math 731: partial di erential equations

Byeongho Ban [email protected]

MATH 731: Partial differential equations

:Taught by Andrea Nahmod

Mathematics & Statistics

University of Massachusetts, Amherst

Byeong Ho Ban

The solutions here are mainly solutions of the exercise problems from the book titled ”Partial Differential Equations : Methods andApplications” written by Robert C. Mcowen. The additional problems appearing sometimes are problem added by A.Nahmod duringthe course of MATH 731.

1

MATH 731 Editor : ByeongHo Ban

Due date : September 26th, 2019 Byeongho Ban

1. Show that if z = u(x, y) is an integral surface of V = (a, b, c) containing a point P , then the surface contains the characteristiccurve χ passing through P . (Assume the vector field V is C1.)

Proof. (Byeongho Ban)

Recall that χ = (x(t), y(t), z(t)) is a curve such that

dx

dt= a

dy

dt= b

dz

dt= c

which means χ is tangent to V . Also recall that the integral surface is a surface tangent to V . Thereforem, if χ passes through P ,χ should be contained in the integral surface containing P . Since the given integral surface z = u(x, y) contains P , χ is containedin z = u(x, y) so we are done.

2. If S1 and S2 are two graphs [i.e. S1 is given by z = ui(x, y), i = 1, 2] that are integral surfaces of V = (a, b, c) and intersect ina curve χ, show that χ is a characteristic curve.


Let P ∈ χ be a given point. Suppose that ξ is a characteristic curve to V passing through P . Since S1 and S2 contains P , byProblem 1, ξ is contained in S1 and S2 at the same time which means ξ is an intersection curve between S1 and S2. However,since intersection curve between S1 and S2 passing through P is χ, we conclude that χ = ξ so χ is a characteristic curve by oursetting.

3. If Γ is a characteristic curve of V , show that there is an infinite number of solutions u of (1) containing Γ in the graph of u.


Firstly, note that any integral surface to the vector field V is a solution u of (1). Thus, it suffices to show that there are infinitelymany integral surfaces containing Γ. Let P ∈ Γ be a given point and let r ∈ R \ 0 be given. Now, let’s choose a curveχ = (x(t), y(y), z(t)) such that

dx

dt= −b− rc dy

dt= a

dz

dt= ra.

and it passes P . Note that χ is noncharacteristic curve to V since it always orthogonal to V . Then V admits a unique integralsurface Sr containing χ. Now note that, by problem 1, Sr contains Γ. Also, note that for any r, s ∈ R \ 0, Sr 6= Ss and it hasonly one curve intersection χ. Since r ∈ R \ 0 was arbitrary, Sr is an integral surface contains Γ for any r ∈ R \ 0. SinceR \ 0 has infinite cardinality, there are infinitely many integral surfaces to V so we are done.


4. Solve the given initial value problem and determine the value of x and y for which it exists:

a) xux + uy = y, u(x, 0) = x2

b) ux − 2uy = u, u(0, y) = y

c) y−1ux + uy = u2, u(x, 1) = x2

Proof. (Byeongho Ban)a) Note that the parametrization of given initial condition is Γ(s) = (s, 0, s2) = (f(s), g(s), h(s)) and that we have the vector fieldV = (x, 1, y) = (a, b, c). Now observe that

f ′(s)b(f(s), g(s))− g′(s)a(f(s), g(s)) = 1− 0 = 1 6= 0.

Thus, γ = (f(s), g(s)) is a non-characteristic curve to v = (a, b). Now, solve the ODE

dx

dt= x

dy

dt= 1

dz

dt= y.

Then we get

x(s, t) = c1(s)et, y(s, t) = t+ c2(s), z(s, t) =1

2t2 + c2(s)t+ c3(s).

By applying the initial condition Γ(s), we have

x(s, t) = set, y(s, t) = t, z(s, t) =1

2t2 + s2.

Note that

det

[xs ysxt yt

]= et − 0 = et > 0

for any s and t. Thus, we have

t(x, y) = y, s(x, y) = xe−y.

Then, by applying it to z(s, t), we have

u(x, y) = z(x, y) =1

2y2 + x2e−2y.

This is the solution to IVP and it exists for any (x, y) ∈ R2 since there is no singularity.

b) Note that the parametrization of given initial condition is Γ(s) = (0, s, s) = (f(s), g(s), h(s)) and that we have the vector fieldV = (1,−2, z) = (a, b, c). Now observe that

f ′(s)b(f(s), g(s))− g′(s)a(f(s), g(s)) = 0− 1 = −1 6= 0.


dx

dt= 1

dy

dt= −2

dz

dt= z.

Then we get

x(s, t) = t+ c1(s), y(s, t) = −2t+ c2(s), z(s, t) = c3(s)et

for some function c1(s), c2(s), c3(s).By applying the initial condition Γ(s), we have

x(s, t) = t, y(s, t) = −2t+ s, z(s, t) = set.

Note that

det

[xs ysxt yt

]= 0− 1 = −1 < 0

for any s and t. Thus, we have

t(x, y) = x, s(x, y) = 2x+ y.


u(x, y) = z(x, y) = (2x+ y)ex.

This is the solution to IVP and it exists for any (x, y) ∈ R2 since there is no singularity.


6. Solve the initial value problem and determine the values of x and y for which it exists:

a) ux + u2uy = 1, u(x, 0) = 1.

b) ux +√uuy = 0, u(x, 0) = x2 + 1


a)Note that the parametrization of given initial condition is Γ(s) = (s, 0, 1) = (f(s), g(s), h(s)) and that we have the vector fieldV = (1, z2, 1) = (a, b, c). Now observe that

f ′(s)b(f(s), g(s))− g′(s)a(f(s), g(s)) = z2 − 0 = z2 6= 0.


dx

dt= 1

dy

dt= z2 dz

dt= 1.

Then we get

x(s, t) = t+ c1(s), y(s, t) =1

3t3 + t2c3(s) + c23(s)t+ c2(s), z(s, t) = t+ c3(s)

for some function c1(s), c2(s), c3(s) and where y(s, t) is obtained by using z(s, t).By applying the initial condition Γ(s), we have

x(s, t) = t+ s, y(s, t) =1

3t3 + t2 + t, z(s, t) = t+ 1.

Note that

det

[xs ysxt yt

]= (t2 + 2t+ 1)− 0 = (t+ 1)2 6= 0

for any s ∈ R and t 6= −1. Since our situation is when t = 0, we can solve this problem. Thus, we have

t(x, y) = 3√

3y + 1− 1 s(x, y) = x− 3√

3y + 1 + 1


u(x, y) = z(x, y) = t(x, y) + 1 = 3√

3y + 1.

This is the solution to IVP and it exists for any x and y.


9. Consider the equation y2ux + xuy = sinu2.

a) Describe all projected characteristic curves in the xy-plane.

b) For the solution u of the initial value problem with u(x, 0) = x, determine the values of ux, uy, uxx, uxy, uyy on the x-axis.


a) Note that the characteristic curve χ(t) = (x(t), y(t), z(t)) satisfies

dx

dt= y2 dy

dt= x

dz

dt= sinu2.

From the first two equation, we have

dx

y2=dy

x⇐⇒ 1

2x2 =

1

3y3 + C ⇐⇒ 1

2x2 − 1

3y3 = C.

Thus, all the projected characteristic curves in xy-plane are of the form 12x

2 − 13y

3 = C for any constant C.

b)Firstly, by parametrizing the initial condition into Γ(s) = (s, 0, s), we can note that, due to a), the projected curve γ(s) = (s, 0)is non-characteristic. Thus, there is a unique solution to the IVP.

Now, by taking the derivative of the initial condition wrt x, we have ux(x, 0) = 1. Nextly, by the PDE,

y2ux + xuy = sinu2 =⇒ 02ux(x, 0) + xuy(x, 0) = sinu2(x, 0) =⇒ uy(x, 0) =sinu2(x, 0)

x=

sinx2

xfor any x 6= 0.

When x = 0, since uy is continuous, observe that, by L’Hospital’s rule,

uy(0, 0) = limx→0

sinu2(x, 0)

x= limx→0

2ux(x, 0)u(x, 0) cosu2(x, 0) = limx→

2x cosx2 = 0.

Again, by taking the derivative of ux(x, 0) and uy(x, 0) wrt x, we have

uxx(x, 0) = 0, uxy(x, 0) =2x2 cosx2 − sinx2

x2.

As for uxy, when x = 0, since uxy is continuous, by L’Hospital’s rule,

uxy(0, 0) = limx→0

2x2 cosx2 − sinx2

x2= 2− lim

x→0

sinx2

x2= 1.

Lastly, in order to find uyy(x, 0), we take the derivative of the PDE, then

2yux + y2uxy + xuyy = 2uyu cosu2 =⇒ xuyy(x, 0) = 2uy(x, 0)u(x, 0) cosu2(x, 0) = 2 sinx2 cosx2 = sin 2x2

=⇒ uyy(x, 0) =sin 2x2

x.

And, when x = 0, since uyy is continuous, by L’Hospital’s rule,

uyy(0, 0) = limx→0

sin 2x2

x= 0.

We are done.


Additional Problem 1Write down an explicit formula for a function u solving the inhomogeneous initial value problem

ut + b · ∇u = f on Rn × (0,∞)

u = g on Rn × t = 0

where f = f(x, t), f : Rn × [0,∞)→ R, c ∈ R, b ∈ Rn and g : Rn → R are all given. (Hint : Solve first the homogeneous problem(i.e. f = 0) using the method of characteristics (as in class or section 1.1a)). Then use the Fundamental Theorem of Calculus towrite u(x, t) − g(x − bt) for (x, t) along a characteristic (g(x − bt) is the homogeneous solution) as an appropriate integral of f .At the end u will be the sum of the homogeneous solution plus an integral term.)


Note that, if u was the solution, the PDE indicates that the directional derivative of u in the direction of (b, 1) ∈ Rn+1 is f .Firstly, let’s consider the homogeneous case where f = 0. In this case the directional derivative should be zero so we get

d

dsu(x+ sb, t+ s) = 0

which implies that u(x + sb, t + s) = u(x − tb, 0) = g(x − tb) by the initial condition. Thus, we get the solution of IVP forhomogeneous case.

Now, let’s come back to original inhomogeneous problem. Then we have

d

dsu(x+ sb, t+ s) = f(x+ sb, t+ s).

Then by the fundamental theorem of calculus, we have

u(x, t)− g(x− tb) = u(x, t)− u(x− tb, 0) =

∫ 0

−t

d

dsu(x+ sb, t+ s)ds =

∫ 0

−tf(x+ sb, t+ s)ds.

By moving terms around and using change of variables s→ −s, we get

u(x, t) = g(x− tb) +

∫ t

0

f(x− sb, t− s)ds.

Furthermore, if we use the change of variable t− s→ τ , since ds = −dτ , we get

u(x, t) = g(x− tb) +

∫ 0

t

f(x+ (τ − t)b, τ)(−dτ) = g(x− tb) +

∫ t

0

f(x+ (τ − t)b, τ)dτ.


Additional Problem 2Write down an explicit formula for a function u solving the initial value problem

ut + b · ∇u+ cu = 0 on Rn × (0,∞)

u = g on Rn × t = 0where c ∈ R, b ∈ Rn and g : Rn → R are given.


Let v(x, t) = ectu(x, t) then observe that

vt + b · ∇v = cectu+ ectut + ect(b · ∇u) = ect(ut + b · ∇u+ cu) = 0 on Rn × (0,∞)

v = e0u = u = g on Rn × t = 0.Thus, v satisfies the homogeneous transport equation with the initial condition v = g on Rn × t = 0. Note that, from theequation, we have

vt + b · ∇v = 0 =⇒ ∇(b,1)v = 0 =⇒ dv

ds(x+ bs, t+ s) = 0

Thus, v(x+ bs, t+ s) is constant in s so observe that

v(x, t) = v(x+ bs, t+ s) = v(x− bt, 0) = g(x− bt) ∀(x, t) ∈ Rn × (0,∞).

Therefore, we have ectu(x, t) = g(x− bt) which implies that u(x, t) = e−ctg(x− bt).

For making sure it is the solution, observe that

ut + b · ∇u+ cu = (−ce−ctg(x− bt) + e−ctb · ∇g(x− bt)) + (b · ∇e−ctg(x− ct)) + (ce−ctg(x− bt)) = 0.

Therefore, it is the solution to the IVP.


Additional Problem 3Let f be a continuous function on an open set D ⊂ Rn such that∫

D0

f(x)dx = 0 D0 ⊂ D

Prove that then f ≡ 0 on D.


Assume that f 6≡ 0 on D. And assume that there exists a ∈ D such that f(a) > 0. Since f is continuous on D, there exists anopen neighborhood N of a such that

|f(x)− f(a)| < f(a)

2∀x ∈ N.

Then it implies that f(a)2 < f(x) for all x ∈ N . Then observe that

0 <

∫N

f(a)

2dx <

∫N

f(x)dx = 0

which is a contradiction. If there is no such a ∈ R, then there should have been b ∈ R such that f(b) < 0. Then by definingg(x) = −f(x) we conclude that, by same reasoning, with appropriate open neighborhood N of b,

0 <

∫N

g(x)dx = −∫N

f(x)dx =⇒ 0 =

∫N

f(x)dx < 0

which is again a contradiction. Therefore, it should have been f ≡ 0.


Due date : October 10th, 2019 Byeongho Ban

2.1.1Consider the initial value problem

uzz = u2ux + (uxy)2,

u(x, y, 0) = x− y,uz(x, y, 0) = sinx.

Find the values of uxz, uyz, and uzz when z = 0.


Observe that

uxz(x, y, 0) =∂

∂xuz(x, y, 0) = cosx,

uyz(x, y, 0) =∂

∂yuz(x, y, 0) =

∂

∂ysinx = 0.

As for uzz, we need to use first equation, so we need to find u(x, y, 0), ux(x, y, 0) and uxy(x, y, 0). u(x, y, 0) is already given in thesecond equation. Firstly, observe that

ux(x, y, 0) =∂

∂xu(x, y, 0) =

∂

∂x(x− y) = 1.

And then we note that

uxy(x, y, 0) =∂

∂yux(x, y, 0) =

∂x

∂y= 0.

Thus, by the first equation,

uzz(x, y, 0) = (u(x, y, 0))2ux(x, y, 0) + (uxy(x, y, 0))2 = (x− y)2 · 1 + 02 = (x− y)2.

2.1.2 Is the heat equation ut = kuxx in normal form for Cauchy data on the x−axis? On the t-axis? What form would theCauchy data (3) take?


If the Cauchy data is given on the x-axis (when t = 0), this equation is not in normal.If the Cauchy data is given on t-axis(when x = 0), after dividing k from both sides(it is possible since k > 0), uxx = 1

kut, it is in

the normal(canonical) form. And our initial surface would be S = (x, t) ∈ R2 : x = 0. Then Cauchy data should be

u(0, t) = h1(t), ux(0, t) = h2(t).


2.1.7 Consider the Cauchy problem for Laplace’s equationuxx + uyy = 0,

u(x, 0) = 0,

uy(x, 0) = k−1 sin kx,

where k > 0. Use separation of variables to find the solution explicitly. If we let k → ∞, notice that the Cauchy data tendsuniformly to zero, but the solution does not converge to zero for any y 6= 0. Therefore, a small change from zero Cauchy data[which has the solution u(x, y) ≡ 0] induces more than a small change in the solution; this means that the Cauchy problem forthe Laplace equation is not well posed.


Let u(x, y) = X(x)Y (y). Then observe that

X ′′Y = uxx = −uyy = −XY ′′

which implies that

X ′′

X= −Y

′′

YNow, note that, LHS of the equation above only depends on x and that of RHS only depends on y. Since x and y are independentvariables, each sides should be constant, let’s say λ. Then we get two different cases depending on the sign of λ as

X ′′ = λX

Y ′′ = −λY=⇒

X(x) = Ae

√λx +Be−

√λx

Y (y) = Cei√λy +De−i

√λy

or

X(x) = Aei

√λx +Be−i

√λx

Y (y) = Ce√λy +De−

√λy

.

However, the first case is impossible since, by the second Cauchy data,

k−1 sin kx = uy(x, 0) = X(x)Y ′(0) = i√λY (C −D)(Ae

√λx +Be−

√λx)

where RHS can never be a sine function. Thus, λ < 0, let’s say λ = −n2 so we have

u(x, y) = X(x)Y (y) = (Aeinx +Be−inx)(Ceny +De−ny).

Now, by the first Cauchy data, we get

0 = u(x, 0) = X(x)Y (0) = (Aeinx +Be−inx)(C +D)

which implies C = −D since X(x) 6= 0 by second Cauchy data.

Also, by the second Cauchy data,

sin kx

k= uy(x, 0) = X(x)Y ′(y) = (Aeinx +Be−inx)2nC

which implies that 2nkCA = −2nkCB = 12i and k = n by the relation

sin kx =eikx − e−ikx

2i.

Thus, we have

u(x, y) = (Aeikx +Be−ikx)(Ceky − Ce−ky) =eikx − e−ikx

4k2i(eky − e−ky) =

sin kx sinh ky

k2.

As k → ∞, the first Cauchy data is clearly converges to 0 uniformly. As for second Cauchy data, it is also converges uniformlybecause, for any ε > 0, letting K > 1

ε (existence of K is guaranteed due to Archimedean property) observe that, when ever k > K,we have

|uy(x, 0)− 0| =∣∣∣∣ sin kxk

∣∣∣∣ ≤ ∣∣∣∣1k∣∣∣∣ ≤ 1

K< ε ∀x ∈ R.

