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EXERCISES PDE 31.10.12-02.11.12
1. Exercise
Let U ∈ RN be a bounded open set. We say that v ∈ C2(U) is subharmonic iff −∆v ≤ 0 in U .
(a) Prove that subharmonic functions enjoy the following form of the mean-value property: forevery x ∈ U , for every r > 0 such that B(x, r) ⊂ U
v(x) ≤ B(x,r)
v(y) dy
(b) Use the previous property to prove the maximum principle for subharmonic functions:
maxU
v = max∂U
v .
Try also to give an elementary direct proof of this result.Hint: first consider the case −∆v < 0. Then prove the general case by approximating v withthe functions vε(x) := v(x) + ε|x|2 .
(c) Consider φ ∈ C∞(R) convex and let u be harmonic in U . Prove that the composition φ(u)is subharmonic.
(d) Let u be harmonic in U . Prove that |∇u|2 is subharmonic.
Solution: (a) Fix x ∈ U and r > 0 such that B(x, r) ⊂ U . Define the function
ψ(ρ) :=
∂B(x,ρ)
v(ξ) dS(ξ) =
∂B(0,1)
v(x+ ρζ) dS(ζ)
for ρ ∈ (0, r). Observe that
limρ→0+
φ(ρ) = v(x) . (1.1)
Now using the divergence theorem one has
ψ′(ρ) =
∂B(0,1)
∇v(x+ ρζ) · ζ dS(ζ) =
∂B(x,ρ)
∇v(ξ) · ξ − xρ
dS(ξ) =
=
∂B(x,ρ)
∂v
∂ν(ξ) dS(ξ) =
r
N
B(x,ρ)
∆v(y) dy ≥ 0
so that the function ψ(ρ) is increasing. Taking into account (1.1) this gives
ψ(ρ) ≥ ψ(0+) = v(x)
whence, also using Fubini’s theorem, and denoting by αN the voulume of the unit sphere B(x,r)
v(y) dy =1
αNrN
ˆB(x,r)
v(y) dy =1
αNrN
ˆ r
0
(ˆ∂B(x,ρ)
v(ξ) dS(ξ))dρ =
=1
rN
ˆ r
0
NρN−1ψ(ρ) dρ ≥ v(x)
rN
ˆ r
0
NρN−1 dρ = v(x) ,
as required.
(b) Assume that v takes its maximum at an interior point x0 . But for every r > 0 such thatB(x0, r) ⊂ U one has, by part (a)
B(x0,r)
v(y)− v(x0) dy ≥ 0 ;
1
2 EX1
since the integrand must be nonpositive this implies v(y) = v(x0) for every y ∈ B(x0, r). It followsthat v is constant in the connected component of U containing x0 (strong maximum principle) whichimplies the weaker conclusion. Alternative proof will appear here next week.
(c) By the usual rules of differentiation, one has
∆[φ(u)] = φ′(u)∆u+ φ′′(u)|∇u|2 .Since ∆u = 0 and φ′′ ≥ 0 by convexity, one gets ∆[φ(u)] ≥ 0 as required.
(d) For every i = 1, . . . , N the derivatives ∂u∂xi
are still harmonic. By the previous step with
φ(t) = t2 , the functions(∂u∂xi
)2are subharmonic. Then
|∇u|2 =
N∑i=1
(∂u∂xi
)2must be subharmonic as well.
2. Exercise
Let U ∈ RN be a bounded open set. Let v ∈ C2(U) satisfy−∆v = f in U
v = g on ∂U .
Prove that there exist a constant CU only depending on U such that
maxU|v| ≤ CU (max
∂U|g|+ max
U|f |) .
Solution: Let L := maxU |f | , and define v(x) := v(x) + L2N |x|
2 . One has
−∆v = f − L = f −maxU|f | ≤ 0
and therefore by the maximum principle for subharmonic functions and trivial estimates
maxU
v ≤ maxU
v = max∂U
v ≤ max∂U
g + L maxy∈∂U
1
2N|y|2 .
Therefore, setting MU := maxy∈∂U1
2N |y|2 and recalling the expression of L , we get
maxU
v ≤ max∂U
g +MU maxU|f | ≤ max
∂U|g|+MU max
U|f | (2.1)
where the last estimate is obvious. The same argument with −v in place of v gives
maxU
(−v) ≤ max∂U| − g|+MU max
U| − f | = max
∂U|g|+MU max
U|f | . (2.2)
Combining (2.1) and (2.2) gives
maxU|v| ≤ max
∂U|g|+MU max
U|f |
and eventually the conclusion, with CU the larger between 1 and MU .
EXERCISES PDE WS 2012-2013
1. Exercises PDE 07.11.12-09.11.12
Exercise 1. Let U ⊂ RN be a bounded open set. Consider a sequence un of harmonic functionssuch that un ⇒ u uniformly on U . What can we say about u?
Solution: u is a harmonic function, too. Actually, u is continuous since it is the limit of continuousfunctions; moreover, x ∈ U and r > 0 such that B(x, r) ⊂ U one has
u(x) = limn→+∞
un(x) = limn→+∞
∂B(x,r)
un(ξ) dS(ξ) =
∂B(x,r)
u(ξ) dS(ξ) ,
that is, it holds the mean-value property. This one implies first that u ∈ C∞(U) (remember theargument with the mollifiers!), and eventually that u is harmonic, since such is a C∞ function withthe mean value property.
Exercise 2. Fix R > 0 and consider a nonnegative function u ≥ 0 which is harmonic in the ballB(0, R). Prove the following Harnack’s inequality: for every x such that |x| < R one has
RN−2 R− |x|(R+ |x|)N−1
u(0) ≤ u(x) ≤ RN−2 R+ |x|(R− |x|)N−1
u(0) . (1.1)
Deduce that
supB(0,R2 )
u ≤ 3N infB(0,R2 )
u .
Solution: Denote by αN the volume of the unit sphere. By Poisson’s formula, for every x such that|x| < R one has
u(x) =
ˆ∂B(0,R)
KR(x, ξ)u(ξ) dS(ξ)
with KR the Poisson’s kernel
KR(x, ξ) :=R2 − |x|2
NαNR |ξ − x|N.
Since |ξ| = R > |x| by means of elementary triangle inequalities we get
R− |x|NαNR (R+ |x|)N−1
≤ KR(x, ξ) ≤ R+ |x|NαNR (R− |x|)N−1
and since u is nonnegative, substituting into the Poisson’s formula we arrive at
R− |x|NαNR (R+ |x|)N−1
ˆ∂B(0,R)
u(ξ) dS(ξ) ≤ u(x) ≤ R+ |x|NαNR (R− |x|)N−1
ˆ∂B(0,R)
u(ξ) dS(ξ) .
Since by the mean-value formulaˆ∂B(0,R)
u(ξ) dS(ξ) = NαNRN−1 u(0) ,
we eventually get (1.1).Now, for |x| = R
2 , Harnack’s inequality reads, after simple computations,
2N−2
3N−1u(0) ≤ u(x) ≤ 3 · 2N−2u(0) ,
1
2 EXERCISES PDE
which gives
max∂B(0,R2 )
u ≤ 3 · 2N−2u(0) and min∂B(0,R2 )
u ≥ 2N−2
3N−1u(0) .
Combining the two we arrive at
max∂B(0,R2 )
u ≤ 3N min∂B(0,R2 )
u
which entails the conclusion, by the maximum and minimum principles.
Exercise 3. Find an explicit solution for the following Dirichlet problem on the square QL := (0, L)2
associated to the Laplace equation
∆u(x, y) = 0
u(0, y) = f(y)
u(L, y) = 0
u(x, 0) = 0
u(x, L) = 0
in the following cases:
(a) f(y) = sin(nπL y), with n ∈ N ;
(b) f is a generic C1 function satisfying f(0) = f(L) = 0 and thus can be developed in a Fourierseries
f(y) =
+∞∑n=1
an sin(nπ
Ly)
with+∞∑n=1
|an| < +∞ . (1.2)
Hint: first, try to see which functions of the form u(x, y) = v(x)w(y) satisfy the Laplace equationwith 0 boundary conditions on three sides of the square.