Now, note that as k →∞, u(π/2k, y) = sinh kyk2 →∞ 6= 0. If the solution u depends on the Cauchy data continuously, when Φ is

a map from data to solution, then

0 6= limk→∞

sinh ky sin ky

k2= limk→∞

φ(datak) = u( limk→∞

datak) = φ(0) = 0

since zero Cauchy data gives zero solution. But it is contradiction. Therefore, solution does not depends on Cauchy datacontinuously which means this problem is not well posed.


2.2.1 Reduce to canonical form(a) uxx + 5uxy + 6uyy = 0

(b) x2uxx − y2uyy = 0


(a) Firstly, we need to find the characteristics. Observe that

dy

dx=

5±√

25− 24

2= 3 or 2.

Since there are two characteristics, this PDE is hyperbolic. And then we have y = 3x+C1 and y = 2x+C2 where C1 and C2 arearbitrary constants. Then our characteristics are ξ(x, y) = 3x− y and η(x, y) = 2x− y. Then observe that

ux = ξxuξ + ηxuη = 3uξ + 2uη

uy = ξyuξ + ηyuη = −uξ − uηand that

uxx = 3ξxuξξ + 3ηxuξη + 2ξxuξη + 2ηxuηη = 9uξξ + 12uξη + 4uηη

uyy = −ξyuξξ − ηyuξη − ξyuξη − ηyuηη = uξξ + 2uξη + uηη

uxy = (ux)y = 3ξyuξξ + 3ηyuξη + 2ξyuξη + 2ηyuηη = −3uξξ − 5uξη − 2uηη.

By plugging uxx, uxy and uyy we found into the given PDE, we get

uxx + 5uxy + 6uyy = (9uξξ + 12uξη + 4uηη) + 5(−3uξξ − 5uξη − 2uηη) + 6(uξξ + 2uξη + uηη) = uξη = 0.

So our canonical form is uξη = 0.

(b)Firstly, we need to find the characteristics. Observe that

dy

dx=

0±√

0 + 4x2y2

2x2=|y|x

or − |y|x.

Note these two curves are equivalent to two curves ± yx . so we have

dy

dx= ±y

x.

Since there are two characteristics, this PDE is hyperbolic. And then, by using separation of variables, we have y = C1x andxy = C2 where C1 and C2 are arbitrary constants. Then our characteristics are ξ(x, y) = y/x and η(x, y) = xy. Then observethat

ux = ξxuξ + ηxuη = − y

x2uξ + yuη

uy = ξyuξ + ηyuη =1

xuξ + xuη

and that

uxx =2y

x3uξ +

y2

x4uξξ −

y2

x2uξη −

y2

x2uξη + y2uηη

uyy =1

x2uξξ + 2uξη + x2uηη.

By plugging uxx and uyy we found into the given PDE, we get

0 = x2uxx − y2uyy =2y

xuξ +

y2

x2uξξ − y2uξη − y2uξη + x2y2uηη −

y2

x2uξξ − 2y2uξη − x2y2uηη = 2ξuξ − 4y2uξη

So our canonical form is 4ξηuξη − 2ξuξ = 0.


Additional Problem 1)(a) Find the characteristics of the PDE

y2uxx − 2yuxy + uyy = ux + 6y,

and determine if elliptic, parabolic or hyperbolic.

(b) Then find the canonical form and use it to find the solution u first in the ξ and η variables and then in the x and y variables.


(a) Observe that the characteristic is determined by

dy

dx=−2y ±

√4y2 − 4y2

2y2= −1

y

Then we get y2 = −2x + C for arbitrary constant C. Then the characteristic is ξ(x, y) = y2 + 2x. Since there is only onecharacteristic, this PDE is parabolic.

(b) Now, with the ξ(x, y) found in (a), let η(x, y) = y then note that y = C always transverse the graph ξ(x, y) = Const. Now,observe that

ux = ξxuξ + ηxuη = 2uξ

uy = ξyuξ + ηyuη = 2yuξ + uη

and that

uxx = 4uξξ

uyy = 4y2uξξ + 4yuξη + uηη + 2uξ

uxy = 4yuξξ + 2uξη.

By plugging it back to the PDE, we get

y2uxx − 2yuxy + uyy − ux − 6y = uηη − 6η = 0.

By solving the canonical PDE uηη = 6η, we get

u(ξ, η) = η3 + F (ξ)η +G(ξ)

for arbitrary function F and G which are in C2. By representing it in terms of x and y, we get

u(x, y) = y3 + F (y2 + 2x)y +G(y2 + 2x).

And it is the solution.


Additional Problem 2)(a) Find the characteristics of the PDE

xuxx + (x− y)uxy − yuyy = 0, x > 0, y > 0,

and determine if elliptic, parabolic or hyperbolic.

(b) Then show that it can be transformed into the canonical form

(ξ2 + 4η)uξη + ξuη = 0

for ξ and η are suitably chosen canonical coordinates and use this to obtain the general solution in the ξ and η variables.


(a) Observe that the characteristic is determined by

dy

dx=

(x− y)±√

(x− y)2 + 4yx

2x=

(x− y)± |x+ y|2x

= 1 or − y

x

since x, y > 0. Then note that we have two characteristics ξ(x, y) = y−x and η(x, y) = xy. And since there are two characteristics(Note x+ y > 0. In particular, x+ y 6= 0), this PDE is hyperbolic.

(b) Observe that

ux = ξxuξ + ηxuη = −uξ + yuη

uy = ξyuξ + ηyuη = uξ + xuη

and that

uxx = −(ξxuξξ + ηxuηξ) + y(ξxuξη + ηxuηη) = uξξ − 2yuξη + y2uηη

uyy = (ξyuξξ + ηyuξη) + x(ξyuξη + ηyuηη) = uξξ + 2xuξη + x2uηη

uxy = (ξxuξξ + ηxuηξ) + uη + x(ξxuξη + ηxuηη) = −uξξ + (y − x)uξη + xyuηη + uη.

By plugging these back to the original equation, we get

xuxx + (x− y)uxy − yuyy = x(uξξ − 2yuξη + y2uηη) + (x− y)(−uξξ + (y − x)uξη + xyuηη + uη)− y(uξξ + 2xuξη + x2uηη)

= (−4xy − (x− y)2)uξη + (x− y)uη

= −(4η + ξ2)uξη − ξuη = 0

so we get the desired form (ξ2 + 4η)uξη + ξuη = 0. Now, let’s solve this PDE. Observe that we have

(uη)ξ = − ξ

ξ2 + 4ηuη

so we have uη(x, y) = F (η) exp[

12(ξ2+4η)

]. By integrating both sides with respect to ξ, we have

u(ξ, η) =

∫F (η) exp

[1

2(ξ2 + 4η)

]dη +G(ξ).

for arbitrary C2-functions F and G.


3.1.1Solve the initial value problems:

(a)

utt − c2uxx = 0,

u(x, 0) = x3

ut(x, 0) = sinx.

(b)

utt − c2uxx = 2t

u(x, 0) = x2

ut(x, 0) = 1.


(a) Since the Cauchy date are C2 functions, we can simply use d’Alembert’s formula

u(x, y) =1

2

((x− ct)3 + (x+ ct)3

)+

1

2c

∫ x+ct

x−ctsin ξdξ

=1

2

(2x3 + 6xc2t2

)+

1

2c[cos(x− ct)− cos (x+ ct)]

= x3 + 3xc2t2 +sinx sin ct

cso we found the solution.

(b) We are given a nonhomogeneous PDE. Thus, we will use Duhamel’s principle to solve it. Thus, our solution is

u(x, t) =1

2((x− ct)2 + (x+ ct)2) +

1

2c

∫ x+ct

x−ctdξ +

1

2c

∫ t

0

(∫ x+c(t−s)

x−c(t−s)2sdξ

)ds

= x2 + c2t2 + t+t3

3.


3.1.4Consider initial boundary value problem

utt − c2uxx = 0 x, t > 0,

u(x, 0) = g(x), ut(x, 0) = h(x) x > 0,

u(0, t) = 0 t ≥ 0,

where g(0) = 0 = h(0). If we extend g and h as odd functions on −∞ < x < ∞, show that d’Alembert’s formula (6) gives thesolution.


First, we extend our function g and h to odd functions by defining

G(x) =

g(x) x ≥ 0

−g(−x) x < 0H(x) =

h(x) x ≥ 0

−h(−x) x < 0

Then, let’s solve the same initial and boundary value problem with whole real line R and let v be the solution to the new PDEproblem. If we use the d’Alembert’s formula for the extended functions, we get

v(x, t) =1

2(G(x+ ct) +G(x− ct)) +

1

2c

∫ x+ct

x−ctH(ξ)dξ

Now, let’s restrict our solution to half real line (0,∞), so rename it to u(x, t) = v(x, t) when x, t ∈ (0,∞). Then u(x, t) is clearlysatisfies the PDE and initial conditions. Lastly, we need to check the boundary contion. Since G and H are odd functions, observethat

v(0, t) =1

2(G(ct) +G(−ct)) +

1

2c

∫ ct

−ctH(ξ)dξ

=1

2(g(ct)− g(ct)) +

1

2c

[∫ ct

0

h(ξ)dξ −∫ 0

−cth(−ξ)dξ

]=

1

2c

(∫ ct

0

h(ξ)dξ −∫ 0

ct

h(η)d(−η)

)η = −ξ

=1

2c

(∫ ct

0

h(ξ)dξ −∫ ct

0

h(η)dη

)= 0

for any t. Therefore, by restricting our solution to the right half real line, it gives the solution for the given initial boundary valueproblem.

Furthermore, let’s unwind this formula. Noting that G and H are odd function, we can categorize it into two cases. Firstly, whenx > |ct|, we have x± ct > 0. Then, we have

v(x, t) =1

2(g(x+ ct) + g(x− ct)) +

1

2c

∫ x+ct

x−cth(ξ)dξ.

Otherwise, if 0 < x < |ct|, then observe that

v(x, t) =1

2(G(x+ ct) +G(x− ct)) +

1

2c

∫ x+ct

x−ctH(ξ)dξ

=1

2(g(x+ ct)− g(ct− x))− 1

2c

∫ 0

x−cth(−ξ)dξ +

1

2c

∫ x+ct

0

h(ξ)dξ

=1

2(g(x+ ct)− g(ct− x)) +

1

2c

∫ x+ct

ct−xh(ξ)dξ.

Therefore,

v(x, t) =

12 (g(x+ ct) + g(x− ct)) + 1

2c

∫ x+ct

x−ct h(ξ)dξ x > |ct|12 (g(x+ ct)− g(ct− x)) + 1

2c

∫ x+ct

ct−x h(ξ)dξ 0 < x < |ct|.


3.1.5Find in closed form (similar to d’Alembert’s formula ) the solution u(x, t) of

utt − c2uxx = 0 x, t > 0,

u(x, 0) = g(x), ut(x, 0) = h(x) x > 0

u(0, t) = α(t) t ≥ 0,

where g, h, α ∈ C2 satisfy α(0) = g(0), α′(0) = h(0), and α′′(0) = c2g′′(0). Verify that u ∈ C2, even on the characteristic x = ct.


Firstly, observe that, when x > ct,

u(x, t) =1

2[g(x− ct) + g(x+ ct)] +

1

2c

∫ x+ct

x−cth(ξ)dξ.

It is clearly satisfy IBVP. As for the boundary condition, since in this case we have x > ct > 0, when x = 0, it is necessary tohave t = so observe that

u(0, t) = u(0, 0) = g(0) = α(0) = α(t).

Now, we want to investigate when 0 < x < ct.Firstly, similar to previous exercise, we extend our function g(x)− α(0) and h(x)− α′(0) to odd functions by defining

G(x) =

g(x)− α(0) x ≥ 0

−g(−x) + α(0) x < 0H(x) =

h(x)− α′(0) x ≥ 0

−h(−x) + α′(0) x < 0

Let v(x, t) = u(x, t)− α(t). Then observe that v satisfiesvtt − c2vxx = A(x, t) t > 0

v(x, 0) = G(x), vt(x, 0) = H(x)

v(0, t) = 0

where

A(x, t) =

−α′′(t) x > 0

α′′(t) x < 0

which is similar problem with 3.1.4 but note now it is non-homogeneous.Then by using d’Alembert’s formula and Duhamel’s principle, our solution would be

v(x, t) =1

2(G(x+ ct) +G(x− ct)) +

1

2c

∫ x+ct

x−ctH(ξ)dξ +

1

2c

∫ ∫∆

A(ξ, s)dξds

where ∆ is the domain of dependence of the point (x, t). According to the reflection point, we can divide the domain of dependenceinto two parts. One part is the region in ξs-plane bounded by three curves : ξ = x + c(t − s), ξ = x − c(t − s) and s = t − x

c .Another one is the region in ξs-pla-ne bounded by four curves: ξ = c(t− s) + x, ξ = c(t− s)− x, s = 0 and s = t− x

c . Note that,A(ξ, s) = −α′′(s) in the first region and A(ξ, s) = α′′(s) in another region. Then observe that the Duhamel term is

1

2c

∫ ∫∆

A(ξ, s)dξds =1

2c

(∫ t

t− xc

∫ x+c(t−s)

x−c(t−s)−α′′(s)dξds

)+

1

2c

(∫ t− xc

0

∫ c(t−s)+x

c(t−s)−xα′′(s)dξds

)

=

∫ t

t− xcα′′(s)(s− t)ds+

x

c

∫ t− xc

0

α′′(s)ds

= α′(s)(s− t)∣∣∣∣tt− xc

−∫ t

t− xcα′(s)ds+

x

c

[α′(t− x

c

)− α′(0)

]= −x

cα′(t− x

c

)− α(t) + α

(t− x

c

)+x

cα′(t− x

c

)− x

cα′(0)

= −α(t) + α(t− x

c

)− x

cα′(0)



Then by using the result above, observe that

v(0, t) =1

2[G(ct) +G(−ct)] +

1

2c

∫ ct

−ctH(ξ)dξ +

1

2c

∫ ∫∆

A(ξ, s)dξds

= 0 + 0− α(t) + α(t) + 0 = 0

since G and H are odd functions, so it satisfies the boundary condition. Also, since we are just dealing with the case when0 < x < ct, when t = 0 we should have x = 0. Thus observe that

v(x, 0) = v(0, 0) = G(0) + 0− α(0) + α(0)− 0 = G(0) = G(x)

vt(x, 0) = vt(0, 0) = 0 +H(0)− α′(0) + α′(0) = H(0) = H(x)

so it satisfies the initial condition. Thus, the v gives us the solution to the IBVP. Then letting u(x, t) = v(x, t) + α(t), and byrestricting our domain from R to (0,∞), it gives us the solution to the original IBVP.

Now, we want to unwind the formula in terms of g and h. Since we only consider then 0 < x < ct, similarly to the previousexercise, we have

u(x, t) = v(x, t) + α(t) =1

2[g(ct+ x)− α(0)− g(ct− x) + α(0)] +

1

2c

∫ ct+x

ct−xh(ξ)− α′(0)dξ + α

(t− x

c

)− x

cα′(0)

=1

2[g(ct+ x)− g(ct− x)] +

1

2c

∫ ct+x

ct−xh(ξ)dξ + α

(t− x

c

).

Therefore, out solution would be

u(x, t) =

12 [g(x− ct) + g(x+ ct)] + 1

2c

∫ x+ct

x−ct h(ξ)dξ, x > ct12 [g(ct+ x)− g(ct− x)] + 1

2c

∫ ct+xct−x h(ξ)dξ + α

(t− x

c

), 0 < x < ct.

This u is clearly in C2 since g, h, α ∈ C2.

When x = ct, letting u1 and u2 be the solution in the domain of x > ct and 0 < x < ct respectively, observe that

u1(ct, t) =1

2[g(0) + g(2ct)] +

1

2c

∫ 2ct

0

h(ξ)dξ

=1

2[g(2ct)− g(0)] +

1

2c

∫ 2ct

0

h(ξ)dξ + g (0)

=1

2[g(2ct)− g(0)] +

1

2c

∫ 2ct

0

h(ξ)dξ + α (0)

= u2(ct, t)

which implies that u is in C on x = ct.Also, observe that

u1x(ct, t) =

1

2[g′(0) + g′(2ct)] +

1

2c[h(2ct)− h(0)]

=1

2[g′(2ct) + g′(0)] +

1

2c[h(2ct) + h(0)]− 1

ch (0)

=1

2[g′(2ct) + g′(0)] +

1

2c[h(2ct) + h(0)]− 1

cα′ (0)

= u2x(ct, t)

and that

u1t (ct, t) =

1

2[−cg′(0) + cg′(2ct)] +

1

2c[ch(2ct) + ch(0)]

=1

2[cg′(2ct)− cg′(0)] +

1

2c[ch(2ct)− ch(0)] + h (0)

=1

2[cg′(2ct) + cg′(0)] +

1

2c[ch(2ct)− ch(0)] + α′ (0)

= u2t (ct, t)

so u is in C1 on x = ct.



Lastly observe that

u1xx(ct, t) =

1

2[g′′(0) + g′′(2ct)] +

1

2c[h′(2ct)− h′(0)]

=1

2[g′′(2ct)− g′′(0)] +

1

2c[h′(2ct)− h(0)] + g′′(0)

=1

2[g′′(2ct)− g′′(0)] +

1

2c[h′(2ct)− h(0)] +

1

c2α′′ (0)

= u2xx(ct, t)

and that

u1xt(ct, t) =

1

2[−cg′′(0) + cg′′(2ct)] +

1

2c[ch′(2ct) + ch′(0)]

=1

2[cg′′(2ct) + cg′′(0)] +

1

2c[ch′(2ct) + ch′(0)]− cg′′(0)

=1

2[cg′′(2ct) + cg′′(0)] +

1

2c[ch′(2ct) + ch′(0)]− 1

cα′′ (0)

= u2xt(ct, t)

since c2g′′(0) = α′′(0). Thus, u is in C2 on the line x = ct. As for u1tt = u2

tt, we can easily see this from u1xx = u2

tt by the relationof utt = c2uxx.


Additional Problem 1)(a) Show that the general solution to the PDE uxy = 0 is

u(x, y) = F (x) +G(y)

for arbitrary functions F,G.

(b) Using a change of variables ξ = x+ t and η = x− t, show that

utt − uxx = 0 if and only if uξη = 0.

(c) Use parts (a) and (b) to rederive D’Alembert’s formula.


(a) Observe that

uxy = 0 =⇒ ux(x, y) = f(x)

for arbitrary function f . Then observe that

ux(x, y) = f(x) =⇒ u(x, y) =

∫ x

0

f(ξ)dξ +G(y) + C

for arbitrary function G and f and for some constant C.Letting

∫ x0f(ξ)dξ + C = F (x), we have u(x, y) = F (x) +G(y) for arbitrary function F and G.

(b) Let’s reset ξ = x + ct and η = x − ct with c > 0. And we will prove that utt − c2uxx = 0 if and only if uξη. Firstly, observethat

ut = ξtuξ + ηtuη = cuξ − cuηutt = c(ξtuξ + ηtuξη)− c(ξtuξη + ηtuηη) = c2uξξ − 2c2uξη + c2uηη

and that

ux = ξxuξ + ηxuη = uξ + uη

uxx = (ξxuξξ + ηxuηξ) + (ξxuξη + ηxuηη) = uξξ + 2uξη + uηη.

Then we have the following equivalence

0 = utt − c2uxx ⇐⇒ 0 = (c2uξξ − 2c2uξη + c2uηη)− c2(uξξ + 2uξη + uηη) = −4c2uξη ⇐⇒ uξη = 0.

Therefore, utt − c2uxx = 0 if and only if uξη = 0. And by setting c = 1, we get the desired result.

(c) We want to find the solution for the wave equation utt − c2uxx = 0 with initial conditions u(x, 0) = g(x) and ut(x, 0) = h(x).Note that by part (b), by having same ξ and eta, we get uηξ = 0. And by part (a), we get

u(ξ, η) = F (ξ) +G(η) ⇐⇒ u(x, t) = F (x+ ct) +G(x− ct).for some function F and G. Now, by the given initial condition, we get

g(x) = F (x) +G(x) and h(x) = cF ′(x)− cG′(x).

And note that, by the fundamental theorem of calculus,∫ x

0

h(s)ds = c(F (x)− F (0))− c(G(x)−G(0)) and cg(x) = cF (x) + cG(x).

By solving this simultaneous equation, we get 2cF (x) = cF (0)−cG(0)+∫ x

0h(s)ds+cg(x) and 2cG(x) = −cF (0)+cG(0)+cg(x)−∫ x

0h(s)ds which gives

F (x) =1

2(F (0)−G(0) + g(x)) +

1

2c

∫ x

0

h(s)ds and G(x) =1

2(−F (0) +G(0) + g(x))− 1

2c

∫ x

0

h(s)ds.

Then, by plugging it back to u(x, t), we get

u(x, t) = F (x+ ct) +G(x− ct)

=1

2(F (0)−G(0) + g(x+ ct)) +

1

2c

∫ x+ct

0

h(s)ds+1

2(−F (0) +G(0) + g(x− ct))− 1

2c

∫ x−ct

0

h(s)d

=1

2(g(x+ ct) + g(x− ct)) +

1

2c

∫ x+ct

x−cth(s)ds.

which is D’Alembert’s formula.


Additional Problem 2)Let u ∈ C2(R× [0,∞)) be a solution to the Cauchy initial value problem

utt − uxx = 0 in R× (0,∞)

u(0, x) = g(x) in Rut(0, x) = h(x) in R.

Suppose that g and h are smooth and have compact support. The kinetic energy is

k(t) =1

2

∫Ru2t (t, x)dx

and the potential energy is

p(t) =1

2

∫Ru2x(t, x)dx.

Prove :(a) k(t) + p(t) is constant in t.(b) k(t) = p(t) for all large enough times t.



(a) Firstly, note that by d’Alembert’s formula,

u(x, t) =1

2(g(x+ t) + g(x− t)) +

1

2

∫ x+t

x−th(ξ)dξ.

Then, by the fundamental theorem of calculus,

ux(x, t) =1

2(g′(x+ t) + g′(x− t)) +

1

2[h(x+ t)− h(x− t)]

ut(x, y) =1

2(g′(x+ t)− g′(x− t)) +

1

2(h(x+ t) + h(x− t))

Now, note that, since h has compact support, h(x ± t) → 0 as x → ∞. Since g(x) = 0 for large enough x, g′(x) = 0 for largeenough x which implies limx→∞ g′(x± t) = 0. Thus, lim|x|→∞ ux(x, t) = 0.Let

E(t) = k(t) + p(t) =1

2

∫Ru2t (t, x) + u2

x(t, x)dx.

Then observe thatdE

dt(t) =

∫Rututt + uxuxtdx

=

∫Rututtdx+

[uxut

∣∣∞x=−∞ −

∫Ruxxutdx

]∵ by the integration by parts

=

∫Rut(utt − uxx)dx ∵ since lim

|x|→∞ux(x, t) = 0

= 0 ∵ utt − uxx = 0.

Therefore, E(t) is constant in time t.

(b) Note from (a) that, since g′ and h are compactly supported. Therefore, ∃C,D ∈ R such that

supp(f) ⊂ [−C,C] supp(g) ⊂ [−D,D].

So, let R = maxC,D, then

supp(f) ⊂ [−R,R] supp(g) ⊂ [−R,R].

And now, note that

k(t)− p(t) =

∫ ∞−∞

u2t − u2

x

dx = 0

=

∫ ∞−∞

h(x+ t)h(x− t)− g′(x+ t)g′(x− t) +

1

2g′(x+ t)h(x− t)− 1

2g′(x− t)h(x+ t)

dx.

Observe that, if t > R,

h(x+ t) 6= 0 =⇒ x+ t ∈ [−R,R] =⇒ x− t ∈ [−R− 2t, R− 2t]

=⇒ x− t 6∈ [−R,R] =⇒ h(x− t) = 0 and g′(x− t) = 0

h(x− t) 6= 0 =⇒ x− t ∈ [−R,R] =⇒ x+ t ∈ [−R+ 2t, R+ 2t]

=⇒ x+ t 6∈ [−R,R] =⇒ h(x+ t) = 0 and g′(x+ t) = 0

g′(x+ t) 6= 0 =⇒ x+ t ∈ [−R,R] =⇒ x− t ∈ [−R− 2t, R− 2t]

=⇒ x− t 6∈ [−R,R] =⇒ g′(x− t) = 0 and h(x− t) = 0

g′(x− t) 6= 0 =⇒ x− t ∈ [−R,R] =⇒ x+ t ∈ [−R+ 2t, R+ 2t]

=⇒ x+ t 6∈ [−R,R] =⇒ g′(x+ t) = 0 and h(x+ t) = 0.

Then, by the argument above, for t > R, all the terms in the integrand are always zero since one of the factors should be zero foreach terms. Therefore, we conclude that

k(t)− p(t) = 0 ∀t > R.

Therefore, k(t) = p(t) for all t > R.


Due date : October 29th, 2019 Byeongho Ban

2.3.4

Let a, b ∈ R and let v, w ∈ D be given. Then observe that

Ff (av + bw) =

∫Ω

f(x)(av(x) + bw(x))dx = a

∫Ω

f(x)v(x)dx+ b

∫Ω

f(x)w(x)dx = aFf (v) + bFf (w).

Thus, Ff is clearly a linear mapping. And now, suppose that vi ⊆ D be a sequence of functions such that supp(vi) ⊆ K forsome compact set K ⊆ Ω and Dαvi → 0 uniformly in K. Then note that, for given ε > 0, there is N ∈ N such that |vn(x)| < ε

A

for all n > N and all x ∈ Ω where A =∫K|f(x)|dx. Now, observe that, for n > N ,

|Ff (vn)− 0| =∣∣∣∣ ∫

Ω

f(x)vn(x)dx

∣∣∣∣ ≤ ∫K

|f(x)||vn(x)|dx ≤ ε

A

∫Ω

|f(x)|dx = ε

Therefore, Ff (vi)→ 0 as i→∞ which implies that Ff is continuous on D so Ff is a distribution.

2.3.8

Let v ∈ D be given. Firstly, note that v is continuous so for a given ε > 0, there is δ > 0 such that |v(x) − v(0)| < ε whenever|x| < δ. Secondly, observe that there exists N ∈ N such that 1

n < δ for any n > N . Then observe that, for n > N ,∣∣∣∣ ∫Rfn(x)v(x)dx−

∫Rδ(x)v(x)dx

∣∣∣∣ =

∣∣∣∣ ∫Rfn(x)v(x)dx− v(0)

∣∣∣∣=

∣∣∣∣ ∫Rfn(x)v(x)dx− v(0)

∫Rfn(x)dx

∣∣∣∣≤∫R|fn(x)||v(x)− v(0)|dx

=

∫|x|< 1

n

n

2|v(x)− v(0)|dx

< ε

∫|x|< 1

n

n

2dx

= ε.

Therefore, fn converges to δ in distribution.

2.3.10

(a)Observe that, if F is the fundamental, then we have, for any v ∈ D,

v(0) = 〈δ, v〉 =

⟨dF

dx− aF, v

⟩=

⟨dF

dx, v

⟩− 〈aF, v〉 = −

⟨F,dv

dx

⟩− 〈F, av〉 = −

⟨F,dv

dx+ av

⟩.

We claim that

F (x) =

eax x > 0

0 x < 0

is the fundamental solution. Observe that, since v has compact support,⟨F,dv

dx

⟩=

∫ ∞0

eaxdv

dx(x)dx = eaxv(x)

∣∣∣∣∞0

−∫ ∞

0

aeaxv(x)dx = −v(0)− 〈F, av〉

⇐⇒ −⟨F,dv

dx+ aF

⟩= v(0)

as we desired.


2.3.10 (b)

Observe that, for any v ∈ D,⟨d2F

dx2, v

⟩=

⟨F,d2v

dx2

⟩=

∫ ∞0

1

asinh ax

d2v

dx2(x)dx

=1

asinh ax

dv

dx

∣∣∣∣∞0

−∫ ∞

0

cosh axdv

dx(x)dx ∵ integration by parts

= −∫ ∞

0

cosh axdv

dx(x)dx ∵ sinh 0 = 0 and

dv

dx∈ D

= cosh (ax)v(x)

∣∣∣∣∞0

+

∫ ∞0

a sinh (ax)v(x)dx ∵ integration by parts

= v(0) + a2

∫ ∞0

1

asinh (ax)v(x)dx ∵ v ∈ D and cosh 0 = 1

= 〈δ, v〉+⟨a2F, v

⟩.

Therefore, since v ∈ D was arbitrary, we have ⟨d2F

dx2− a2F, v

⟩= 〈δ, v〉

which implies that F is a fundamental solution for the differential operator.

2.3.11 (c)Suppose that f ∈ L1(R) has compact support and is continuous. Now, let

v(x) =1

2

(∫ x

−∞f(y)dy −

∫ ∞x

f(y)dy

).

Note that v(x) is continuous and differentiable with respect to x. Observe that, by the fundamental theorem of calculus,

v′(x) =1

2(f(x) + f(x)) = f(x)

and since f is continuous, v ∈ C1(R). Also, since v′ = f , we have v = u′ which implies that u ∈ C2(R) and u′′ = f .

Additional Problem (1)

Let v ∈ D be given. Then for any η > 0 there is γ > 0 such that |v(x) − v(0)| < ηA whenever |x| < γ. Also, there is R > 0 such

that∫

Ωg(x)dx =

∫BR

g(x)dx since g ∈ L1(Ω). Now, note that supp(gε) ⊆ BεR. Let ε < γR , then observe that∣∣∣∣ 〈gε, v〉 − 〈Aδ, v〉 ∣∣∣∣ =

∣∣∣∣ ∫Ω

gε(x)v(x)dx−Av(0)

∣∣∣∣=

∣∣∣∣ ∫Ω

gε(x)v(x)dx− v(0)

∫Ω

gε(x)dx

∣∣∣∣≤∫

Ω

|gε(x)||v(x)− v(0)|dx

=

∫BεR

|gε(x)||v(x)− v(0)|dx

≤∫BεR

|gε(x)| ηAdx

< η.

Since η was arbitrary, we proved that gε → δ in distribution when ε→ 0.


Additional Problem (1’)

Let ε > 0 be given and v ∈ D be given. Firstly, note that there is N ∈ N such that∣∣∣∣ ∫Rnfj(x)dx− 1

∣∣∣∣ < ε

|v(0)|∀j > N.

Or equivalently,∫Rn

fj(x) < ε|v(0)| + 1 for all j > N since fj is non-negative. Also, note that, since v is continuous at 0, there is

δ > 0 such that |v(x)− v(0)| < ε whenever |x| < δ. Additionally, note that, there is N ′ ∈ N such that∣∣∣∣ ∫|x|>δ

fj(x)dx

∣∣∣∣ < ε

2A∀j > N ′

where A is a maximum of v over all Rn. (possible because v has compact support.)Then observe that, for any j > M = max(N,N ′),

|⟨Ffj , v

⟩− 〈δ, v〉 | =

∣∣∣∣ ∫Rnfj(x)v(x)dx− v(0)

∣∣∣∣ =

∣∣∣∣ ∫Rnfj(x)v(x)dx−

∫Rn

fj(x)v(0)dx+

∫Rnfj(x)v(0)dx− v(0)

∣∣∣∣≤∫Rnfj(x)|v(x)− v(0)|dx+ |v(0)|

∣∣∣∣ ∫Rnfj(x)dx− 1

∣∣∣∣<

∫|x|<δ

fj(x)|v(x)− v(0)|dx+

∫|x|>δ

fj(x)|v(x)− v(0)|dx+ ε

≤ ε∫|x|<δ

fj(x)dx+ 2A

∫|x|>δ

fj(x)dx+ ε

< ε

(ε

|v(0)|+ 1

)+ 2ε

=ε2

|v(0)|+ 3ε.

Therefore, we proved that

limj→∞

⟨Ffj , v

⟩= 〈δ, v〉 ∀v ∈ D.

Thus, Ffj → δ in D′.


Observe that, for any v ∈ D ⟨δ(k), v

⟩= (−1)k

⟨δ,dkv

dxk

⟩= (−1)k

dkv

dxk(0)

by integration by parts and since any derivatives of v has compact support.Thus,

δ(k)(v) = (−1)kdk

dxk

∣∣∣∣x=0

as a distribution.


Note that (sgn(x) + 1)′ = (sgn(x))′. Thus, observe that, for any v ∈ D,

〈(sgn(x))′, v〉 = 〈(sgn(x) + 1)′, v〉 = −〈sgn(x) + 1, v′〉 = −∫ ∞

0

2v′(x)dx = 2v(0) = 〈2δ, v〉

since v has compact support and by fundamental theorem of calculus.Therefore, (sgn(x))′ = 2δ as a distribution.



Let v ∈ D be given. And observe that⟨d

dx(log |x|), v

⟩= −〈log |x|, v′〉 = −

∫ ∞0

log (x)v′(x)dx−∫ 0

−∞log(−x)v′(x)dx

= −∫ ∞

0

log (x)v′(x)dx−∫ ∞

0

log(x)v′(−x)dx

= −∫ ∞

0

log(x)(v′(x) + v′(−x))dx

= − log(x)(v(x)− v(−x))

∣∣∣∣∞0

+

∫ ∞0

v(x)− v(−x)

xdx

= limε→0+

log(ε)(v(ε)− v(−ε)) +

∫ ∞0

v(x)

xdx−

∫ ∞0

v(−x)

xdx

= limε→0+

log(ε)(v(ε)− v(−ε)) +

∫ ∞0

v(x)

xdx+

∫ 0

−∞

v(x)

xdx

= limε→0+

log(ε)(v(ε)− v(−ε)) +

∫ ∞−∞

v(x)

xdx

= limε→0+

log(ε)(v(ε)− v(−ε)) + pv

⟨1

x, v

⟩=

⟨pv

1

x, v

⟩since

limε→0+

log(ε)(v(ε)− v(−ε)) = limε→0+

1ε

−(v′(ε)+v′(−ε))(v(ε)−v(−ε))2

= − limε→0+

(v(ε)− v(−ε))2

ε(v′(ε) + v′(−ε))

= limε→0+

2(v′(ε) + v′(−ε))(v(ε)− v(−ε))(v′(ε) + v′(−ε)) + ε(v′′(ε)− v′′(−ε))

= 0.

Therefore, ddx log |x| = pv 1

x as a distribution.


Assume that there exists one f ∈ L1loc(Rn) such that

〈δx0, v〉 = 〈f, v〉 ∀v ∈ D.

Let

vε(x) =

exp

(− ‖x−x0‖ε−‖x−x0‖

)x ∈ Bε(x0)

0 Otherwise

where ε > 0. vε is clearly in D. Then observe that

0 < 1 = |vε(x0)| =

∣∣∣∣∣∫Bε(x0)

f(x)vε(x)dx

∣∣∣∣∣ ≤∫Bε(x0)

|f(x)||vε(x)|dx ≤ |vε(x0)|∫Bε(x0)

|f(x)|dx ≤ |vε(x)|Aεnα(n)

where A is the essential sup over B1(x0) and α(n) is a volume of n−dimensional unit ball. Then it is clear that we can make εsmall enough so that Aεnα(n) < 1

2 . Then we get

1 = |v(x0)| < 1

2|v(x0)| = 1

2

which is a contradiction. Therefore, there does not exists such f ∈ L1loc(Rn).