Solution: We first observe that due to the boundary conditions, v and w must satisfy
w(0) = w(L) = 0 (1.3)
and
v(L) = 0 . (1.4)
Moreover, the Laplace equation givesv′′(x)
v(x)= −w
′′(y)
w(y)
for every x and y ; but this is possible only if both sides are equal to a constant, and thus there mustexist λ ∈ R such that
v′′(x)− λv(x) = w′′(y) + λw(y) = 0 (1.5)
for every x and y . Taking into account the boundary conditions (1.3), the function w has to solve aboundary-value problem of the form
w′′(y) + λw(y) = 0
w(0) = w(L) = 0 .(1.6)
Without entering the details of the related (beautiful) theory, we only observe that by solving explicitlyand imposing the boundary conditions, one sees that the problem (1.6) has solution only for a discreteset of values of λ (no wonder: this is not a Cauchy problem)! Precisely, it must be
λ = λn :=(nπL
)2
, n ∈ N (1.7)
EXERCISES PDE 3
and for given n the solutions of (1.6) are all multiples of the functions
wn(y) := sin(nπLy), n ∈ N .
For fixed n , now, and thus for fixed λn , by (1.5) and (1.4) we can recover the functions vn(x) assolutions of the problem
v′′n(y)− λnvn(y) = 0
vn(L) = 0 .
The solutions of this problem are all multiples of the functions
vn(x) := sinh(nπL
(x− L)).
We recall that the function sinh(t) := 12 (et − e−t) is an odd increasing function, sinh(t) ≥ 0 ⇐⇒
t ≥ 0.We can now take a small breath by observing that we have answered the question posed by the
hint: the harmonic functions we searched are of the form
un(x, y) = µ sinh(nπL
(x− L))
sin(nπLy), µ ∈ R, n ∈ N . (1.8)
Now for f(y) = sin(nπL y) we immediately get that the solution can be recovered from (1.8) byimposing the boundary condition (namely, recover µ by imposing equality for x = 0) and we get thesolutions
un(x, y) = − 1
sinh(nπ)sinh
(nπL
(x− L))
sin(nπLy).
In the general case, we first proceed formally. Observe that finite linear combinations of the ungiven by (1.8) are still harmonic and 0 on the three sides of the square; but since f is given by aninfinite sum, we look for a candidate solution in the form of an infinite linear combination
u(x, y) =
+∞∑n=1
bn sinh(nπL
(x− L))
sin(nπLy).
By imposing equality for x = 0 with the Fourier development of f we get that our candidate solutionis
u(x, y) =
+∞∑n=1
− ansinh(nπ)
sinh(nπL
(x− L))
sin(nπLy). (1.9)
But since the solution is given by an infinite series, we are not done. We must first show that the seriesconverges, to have a well defined function! And actually, we want to show that it converges uniformlyon QL : thus, the limit will be a harmonic function by Exercise 1 (by construction the partial sumsare harmonic functions), while the boundary conditions are attained by the choice of the coefficients.By the properties of sinh, we now observe that for every x ∈ [0, L] we have
0 ≤ −sinh
(nπL (x− L)
)sinh(nπ)
≤ 1
and thus, for every (x, y) ∈ QL∣∣∣− ansinh(nπ)
sinh(nπL
(x− L))
sin(nπLy)∣∣∣ ≤ |an| ;
the required uniform convergence follows now by (1.2).
4 EXERCISES PDE
2. Exercises PDE 14.11.12-16.11.12
Exercise 1. Harnack’s Theorem: consider an increasing sequence un : B(0, R) → R of harmonicfunctions, that is un(x) ≤ un+1(x) for every x ∈ B(0, R). Assume that un(0) is a Cauchy sequence.Then there exists a harmonic function u such that un ⇒ u uniformly in B(0, r) for every r < R .
Solution: It suffices to show that un(x) is a Cauchy sequence uniformly for x ∈ B(0, r). To this aim,fix h ∈ N , and observe that, by the hypothesis, for every k ≤ h we have that uk − uh is a positiveharmonic function. Thus, by Harnack’s inequality there exists a constant C(N, r,R) such that
supx∈B(0,r)
|uk(x)− uh(x)| = supx∈B(0,r)
(uk(x)− uh(x))
≤ C(N, r,R) infx∈B(0,r)
(uk(x)− uh(x)) ≤ C(N, r,R)(uk(0)− uh(0)) .
Since the last term of this chain of inequalities is by the hypothesis a Cauchy sequence, we get theconclusion.
Exercise 2. Let u : RN → R be a harmonic function. Prove that:
(a)
ˆRnu2(x) dx < +∞⇒ u ≡ 0;
(b)
ˆRn|∇u|2(x) dx < +∞⇒ u ≡ const.
Solution: For part (a), observe that u2 is subharmonic (Ex.1 (c), 31.10.12). Then by the mean-valueproperty one has, for every x ∈ Rn and every R > 0
u2(x) ≤ 1
αN RN
ˆB(x,R)
u2(x) dx ≤ 1
αN RN
ˆRnu2(x) dx =
C
αN RN.
Letting R to +∞ , we get u2(x) ≤ 0 for every x , which implies the conclusion.For part (b), apply part (a) to the harmonic functions ∂
∂xiu with i = 1, . . . , N to obtain ∇u ≡ 0
and thus the conclusion.
Exercise 3.
• Consider a holomorphic function f : C→ C and its real and imaginary part u : R2 → R andv : R2 → R , that is
f(z) = u(x, y) + i v(x, y) (2.1)
with z = x+ iy . Prove that u and v are harmonic.• Prove the following converse. Let u : B(0, R) ⊂ R2 → R be a harmonic function. Then,
there exists v : B(0, R) ⊂ R2 → R harmonic such that f(z) defined by (2.1) is a holomorphicfunction. Is v unique?
Solution: We need to recall the following preliminaries about holomorphic functions. By the Cauchy-Riemann theorem, f is holomorphic in region of the complex plane D if and only if the followingconditions are satisfied
∂
∂xu =
∂
∂yv
∂
∂yu = − ∂
∂xv
(2.2)
and if this happens one has
f ′(z0) =∂
∂xu(x0, y0) + i
∂
∂xv(x0, y0) (2.3)
for every z0 ∈ D , with z0 = x0 + iy0 .It also follows that if Γ is a Jordan curve surrounding a domain D and f is holomorphic in D ,
then the line integral of f over Γ vanishes: in formulasˆΓ
f(z) dz = 0 (2.4)
EXERCISES PDE 5
Indeed, by (2.1), dz = dx+ i dy we getˆΓ
f(z) dz =
ˆΓ
P (x, y) · ds+ i
ˆΓ
Q(x, y) · ds ,
where the vector fields P and Q are given by P = (u,−v) and Q = (v, u) respectively; by Stokes’Theorem and (2.2) we get ˆ
Γ
P (x, y) · ds =
ˆD
curlP (x, y) dxdy = 0
ˆΓ
Q(x, y) · ds =
ˆD
curlQ(x, y) dxdy = 0 ,
proving (2.4). With these preliminaries we can solve the exercise.First, implication (a) is quite easy. Indeed, if f is holomorphic, by (2.2) we get
∆u =∂2
∂x2u+
∂2
∂y2u =
∂2
∂x∂yv − ∂2
∂y∂xv = 0
since we can exchange the order or derivations by smoothness of u . A similar proof shows that v isharmonic, too.
For part (b), we start by observing that if f holomorphic having u as its real part exists, then by(2.3) and (2.2) it must be
f ′(z) =∂
∂xu(x, y)− i ∂
∂yu . (2.5)
Then, defining g(z) as the right-hand side, which depends only on u , it only suffices to show thatthere exists a holomorphic primitive f of g satisfying f(0) = u(0). If it exists, its imaginary part vwill be harmonic by part (a), and uniquely determined up to a constant, thus concluding the exercise.