Due date : November 5th, 2019 Byeongho Ban

3.2.1(a) Observe that, for a given F ∈ C2(R), in order for the formula to be a solution to utt = c2∆u, letting x = (x1, x2, x3),

F ′′(α · x− t) = ∂ttF (α · x− t) = ∂ttu(x, t) = c2∆u(x, t) = c2∆F (α · x− t) = c2(α21 + α2

2 + α23)F ′′(α · x− t).

If F ′′ is nonzero, then we have |α|2 = c−2. Thus, the condition for α would be |α| = 1/c.

(b) Observe that, by the condition found from (a),

c2α · ∇g(x) = c2α · ∇F (α · x) = c2(α21 + α2

2 + α23)F ′(α · x) = F ′(α · x) = −∂tF (α · x− t)|t=0 = ut(x, 0) = −h(x).

Therefore, the relation is h = −c2α · ∇g.

(c) By the Kirchhoff’s formula and the results found from (a) and (b), observe that,

u(x, t) =1

4π∂t

(t

∫|ξ|=1

g(x+ ctξ)dSξ

)+

t

4π

∫|ξ|=1

h(x+ ctξ)dSξ

=1

4π∂t

(t

∫|ξ|=1

[(x1 + ctξ1)− (x2 + ctξ2) + 1] dSξ

)− c2t

4π

∫|ξ|=1

[α1 − α2]dSξ

=1

4π∂t

(t

[4π(x1 − x2 + 1) + ct

(∫|ξ|=1

[ξ1 − ξ2]dSξ

)])− c2t(α1 − α2)

=1

4π

[4π(x1 − x2 + 1) + ct

(∫|ξ|=1

[ξ1 − ξ2]dSξ

)]+

1

4π

[ct

(∫|ξ|=1

[ξ1 − ξ2]dSξ

)]− c2t(α1 − α2)

= (x1 − x2 + 1)− c2t(α1 − α2)

because∫|ξ|=1

[ξ1 − ξ2]dSξ =

∫ π

0

(∫ 2π

0

(sin θ cosφ− sin θ sinφ)dφ

)sin θdθ =

∫ π

0

(∫ 2π

0

(cosφ− sinφ)dφ

)sin2 θdθ =

∫ π

0

0 · sin2 θdθ = 0.

Therefore, our solution would be u(x1, x2, x3, t) = (x1 − x2 + 1)− c2t(α1 − α2).

3.2.2(a) Noting that c = 1, observe that, by using (37),

u(x, y, z, t) =1

4π∂t

(t

∫|ξ|=1

g(x+ tξ)dSξ

)+

t

4π

∫|ξ|=1

h(x+ tξ)dSξ

=1

4π∂t

(t

∫|ξ|=1

[(x+ tξ1)2 + (y + tξ2)2]dSξ

)+

t

4π

∫|ξ|=1

0dSξ

=1

4π∂t

(t

∫|ξ|=1

[(x+ tξ1)2 + (y + tξ2)2]dSξ

)

=1

4π∂t

[t

(4π(x2 + y2) + t2

∫|ξ|=1

(ξ21 + ξ2

2)dSξ

)]∵∫|ξ|=1

ξidSξ = 0, i = 1, 2 as 3.2.1(c)

=1

4π∂t

[t

(4π(x2 + y2) + t2

∫ π

0

∫ 2π

0

(sin2 θ cos2 φ+ sin2 θ sin2 φ) sin θdφdθ

)]=

1

4π∂t

[t

(4π(x2 + y2) + 2πt2

∫ π

0

sin3 θdθ

)]=

1

4π∂t

[t

(4π(x2 + y2) + 2πt2

4

3

)]∵∫ π

0

sin3 θdθ =4

3

= ∂t

[t(x2 + y2) +

2

3t3]

= (x2 + y2) + 2t2.

It clearly satisfies our PDE since

∂tt((x2 + y2) + 2t2) = 4 = 2 + 2 + 0 = ∂xx[(x2 + y2) + 2t2] + ∂yy[(x2 + y2) + 2t2] + ∂zz[(x

2 + y2) + 2t2]

and u(x, y, z, 0) = x2 + y2 and ut(x, y, z, t) = 4t so ut(x, y, z, 0) = 0.


3.2.2 (b)Noting that c = 1, observe that, using (39),

u(x, y, t) =1

4π∂t

[2t

∫ξ21+ξ22<1

g(x+ ctξ1, y + ctξ2)√1− ξ2

1 − ξ22

dξ1dξ2

]+

t

4π

[2

∫ξ21+ξ22<1

h(x+ ctξ1, y + ctξ2)√1− ξ2

1 − ξ22

dξ1dξ2

]

=1

4π∂t

[2t

∫ξ21+ξ22<1

(x+ ctξ1)2 + (y + ctξ2)2√1− ξ2

1 − ξ22

dξ1dξ2

]∵ h(x, y, z) = ut(x, y, z, 0) = 0.

=1

4π∂t

[2t

∫ 1

0

∫ 2π

0

x2 + y2 + 2ctxr cos θ + 2ctyr sin θ + c2t2r2

√1− r2

rdθdr

]=

1

4π∂t

[4πt

∫ 1

0

(x2 + y2)r + c2t2r3

√1− r2

dr

]∵∫ 2π

0

sin θdθ = 0 =

∫ 2π

0

cos θdθ

= ∂t

[t

((x2 + y2) + c2t2

∫ 1

0

r3

√1− r2

dr

)]∵∫

r√1− x2

dx = −√

1− r2 + C

= ∂t

[t

((x2 + y2) +

2

3c2t2

)]∵∫

r3

√1− r3

dr = −1

3

√1− r2(r2 + 2) + C

= (x2 + y2) + 2c2t2

= (x2 + y2) + 2t2

which is same with (a).

3.2.2’ (a)Noting that c = 1, observe that, by using (37),

u(x, y, z, t) =1

4π∂t

(t

∫|ξ|=1

g(x+ tξ)dSξ

)+

t

4π

∫|ξ|=1

h(x+ tξ)dSξ

=1

4π∂t

(t

∫|ξ|=1

0dSξ

)+

t

4π

∫|ξ|=1

(y + tξ2)dSξ

=t

4π

∫|ξ|=1

(y + tξ2)dSξ

= ty +t2

4π

∫ 1

0

∫ π

0

∫ 2π

0

sin θ sinφr2 sin θdφdθdr

= ty ∵∫ 2π

0

sinφdφ = 0.

which is clearly satisfying our PDE and initial condition.

3.2.2’ (b)Noting that c = 1, observe that, using (39),

u(x, y, t) =1

4π∂t

[2t

∫ξ21+ξ22<1

g(x+ ctξ1, y + ctξ2)√1− ξ2

1 − ξ22

dξ1dξ2

]+

t

4π

[2

∫ξ21+ξ22<1

h(x+ ctξ1, y + ctξ2)√1− ξ2

1 − ξ22

dξ1dξ2

]

=1

4π∂t

[2t

∫ξ21+ξ22<1

0√1− ξ2

1 − ξ22

dξ1dξ2

]+

t

4π

[2

∫ξ21+ξ22<1

y + ctξ2√1− ξ2

1 − ξ22

dξ1dξ2

]

=t

4π

[2

∫ξ21+ξ22<1

y + ctξ2√1− ξ2

1 − ξ22

dξ1dξ2

]

=t

4π

[2

∫ 1

0

∫ 2π

0

y + ctr sin θ√1− r2

rdθdr

]=

t

4π

[2

∫ 1

0

∫ 2π

0

yr√1− r2

dθdr

]∵∫ 2π

0

sin θdθ = 0

= t

∫ 1

0

yr√1− r2

dr

= ty ∵∫

r√1− x2

dx = −√

1− r2 + C

which is same with (a).


3.2.3

Our problem is utt − c2∆u = f(x, t) , (x, t) ∈ R3 × (0,∞)

u(x, 0) = 0 = ut(x, 0).

The solution would be the sum of homogeneous part, uh, and particular part. Observe that, by Kirchhoff, the homogeneous partwould be

uh(x, t) =1

4π∂t

(t

∫|ξ|=1

g(x+ tξ)dSξ

)+

t

4π

∫|ξ|=1

h(x+ tξ)dSξ =1

4π∂t

(t

∫|ξ|=1

0dSξ

)+

t

4π

∫|ξ|=1

0dSξ = 0.

Thus, we are only interested in particular part up which is also a Duhamel part. Let’s consider the following problem.vtt − c2∆v = 0 , (x, t) ∈ R3 × (0,∞)

v(x, 0, s) = 0 vt(x, 0, s) = f(x, s).

where s > 0. Then by Duhamel principle,

u(x, t) =

∫ t

0

v(x, t− s, s)ds

Note that, by Kirchhoff,

v(x, t, s) =t

4π

∫|ξ|=1

f(x+ ctξ, s)dSξ

Then, combining the results, we have

u(x, t) =

∫ t

0

v(x, t, s)ds =

∫ t

0

t− s4π

∫|ξ|=1

f(x+ c(t− s)ξ, s)dSξds.

Provided that f(x, t) is C1 in x and C0 in t, v(x, t, s) is C2 in x and t and C0 in s. Then u ∈ C2.

3.2.5Our problem is

vtt − c2∆v = −m2v, (x, y, t) ∈ R2 × (0,∞)

v(x, y, 0) = g(x, y) vt(x, , y0) = h(x, y).

Now, let u(x, y, z, t) = cos(mc z)v(x, y, t). Then observe that

utt(x, y, z, t)− c2∆u(x, y, z, t) = cos(mcz)vtt(x, y, t)

− c2[cos(mcz)vxx + cos

(mcz)vyy + cos

(mcz)vzz −

m2

c2cos(mcz)v − 2

m

csin(mcz)vz

]= cos

(mcz) [vtt − c2∆v +m2v

]= 0

since vz = 0 and v satisfies the Klein-Gordon equation. Therefore, u satisfies the three dimensional homogeneous wave equation.Also, observe that

u(x, y, z, 0) = cos(mcz)v(x, y, 0) = cos

(mcz)g(x, y)

ut(x, y, z, 0) = cos(mcz)vt(x, y, 0) = cos

(mcz)h(x, y)

Thus, by using Kirchhoff, we get

u(x, y, z, t) =1

4π∂t

(t

∫|ξ|=1

cos(mc

(z + ctξ3))g(x+ ctξ1, y + ctξ2)dSξ

)+

t

4π

∫|ξ|=1

cos(mc

(z + ctξ3))h(x+ ctξ1, y + ctξ3)dSξ

Now, observe that v(x, y, t) = cos (0) v(x, y, t) = u(x, y, 0, t). Therefore, we can note that, by using method of descent,

v(x, y, t) = u(x, y, 0, t) =1

4π∂t

(t

∫|ξ|=1

cos (mtξ3) g(x+ ctξ1, y + ctξ2)dSξ

)+

t

4π

∫|ξ|=1

cos (mtξ3)h(x+ ctξ1, y + ctξ3)dSξ

=1

4π∂t

(2t

∫ξ21+ξ

22<1

cos(mt√

1− ξ21 − ξ22)g(x+ ctξ1, y + ctξ2)√1− ξ21 − ξ22

dξ1dξ2

)+

t

4π

(2

∫ξ21+ξ

22<1

cos(mt√

1− ξ21 − ξ22)h(x+ ctξ1, y + ctξ2)√1− ξ21 − ξ22

dξ1dξ2

).

It is the formula we have found.


3.2.6 (a)

Note that there are R1, R2 > 0 such that supp(g) ⊂ B(0, R1) and supp(h) ⊂ B(0, R2). Since g and h are continuous in compactregion, it is also bounded. Thus, there are M1,M2 > 0 such that |g| ≤ M1 and |h| ≤ M2. Also, note that ∇g is also continuousand have compact support. It is clear that supp(∇g) ⊂ supp(g) and there is M3 > 0 such that |∇g| ≤M3.Then, observe that, by Kirchhoff and the change of variables,

|u(x, y, z, t)| =

∣∣∣∣∣ 1

4π∂t

(t

∫|ξ|=1

g(x+ ctξ)dSξ

)+

t

4π(ct)2

∫∂B(x,ct)

h(ξ)dSξ

∣∣∣∣∣≤

∣∣∣∣∣ 1

4π∂t

(t

∫|ξ|=1

g(x+ ctξ)dSξ

)∣∣∣∣∣+

∣∣∣∣∣ 1

4πc2t

∫∂B(x,ct)

h(ξ)dSξ

∣∣∣∣∣≤ 1

4π

∣∣∣∣∣(∫|ξ|=1

g(x+ ctξ)dSξ +

∫|ξ|=1

tcξ · ∇g(x+ ctξ)dSξ

)∣∣∣∣∣+1

4πc2t

∫∂B(x,ct)

|h(ξ)|dSξ

≤ 1

4π(ct)2

∫∂B(x,ct)

|g(ξ)|dSξ +1

4π(ct)2

∫∂B(x,ct)

|ξ − x||∇g(ξ)|dSξ +1

4πc2t

∫∂B(x,ct)

|h(ξ)| dSξ

∵ |ξ − x| = ct and Cauchy Swartz inequality

=1

4π(ct)2

∫∂B(x,ct)∩B(0,R1)

|g(ξ)|dSξ +1

4πct

∫∂B(x,ct)∩B(0,R1)

|∇g(ξ)|dSξ +1

4πc2t

∫∂B(x,ct)∩B(0,R2)

|h(ξ)| dSξ

≤ M1

4πc2t2|∂B(x, ct) ∩B(0, R1)|+ M3

4πct|∂B(x, ct) ∩B(0, R1)|+ M2

4πc2t|∂B(x, ct) ∩B(0, R2)|

Now, let R ≥ max(1, R1, R2) and M = max(M1,M2,M3). Now, we have two situations. One is when t ≥ 1. In this case, we have1t2 ≤

1t . Thus, we have bound

|u(x, y, z, t)| ≤ M1

4πc2t2|∂B(x, ct) ∩B(0, R1)|+ M3

4πct|∂B(x, ct) ∩B(0, R1)|+ M2

4πc2t|∂B(x, ct) ∩B(0, R2)|

≤ M

4πc2t2|∂B(x, ct) ∩B(0, R)|+ M

4πct|∂B(x, ct) ∩B(0, )|+ M

4πc2t|∂B(x, ct) ∩B(0, R)|

≤ M

4πc2t2|∂B(0, R)|+ M

4πct|∂B(0, R)|+ M

4πc2t|∂B(0, R)|

≤ M

4πc2t4πR2 +

M

4πct4πR2 +

M

4πc2t4πR2

=MR2(2 + c)/c2

t.

It is clear that |∂B(x, ct)∩B(0, R)| ≤ |∂B(0, R)| since ∂B(x, ct)∩B(0, R) is a 2-dimensional subset of B(0, R). If ct ≤ R, it is clearsince 4π(ct)2 ≤ 4πR2. If ct > R, then |∂B(x, ct)∩B(0, R)| has maximum when ∂B(x, ct) passes the latitude of B(0, R). Also, notethat, as ct gets bigger the intersection gets planner so the the intersection area gets smaller. It implies that |∂B(x, ct) ∩B(0, R)|gets maximum when the sphere passes the latitude of the ball and it gets another maximum when ct is small as much as it canwhich is R. And note that, when the sphere passes the latitude and has radius R, ∂B(x, ct) = ∂B(0, R), thus, we get

|∂B(x, ct) ∩B(0, R)| ≤ |∂B(0, R)| = 4πR2.

So we are done for one case.

On the other hands, if 0 < t < 1, we have 4π(ct)2 ≤ 4πc2tR since R ≥ 1. Thus, observe that

|u(x, y, z, t)| ≤ M1

4πc2t2|∂B(x, ct) ∩B(0, R1)|+ M3

4πct|∂B(x, ct) ∩B(0, R1)|+ M2

4πc2t|∂B(x, ct) ∩B(0, R2)|

≤ M

4πc2t2|∂B(x, ct)|+ M

4πct|∂B(x, ct)|+ M

4πc2t|∂B(x, ct)|

≤ M

4πc2t24π(ct)2 +

M

4πct4π(ct)2 +

M

4πc2t4π(ct)2

≤ M

4πc2t24πc2tR+

M

4πct4πc2R2 +

M

4πc2t4πc2R2

≤ MR(1 + cR+R)

t.


3.3.1

Note

EΩ(t) =1

2

∫Ω

(u2t + c2|∇u|2)dx.

Then observe that, if either u(x, t) = 0 or dudν (x, t) = 0 in ∂Ω,

d

dtEΩ(t) =

∫Ω

(ututt + c2∇ut · ∇u)dx

=

∫Ω

ututtdx+

∫Ω

c2∇ut · ∇udx

=

∫Ω

ututtdx+ c2(∫

∂Ω

ut∇u · νdS −∫

Ω

ut∆udx

)∵ Integration by parts

==

∫Ω

ututtdx+ c2(∫

∂Ω

ut∂u

∂νdS −

∫Ω

ut∆udx

)=

∫Ω

ut(utt − c2∆u)dx Either u(x, t) = 0 ordu

dν(x, t) = 0 in ∂Ω

=

∫Ω

ut · (0)dx ∵ u satisfies the PDE

= 0.

(Note u(x, t) = 0 in ∂Ω implies ut(x, t) = 0 in ∂Ω.)Therefore, EΩ is constant in time t.

3.3.2

Note that our problem is utt −∆u = f(x, t) , x ∈ Ω ⊆ Rn, t > 0.

In addition, suppose we have either u ≡ g or dudν ≡ h in ∂Ω.

Now, assume there are two solutions u1 and u2 for the problem. Then, defining w = u1 − u2, observe thatwtt −∆w = 0 , x ∈ Ω ⊆ Rn, t > 0

and w ≡ 0 or dwdν ≡ 0 in ∂Ω. Then, by previous problem 3.3.1, we know EΩ is constant in time. Thus, EΩ(t) = EΩ(0) for all t.