By analogy with integral calculus in one variable, we set
f(z) = u(0) +
ˆ[0,z]
g(w) dw
where [0, z] is the line segment connecting 0 and z , oriented from 0 to z . We observe that g satisfiesthe Cauchy-Riemann conditions; indeed, since u is smooth and harmonic
∂
∂xReg =
∂2
∂x2u = − ∂2
∂y2u =
∂
∂yImg
∂
∂xImg = − ∂2
∂x∂yu = − ∂2
∂y∂xu = − ∂
∂xReg .
From this, by (2.4) with Γ the union of the three segments [0, z0] , [z0, z] and [z, 0] all oriented fromthe first to the second endpoint (observe that this curve is the boundary of a domain interely containedin the ball, thus g is holomorphic inside!) we easily getˆ
[0,z]
g(w) dw −ˆ
[0,z0]
g(w) dw =
ˆ[z0,z]
g(w) dw ,
so that
f(z)− f(z0)
z − z0=
´[z0,z]
g(w) dw
z − z0
for every z and z0 in B(0, R); and now, only continuity of g suffices to say that the righ-hand sidehas a limit when z goes to z0 , which is equal to g(z0), as we wanted.
Remark 2.1. In Exercise 3 part (a) it was used the fact that when f is holomorphic, then u andv are C∞ . Actually, this is a consequence of the fact that the derivative of a holomorphic functionis still holomorphic, usually recovered as a consequence of Cauchy’s formula for the derivatives. It isinteresting to show that this fact can be also derived only using the Cauchy-Riemann conditions andthe theory of harmonic functions presented so far. I give a sketch of the proof.
6 EXERCISES PDE
Let f , u and v as in (2.1) and consider fn = ρn ? u+ i(ρn ? v), where ρn are the usual mollifiers.It is easy to see that ρn ? u and ρn ? v (which are now C∞ ) still satisfy (2.2) (why?) so that fn areholomorphic and ρn ? u and ρn ? v harmonic. As ρn ? u and ρn ? v converge locally uniformly to uand v with n going to +∞ , we get that u and v are harmonic, and in particular C∞ . Now, (2.5)holds, and since u is harmonic we get (how?) that f ′ is holomorphic, as required.
3. Exercises PDE 21.11.12-23.11.12
Exercise 1. Assume N = 1 and u(x, t) = v( x√t).
(a) Show that ut = uxx if and only if v solves the differential equation
v”(z) +z
2v′(z) = 0
and compute the solutions of it in dependance of two arbitrary constants c and d .(b) Differentiate with respect to x and select the constant c properly, to obtain the fundamental
solution for N = 1. Why this procedure gives the fundamental solution? (Hint: What is theinitial condition for u?)
Solution: (a) It holds, denoting with primes the derivatives of v with respect to its argument z :
∂
∂tu(x, t) =
−x2t√tv′(z) = −zv
′(z)
2t
∂2
∂x2u(x, t) =
1
2tv′′(z)
so that ut = uxx if and only if v solves the differential equation
v”(z) +z
2v′(z) = 0 .
(b) Setting w = v′ one has that it must be
w′(z) = ce−z24
for some arbitrary constant c ∈ R , whence
v(z) = c
ˆ z
0
e−s24 ds+ d
with d ∈ R again arbitrary. It is easy to check that
U(x, t) :=∂
∂xu(x, t) =
c√te−x24t
solves again the heat equation. Moreover we can impose to this function the “initial condition” forthe fundamental solution, that is it must approximate the Dirac delta when t tends to 0. This resultsin the coupling of the two conditions
limt→0
U(x, t) = 0 if |x| 6= 0
and ˆRU(x, t) dx = 1 (3.1)
for t in a neighborhood of 0. The first is automatically satisfied, while the second gives the requiredconstant c = 1
2√π
.
Also, observe that (3.1) asserts conservation of the total heat, a property that follows from theequation for a solution decaying at infinity with its derivative. It is actually, for t 6= 0
− d
dt
ˆRU(x, t) dx =
ˆRUxx(x, t) dx = Ux(+∞, t)− Ux(−∞, t) = 0
which implies (3.1) possibly with a constant different from 1, provided that the total heat keepsbounded for t going to 0 (if not we have identically +∞ , as it happens for u(x, t)!)
EXERCISES PDE 7
Exercise 2. Write down an explicit formula for a solution ofut −∆u+ cu = f in Rn × (0,+∞)
u = g in Rn × t = 0(3.2)
where c ∈ R .
Solution: Set v(t, x) = ect u(t, x) where u solves (3.2). By a direct computation one gets
vt −∆v = ect(ut −∆u+ cu)
in Rn × (0,+∞) therefore v solvesvt −∆v = ectf(x, t) in Rn × (0,+∞)
v = g in Rn × t = 0 .
A solution of the previous problem is given via Duhamel’s principle by
v(x, t) =1
(4πt)N2
ˆRne−|x−y|2
4t g(y) dy +
ˆ t
0
ˆRn
ecs
(4π(t− s))N2e−|x−y|24(t−s) f(y, s) dsdy
Now, setting u(x, t) := e−ctv(t, x), with v as above, gives a solution to (3.2).
Exercise 3. Tychonov’s counterexample: Consider the holomorphic function g(z) = e−1z2 for z ∈
C\0 and denoting by g(k) its k -th derivative, define the function
u(x, t) =
+∞∑k=0
g(k)(t)
(2k)!x2k if t > 0 , x ∈ R
0 if t = 0 , x ∈ RRigorously justify that this is a solution of the heat equation with 0 Cauchy datum by showinguniform convergence of the series (and of the series of its time and space derivatives involved in theequation) on any semi-strip of the type
(x, t) ∈ [−a, a]× [δ,+∞)
with a, δ > 0.Hint: apply Cauchy’s formula for the derivatives of holomorphic functions in this form
g(k)(t) =k!
2π i
ˆ∂B(t, t2 )
g(z)
(z − t)k+1dz (3.3)
to estimate g(k)(t). Obviously, if you find a method not using complex analysis, it is also valid!
Solution: For every z ∈ Ct one has that there exists ω ∈ [0, 2π) such that
z = t(1 +1
2eiω)
whence1
z=
1
t
1
1 + 12eiω
=1
t
2(2 + e−iω)
5 + 4 cosω.
Therefore1
z2=
4
t2(2 + e−iω)2
(5 + 4 cosω)2.
so that
Re1
z2=
4
t24 + cos(2ω)− 4 cos(ω)
(5 + 4 cosω)2≥ 4
81t2,
where the last estimate is simply obtained by observing that 0 is a minimizer of the function at thenumerator, plus the trivial fact that (5 + 4 cosω)2 ≤ 81.
With this, we get
|e−1z2 | = e−Re
(1z2
)≤ e−
481t2
8 EXERCISES PDE
for every z ∈ Ct , therefore (3.3) gives
|g(k)(t)| ≤ k!
2π
ˆ∂B(t, t2 )
|g(z)||z − t|k+1
|dz| ≤ k!
2π
ˆ 2π
0
2k+1 e−Re
(1z2
)tk+1
t
2dθ ≤ k!
(2
t
)ke−
481t2 .
By recalling the easy estimate2kk!
(2k)!≤ 1
k!
for every k ∈ N , we conclude that ∣∣∣g(k)(t)
(2k)!x2k∣∣∣ ≤ e− 4
81t2
(x2
t
)k 1
k!
for every k ∈ N and every (x, t). Therefore the series defining u(x, t) is totally and thus uniformlyconvergent in [−a, a]× [δ,+∞); moreover, we have the estimate
|u(x, t)| ≤ e−4
81t2+ a2
t
for every x ∈ [−a, a] and t > 0. This implies
limt→0+
|u(x, t)| = 0
uniformly in x ∈ [−a, a] , which means that u ∈ C(R× [0,+∞)).A similar proof shows that the series
+∞∑k=0
g(k+1)(t)
(2k)!x2k , (3.4)
formally giving ut(x, t), as well as+∞∑k=1
g(k)(t)
(2k − 1)!x2k−1
and+∞∑k=1
g(k)(t)
(2(k − 1))!x2(k−1) (3.5)
formally giving ux(x, t), and uxx(x, t), respectively, uniformly converge in [−a, a]×[δ,+∞). Thereforeu(x, t) is twice differentiable in x and once t ; a trivial algebraic computation using (3.4) and (3.5)shows that it solves the heat equation, as required.