Then observe that

EΩ(0) =1

2

∫Ω

(w2t (x, 0) + c2|∇w|2(x, 0))dx =

1

2

∫Ω

(0 + 0)dx = 0.

Therefore, EΩ(t) = 0 ∀t. Then, since the integrands of EΩ is positive, we conclude that wt ≡ 0 and ∇w ≡ 0 in Ω which means thatall partial derivative of w is zero. Therefore, w ≡ const. Furthermore, note that, if either w ≡ 0 or dw

dν ≡ 0 in ∂Ω, then w(x, t) = 0for all x ∈ ∂Ω and t. Therefore, since w is constant over Ω, w(x, t) = 0 in Ω which implies u1 ≡ u2 in Ω so the solution is unique.

3.3.4(a) Let’s define our energy as E(t) = 1

2

∫Ω

[u2t + c2|∇u|2 + q(x)u2]dx where Ω is the domain of our interest. Now, observe that

d

dtE(t) =

∫Ω

[ututt + c2∇ut · ∇u+ q(x)utu]

Assuming that u vanishes in ∂Ω, since then ut vanishes in the boundary, by the integraion by parts,∫Ω

[ututt + c2∇ut · ∇u+ q(x)utu]dx =

∫Ω

[ututt + q(x)utu]dx+ c2∫

Ω

∇ut · ∇udx

=

∫Ω

[utttt + q(x)uuu]dx+ c2(∫

∂Ω

ut∇u · νdσ(x)−∫

Ω

ut∆udx

)=

∫Ω

[ututt + q(x)utu]dx− c2∫

Ω

ut∆udx ∵ ut = 0 in ∂Ω

=

∫Ω

ut[utt − c2∆u+ q(x)u]dx

= 0 ∵ utt = c2∆u− q(x)u.

Thus, E(t) = const and our new energy formula gives the global conservation of energy. It implies the energy formula is appropriate.


3.3.4 (b)(b) We want to show that, for given x0 ∈ Rn and t0 > 0,

Ex0,t0(τ) ≤ Ex0,t0(0) for 0 ≤ τ ≤ t0where

Ex0,t0 =1

2

∫Bτ

(u2t + c2|∇u|2 + qu2)

∣∣∣∣t=τ

dx

where

Bτ = x ∈ Rn : |x− x0| < c(t0 − τ).We set

Ωτ = (x, t) : |x− x0| < c(t0 − t), 0 < t < τCτ = (x, t) : |x− x0| = c(t0 − t), 0 < t < τB0 = x ∈ Rn : |x− x0| ≤ ct0

Note that ∂Ωτ = Cτ ∪(B0×0)∪(Bτ ×τ) where the union is disjoint. And note that the normal vector ν would be (0, . . . , 0, 1)on Bτ × τ, (0, . . . , 0,−1) on ]B0 × 0. And note that, on Cτ , we have the relation

c2(ν21 + · · ·+ ν2

n) = ν2n+1

and so

ν21 + · · ·+ ν2

n =ν2n+1

c2=

1

1 + c2.

Now, let us define a vector field−→V = (2c2utux1

, . . . , 2c2utuxn ,−(c2|∇u|2 + u2t + qu2))

Now, observe that

div−→V = 2c2(utx1

ux1+ utux1x1

+ · · ·+ utxnuxn + utuxnxn)− 2c2(utx1ux1

+ · · ·+ utxnuxn)− 2ututt − 2qutu

= 2ut(c2∆u− qu− utt) = 0.

Therefore, by the divergence theorem, we have ∫∂Ωτ

−→V · νdS =

∫Ωτ

div−→V dV = 0.

Actually, we have following inequality on Cτ ,

2ut(ux1ν1 + · · ·+ uxnνn) ≤ 2ut|∇u|

√ν2

1 + · · ·+ ν2n ∵ Cauchy Swartz inequality

=2utc|∇u|c√

1 + c2

≤ u2t + c2|∇u|2

c√

1 + c2∵ (ut − |∇u|)2 ≥ 0

Thus, on Cτ , we have−→V · ν = 2c2ut(ux1ν1 + · · ·+ uxnνn)− (c2|∇u|2 + u2

t + qu2)νn+1

≤ c(u2t + c2|∇u|2)√

1 + c2− (c2|∇u|2 + u2

t + qu2)c√

1 + c2

= − qu2c√1 + c2

≤ 0 ∵ q is nonnegative function and c > 0.

Therefore,

0 ≤∫B0×0

−→V · νdS +

∫Bτ×τ

−→V · νdS =

∫B0

(c2|∇u|2 + u2t + qu2)

∣∣t=0

dS −∫Bτ

(c2|∇u|2 + u2t + qu2)

∣∣t=τ

dS

and so

Ex0,t0(τ) =

∫Bτ

(c2|∇u|2 + u2t + qu2)

∣∣t=τ

dS ≤∫B0

(c2|∇u|2 + u2t + qu2)

∣∣t=0

dS = Ex0,t0(0).

Thus, we prove the local every inequality.


3.3.4 (c)

Let’s formally state our problem as utt = ∆u− q(x)u x ∈ Ω ⊂ Rn, t > 0

u(x, 0) = g(x) ut(x, 0) = h(x).

Now, suppose that there are two different solutions u and v of the Cauchy problem above. Then, letting w = u− v, observe that

wtt = utt − vtt = ∆u− qu−∆v + qv = ∆(u− v)− q(u− v) = ∆w − qwwhich implies that w satisfies

wtt = ∆w − q(x)w x ∈ Ω, t > 0

w(x, 0) = 0 wt(x, 0) = 0.

Then observe that, the energy of w,

Ew(0) =1

2

∫Ω

[w2t (x, 0) + c2|∇w(x, 0)|2 + q(x)w2(x, 0)]dx =

1

2

∫Ω

0dx = 0.

Therefore, by the consequence of (a), we have Ew(t) = 0 for all t which means

Ew(t) =1

2

∫Ω

[w2t + c2|∇w|2 + q(x)w2]dx = 0 ∀t > 0.

Note that the integrand is positive because q is non-negative function. Thus, w2t + c2|∇w|2 + q(x)w2 = 0 ∀t > 0 ∀x ∈ Ω which

means wt = 0, |∇w| = 0 and w = 0 for all t > 0 and x ∈ Ω. Therefore, our solution is u − c ≡ w ≡ 0 so u ≡ v. Therefore, thesolutions is uniquely determined.

3.3.5

We have initial value problem utt − c2∆u+ αut = 0

u(x, 0) = g(x) ut(x, 0) = h(x),

where g and h have compact support and α ≥ 0 is a constant. And we define the energy as

E(t) =1

2

∫Rn

(u2t + c2|∇u|2)dx.

(a) We need to repeat the proof of Theorem 2 in page 92. Following all defined notation in the proof, note that the only place

where the PDE utt − c2∆u = 0 was used is when we prove div−→V = 0. Now, with different PDE, we have

div−→V = 2c2(utx1ux1 + utux1x1 + · · ·+ utxnuxn + utuxnxn)− 2c2(utx1ux1 + · · ·+ utxnuxn)− 2ututt

= 2ut(c2∆u− utt) = 2ut(αut) = 2u2

tα ≥ 0

since α ≥ 0. Then this gives us that, by the divergence theorem,∫∂Ωτ

−→V · νdS =

∫Ωτ

div−→V dV ≥ 0.

Since we have ∫Cτ

−→V · νdS ≤ 0,

we should have the inequality

0 ≤∫B0×0

−→V · νdS +

∫Bτ×τ

−→V · νdS =

∫B0

(c2|∇u|2 + u2t )∣∣t=0

dx−∫Bτ

(c2|∇u|2 + u2t )∣∣t=τ

dx

and so

Ex0,t0(τ) =

∫Bτ

(c2|∇u|2 + u2t )∣∣t=τ

dx ≤∫B0

(c2|∇u|2 + u2t )∣∣t=0

dx = Ex0,t0(0).

Note that

Ex0,t0(τ) ≤ Ex0,t0(0) =

∫B0

(c2|∇u|2 + u2t )∣∣t=0

dx =

∫B0

(c2|∇g|2 + h2)dx = 0

since g and h are zero in B0. Therefore, we conclude that ut(x, t) = 0 and ∇u(x, t) = 0 in Ωt0 so u is constant in Ωt0 . In addition,since u(x, t) = 0 in B0 by Kirchhoff, u ≡ 0 in Ωt0 so u(x0, t0) = 0. It proves the domain of dependence. And it also implies thefinite propagation speed since the domain of dependence result says the impact of the initial condition cannot exceed the speed c.


3.3.5 (b)

Now, observe that, by the integration by parts,

d

dtE(t) =

d

dt

1

2

∫Rn

(u2t + c2|∇u|2)dx =

∫Rn

(ututt + c2∇u · ∇ut)dx

=

∫Rnututtdx+ c2

[−∫Rnut∆u

]dx

=

∫Rnut(utt − c2∆u)dx

=

∫Rnut(−αut)dx

= −∫Rnαu2

tdx ≤ 0

since α ≥ 0. Note that, in the computation of integration by parts, the boundary term is disappeared since g and h havecompact support and the propagation of impact of g and h has finite speed, u gets zero at infinity. Therefore, E(t) is nonincreasing.

3.3.5. (c)

Assume that there are two solutions u and v to the given IVP. Then, letting w = u− v, observe that

wtt − c2∆w + αwt = (utt − c2∆u+ αut)− (vtt − c2∆v + αvt) = 0

and that w(x, 0) = u(x, 0)− v(x, 0) = g(x)− g(x) = 0 and wt(x, 0) = ut(x, 0)− vt(x, 0) = h(x)− h(x) = 0. Now, observe that, forany t > 0 by (b),

0 ≤ E(t) ≤ E(0) =1

2

∫Rn

(wt(x, 0)2 + c2|∇w(x, 0)|2)dx = 0.

It implies that E(t) = 0 for all t > 0. Since the integrand in the energy is positive everywhere, we should have wt = 0 and ∇w = 0everywhere which means that w is constant in Rn. Note that w(x, 0) = 0. Therefore, w ≡ 0 which implies that u ≡ v so thesolution to the IVP is unique.


Additional Problem 1

(a) Observe that, by the fundamental theorem of calculus and change of variables,

d

dr

∫Br(x)

f(y)dy =d

dr

(∫ r

0

∫Sn−1

f(x+ ρz)dσ(z)ρn−1dρ

)=

∫Sn−1

f(x+ rz)dσ(z)rn−1 =

∫∂Br(x)

f(y)dσ(y).

(b)Observe that

φ(r + h)− φ(r)

h=

∫Br+h(x)

f(r + h, y)− f(r, y)

hdy +

1

h

[∫Br+h(x)

f(r, y)dy −∫Br(x)

f(r, y)dy

].

Firstly, observe that, by part (a),

limh→0

1

h

[∫Br+h(x)


f(r, y)dy

]=

d

dr

∫Br(x)

f(r, y)dy =

∫∂Br(x)

f(r, y)dσ.

Secondly, let hn be a given sequence converging to 0. Then the sequence is bounded. Let |hn| ≤ H for all n. Then let

Fn(r, y) =f(r + hn, y)− f(r, y)

hnχBr+hn (x)

Clearly, Fn(r, y)→ ∂∂rf(r, y)χBr(x) as n→∞. And, also, note that, by mean value theorem, for any n, there is cn ∈ [0, hn] such

that

f(r + hn, y)− f(r, y)

hn= ∂rf(r + cn, y).

Now, observe that, for all n,

f(r + hn, y)− f(r, y)

hnχBr+hn (x) ≤

f(r + hn, y)− f(r, y)

hnχBr+H(x) = ∂rf(r + cn, y)χBr+H(x) ≤ χBr+H(x) sup

t∈[r−H,r+H]

|∂rf(t, y)|

Since |∂rf | is continuous, over the compact set, [r −H, r + H] × Br+H(x), it has a maximum value, let’s say L. Then we have,for any n,

|Fn(r, y)| ≤ LχBr+H(x)(y).

Note that the latter function is integrable. Therefore, by the dominated convergence theorem, we have

limh→0

∫Br+h(x)

f(r + h, y)− f(r, y)

hdy = lim

n→∞

∫RnχBr+hn (x)

f(r + hn, y)− f(r, y)

hndy =

∫Rn∂rf(r, y)χBr(x)dy =

∫Br(x)

∂rf(r, y)dy.

Therefore, by combining our results,

d

drφ(r) = lim

h→∞

φ(r + h)− φ(r)

h

= limh→0

∫Br+h(x)

f(r + h, y)− f(r, y)

hdy + lim

h→0

1

h

[∫Br+h(x)


f(r, y)dy

]

=

∫Br(x)

∂rf(r, y)dy +

∫∂Br(x)

f(r, y)dσ.

We are done.


Due date : November 12th, 2019 ByeongHo Ban

4.1.1Let Ω = (x, y) ∈ R2 : x2 + y2 = 1 = (r, θ) : 0 ≤ r < 1, 0 ≤ θ < 2π, and use separation of the variables (r, θ) to solve theDirichlet problem

∆u = 0 in Ω

u(1, θ) = g(θ) for 0 ≤ θ < 2π


We consider the situation when u(r, θ) = R(r)Θ(θ). Then observe that

0 = ∆u = ∂2ru+

1

r∂ru+

1

r2∂2θu = R′′Θ +

1

rR′Θ +

1

r2RΘ′′.

Also,note that, by the change of variables r = e−t and letting R(r) = T (t), since ∂t∂r = −et, we get

R′′ +1

rR′ =

∂

∂r

[∂

∂rR

]+

1

r

∂

∂rR = −et ∂

∂t

[−et ∂

∂tT

]+ et(−et) ∂

∂tT

= et∂

∂t

[et∂

∂tT

]− e2t ∂

∂tT = et

[et∂2

∂t2T + et

∂

∂tT

]− e2t ∂

∂tT = e2tT ′′.

Thus, observe that

0 = R′′Θ +1

rR′Θ +

1

r2RΘ′′ =

(e2tT ′′

)Θ + e2tRΘ′′

which implies

T ′′

T= −Θ′′

Θ= λ

for some constant λ since each side only depends on different independent variables. Then we have three different cases whenλ = 0, λ > 0 or λ < 0.Case 1. λ = 0.When λ = 0, we get T ′′ = 0 = Θ′′ which implies that T (t) = at + b and Θ(θ) = cθ + d for some constants a, b, c, d. SinceΘ(0) = Θ(2π), we have d = 2πc+ d which says c = 0. Also, Note R(r) = −a ln r + b which implies a = 0 since u, so R should befinite as r → 0. Thus, in this case, our solution to the equation would be u(r, θ) = R(r)Θ(θ) = bd.Case 2. λ < 0.Let’s notate λ = −n2 for some n > 0. Then we get Θ′′ − n2Θ = 0 and T ′′ + n2T = 0. Then the solution to the two differentequation would be Θ(θ) = a sinh θ + b cosh θ and T (t) = ce−nt + dent so R(r) = crn + dr−n. However, we know that Θ should beperiodic but the Θ we found is not periodic if at least one of a or b is not zero. Thus, Θ ≡ 0 which implies that u(r, θ) = RΘ = 0.It is a solution when g(x) = 0.Case 3. λ > 0.Now, suppose that λ = n2 for some n > 0. Then we get T ′′ = n2T and Θ′′ + n2Θ. Then we get T (t) = ae−nt + bent andΘ(θ) = ce−inθ + deinθ. Especially, R(r) = arn + br−n. Here, we should have b = 0 since the solution should be finite asr → 0. Therefore, we have, for any n, u(r, θ) = rn(Ane

−inθ + Bneinθ). Also, if we focus on the fact that Θ is periodic, we have

An + Bn = Ane−2πin + Bne

2πin and An + Bn = Ane−4πin + Bne

4πin which gives us An(1 − e−2πin) = Bne2πin(1 − e−2πin) and

An(1− e−2πin) = Bne4πin(1− e−4πin) so Bne

2πin = An = Bne4πin and so cos 2πn+ i sin 2πn = e2πin = 1. It implies that n is an

integer and An = Bn.

By superposition principle, by summing up all the solutions, we get

u(r, θ) =

∞∑n=−∞

rn(Ane−inθ +Bne

inθ) =

∞∑n=−∞

rnA′n cosnθ

where the constant cases is counted in when n = 0 and A′n = 2An.Now, by the boundary condition and orthogonality condition we have∫ 2π

0

g(θ) cosnθdθ =

∫ 2π

0

u(1, θ) cosnθdθ =

∞∑m=−∞

∫ 2π

0

A′n cosmθ cosnθdθ = A′nπ.

It gives us the solution

u(r, θ) =

∞∑n=0

rn

π

[∫ 2π

0

g(θ) cosnθdθ

]cosnθ.


4.1.2Let Ω = (0, π)× (0, π), and use separation of the variables to solve the mixed boundary value problem

∆u = 0 in Ω

ux(0, y) = 0 = ux(π, y) 0 < y < π

u(x, 0) = 0, u(x, π) = g(x) 0 < x < π.