4. Exercises PDE 28.11.12-30.11.12
Exercise 1.
(a) Show that the general solution of the PDE uxy = 0 is
u(x, y) = F (x) +G(y) (4.1)
for arbitrary functions F and G .(b) Using the change of variables ξ = x + t , η = x − t , show that utt − uxx = 0 if and only if
uξη = 0.(c) Rederive D’Alembert’s formula.(d) Under which conditions on the initial data g, h is the solution a right-moving wave? A left-
moving wave?
Solution: (a) Clearly any function of the form (4.1) solves the equation, and it must be only shownthe converse. Take a solution u of the PDE uxy = 0 and, for fixed x , set fx(y) := ux(x, y). One hasthat
d
dyfx(y) = 0
EXERCISES PDE 9
therefore
fx(y) ≡ Cx , (4.2)
a constant depending only on x . We now set f(x) = Cx for every x , we call F (x) a primitive of f ,and we observe that by definition of fx we can rewrite (4.2) as follows: for every x and y , it holds
ux(x, y) =d
dxF (x) .
We now set, for fixed y , gy(x) := u(x, y). We have
d
dxgy(x) = ux(x, y) =
d
dxF (x)
for every x and y . This gives
gy(x)− F (x) ≡ Gy ,
a constant depending only on y . Setting G(y) := Gy , the previous equality gives the conclusion.(b) Setting ξ = x+ t , η = x− t is equivalent to t = 1
2 (ξ − η) and x = 12 (ξ + η). Consequently, let
v(ξ, η) := u( 12 (ξ − η), 1
2 (ξ + η)). denoting with ut differentiation with respect to the first argumentand with ux differentiation with respect to the second argument, we have
∂
∂ηv(ξ, η) = −1
2ut
(1
2(ξ − η),
1
2(ξ + η)
)+
1
2ux
(1
2(ξ − η),
1
2(ξ + η)
)and therefore
∂2
∂ξ∂ηv(ξ, η) = −1
4utt
(1
2(ξ − η),
1
2(ξ + η)
)− 1
4uxt
(1
2(ξ − η),
1
2(ξ + η)
)+
+1
4utx
(1
2(ξ − η),
1
2(ξ + η)
)+
1
4uxx
(1
2(ξ − η),
1
2(ξ + η)
)so that utt − uxx = 0 if and only if vξη = 0 as we wanted.
(c) By part (a) and (b) we have, denoting with u a solution of the wave equation, and for ξ = x+t ,η = x− t :
u(1
2(ξ − η),
1
2(ξ + η)
)= F (ξ) +G(η)
for arbitrary functions F and G , that is
u(t, x) = F (x+ t) +G(x− t) ,
the sum of the left-travelling wave F (x+ t) and of the right-travelling wave G(x− t).(d) A pure left-travelling wave corresponds to G = const. , and similarly a pure right-travelling
wave corresponds to F = const . Imposing the initial conditions g(x) = u(0, x) and h(x) = ut(0, x)we get that it must be
F (x) +G(x) = g(x)
F ′(x)−G′(x) = h(x)
for every x , that is F (x) +G(x) = g(x)
F (x)−G(x) =
ˆ x
0
h(s) ds+ const. .
We then have G(x) ≡ const. if and only if g(x) −´ x
0h(s) ds ≡ const. , which is to say that we have
a left-traveling wave if and only if g′ = h . Similarly we have a right-traveling wave if and only ifg′ = −h .
10 EXERCISES PDE
Exercise 2.
(a) Let E := (E1, E2, E3) and B := (B1, B2, B3) be solutions of the Maxwell equationsEt = curlB, Bt = −curlE
divE = divB = 0 .
Show
Ett −∆E = 0, Btt −∆B = 0 .
(b) Assume that u := (u1, u2.u3) solves the evolution equation of linear elasticity
utt − µ∆u− (λ+ µ)∇(divu) = 0 in R3 × (0,+∞) . (4.3)
Show that w := divu and w := curlu each solve wave equation, but with different speed ofpropagation.
Solution: (a) We use the following vector calculus identity:
curl(curlF) = ∇(divF)−∆F (4.4)
for every C2 vector field F . Taking into account that divE = 0, (4.4) yields
curl(curlE) = −∆E .
Observing that by Maxwell’s equations Ett = curlBt = −curl(curlE) we get that Ett−∆E = 0. Theproof of the other case is similar.
(b) Using (4.3), one has
wtt = curlutt = µ curl∆u + (λ+ µ) curl(∇(divu)) .
The curl of a gradient vector field being zero, we arrive at
wtt = µ curl ∆u = µ∆w
by simply exchanging the order of differentiation. This is a wave equation with speed of propagationµ .
Analogously
wtt = div (utt) = µdiv (∆u) + (λ+ µ) div (∇(divu)) = µ∆w + (λ+ µ) div (∇w)
again exchanging the order of differentiation and recalling that w = divu . Since div (∇w) = ∆w weget
wtt = (λ+ 2µ) ∆w ,
that is a wave equation with speed of propagation λ+ 2µ .
Exercise 3. Let V be a (complex) pre-Hilbertian space and A a linear mapping from V to V .Suppose preliminarly that the dimension of V is finite and equal to n ∈ N and suppose that thereexists an orthonormal basis v1, . . . , vn of V consisting of eigenvectors of A relative to the eigenvalues(λ1, . . . , λn).
(a) Consider the following Cauchy problem associated to a linear ODE in V :
d
dtw = Aw, w(0) = g (4.5)
and show that
w(t) :=
n∑i=1
〈g, vi〉eλitvi
is its only solution.(b) Give an analogous statement for this other initial value problem
d2
dt2w = Aw, w(0) = g,
d
dtw(0) = h . (4.6)
EXERCISES PDE 11
We now want to generalise this method to an infinite-dimensional situation where we look to thesolution of the wave equation with periodic boundary conditions, that is
utt − uxx = 0 (x, t) ∈ (0, 2π)× (0,+∞)
u(0, x) = g(x), ut(0, x) = h(x)
u(t, 0)− u(t, 2π) = ux(t, 0)− ux(t, 2π) = uxx(t, 0)− uxx(t, 2π) = 0 for every t .
(4.7)
To this end, let
V := v ∈ C2([0, 2π],C) : v(0)− v(2π) = vx(0)− vx(2π) = vxx(0)− vxx(2π) = 0 . (4.8)
(c) Interpret (4.7) as a linear ODE of the type (4.6) in the pre-Hilbertian space V , endowed withthe usual L2 scalar product
〈u, v〉 :=1
2π
ˆ 2π
0
u(x)v(x) dx ,
where the operator A is given by
A :=d2
dx2: V ⊂ L2((0, 2π))→ L2((0, 2π)) . (4.9)
What are the initial conditions? What happens of the boundary value conditions?(d) Show that (eikx)k∈Z is an orthonormal basis of V consisting of eigenvectors of A and compute
the relative eigenvalues (hint: Fourier series).(e) Consider now (4.7) with g := x2(x − 2π)2 and h ≡ 0. Verify that g and h belong to V ,
then find a formal solution to (4.7) assuming that the statement you found in (b) can begeneralised to an infinite-dimensional situation, and using the basis (eikx)k∈Z . Then showthat you actually found a solution, in this way!
Solution: (a) Uniqueness is a well-known fact. By a direct calculation, since λivi = Avi we get
d
dtw(t) :=
n∑i=1
〈g, vi〉eλitλivi =
n∑i=1
〈g, vi〉eλitAvi = A( n∑i=1
〈g, vi〉eλitvi),
so that ddtw = Aw as we wanted. On the other hand w(0) =
∑ni=1〈g, vi〉vi = g since vi is a
orthonormal basis. In view of point (b) it is important to remark that if vi is a basis of eigenvectors,but no orthonormal basis, then, denoting with gi the components of g with respect to vi , thesolution to (4.5) is given by
w(t) :=
n∑i=1
gieλitvi ; (4.10)
only, gi 6= 〈g, vi〉 , in general!(b) Setting v(t) := d
dtw and W := (v, w) ∈ V × V the problem is equivalent to the following one
d
dtW = BW, W (0) = (g, h)
where the linear operator B acts on V × V as follows:
B(w, v) := (v,Aw) .