Proof. (Byeongho Ban)Here, we consider when u(x, y) = X(x)Y (y). Then we get X ′′Y +XY ′′ = 0 which gives us

X ′′

X= −Y

′′

Y= λ

since first equality means two functions with different independent variables are equal so they should be a constant. Then, wehave the three cases when λ = 0, λ > 0 and λ < 0.Case 1. λ = 0.When λ = 0, then we have X ′′ = 0 = Y ′′ which gives us X(x) = ax + b and Y (y) = cy + d for some constants a, b, c, d.Then by the given condition in the problem, we have u(x, 0) = X(x)Y (0) = (ax + b)d = 0 which says d = 0 or a = b = 0.If a = b = 0, then g(x) = u(x, π) = (ax + b)(cπ + d) = 0. Then the uniqueness of the solution u ≡ 0. If d = 0, thena(cy) = ux(0, y) = 0 = ux(π, y) = a(cy). It again tells us that a = 0 or c = 0. If a = 0, then u(x, y) = bcy which is a solution tothe problem when g(x) = bcπ which is a contradiction if we assume the function g is not like that. If c = 0, then, similarly, u ≡ 0.Case 2. λ < 0.When λ < 0, let λ = −n2 for some n > 0. Then we get X ′′ + n2X = 0 and Y ′′ − n2Y = 0 which gives us X(x) = Aeinx +Be−inx

and Y (y) = Ceny + De−ny for some constants A,B,C and D. Then note that 0 = u(x, 0) = (Aeinx + Be−inx)(C + D). Thenwe have C + D = 0 or A = B = 0. If A = B = 0, then u ≡ 0 which is a solution to the problem when g ≡ 0. If C + D = 0,then u(x, y) = (Aeinx +Be−inx)(2C sinhny) and ux(x, y) = in(Aeinx−Be−inx)C sinhny so ux(0, y) = inC(A−B) sinhny = 0 =ux(π, y) = inC(Aeinπ − Be−inπ) sinhny. It says A − B = 0 or C = 0. If A = B, then u(x, y) = 4iAC cosnx sinhny . If C = 0,then D = 0 since we are assuming that C +D = 0. And it again gives us that u(x, y) = 0.In this case, renaming A = An and C = Cn since each constants depends on n, by the superposition principle, we have

u(x, y) =

∞∑n=−∞

4iAnCn cosnx sinhny =

∞∑n=−∞

A′n cosnx sinhny

where A′n = 4iAnCn. From here, by we have

u(x, π) =

∞∑n=−∞

A′n cosnx sinhnπ =⇒ π sinhnπA′n =

∫ 2π

0

g(x) cosnxdx

by the orthogonality. Thus,

u(x, y) =

∞∑n=−∞

1

π sinhnπ

(∫ 2π

0

g(x) cosnxdx

)cosnx sinhny

Case 3. λ > 0.When λ > 0, let λ = n2 for some n > 0. Then X ′′ − n2X = 0 and Y ′′ + n2Y = 0. So u(x, y) = X(x)Y (y) =(Aenx + Be−nx)(Ceiny + De−iny). Then note that 0 = u(x, 0) = (Aenx + Be−nx)(C + D) = 0 which implies thatC + D = 0 or A = 0 = B. If A = 0 = B, then u(x, y) = 0 which is a solution when g(x) = 0. If C + D = 0, thenu(x, y) = (Aenx +Be−nx)2iC sinny. Note that ux(x, y) = n(Aenx −Be−nx)2iC sinny and so 0 = ux(0, y) = n(A−B)2iC sinny.Then C = 0 or A − B = 0. If C = 0, then D = 0 so u(x, y) = 0 which is a solution if g ≡ 0. If A = B thenu(x, y) = 4iAC coshnx sinny. Also, it satisfies 0 = ux(π, y) = 4iACn sinhnπ sinny. But it is a contradiction since n > 0 andsinhx = 0 if and only if x = 0.

Therefore, our solution would be

u(x, y) =

∞∑n=−∞

1

π sinhnπ

(∫ 2π

0

g(x) cosnxdx

)cosnx sinhny.


4.1.3Prove that the solution of the Robin or third boundary value problem (5) for the Laplace equation is unique when α > 0 is aconstant.


Consider the problem ∆u = 0 in Ω∂u∂ν + αu = β in ∂Ω.

And suppose that u and v are two solutions of the problem. Then, letting w = u− v, observe that ∆w = ∆u−∆v = 0 and

∂w

∂ν+ αw = ∇w · ν + αw = ∇u · ν + αu−∇v · ν − αv = β − β = 0.

Thus, w satisfies ∆w = 0 in Ω∂w∂ν + αw = 0 in ∂Ω.

Now, observe that, by the Green’s identity,∫Ω

|∇w|2dx =

∫Ω

(w∆w + |∇w|2)dx =

∫∂Ω

w∂w

∂ν= −α

∫∂Ω

w2dS

since ∆w = 0 and ∂w∂ν = −αw. Then note that left most hand side is non-negative and right most hand side is non-positive since

α > 0. Therefore, only possibility is when∫∂Ωw2dS = 0 which implies that, since w is continuous, w ≡ 0. It proves that u ≡ v

which tells us that the solution to the Robin problem is unique.


4.1.5Suppose q(x) ≥ 0 for x ∈ Ω and consider solutions u ∈ C2(Ω) ∩C1(Ω) of ∆u− q(x)u = 0 in Ω. Establish uniqueness theorem for(a) the Dirichlet problem, and (b) the Neumann problem.


(a)Consider a Dirichlet problem

∆u− q(x)u = 0 in Ω

u(x) = g(x) in ∂Ω.

for some function g.Suppose that there are two solutions u and v to the problem above. Letting w = u−v, observe that ∆w−q(x)w = (∆u−q(x)u)−(∆v − q(x)v) = 0 in Ω and that w(x) = u(x)− v(x) = g(x)− g(x) = 0 in ∂Ω. Thus, w satisfies

∆w − q(x)w = 0 in Ω

w(x) = 0 in ∂Ω.

Then, by the Green’s identity,

0 =

∫∂Ω

w∂w

∂νdS =

∫Ω

(w∆w + |∇w|2)dx =

∫Ω

(q(x)w2 + |∇w|2)dx

since w = 0 over ∂Ω and ∆w = q(x)w in Ω. Since q(x)w2 + |∇w|2 is non-negative over Ω and continuous, we conclude thatq(x)w2 + |∇w|2 ≡ 0 which means q(x)w2 ≡ 0 and |∇w| ≡ 0. Then from ∇w ≡ 0, we get w is constant. And from q(x)w ≡ 0, wehave q ≡ 0 or w ≡ 0. If w ≡ 0, when u− v ≡ 0 so u ≡ v and we have unique solution. If q ≡ 0, then the given problem is Laplaceequation with zero boundary condition. Since 0 is a solution to the problem and the problem has unique solution, it must bew ≡ 0 so again u ≡ v so we have unique for the given Dirichlet problem.

(b)Consider a Neumann problem

∆u− q(x)u = 0 in Ω∂u∂ν = g(x) in ∂Ω.

for some given function g. Now, assume that there are two solutions u and v to the given problem. Letting w = u − v, observethat ∆w − q(x)w = 0 by argument in (a) and ∂w

∂ν = ∇w · ν = ∇u · ν −∇v · ν = ∂u∂ν −

∂v∂ν = g(x)− g(x) = 0. Thus, w satisfies

∆w − q(x)w = 0 in Ω∂w∂ν = 0 in ∂Ω.

Then, by the Green’s identity,

0 =

∫∂Ω

w∂w

∂νdS =

∫Ω

(w∆w + |∇w|2)dx =

∫Ω

(q(x)w2 + |∇w|2)dx

since ∂w∂ν = 0 over ∂Ω and ∆w = q(x)w in Ω. Since q(x)w2 + |∇w|2 is non-negative over Ω and continuous, we conclude that

q(x)w2 + |∇w|2 ≡ 0 which means q(x)w2 ≡ 0 and |∇w| ≡ 0. Then from ∇w ≡ 0, we get w is constant. And from q(x)w ≡ 0, wehave q ≡ 0 or w ≡ 0. If w ≡ 0, when u− v ≡ 0 so u ≡ v and we have unique solution. If q ≡ 0, then the given problem is Laplaceequation with zero boundary condition. Since 0 is a solution to the problem and the problem has unique solution, it must bew ≡ 0 so again u ≡ v so we have unique for the given Neumann problem.


4.1.6By direct calculation, show that v(x) = |x − x0|2−n is harmonic in Rn \ x0 for n ≥ 3. Do the same for v(x) = log |x − x0| ifn = 2.


Note that

v(x) = |x− x0|2−n =

(n∑i=1

(xi − xi0)2

) 2−n2

Now, observe that

∂

∂xiv(x) =

2− n2

2(xi − xi0)

(n∑i=1

(xi − xi0)2

)−n2and that

∂2

∂(xi)2v(x) =

∂

∂xi

(∂

∂xiv(x)

)=

∂

∂xi

2− n2

2(xi − xi0)

(n∑i=1

(xi − xi0)2

)−n2 = (2− n)

∂

∂xi

(xi − xi0)

(n∑i=1

(xi − xi0)2

)−n2 = (2− n)

( n∑i=1

(xi − xi0)2

)−n2− n(xi − xi0)2

(n∑i=1

(xi − xi0)2

)−n+22

= (2− n)

(n∑i=1

(xi − xi0)2

)−n2 1− n(xi − xi0)2

(n∑i=1

(xi − xi0)2

)−1 .

Then we have

∆v(x) =

n∑k=1

vxkxk(x) = (2− n)

(n∑i=1

(xi − xi0)2

)−n2 n∑k=1

1− n(xk − xk0)2

(n∑i=1

(xi − xi0)2

)−1

= (2− n)

(n∑i=1

(xi − xi0)2

)−n2 n− n n∑k=1

(xk − xk0)2

(n∑i=1

(xi − xi0)2

)−1

= (2− n)

(n∑i=1

(xi − xi0)2

)−n2 n− n( n∑i=1

(xi − xi0)2

)−1( n∑k=1

(xk − xk0)2

)= (2− n)

(n∑i=1

(xi − xi0)2

)−n2[n− n]

= 0.

Therefore, v(x) is harmonic in Rn \ x0 with n ≥ 3.

When n = 2, setting v(x) = log |x− x0|, observe that

∂xiv(x) = ∂xi log√

(x1 − x10)2 + (x2 − x2

0)2 =122(xi − xi0)((x1 − x1

0)2 + (x2 − x20)2)−

12√

(x1 − x10)2 + (x2 − x2

0)2=

xi − xi0(x1 − x1

0)2 + (x2 − x20)2

and that

∂xixi = ∂xi

(xi − xi0

(x1 − x10)2 + (x2 − x2

0)2

)=

[(x1 − x10)2 + (x2 − x2

0)2]− 2(xi − xi0)2

[(x1 − x10)2 + (x2 − x2

0)2]2

Thus,

∆v(x) = ∂x1x1v(x) + ∂x2x2v(x) =2[(x1 − x1

0)2 + (x2 − x20)2]− 2(x1 − x1

0)2 − 2(x2 − x20)2

[(x1 − x10)2 + (x2 − x2

0)2]2 = 0.

Therefore, v(x) is harmonic in R2 \ x0.


4.2.3The symmetry of the Green’s function [i.e, G(x, ξ) = G(ξ, x) for x, ξ ∈ Ω] is an important fact connected with the self-adjointnessof ∆.(a) Verify the symmetry of G(x, ξ) by direct calculation when Ω = Rn+ and Ω = Ba(0).(b) Prove the symmetry of G(x, ξ) when Ω is any smooth, bounded domain.


(a) First, we consider when Ω = Rn+. Observe that

|x− ξ∗| =

√√√√ n∑k=1

(x1 − ξ∗i )2 =

√√√√n−1∑k=1

(x1 − ξi)2 + (xn + ξn)2 =

√√√√n−1∑k=1

(ξ1 − xi)2 + (ξn + xn)2 = |ξ − x∗|.

Then, observe that, when n = 2,

G(x, ξ) = K(x− ξ)−K(x− ξ∗) =1

2πlog |x− ξ| − 1

2πlog |x− ξ∗| = 1

2πlog |ξ − x| − 1

2πlog |ξ − x∗| = G(ξ, x)

and that, when n ≥ 3,

G(x, ξ) = K(x− ξ)−K(x− ξ∗) =1

(2− n)ωn|x− ξ|2−n − 1

(2− n)ωn|x− ξ∗|2−n

=1

(2− n)ωn|ξ − x|2−n − 1

(2− n)ωn|ξ − x∗|2−n

= K(ξ − x)−K(ξ − x∗)= G(ξ, x).

So we are done for when Ω = Rn+.Now, consider when Ω = Ba(0). Note that

G(x, ξ) =

K(x− ξ)− 12π log

(|ξ|a |x− ξ

∗|)

n = 2

K(x− ξ)−(a|ξ|

)n−2

K(x− ξ∗) n ≥ 3.

Recalling that K is radial symmetry, observe that, by the observation above, when n = 2,

G(x, ξ) = K(x− ξ)− 1

2πlog

(|ξ|a|x− ξ∗|

)= K(ξ − x)− 1

2πlog

(|x||ξ||ξ − x∗|

)= G(ξ, x)

since |ξ|x |x− ξ∗| = |x− ξ| = |ξ − x| = |x|

ξ |ξ − x∗| and that, similarly, when n ≥ 3,

G(x, ξ) = K(x− ξ)−(a

|ξ|

)n−2

K(x− ξ∗) = K(ξ − x)−(|ξ||x|

)n−2

K(ξ − x∗) = G(ξ, x)

since K(x− ξ∗) = 1(2−n)ωn

1|x−ξ∗|n−2 and

(a|ξ|

)n−21

|x−ξ∗|n−2 = 1|x−ξ|n−2 = 1

|ξ−x|n−2 =(|ξ||x|

1|ξ−x∗|

)n−2

.

So we are done.


4.2.4(a) Use the weak maximum principle (16) to prove that G(x, ξ) ≤ 0 for x, ξ ∈ Ω with x 6= ξ.(b) Use the strong maximum principle (15) to prove that G(x, ξ) < 0 for x, ξ ∈ Ω with x 6= ξ.


(a)Let x, ξ ∈ Ω with x 6= ξ be given. And note that G(x, ξ) = K(x− ξ) + ωξ(x) where ωξ is a C2 solution to the problem

∆ωξ(y) = 0 y, ξ ∈ Ω

ωξ(y) = −K(y − ξ) y ∈ ∂Ω.

Since ωξ is a harmonic function, by the weak maximum principle,

maxy∈∂Ω

ωξ = maxy∈Ω

ωξ.

Let’s call the maximum value above M . And also note that, K(x− ξ)→ −∞ as x→ ξ. Thus, since Ω is a Hausdorff space, thereis r > 0 such that K(y − ξ) < −M ∀y ∈ Br(ξ) ⊂ Ω and x 6∈ Br(ξ). Then, we have G(y, ξ) = K(y − ξ) + ωξ(y) < −M + M = 0for all y ∈ ∂Br(ξ).Now, observe that G(·, ξ) is a harmonic function over Ωr := Ω \ Br(ξ) and G(y, ξ) ≤ 0 for all y ∈ ∂Ωr = ∂Ω ∪ ∂Br(ξ) so, by theweak maximum principle, we have

G(x, ξ) ≤ maxy∈Ωr

G(y, ξ) = maxy∈∂Ωr

G(y, ξ) ≤ 0

(b)Now, we follow all the settings from (a) and note that G(·, ξ) is harmonic in Ωr and is not constant since G(y, ξ) = 0 for y ∈ ∂Ωand G(y, ξ) < 0 for y ∈ ∂Br(ξ). Therefore, by the strong maximum principle, we have

G(x, ξ) < supz∈Ωr

G(z, ξ)

And note that, since G(z, ξ) ≤ 0 for all z ∈ Ωz by weak maximum principle, we also have supz∈Ωr G(z, ξ) ≤ 0 due to inequalitypreserving property of limit. Therefore, we have G(x, ξ) < supz∈Ωr G(z, ξ) ≤ 0 and we are done.


4.2.5Use (33) to prove (38).


Recall that (33) is

u(ξ) =

∫Ω

G(x, ξ)∆udx+

∫∂Ω

u(x)∂G(x, ξ)

∂νxdSx

and (38) is that ∫Rn−1

H(x′, ξ)dx′ = 1 ∀ξ ∈ Rn+.

Since the Ω in (33) is assumed to be bounded, we cannot replace it with Rn+. Thus, we start from ΩR := BR ∩Rn+. Then, lettingξ ∈ Rn+ be given and, by using as (31), we have

u(ξ) =

∫ΩR

G(x, ξ)∆u+

∫∂ΩR

[u(x)

∂G(x, ξ)

∂νx−G(x, ξ)

∂u(x)

∂ν

]dSx

where G is the Green’s function for the upper half plane. Now, let’s set u ≡ 1, then the first and last term disappear and we get

1 =

∫∂ΩR

∂G(x, ξ)

∂νxdSx =

∫∂ΩR

H(x, ξ)dSx =

∫∂ΩR∩Rn−1

H(x, ξ)dSx +

∫∂ΩR\Ω0

H(x, ξ)dSx

where

H(x, ξ) =2ξnωn|x− ξ|−n.

It is clear that H is integrable over n− 1 dimensional subspace since its power of denominator is n > n− 1. In fact, as explainedin the above of (38) in the book, H(x, ξ) = O(|x|−n) as |x| → ∞ and the decay O(|x|−n) is enough for an integral over Rn−1 toconverge. Therefore, by Dominated convergence theorem, since χ∂ΩR∩Rn−1H(x, ξ) ≤ H(x, ξ), we have

limR→∞

∫Rn−1

χ∂ΩR∩Rn−1H(x, ξ)dx =

∫Rn−1

H(x, ξ)dx.

And, since H(x, ξ)→ 0 as |x| → 0, the last term also converges to 0. Therefore, we have

1 =

∫Rn−1

H(x, ξ)dx.

Then we are done.


4.2.6For n = 2, use the method of reflections to find the Green’s function for the first quadrant Ω = (x, y) : x, y > 0.