Observe that the only hypothesis that A is diagonalisable in V does not assure that B is diagonalisablein V × V ! We have actually to distinguish two cases.
(b1) ker A = 0 : this is the case where B possesses a basis of eigenvectors. We make the following
convention: for every λ ∈ C with λ 6= 0 we denote with√λ the complex square root of λ
with argument in [0, π). Then, a basis in V × V made of eigenvectors of B is easily given bythe set of vectors
(W1)+, (W1)−, . . . , (Wn)+, (Wn)−
12 EXERCISES PDE
where for every i = 1, . . . , n we have set
(Wi)+ =1√
1 + |λi|(vi,√λivi)
and
(Wi)− =1√
1 + |λi|(vi,−
√λivi) .
The corresponding eigenvalues are√λi and −
√λi , respectively. But since this basis is no
orthonormal one, we can apply case (a) only in the form (4.10), to get
W (t) =
n∑i=1
(wi)+e√λit(Wi)+ +
n∑i=1
(wi)−e−√λit(Wi)− .
where (wi)+ and (wi)− are the coefficients of W (0) with respect to the basis
(Wi)+, (Wi)−i=1,...,n
. Projecting on the first factor we obtain the expression of w in this form
w(t) =
n∑i=1
aie√λitvi +
n∑i=1
bie−√λitvi (4.11)
with ai = (wi)+√1+|λi|
and bi = (wi)+√1+|λi|
. But the key point is that now we can easily determine
ai and bi by imposing the initial conditions, since vi is an orthonormal basis! Precisely, itmust be
ai + bi = 〈g, vi〉√λiai −
√λibi = 〈h, vi〉
for every i . Solving the system gives that the coefficients ai and bi in (4.11) are given by
ai =1
2
(〈g, vi〉+
〈h, vi〉√λi
), bi =
1
2
(〈g, vi〉 −
〈h, vi〉√λi
). (4.12)
(b2) dim ker A = m 6= 0. Denoting with vin−mi=1 the eigenvectors of A relative to the nonzeroeigenvalues λin−mi=1 we see that the analogous of (4.11), that is the function
w1(t) =
n−m∑i=1
aie√λitvi +
n−m∑i=1
bie−√λitvi , (4.13)
wtih ai and bi given by (4.12) still solves the equation in (4.6), but with a different initialdatum: precisely w1(0) = g − Pker A g and d
dtw1(0) = h− Pker A h where Pker A denotes the
orthogonal projection onto ker A . But now, given an orthonormal basis vker1 , . . . , vkerm ofker A it is easy to check that the unique solution of (4.6) is w(t) = w1(t) + z(t) where w1(t)is given by (4.13) and (4.12) and z(t) is given by:
z(t) :=
m∑j=1
[〈g, vkerj 〉+ 〈h, vkerj 〉t]vkerj . (4.14)
Indeed z(t) ∈ ker A for every t , d2
dt2 z(t) ≡ 0 = Az(t), and z(0) = Pker A g as well asddtz(0) = Pker A h so that we conclude by linearity.
(c) Simply interpret (4.7) as a generalised Cauchy problem
d2
dt2w = Aw, w(0) = g,
d
dtw(0) = h .
with w : R→ V , where this one is given by (4.8), and A is given by (4.9). Setting u(t, ·) = w(t) theboundary conditions ar already incorporated in the request that w(t) must belong to V . Obviously,for the problem to have sense, it must be g and h in V , too!
EXERCISES PDE 13
(d) It is a very-well known fact that (eikx)k∈Z is an orthonormal basis of L2 by the theory of Fourierseries; since (eikx)k∈Z ⊂ V for, it constitutes an orthonormal basis of V , too. A direct calculationshows that for every k ∈ Z
A(eikx) = −k2eikx
so that they are eigenvectors of A with eigenvalues −k2 .(e) Checking that g belongs to V is straightforward. Now, for k 6= 0, we have according to our
convention√λk = i|k| , so that, setting
αk := 〈g, eikx〉we get that the coefficients ak and bk in (4.12) satisfy ak = bk = 1
2αk . Therefore the function w1(t)defined in (4.13) reduces in our case to
w1(t) =∑
k∈Z, k 6=0
1
2αk[ei|k|t + e−i|k|t]eikx =
∑k∈Z, k 6=0
αk cos(|k|t)eikx .
By a direct computation, αk = − 24k4 for every k 6= 0, so that
w1(t) = −∑
k∈Z, k 6=0
24
k4cos(|k|t)eikx = −48
∞∑k=1
1
k4cos(kt) cos(kx) .
The eigenvector corresponding to the eigenvalue 0 is the constant 1, so that in our case
z(t) = 〈g, 1〉 =8π4
15
therefore we get to our candidate solution
w(t) =8π4
15− 48
∞∑k=1
1
k4cos(kt) cos(kx) . (4.15)
The verification that (4.15) is actually the solution of (4.7) is an easy exercise in differentiation ofseries whose proof I omit.
5. Exercises PDE 5.12.12-7.12.12
Exercise 1. Let u(t, x) : R× R→ R be a solution of the wave equation utt − uxx = 0. Show that
v(t, x) :=
ˆ +∞
−∞
e−s2
4t
√tu(s, x) ds
satisfies the heat equation ut − uxx = 0 for every t > 0 and x ∈ R .
Solution: By D’Alembert’s formula, u(s, x) = F (x+s)+G(x−s) for two functions of a real variableF and G . With the change of variable z = x+ s one has
ˆ +∞
−∞
e−s2
4t
√tF (x+ s) ds =
ˆ +∞
−∞
e−(x−z)2
4t
√t
F (z) dz
and similarly, with z = x− s we get
ˆ +∞
−∞
e−s2
4t
√tG(x− s) ds =
ˆ +∞
−∞
e−(x−z)2
4t
√t
G(z) dz
so that
v(t, x) =
ˆ +∞
−∞
e−(x−z)2
4t
√t
(F (z) +G(z)) dz
solving the heat equation with initial datum 2√πu(0, x).
14 EXERCISES PDE
Exercise 2. We say that a function u : [0, 1] → R is absolutely continuous if for every ε > 0 thereexists δ > 0 such that if 0 ≤ t0 < s0 < t1 < s1 < . . . tn < sn ≤ 1 are points in [0, 1] satisfyng
n∑i=1
(si − ti) ≤ δ
thenn∑i=1
|u(ti)− u(si)| ≤ ε
An absolutely continuous function u is continuous and u(t) (the pointwise derivative) exists for almostevery t ∈ [0, 1].
Show that if u ∈ W 1,p(0, 1) for 1 ≤ p then there exists v absolutely continuous such that u = valmost everywhere; moreover the weak derivative of u is given exactly by the pointwise derivative vand v ∈ Lp(0, 1). Moreover, if p > 1:
|u(x)− u(y)| ≤ |x− y|1−1p (
ˆ 1
0
|u′(t)|p dt)1p (5.1)
for every x and y in [0, 1].
Solution: The key of the exercise, important in itself, is that u satisfies the fundamental Theoremof calculus, that is
u(x) = u(0) +
ˆ x
0
u′(t) dt (5.2)
for every x ∈ [0, 1], with u′ the weak derivative. Observe that the right-hand side is well defined,if we assume that the weak derivative belongs to Lp . To prove (5.2), no knowledge of absolutelycontinuous function is needed, but only Fubini’s theorem and exercise 4 below. Let us see how. Givena function v ∈ L1([0, 1]), define V (x) :=
´ x0v(y) dy and we claim that
V ′ = v , (5.3)
that is, the weak derivative of V is v . To this aim, take ϕ ∈ C∞c ([0, 1]) and observe that by Fubini’sTheorem
−ˆ 1
0
V (x)ϕ′(x) dx = −ˆ 1
0
( ˆ x
0
v(y) dy)ϕ′(x) dx = −
ˆ 1
0
( ˆ 1
y
ϕ′(x) dx)v(y) dy =
ˆ 1
0
v(y)ϕ(y) dy .