Let ξ = (ξ1, ξ2) ∈ Ω be given. and introduce ξ∗ = (−ξ1, ξ2). Then, observe that when x′ ∈ ∂Ω ∩ (x, 0) : x ∈ R, we have|x′ − ξ| = |x′ + ξ∗| and |x′ − ξ∗| = |x′ + ξ|. Then we have

|x′ − ξ||x′ − ξ∗|

=|x′ + ξ∗||x′ + ξ|

=⇒ |x′ − ξ| = |x′ + ξ∗||x′ − ξ∗||x′ + ξ|

.

Also, if y′ ∈ ∂Ω ∩ (0, y) : y ∈ R, we have |y′ − ξ| = |y′ − ξ∗| and |y′ + ξ| = |y′ + ξ∗|. Then we have

|y′ − ξ||y′ + ξ| = |y′ − ξ∗||y′ + ξ∗| =⇒ |y′ − ξ| = |y′ + ξ∗||y′ − ξ∗||y′ + ξ|

.

Then, by setting

G(x, ξ) = K(x− ξ)−K(|x+ ξ∗||x− ξ∗||x+ ξ|

)we get the Greens’ function G since ξ∗,−ξ,−ξ∗ 6∈ Ω and so the second term is harmonic over Ω and G(x′, ξ) = 0 for all x′ ∈ ∂Ω.


4.2.8Let Ω = Ba(0), Ω+ = Ω ∩ Rn+ and Ω0 = x ∈ Ω : xn = 0. If u ∈ C2(Ω) ∩ C(Ω+ ∪ Ω0) [Typo! It should have beenC2(Ω+) ∩ C(Ω+ ∪ Ω0)! Otherwise, we can find a counter example.] is harmonic in Ω+ and u = 0 on Ω0, prove that u maybe extended to a harmonic function on all of Ω. (This is called a reflection principle.)


Let’s define

u(x) =

u(x1, x2, . . . , xn−1, xn) ,if x ∈ Ω+ ∪ Ω0

−u(x1, x2, . . . , xn−1,−xn) ,if x 6∈ Ω+ ∪ Ω0.

Due to the condition that u ≡ 0 over Ω0, u is continuous over Ω. Now, note that, by Poisson formula, we have

v(ξ) =r2 − |ξ|2

aωn

∫|x|=1

u(x)

|x− ξ|ndSx

which is C∞ harmonic function since u(x) is continuous over the ∂Ω. Now, let ξ ∈ Ω0, then observe that∫|x|=1

u(x)

|x− ξ|ndSx =

∫∂Ω∩Ω+

u(x)

|x− ξ|ndSx +

∫∂Ω∩Ω−

−u(x′)

|x− ξ|ndSx =

∫∂Ω∩Ω+

u(x)

|x− ξ|ndSx −

∫∂Ω∩Ω+

u(x)

|x− ξ|ndSx = 0

where x′ = (x1, x2, . . . , xn−1,−xn) and Ω− = Rn− ∩ Ω and since |x − ξ| = |x′ − ξ| when ξ ∈ Ω0. Thus, v = 0 over Ω0. It implies

that v ≡ u over ∂Ω+ so u − v ≡ 0 over ∂Ω+. Then, by the maximum principle, for given ξ ∈ Ω+ (u − v)(ξ) ≤ maxΩ+(u − v) =

max∂Ω+(u− v) = 0 and (v−u)(ξ) ≤ maxΩ+

(v−u) = max∂Ω+(v−u) = 0. Therefore, u ≡ v over Ω+. Therefore, v is the harmonic

extension of u in the upper half ball to the lower half ball.


4.2.10 a) Suppose u ∈ C2(Ω) is harmonic, u ≥ 0, and Ba(0) ⊂ Ω. Use (44) to show

an−2(a− |ξ|)(a+ |ξ|)n−1

u(0) ≤ u(ξ) ≤ an−2(a+ |ξ|)(a− |ξ|n−1)

u(0)


Recall that (44) is

u(ξ) =a2 − |ξ|2

aωn

∫|x|=a

g(x)

|x− ξ|ndSx =

a2 − |ξ|2

aωn

∫|x|=a

u(x)

|x− ξ|ndSx

where g is the Dirichlet data.Firstly, since |x− ξ| ≥ |x| − |ξ| so 1

|x−ξ| ≤1

|x|−|ξ| (since ξ ∈ Ba(0) so |ξ| ≤ a) and since |x| = a in the integral, observe that

u(ξ) =a2 − |ξ|2

aωn

∫|x|=a

u(x)

|x− ξ|ndSx ≤

a2 − |ξ|2

aωn

∫|x|=a

u(x)

(a− |ξ|)ndSx

=a2 − |ξ|2

aωn

1

(a− |ξ|)n

∫|x|=a

u(x)dSx

=(a− |ξ|)(x+ |ξ|)

(a− |ξ|)nan−2

an−1ωn

∫|x|=a

u(x)dSx

=a+ |ξ|

(a− |ξ|)n−1

an−2

|∂Ba(0)|

∫∂Ba(0)

u(x)dSx

=an−2(a+ |ξ|)(a− |ξ|)n−1

u(0)

by Gauss Mean Value Theorem. Similarly, since |x− ξ| ≤ |x|+ |ξ| = a+ |ξ| so 1|x−ξ| ≥

1a+|ξ| , observe that

u(ξ) =a2 − |ξ|2

aωn

∫|x|=a

u(x)

|x− ξ|ndSx ≥

a2 − |ξ|2

aωn

∫|x|=a

u(x)

(a+ |ξ|)ndSx

=a2 − |ξ|2

aωn

1

(a+ |ξ|)n

∫|x|=a

u(x)dSx

=(a− |ξ|)(x+ |ξ|)

(a+ |ξ|)nan−2

an−1ωn

∫|x|=a

u(x)dSx

=a− |ξ|

(a+ |ξ|)n−1

an−2

|∂Ba(0)|

∫∂Ba(0)

u(x)dSx

=an−2(a− |ξ|)(a+ |ξ|)n−1

u(0)

by Gauss Mean Value Theorem. Therefore, we are done.

P.SWe should recognize that we could have made the inequality due to positivity of u. For example, if u was negative somewhere inthe domain so the integral gets negative, then the inequality above should be reversed.


Due date : December 5th, 2019 ByeongHo Ban

4.1.7(a) If Ω is a bounded domain and u ∈ C2(Ω) ∩ C(Ω) satisfies ∆u = 0, then maxΩ |u| = max∂Ω |u|.

(b) If Ω = x ∈ Rn : |x| > 1 and u ∈ C2(Ω) ∩ C(Ω) satisfies ∆u = 0 and lim|x|→∞ u(x) = 0, then maxΩ |u| = max∂Ω |u|.


(a) Let M = maxΩ |u| and then let A = x ∈ Ω : |u|(x) = M Assume that there is no x ∈ ∂Ω such that |u|(x) = M and assumethat there is ξ ∈ Ω such that |u|(ξ) = maxΩ |u|. Equivalently, we are assuming that maxΩ |u| > max∂Ω |u| Then observe that,

there exists ρ such that B(ξ, ρ) ⊂ Ω, so

M = |u(ξ)| =

∣∣∣∣∣ 1

|B(0, ρ)|

∫B(0,ρ)

u(x)dx

∣∣∣∣∣ ≤ 1

|B(0, ρ)|

∫B(0,ρ)

|u|(x)dx ≤M.

Then note that the inequalities above are all equality. Thus, observe that

1

|B(0, ρ)|

∫B(0,ρ)

|u|(x)− |u|(ξ)dx = 0 =⇒ |u|(x) = |u|(ξ) ∀x ∈ B(0, 1)

since |u|(x) − |u|(ξ) ≤ 0 for all x ∈ B(0, ρ). It implies that A is open since B(ξ, ρ) ⊂ A. Since A = |u|−1(M) is closed, A isopen and closed so by the connectedness of Ω, A is ∅ or Ω. If it cannot be ∅ due to our assumption on ξ, so A = Ω which means|u| is constant. But it is also impossible since we assumed there is no x ∈ ∂Ω such that |u|(x) = M and |u| is continuous in Ω soit should be u ≡M over Ω.

(b) Let R > be given. Note that, by the result of (a), letting AR = B(0, R) \B(0, 1), we have

maxAR

|u| = max∂AR|u|.

Now, let m = min∂B(0,1) |u|, then note that there exists L > 0 such that |u|(x) < m for all |x| > L since lim|x|→∞ |u| = 0. Thenwe have maxAR |u| = max∂AR |u| = max∂B(0,1) |u| for all R > L. Therefore, since limit preserves equality, by sending R→∞, we

have max∂Ω |u| ≤ maxΩ |u| = max∂B(0,1) |u| ≤ max∂Ω |u| which implies that

maxΩ|u| = max

∂Ω|u|.


4.1.9Suppose u ∈ C(Ω) satisfies the mean value property in Ω.(a) Show that u satisfies the maximum principle (15).

(b) Show that boundary values on Br(ξ) ⊂ Ω uniquely determine u: If v ∈ C(Br(ξ)) satisfies the mean values property in Br(ξ)and u = v on ∂Br(ξ), then u ≡ v in Br(ξ).


(a)We need to show (15), which is

u(ξ) < supx∈Ω

u(x) ∀ξ ∈ Ω

supposing that u ∈ C(Ω) satisfies the mean value property. We are gonna modify the proof of Theorem 13 in the book. Sowe can assume that A = supx∈Ω u(x) < ∞ because, otherwise, it is just true with finite value u(ξ) for all ξ ∈ Ω. Now, letB = x ∈ Ω : u(x) = A. Since B = u−1A, u is continuous, and A is closed set, B is closed. Now, let ξ ∈ Ω be an interiorpoint of B. Then there is r > 0 such that Br(ξ) ⊂ Ω. Now, by mean value property, we have

u(ξ) =n

ωn

∫|x|≤1

u(ξ + rx)dSx

and it is possible when u(x) = u(ξ) for all x ∈ Br(ξ) which means Br(ξ) ⊆ B so B is an open set. Therefore, B is open andclosed at the same time. Then, by connectedness of Ω, B is ∅ or Ω. Therefore, if B = Ω, u is constant, and if B = ∅, u(ξ) < Afor all ξ ∈ Ω which is (15).

(b)

Suppose that v ∈ C(Br(ξ)) satisfies the mean values property in Br(ξ) and u = v on ∂Br(ξ). Now, by linearity of integration,we note that w = u − v and −w satisfies mean value property. Firstly, assume that w is not constant. Note that, from (a), weconclude that only possibility that w and −w attain their maximum is when it is on the boundary. However, they are 0 at ∂Br(ξ),so w < 0 and −w < 0 at the same time which is a contradiction. Therefore, w is constant. Since w is zero at the boundary, w ≡ 0over Br(ξ) so u ≡ v in Br(ξ).


4.2.7If u ∈ C(Ω) satisfies the mean value property of Section 4.1.d, then u is harmonic in Ω.


Let x ∈ Ω be given and then let r > 0 be such that Br(x) ⊆ Ω. Then, let v ∈ C2 be the solution to the problem∆v = 0 x ∈ Br(x),

v = u x ∈ ∂Br(x).

Then since v is harmonic so it satisfies mean value property, by the result of 4.1.9(a) which is previous homework problem,

u = v over Br(x). Since v ∈ C2, u is twice continuously differentiable at the point x. Since x ∈ Ω was arbitrary, we have u ∈ C2(Ω).

Now, we are left to show u is harmonic. Assume that ∆u 6≡ 0 over Ω. Then, since ∆u is continuous, without loss of generality, wecan find x ∈ Ω and R > 0 such that ∆u > 0 over BR(x) ⊂ Ω. Then define

φ(r) :=1

ωn

∫∂B1(0)

u(x+ ry)dy.

However, note that, by the mean value property of u, we have φ(r) = u(x) which implies that φ(r) is constant in r. Then observethat, by the divergence theorem,

0 = φ′(r) =1

ωn

∫|y|=1

y · ∇u(x+ ry)dSy =1

ωn

∫|y|<1

∆u(x+ ry)dSy =1

|Br(x)|

∫Br(x)

∆u(y)dy > 0.

And it is a contradiction. Therefore, ∆u ≡ 0 over Ω.(If ∆u < 0 over BR(x), then the last inequality above gets reversed.)


4.2.11Use (46) to prove Liouville’s theorem.


Suppose that u is bounded harmonic function over all Rn. Then there is M > 0 such that |u(x)| ≤M for all x ∈ Rn. Let ξ ∈ Rnbe given and let a > 0 be such that ξ ∈ Ba(0). Then, by the Poisson integral formula, we get

u(ξ) =a2 − |ξ|2

aωn

∫|x|=a

g(x)

|x− ξ|ndSx.

Then observe that ∣∣∣∣ ∂u∂ξi (ξ)∣∣∣∣ =

∣∣∣∣∣−2ξiaωn

∫|x|=a

u(x)

|x− ξ|ndSx +

a2 − |ξ|2

aωn

∫|x|=a

n(xi − ξi)|x− ξ|n+2

u(x)dSx

∣∣∣∣∣≤ 1

aωn

[2ξi

∫|x|=a

∣∣∣∣ u(x)

|x− ξ|n

∣∣∣∣ dSx + n(a2 − |ξ|2)

∫|x|=a

∣∣∣∣ (xi − ξi)|x− ξ|n+2

∣∣∣∣ dSx]

≤ M

aωn

[2|ξ|

∫|x|=a

1

(a− |ξ|)ndSx + n(a2 − |ξ|2)

∫|x|=a

a+ |ξ|(a− |ξ|)n+2

dSx

]

=M

a

[2a

1

(a− |ξ|)nan−1 +

n(a2 − |ξ|2)(a+ |ξ|)(a− |ξ|)n+2

an−1

]−→ 0 as a→∞.

Since we can increase a large enough as long as Ba(0) contains ξ and uξi(ξ) is bounded by an expression converging to 0 as a→∞,we conclude that ∇u = 0 which implies that u is constant.


Additional Problem Let Ω ⊂ Rn, bounded open set and g ∈ C1(∂Ω). Find the PDE for u ∈ C2(Ω) such that u is the minimizerover A of

E(v) =

∫Ω

√1 + |∇v|2dx.

where

A := v ∈ C2(Ω) ∩ C1(∂Ω) : v ≡ g on ∂Ω.


Note that, since u is the minimizer of E(v), for any φ ∈ C∞0 and ε > 0, we should have E(u) ≤ E(u+εφ). Then, fixing φ ∈ C∞0 (Ω)where supp(φ) ⊂ Ω, define

G(ε) := E(u+ εφ) =

∫Ω

√1 + |∇(u+ εφ)|2dx.

Then we want G′(0) = 0. Thus, observe that

G′(ε) =

∫Ω

2∇u · ∇φ+ 2ε|∇φ|2

2√

1 + |∇(u+ εφ)|2dx

because |∇(u+ εφ)|2 = |∇u+ ε∇φ|2 = |∇u|2 + 2ε∇u · ∇φ+ ε2|∇φ|2, so, by the integration by parts, we have

0 = G′(0) =

∫Ω

∇u · ∇φ√1 + |∇u|2

dx =

∫∂Ω

∇u · ν√1 + |∇u|2

φdS −∫

Ω

∇ ·

[∇u√

1 + |∇u|2

]φdx.

Note that the first term in the last expression above vanishes since the support of φ is in Ω. Therefore, we get∫Ω

∇ ·

[∇u√

1 + |∇u|2

]φdx = 0 ∀φ ∈ C∞0 (Ω).

This implies that we have

∇ ·

[∇u√

1 + |∇u|2

]= 0

so u should solve this PDE.


Due date : December 10th, 2019 ByeongHo Ban

5.1.2 Describe how you would solve the initial/boundary value problem (3) with the homogeneous Dirichlet condition replacedby u(x, t) = h(x) for x ∈ ∂Ω and t > 0. What happens to u(x, t) as t→∞?


We are interested in the problem ut = ∆u x ∈ Ω, t > 0

u(x, 0) = g(x) x ∈ Ω

u(x, t) = h(x) x ∈ ∂Ω, t > 0.

Suppose that ω is the solution to the problem ∆ω = 0 x ∈ Ω

ω(x) = h(x) x ∈ ∂Ω.

and suppose that v is the solution to the problemvt = ∆v x ∈ Ω, t > 0

v(x, 0) = g(x)− ω(x) x ∈ Ω

v(x, t) = 0 x ∈ ∂Ω, t > 0.

Let u(x, t) = v(x, t) + ω(x). Now observe that

ut = vt +∂

∂tω(x) = vt = ∆v + 0 = ∆v + ∆ω = ∆u

Also, observe that u(x, 0) = v(x, 0) + ω(x) = g(x) − ω(x) + ω(x) = g(x) for x ∈ Ω and that, for x ∈ ∂Ω with t > 0,u(x, t) = v(x, t) + ω(x) = 0 + h(x) = h(x). Therefore, if we solve the two problems above, we can solve the first PDE.

Now, we investigate what happen if t → ∞. Firstly, note that the term ω does not have to do with t. Therefore, it suffices toinvestigate it only for v. Note that,

v(x, t) =

∫Ω

∞∑n=1

e−λntφn(x)φn(y)(g − ω)(y)dy

and it is clear that, as t→∞, v(x, t)→ 0 which implies that u→ ω as t→∞.


5.1.6 Suppose U = Ω× (0, T ) and u ∈ C2,1(U)∩C(U) satisfies ut ≤ ∆u+ cu in U , where c ≤ 0 is a constant. If u ≥ 0, show that(11) holds for u. Give a counterexample without the condition u ≥ 0.


Let c ≤ 0 be given and let v(x, t) = e−ctu(x, t). And observe that, since ut ≤ ∆u+ cu

vt = −ce−ctu+ e−ctut ≤ −ce−ctu+ e−ct(∆u+ cu) = e−ct∆u = ∆(e−ctu) = ∆v.