This proves (5.3). Now, if you define U(x) := u(0) +´ x
0u′(y) dy , (5.3) gives that (U − u)′ = 0, so
that by Exercise 4 below it must be U(x)−u(x) ≡ const. for every x ∈ [0, 1]. But since U(0) = u(0),the constant is 0 and (5.2) is proved.
Let us also observe that (5.1) is an immediate consequence of (5.2) and the Holder’s (or Jensen’sinequality), when p > 1; it is indeed for every 0 ≤ x < y ≤ 1
|u(y)− u(x)| =∣∣∣ˆ y
x
u′(t) dt∣∣∣ = |y − x|
(∣∣∣ 1
|y − x|
ˆ y
x
u′(t) dt∣∣∣p) 1
p ≤ |y − x|1−1p (
ˆ 1
0
|u′(t)|p dt)1p .
Absolute continuity of u follows from (5.2) with the only observation that for every v ∈ L1([0, 1]),V (x) :=
´ x0v(y) dy is absolutely continuous. To show this, we can suppose without loss of generality
that v ≥ 0. For fixed ε > 0, there exist N ∈ N , depending only on ε , such that defining vN (x) :=minv(x), N , one has ˆ 1
0
|v(x)− vN (x)| dx ≤ ε
2. (5.4)
This is because the sequence vk(x) := minv(x), k is L1 convergent to v . Now, fix δ = ε2N ; for
every finite sequence of points 0 ≤ t0 < s0 < t1 < s1 < . . . tn < sn ≤ 1 satisfyngn∑i=1
(si − ti) ≤ δ
EXERCISES PDE 15
one hasn∑i=1
|V (ti)− V (si)| =ˆBn
v(y) dy
with Bn := ∪ni=0[ti, si] . Since now, using (5.4) and our choice of δˆBn
v(y) dy ≤ˆBn
|v(y)− vN (y)| dy +
ˆBn
vN (y) dy ≤ ε
2+N |Bn| ≤
ε
2+Nδ = ε ,
our claim is proved.Finally, define
Uh(t) :=1
h(u(t+ h)− u(t)) .
By (5.2) and the Lebesgue differentiation Theorem we have that
Uh(t) =1
h
ˆ t+h
t
u′(y) dy → u′(t)
in L1([0, 1]) when h → 0, and therefore almost everywhere. Being Uh(t) exactly the differencequotients of u at t , we have shown that u is pointwise almost everywhere differentiable, and that itspointwise derivative coincides a.e. with u′ .
Exercise 3. Let U := (−1, 1)× (−1, 1). Define u as follows
u(x1, x2) :=
1− x1 in T1 := x1 > 0, |x2| < x11 + x1 in T2 := x1 < 0, |x2| < −x11− x2 in T3 := x2 > 0, |x1| < x21 + x2 in T4 := x2 < 0, |x1| < −x2
For which 1 ≤ p ≤ ∞ does u belong to W 1,p(U)?
Solution: Let us take ϕ := (ϕ1, ϕ2) ∈ C∞c (U ;R2) and define the vectors ν+ := (− 1√2, 1√
2) (unit
normal to the line x1 = x2) and ν− := ( 1√2, 1√
2) (unit normal to the line x1 = −x2). First, we
compute ˆT1
x1 divϕ(x1, x2) dx1dx2 .
Since div (x1ϕ)− (1, 0) ·ϕ = x1divϕ , by the divergence theorem (taking into account the orientationof the exterior normal!) we getˆ
T1
x1 divϕ(x1, x2) dx1dx2 = −ˆT1
(1, 0) · ϕ(x1, x2) dx1dx2 +ˆx1=x2; x1>0
ξ1 ϕ(ξ) · ν+ ds(ξ)−ˆx1=−x2; x1>0
ξ1 ϕ(ξ) · ν− ds(ξ) .(5.5)
Here we took into account that ϕ(1, x2) ≡ (0, 0). In a similar way we getˆT4
x2 divϕ(x1, x2) dx1dx2 = −ˆT4
(0, 1) · ϕ(x1, x2) dx1dx2 +ˆx1=−x2; x2<0
ξ2 ϕ(ξ) · ν− ds(ξ)−ˆx1=x2; x2<0
ξ2 ϕ(ξ) · ν+ ds(ξ) .
But this is clearly equivalent toˆT4
x2 divϕ(x1, x2) dx1dx2 = −ˆT4
(0, 1) · ϕ(x1, x2) dx1dx2 +
−ˆx1=−x2; x1>0
ξ1 ϕ(ξ) · ν− ds(ξ) +
ˆx1=x2; x1<0
ξ1 ϕ(ξ) · ν+ ds(ξ) .(5.6)
16 EXERCISES PDE
The same reasonings lead toˆT2
x1 divϕ(x1, x2) dx1dx2 = −ˆT2
(1, 0) · ϕ(x1, x2) dx1dx2 +ˆx1=−x2; x1<0
ξ1 ϕ(ξ) · ν− ds(ξ)−ˆx1=x2; x1<0
ξ1 ϕ(ξ) · ν+ ds(ξ)(5.7)
and ˆT3
x2 divϕ(x1, x2) dx1dx2 = −ˆT3
(0, 1) · ϕ(x1, x2) dx1dx2 +ˆx1=−x2; x1<0
ξ1 ϕ(ξ) · ν− ds(ξ)−ˆx1=x2; x1>0
ξ1 ϕ(ξ) · ν+ ds(ξ)(5.8)
Since ˆU
divϕ(x1, x2) dx1dx2 = 0
again by the divergence Theorem, we have that
−ˆU
udivϕ(x1, x2) dx1dx2 =
ˆT1
x1 divϕ(x1, x2) dx1dx2 −ˆT2
x1 divϕ(x1, x2) dx1dx2+ˆT3
x2 divϕ(x1, x2) dx1dx2 −ˆT4
x2 divϕ(x1, x2) dx1dx2 .
Using (5.5), (5.7), (5.8), and (5.6) we conclude that
−ˆU
udivϕ(x1, x2) dx1dx2 =
ˆU
v · ϕ(x1, x2) dx1dx2
where v ∈ L∞(U ;R2) is defined as
v(x1, x2) :=
(−1, 0) in T1
(1, 0) in T2
(0,−1) in T3
(0, 1) in T4 .
Therefore u ∈W 1,p(U) for every p .
Exercise 4. Suppose U is connected and u ∈ W 1,p(U) satisfies Du = 0 almost everywhere in U ,with Du the Sobolev gradient. Prove that u is (almost everywhere) constant in U .
Solution: Fix a relatively compact open subset U ′ of U ; it suffices to show u constant in U ′ . For εsufficiently small, now, the convolutions with the simmetric mollifiers ρε are well defined; furthermore,for every ϕ ∈ C∞(U ′;Rn), one has, by simmetry of mollifiers and since ρε ? ϕ ∈ C∞(U ;Rn) for εsmall enough, thatˆ
U ′divϕ(x)(ρε ? u)(x) dx =
ˆU
u(x)(ρε ? divϕ)(x) dx =
ˆU
u(x)div (ρε ? ϕ)(x) dx .
By the hypothesis, the right-hand side is 0, so that D(ρε ? u) = 0 on U ′ . By smoothness of ρε ? uthis implies that ρε ? u is constant in U ′ ; by Lp convergence, when ε is going to 0 we get that u isalmost everywhere equal to a constant, as required.
6. Exercises PDE 12.12.12-14.12.12
Exercise 1. (a) Let Ω be an open bounded subset of Rn . For every u ∈ C∞c (Ω), prove that thefollowing interpolation inequality holds:(ˆ
Ω
|Du(x)|2 dx)2
≤ C(ˆ
Ω
u(x)2 dx)( ˆ
Ω
|D2u(x)|2 dx), (6.1)
with D2u the Hessian matrix.(b) Assume ∂Ω ∈ C1 , let u ∈W 2,2(Ω) ∩W 1,2
0 (Ω). Prove that u satisfies (6.1).