Therefore, by the weak maximum principle for heat equation, we have

max(x,t)∈U

u(x, t) ≤ max(x,t)∈U

e−ctu(x, t) ∵ u ≥ 0, e−ct ≥ 0

= max(x,t)∈U

v(x, t)

= max(x,t)∈Γ

v(x, t) ∵ weak maximum principle for heat eq

= max(x,t)∈Γ

e−ctu(x, t).

Then, since c ≤ 0 is arbitrary, we have

max(x,t)∈U

u(x, t) ≤ max(x,t)∈Γ

u(x, t).

Since Γ ⊂ U , we have

max(x,t)∈U

u(x, t) = max(x,t)∈Γ

u(x, t)

which says that (11) holds.QuestionI cannot find good counterexample... What I tried so fart does not provide the bad behavior.

Now, assume that we don’t have the condition u ≥ 0. Then let Ω = (−π/2, π/2) and u(x, t) = t cosx with c = −1. Then observethat

ut = cosx ≤ t sinx+ t sinx = ∆u− u = ∆u+ cu

because sinx ≥ 0 over [0, π] and t ≥ 0.Now, note that

ut(x, t) = − sinx = 0 ⇐⇒ x = 0 or π

∇xu(x, t) =⇐⇒ x =π

2or

3π

26∈ ∂Ω.

Note that u(π/2, t) = −3et and u(3π/2, t) = −et for each t. And, when x ∈ ∂Ω which means x = −π or π, u(x, t) = −2et for eacht. Then, for fixed T > 0,


5.1.7If u satisfies (1), define its heat energy by E(t) =

∫Ωu2(x, t)dx.

(a) If u = Ω× (0,∞) and u ∈ C2,1(U) satisfies (1)and either (i) u = 0 on ∂Ω, or (ii) ∂u∂ν = 0 on ∂Ω, then E(t) is nonincreasing in t.

(b) Use (a) to conclude the uniqueness of a solution u ∈ C2,1(U) for either the nonhomogeneous Dirichlet problem (9) or thecorresponding nonhomogeneous Neumann problem.


(a) Observe that, since ut = k∆u,

dEdt

(t) =

∫Ω

2utudx = 2k

∫Ω

u∆udx = 2k

∫∂Ω

u∂u

∂νdSx − 2k

∫Ω

|∇u|2dx

because of the integration by parts and where ν is a unit normal vector of ∂Ω. If either (i) or (ii) is true, then the first term inthe left most hand side vanish. Therefore,

dEdt

(t) = −2k

∫Ω

|∇u|2dx.

Note that k > 0 and |∇u|2 ≥ 0 which implies that dE/dt ≤ 0 so E is nonincreasing in t.

(b) Suppose that u ∈ C2,1(U) and v ∈ C2,1(U) are either nonhomogeneous Dirichlet problem or corresponding nonhomogeneousNeumann problem. Then observe that, by the linearity of heat equation, w = u− v satisfies

wt = ut − vt = ∆u+ f −∆v − f = ∆w

and w(x, 0) = u(x, 0)− v(x, 0) = g − g = 0 in Ω.If it is for Dirichlet, then w(x, t) = u(x, t) − v(x, t) = h(x, t) − h(x, t) = 0. If it is for corresponding Neumann, then ∂w

∂ν (x, t) =∂u∂ν (x, t) − ∂v

∂ν (x, t) = 0. Therefore, w satisfies either (i) or (ii) in part (a) which means that E(t) =∫

Ωw2dx is nonincreasing

function. However, observe that

0 =

∫Ω

0dx =

∫Ω

w2(x, 0)dx = E(0) ≥ E(t) ∀t > 0.

Since E is nonnegative due to nonnegativity of the integrand, E ≡ 0. It implies that, due to nonnegativity of the integrand, w2 ≡ 0so w ≡ 0. Therefore, u ≡ v so the solution should be unique.


5.2.1Proof of Theorem 1. Let K(x, y, t) be the Gaussian kernel (18).(a) Show that K(x, y, t) is C∞ and satisfies (∂t −∆x)K(x, y, t) = 0 for x, y ∈ Rn and t > 0.(b) Prove that for every x ∈ Rn and t > 0,

∫Rn K(x, y, t)dy = 1.

(c) For any δ > 0, show that

limt→0+

∫|x−y|>δ

K(x, y, t)dy = 0

uniformly for x ∈ Rn.(d) Use (a)-(c) and g ∈ CB(Rn) to show u(x, t)→ g(x) as t→ 0+, proving the theorem.


(a) Recall that

K(x, y, t) =1

(4πt)n/2e−|x−y|2

4t

and it is clear that K is in C∞ when x, y ∈ Rn and t > 0 since all the functions composing K is C∞. Next, observe that

(∂t −∆x)K(x, y, t) = −2πn1

(4πt)n2 +1

e−|x−y|2

4t +1

(4πt)n/2e−|x−y|2

4t

[|x− y|2

4t2

]−∇ ·

1

(4πt)n/2e−|x−y|2

4t

[−2(x− y)

4t

]= −2πn

1

(4πt)n2 +1

e−|x−y|2

4t +1

(4πt)n/2e−|x−y|2

4t

[|x− y|2

4t2

]−

1

(4πt)n/2e−|x−y|2

4t

[−2|x− y|

4t

]2

+1

(4πt)n/2e−|x−y|2

4t

[−2n

4t

]

= K(x, y, t)

[−2πn

4πt+|x− y|2

4t2− |x− y|

2

4t2+−2n

4t

]= 0.

Thus, we are done for (a).

(b) With simple computation, observe that∫RnK(x, y, t)dy =

∫Rn

1

(4πt)n/2e−|x−y|2

4t dy =1

(4πt)n/2

∫Rne−|x−y|2

4t dy =1

(4πt)n/2

n∏k=1

∫Re−

(xk−yk)2

4t dyk.

Here, recall that, with r2 = t,[∫Re−x

2

dx

]2

=

∫R2

e−x2−y2dxdy =

∫ ∞0

∫ 2π

0

e−r2

rdθdr = 2π

∫ ∞0

e−r2

rdr = π

∫ ∞0

e−tdt = π.

Then, by the substitution (xk−yk)

2√t

= zk with dzk = −dyk2√t

for each k, observe that∫RnK(x, y, t)dy =

1

(4πt)n/2

n∏k=1

∫Re−

(xk−yk)2

4t dyk =1

(4πt)n/2

n∏k=1

(2√t)

∫Re−z

2kdzk =

1

(4πt)n/2(2√t√π)n = 1

where the minus sign in the relation of dzk and dyk vanish since it is used for reversing the integral bound and the integrand iseven function.

(c) For given δ > 0, with the substitution of z = (x−y)

2√t

and (2√t)ndz = dy, observe that∫

|x−y|>δK(x, y, t)dy =

1

(4πt)n/2

∫|z|>δ

e−|z|2

(4t)n/2dz =1

(π)n/2

∫|z|> δ

2√t

ez|z|2

dz =1

πn/2

∫Rnχte−|z|2dz

where χt is a characteristic function of set z ∈ Rn : |z| > δ2√t. Since χt → 0 as t→ +0 and since the dominant function e−|z|

2

is integrable, by Dominated convergence theorem, we have

limt→0+

∫|x−y|>δ

K(x, y, t)dy = limt→0+

1

πn/2

∫Rnχte−|z|2dz =

1

πn/2

∫Rn

0dy = 0 ∀x.

(To be continued)



(d) Let ε > 0 be given. Since g ∈ CB(Rn), for any x ∈ Rn there exists δ > 0 such that |g(x)− g(y)| < ε/2 whenever |x− y| < δ.Also, since g is bounded, there exists M > 0 such that |g(x)| < M ∀x ∈ Rn.And, by (c), there exists η > 0 such that ∫

|x−y|>δK(x, y, t)dy <

ε

4M

whenever 0 < t < η.Now, by (b), for any 0 < t < η, observe that

|u(x, t)− g(x)| =∣∣∣∣∫

RnK(x, y, t)g(y)dy − g(x)

∣∣∣∣=

∣∣∣∣∫RnK(x, y, t)g(y)dy −

∫RnK(x, y, t)g(x)dy

∣∣∣∣ ∵ (b)

=

∣∣∣∣∫RnK(x, y, t)(g(y)− g(x))dy

∣∣∣∣≤∫Rn|K(x, y, t)||g(x)− g(y)|dy

=

∫|x−y|<δ

|K(x, y, t)| |g(x)− g(y)|dy +

∫|x−y|>δ

|K(x, y, t)| |g(x)− g(y)|dy

<ε

2

∫|x−y|<δ

K(x, y, t)dy + 2M

∫|x−y|>δ

K(x, y, t)dy

<ε

2

∫RnK(x, y, t)dy + 2M

ε

4M

=ε

2+ε

2= ε.

Therefore, in conclusion, for any ε > 0 there is η > 0 such that |u(x, t) − g(x)| < ε whenever t ∈ (0, η). In other words,u(x, t)→ g(x) as t→ 0+.


5.2.2Let g(x) be bounded and continuous for x ∈ Rn and define u by (19).(a) Show |u(x, t)| ≤ sup|g(y)| : y ∈ Rn.(b) If, in addition,

∫Rn |g(y)|dy <∞, show that limt→∞ u(x, t) = 0 uniformly in x ∈ Rn.


(a) Note that A := supy∈Rn |g(y)| <∞ since g is bounded function. Then observe that, due to positivity of K,

|u(x, t)| =∣∣∣∣∫

RnK(x, y, t)g(y)dy

∣∣∣∣ ≤ ∫RnK(x, y, t)|g(y)|dy ≤ A

∫RnK(x, y, t)dy = A = sup

y∈Rn|g(y)|

by the part (b) of the previous problem.

(b) Observe that

|u(x, t)| =∣∣∣∣∫

RnK(x, y, t)g(y)dy

∣∣∣∣ ≤ ∫RnK(x, y, t)|g(y)|dy =

1

(4πt)n/2

∫Rne−|x−y|2

4t |g(y)|dy ≤ 1

(4πt)n/2

∫Rn|g(y)|dy → 0

as t→∞ since∫Rn |g(y)|dy <∞ and e−

|x−y|24t ≤ 1 for any x, y ∈ Rn and t > 0.


5.2.3Let g(x) ∈ Ck(Rn) with Dαg uniformly bounded on Rn for each |α| ≤ k. Show that u defined by (19) satisfies u ∈ Ck(Rn×[0,∞)).


Note that, by Theorem 1 of Section 5.2, we know that if g is continuous and bounded which is true from assumption of thisproblem, then u is in C∞(Rn × (0,∞)). And with the same theorem, we can extend u to the domain Rn × [0,∞) continuouslysuch that u(x, 0) = g(x).

We are gonna use mathematical induction on k for this problem. When k = 0, it is true from the argument in the previousparagraph. So, now assume that it is the case when k = m. And suppose that g ∈ Cm+1 with Dαg is uniformly boundedon Rn for each |α| ≤ m + 1. Then, by our induction assumption, we know u ∈ Cm(Rn × [0,∞)) ∩ Cm+1(Rn × (0,∞)) ⊂Cm(Rn× [0,∞))∪C∞(Rn× (0,∞)). Now, we want to check that Dαu with |α| = m+1 can be continuously extend to Rn× [0,∞)from Rn × (0,∞).Actually, observe that

∆(Dαu) =

n∑i=1

∂2

∂x2i

Dαu =

n∑i=1

Dα ∂2

∂x2i

u = Dαn∑i=1

∂2

∂x2i

u = Dα∆u = Dαut = (Dαu)t

so Dαu satisfies the heat equation. Also, note that u ∈ Cβ(Rn × [0,∞)) with |β| = m, so Dβu(x, 0) exists. Therefore, byuniqueness, we have

Dβu(x, t) =1

(4πt)n/2

∫Rne−|x−y|2

4t Dβu(y, 0)dy.

And also observe that, when Dα does not contain the derivative with respect to t, by (19)

Dαu(x, t) = Dα 1

(4πt)n/2

∫Rne−|x−y|2

4t g(y)dy

=1

(4πt)n/2Dα

∫Rne−|y|24t g(x− y)dy

=1

(4πt)n/2

∫Rne−|y|24t Dαg(x− y)dy

=1

(4πt)n/2

∫Rne−|x−y|2

4t Dαg(y)dy

since Dαg is uniformly bounded and continuous. And also note that, since heat kernel, Φ satisfies the heat equation, Φt = ∆Φ,observe that, when Dα = ∂tD

β with |β| = m,

Dαu(x, t) = Φt ∗Dβu = ∆Φ ∗Dβu = Φ ∗Dβ(∆u)

=1

(4πt)n/2

∫Rne−|x−y|2

4t Dβ(∆u)(y, 0)dy

=1

(4πt)n/2

∫Rne−|x−y|2

4t Dβ(∆g)(x, 0)dy..

sinceDαg is uniformly bounded and continuous(the last equality is due to the same reason with the previous argument.). Therefore,by Theorem 1, we can extend Dαu to the domain Rn × [0,∞) continuously such that Dαg(x) or Dβ∆g(x). Since this argumenthold for all α with |α| ≤ k, by mathematical induction, u ∈ Ck(Rn × [0,∞)).


5.2.4Formally, check that

u(x, t) =

∞∑k=0

1

(2k)!x2k d

k

dtke−

1t2

satisfies ut = uxx for t > 0, x ∈ R with u(x, 0) = 0 for x ∈ R. (Proving the convergence of the infinite series is not easy; see [John,1] or [Widder])


We formally check it, so we assume that we can interchange the differentiation and summation.Observe that

uxx(x, t) =∂2

∂x2

∞∑k=0

1

(2k)!x2k d

k

dtke−

1t2

=

∞∑k=1

1

(2k)!(2k)(2k − 1)x2k−2 d

k

dtke−

1t2 ∵ only the first term die

=

∞∑k=1

1

(2k − 2)!x2k−2 d

k

dtke−

1t2

=

∞∑k=0

1

(2k)!x2k d

k+1

dtk+1e−

1t2

=∂

∂t

∞∑k=0

1

(2k)!x2k d

k

dtke−

1t2

=∂

∂tu(x, t)

= ut(x, t).

Thus, we have uxx = ut. Now, we want to show that u(x, 0) = 0 for all x ∈ R. Since 1t2 is not defined at t = 0, we should take

the limit as t→ 0+ to figure out the value of u(x, 0). Aagain, we assume that we can interchange the summation and limit as theproblem asked. Therefore, observe that

u(x, 0) = limt→0+

u(x, t)

= limt→0+

∞∑k=0

1

(2k)!x2k d

k

dtke−

1t2

=

∞∑k=0

1

(2k)!x2k lim

t→0+

dk

dtke−

1t2 .

Note that ∂k

∂tke−

1t2 = p(1/t)e−

1t2 where p(1/t) is a polynomial of 1/t. And also note that e−

1t2 decreases faster than the rate of

increasing of any power of 1/t as t→∞. For example, we can see that

limt→0+

e−1t2 = e− limt→0+

1t2 = 0

and that

limt→0+

lne−

1t2

tk= limt→0+

[− 1

t2− ln

1

tk

]= −∞ =⇒ lim

t→0+

e−1t2

tk= 0

since limx→0+ lnx = −∞.These argument shows that

u(x, 0) = limt→0+

u(x, t) =

∞∑k=0

0 = 0.

Therefore, we are done.


5.2.11Find a formula for the solution of the initial value problem

ut = ∆u− u t > 0, x ∈ Rn

u(x, 0) = g(x) x ∈ Rn

where g is continuous and bounded. Is the solution bounded? Is it the only bounded solution?


Let v(x, t) = etu(x, t). Then observe that, since ut = ∆u− u,

vt = etu+ etut = etu+ et(∆u− u) = et∆u = ∆(etu) = ∆v.

And observe that v(x, 0) = e0u(x, 0) = g(x) for all x ∈ Rn. Then, since g is bounded and continuous, by Theorem 1 (or 5.2.1),we have

etu(x, t) = v(x, t) =

∫RnK(x, y, t)g(y)dy =

1

(4πt)n/2

∫Rne−|x−y|2

4t g(y)dy =⇒ u(x, t) =1

(4πt)n/2

∫Rne−[|x−y|2

4t +t

]g(y)dy.

Then, due to boundedness of g and by part (b) of exercise 5.2.1, observe that

|u(x, t)| ≤ e−t∫RnK(x, y, t)|g(y)|dy ≤ e−t sup

z∈Rn|g(z)|

∫RnK(x, y, t)dy = sup

x∈Rn|g(z)| <∞

because e−t ≤ 1 for all t > 0. Thus, u is bounded.

Assume that we have two different bounded solutions u, q to the original problem. Then, letting w = u− q, observe thatwt = ∆w − w t > 0, x ∈ Rn

w(x, 0) = 0 x ∈ Rn.

Then letting w′ := etw, with similar computation with above argument, we havew′t = ∆w′ t > 0, x ∈ Rn

w′(x, 0) = 0 x ∈ Rn.

Note that 0 is a solution to the problem. Note that w′ is non-trivial solution to the problem above. Then, we have

supRn×[0,∞)

|w′(x, t)| 6= 0 = supx∈Rn

w′(x, 0)

since if w′ is a solution, then −w′ is also a solution so we can argue for |w′|.Therefore, by the contrapositive of maximum principle on Rn for heat equation, the growth estimate

w′(x, t) ≤ Aea|x|2

x ∈ Rn

does not hold for any A and a which means that w′ is unbounded in x. Now, note that w′ is bounded in x if and only if w isbounded in x. Thus, w is unbounded which means at least one of q and u is unbounded which is a contradiction. Therefore, Itshould be w ≡ 0 so u ≡ q which proves the uniqueness of bounded solution to the original problem.

math 731: partial di erential equations

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