EXERCISES PDE 17
Solution: (a) For every v ∈ C∞(Ω), it holds
v∆v + |Dv|2 = div (v ·Dv) . (6.2)
For v = u , integrating on Ω, by the divergence theorem we getˆΩ
|Du(x)|2 dx = −ˆ
Ω
u(x)∆u(x) dx
since u = 0 on ∂Ω. By the Cauchy-Schwarz inequality we then have(ˆΩ
|Du(x)|2 dx)2
≤(ˆ
Ω
u(x)2 dx)( ˆ
Ω
|∆u(x)|2 dx)≤ C
(ˆΩ
u(x)2 dx)( ˆ
Ω
|D2u(x)|2 dx),
where the last estimate is obvious. This proves (6.1).(b) By (6.2) and the divergence Theorem, we getˆ
Ω
|Dvn(x)|2 dx = −ˆ
Ω
vn(x)∆vn(x) dx+
ˆ∂Ω
vn(ξ)∂
∂νvn(ξ) dS(ξ) . (6.3)
Now, considering wn as in the hint, we obviously haveˆ∂Ω
vn(ξ)∂
∂νvn(ξ) dS(ξ) =
ˆ∂Ω
(vn − wn)(ξ)∂
∂νvn(ξ) dS(ξ)
since wn(ξ) ≡ 0. By the divergence Theorem, we then getˆ∂Ω
vn(ξ)∂
∂νvn(ξ) dS(ξ) =
ˆΩ
div [(vn(x)− wn(x))Dvn(x)] dx =ˆΩ
(vn(x)− wn(x))∆vn(x) dx+
ˆΩ
D(vn(x)− wn(x)) ·Dvn(x) dx .
Now, by W 2,2 convergence of vn to u , ∆vn and Dvn are bounded in L2 , while ‖vn−wn‖W 1,2(Ω) →‖u− u‖W 1,2(Ω) = 0 as n goes to +∞ , therefore
limn→+∞
ˆ∂Ω
vn(ξ)∂
∂νvn(ξ) dS(ξ) = 0 . (6.4)
By (6.4) and the W 2,2 convergence of vn to u , taking the limit in (6.3) we getˆΩ
|Du(x)|2 dx = −ˆ
Ω
u(x)∆u(x) dx
and we conclude as in part (a).
Exercise 2. Poincare’s inequality:
• Let u ∈W 1,20 ((0, d)). Prove that
ˆ d
0
u(x)2 dx ≤ d2
ˆ d
0
u′(x)2 dx . (6.5)
• Let u ∈W 1,20 ((0, d)× RN−1). Prove thatˆ
[0,d]×RN−1
u(x)2 dx ≤ d2
ˆ[0,d]×RN−1
|Du(x)|2 dx . (6.6)
Hint: Use density of C∞c functions.• Does the inequality hold for u ∈W 1,2((0, d)× RN−1)?
Solution: Let x and y ∈ [0, d] . We have already seen (where?) that
|u(x)− u(y)| ≤ |x− y| 12 (
ˆ d
0
|u′(t)|p dt) 12
18 EXERCISES PDE
and that u is absolutely continuous (in particular, continuous). Therefore u ∈ W 1,20 ([0, d]) implies
u(0) = 0 and then for y = 0 we have
|u(x)| ≤ |x| 12 (
ˆ d
0
|u′(t)|p dt) 12 ≤ d 1
2 (
ˆ d
0
|u′(t)|p dt) 12 ,
where the last estimate is trivial. Taking squares and integrating in [0, d] with respect to x , we get(6.5).
Now, let u ∈ C∞c ((0, d)×RN−1). Writing x = (x1, x) for every x ∈ [0, d]×RN−1 , with x1 ∈ [0, d]and x ∈ RN−1 , we have that for fixed x the function u(·, x) belongs to C∞c ((0, d)). We thereforehave, by (6.5), thatˆ d
0
u(x1, x)2 dx1 ≤ d2
ˆ d
0
( d
dx1u(x1, x)
)2
dx1 = d2
ˆ d
0
( ∂
∂x1u(x)
)2
dx1
for every x ∈ RN−1 . Now, by Fubini’s Theoremˆ[0,d]×RN−1
u(x)2 dx =
ˆRN−1
(ˆ d
0
u(x1, x)2 dx1
)dx ≤
d2
ˆRN−1
( ˆ d
0
( ∂
∂x1u(x)
)2
dx1
)dx ≤ d2
ˆ[0,d]×RN−1
|Du(x)|2 dx .
This proves (6.6) when u ∈ C∞c ((0, d) × RN−1). In the general case, there exists un ∈ C∞c (Ω) suchthat un → u in W 1,2(Ω). In particular
limn→∞
ˆ[0,d]×RN−1
un(x)2 dx =
ˆ[0,d]×RN−1
u(x)2 dx
and
limn→∞
ˆ[0,d]×RN−1
|Dun(x)|2 dx =
ˆ[0,d]×RN−1
|Du(x)|2 dx .
Since ˆ[0,d]×RN−1
un(x)2 dx ≤ d2
ˆ[0,d]×RN−1
|Dun(x)|2 dx ,
(6.6) follows by simply taking the limit.Such an inequality cannot hold in general in W 1,2(Ω). Constant functions in dimension N = 1
provide an easy counterexample.
Exercise 3. Let Ω be an open subset of Rn . We denote by C∞(Ω)W 1,∞(Ω)
the set of functions vin W 1,∞(Ω) such that there exists a sequence vn of C∞(Ω)-functions converging to v in the W 1,∞
norm on compact subsets of Ω.
• C∞(Ω)W 1,∞(Ω)
=?
• In case that C∞(Ω)W 1,∞(Ω)
6= W 1,∞(Ω), find a function v ∈ W 1,∞(Ω) that does not belong
to C∞(Ω)W 1,∞(Ω)
.
Solution: W 1,∞ convergence corresponds to uniform convergence of the sequence of functions andof the sequence of gradients. Therefore, since the limit of continuous functions must be continuous,one easily has that
C∞(Ω)W 1,∞(Ω)
⊆ C1(Ω) .
But also the converse inclusion holds! To prove this, it simply suffices to follow exactly the steps ofthe Meyers-Serrin Theorem (Theorem 2, Section 5.3.2 in the book of Evans); observe actually that instep 2, since now the functions ζiu are C1
c (U), by mollification one can find ui ∈ C∞c (Ω) such thatsuppui ⊂Wi and ‖ui − ζiu‖W 1,∞(Ω) ≤ δ
2i (why?). Therefore
C∞(Ω)W 1,∞(Ω)
= C1(Ω) .
EXERCISES PDE 19
To find the required example, given for instance Ω = B(0, 1), it suffices to find v ∈W 1,∞(B(0, 1))that does not belong to C1(B(0, 1)). Consider for instance v(x) := |x| . This is clearly no C1
function in B(0, 1), since it is not differntiable at 0. On the other hand, for every ε > 0 one has
v ∈ C1(B(0, 1) \B(0, ε)), so that, given ϕ ∈ C∞c (B(0, 1);Rn) one hasˆB(0,1)\B(0,ε)
v(x) divϕ(x) dx = −ˆB(0,1)\B(0,ε)
x
|x|· ϕ(x) dx−
ˆ∂B(0,ε)
|ξ|ϕ(ξ) · ν(ξ) dS(ξ) .
Since ∣∣∣ ˆ∂B(0,ε)
|ξ|ϕ(ξ) · ν(ξ) dS(ξ)∣∣∣ ≤ Nα(N)εN‖ϕ‖L∞(B(0,1);Rn)
and x|x| ∈ L
∞(B(0, 1);Rn), by taking the limit as ε→ 0 we get by dominated convergenceˆB(0,1)
v(x) divϕ(x) dx = −ˆB(0,1)
x
|x|· ϕ(x) dx ,
which proves that v ∈W 1,∞(B(0, 1)).
7. Exercises PDE 19.12.12
Exercise 1. Let Ω be the open subset of RN defined by Ω := B(0, 1) \ xN = 0 . Show that thefunction
u(x1, x2, . . . , xN ) :=
1 if xN > 0
0 if xN < 0
belongs to W 1,p(Ω) for every 1 ≤ p ≤ ∞ , but there exists no sequence vn of C∞(Ω) functions suchthat vn → u in W 1,p(Ω).
Solution: Being u a C∞ function on each connected component of Ω, it is certainly in W 1,p(Ω)for every 1 ≤ p ≤ ∞ . Also observe that u is defined almost everywhere in B(0, 1), and actuallybelongs to L∞(B(0, 1)), but does not belong to W 1,1(B(0, 1))! Indeed, suppose by contradiction thatW 1,1(B(0, 1)). Then, it exists g ∈ L1(B(0, 1);Rn) such thatˆ
B(0,1)
u(x)divϕ(x) dx = −ˆB(0,1)
g(x) · ϕ(x) dx
for every ϕ ∈ C∞c (B(0, 1);Rn). Now, let us define B(0, 1)+ := B(0, 1) ∩ xN > 0 and similarlyB(0, 1)− := B(0, 1)∩xN < 0 . Since Du = 0 in B(0, 1)+ , and obviously every ϕ ∈ C∞c (B(0, 1)+;Rn)also belongs to C∞c (B(0, 1);Rn), it must beˆ
B(0,1)+g(x) · ϕ(x) dx = 0
for every ϕ ∈ C∞c (B(0, 1)+;Rn). Thus, g = 0 a.e. in B(0, 1)+ . Arguing similarly in B(0, 1)− , weget g = 0 a.e. in B(0, 1). We should therefore haveˆ
B(0,1)
u(x)divϕ(x) dx = 0 (7.1)
for every ϕ ∈ C∞c (B(0, 1);Rn). But on the other hand, by definition of u and the divergence theoremwe have ˆ
B(0,1)
u(x)divϕ(x) dx =
ˆB(0,1)+
divϕ(x) dx = −ˆB(0,1)∩xN=0
ϕ(ξ) · eN (ξ) dS(ξ)
for every ϕ ∈ C∞c (B(0, 1);Rn), and this has clearly no reason of being 0! So, u /∈W 1,1(B(0, 1)).Now, if a sequence vn of C∞(Ω) functions such that vn → u in W 1,p(Ω) exists, we have that vn
and Dvn are Cauchy sequences in Lp(Ω) and Lp(Ω;Rn) respectively. Since B(0, 1) and Ω differ bya set of Lebesgue measure 0, we also have that vn and Dvn are Cauchy sequences in Lp(B(0, 1)) andLp(B(0, 1);Rn), respectively. Thus, vn has a limit u in W 1,p(B(0, 1)), and it must be u = u almost
20 EXERCISES PDE
everywhere in B(0, 1). But since the definition of weak derivative does not depend on the Lebesguerepresentative, this would imply u ∈W 1,p(B(0, 1)), a contradiction.
Exercise 2. Let Ω be a bounded open subset of Rn .(a) Let f ∈W 1,p(Ω) and g ∈W 1,q(Ω) with 1 ≤ p, q ≤ ∞ . Find a sufficient condition on p and q
such that fg ∈W 1,1(Ω) and compute the weak gradient D(fg).(b) Show that when N = 1 it suffices to take p = q = 1.(c) Kettenregel: assume that F : R → R ∈ C1 ∩W 1,∞(R). Show that for every u ∈ W 1,p(Ω) the
composition F (u) ∈W 1,p(Ω) and compute the weak gradient.
Solution: (a) Assume that p and q are conjugate exponents, that is 1p + 1
q = 1. By Meyers-Serrin’s
Theorem, there exist two sequences fn and gn of C∞(Ω) functions such that fn → f in W 1,p(Ω)and gn → g in W 1,q(Ω). Now, by Holder’s and Minkowsky’s inequalities
‖fngn − fmgm‖1 = ‖fn(gn − gm) + (fn − fm)gm‖1 ≤ ‖fn‖p‖gn − gm‖q + ‖gm‖q‖fn − fm‖pfor every n and m in N . Thus, fngn is a Cauchy sequence in L1(Ω), and similarly we can provethat D(fngn) = fnDgn + gnDfn is a Cauchy sequence in L1(Ω;Rn). So, fngn is Cauchy in W 1,1(Ω)and is converging to fg in L1 . It follows that fg is in W 1,1(Ω) and that fngn is converging to fgin W 1,1(Ω). In particular,
D(fg) = fDg + gDf .
(b) It suffices to consider the case of an interval, say (0, 1). So, let f and g ∈ W 1,1((0, 1)). We
begin with the case where f and g ∈ W 1,10 ((0, 1)). Take fn and gn sequences of C∞c functions
converging to f , and g , respectively, in W 1,1((0, 1)). Since for every x ∈ [0, 1], one has
|fn(x)− f(x)| ≤ˆ 1
0
|f ′n(y)− f ′(y)| dy
we get that actually fn converges to f uniformly in [0, 1], and clearly, also gn converges to g uniformlyin [0, 1]! Thus, fngn converges to fg uniformly in [0, 1]. On the other hand
(fngn)′ = fng′n + gnf
′n ;
by the uniform convergences of fn and gn , and the L1 convergences of f ′n and g′n , we easily get that
(fngn)′ → fg′ + gf ′
in L1((0, 1)). This proves that fg belongs to W 1,10 ((0, 1)).
In the general case, we only can take fn and gn sequences of C∞([0, 1]) functions converging tof , and g , respectively, in W 1,1((0, 1)). By the fundamental Theorem of calculus, we easily get, forevery m and n ∈ N
|fn(0)− fm(0)| ≤ |fn(x)− fm(x)|+ˆ 1
0
|f ′n(y)− f ′m(y)| dy
so that integrating from 0 to 1, it is
|fn(0)− fm(0)| ≤ˆ 1
0
|fn(x)− fm(x)| dx+
ˆ 1
0
|f ′n(y)− f ′m(y)| dy
which yields that fn(0) is a Cauchy sequence. Now, again by the fundamental theorem of calculus,we have
|fn(x)− fm(x)| ≤ |fn(0)− fm(0)|+ˆ 1
0
|f ′n(y)− f ′m(y)| dy
for every x ∈ (0, 1), so that fn is uniromly Cauchy. It follows that fn converges to f uniformlyin [0, 1], and similarly gn converges to g uniformly in [0, 1]. Arguing as in the previous step, theconclusion follows.
EXERCISES PDE 21
(c) Since F : R → R ∈ C1 ∩W 1,∞(R), for every u ∈ W 1,p(Ω) the functions F (u), F ′(u), andF ′(u)Du are a.e. well defined and belong to Lp(Ω), Lp(Ω), and Lp(Ω;Rn), respectively. Now, takinga sequence un of C∞(Ω) functions such that un → u in W 1,p(Ω), we first have
‖F (un)− F (u)‖Lp(Ω) ≤ ‖F ′‖L∞(R)‖un − u‖Lp(Ω)
which gives F (un)→ F (u) in Lp(Ω). By boundedness of F ′ , the convergence
F ′(un)Du→ F ′(u)Du in Lp(Ω;Rn)
can be proved by dominated convergence. Finally, one has
limn‖F ′(un)Dun − F ′(u)Du‖Lp(Ω;Rn) = lim
n‖F ′(un)Dun − F ′(un)Du‖Lp(Ω;Rn)
≤ ‖F ′‖L∞(R) limn‖Dun −Du‖Lp(Ω;Rn) = 0 ,
whence it follows that F (u) ∈W 1,p(Ω), and that the weak gradient is given by F ′(u)Du .
Exercise 3. Let Ω be a bounded open subset of Rn , and u ∈ W 1,p(Ω). Define as usual u+(x) :=maxu(x), 0 , and u−(x) := max−u(x), 0 . Show that u+ , u− , and |u| ∈ W 1,p(Ω), too, andcompute the corresponding weak gradients.
Hint: u+ = limε→0 Fε(u), with
Fε(z) :=
(z2 + ε2)
12 − ε if z ≥ 0
0 if z < 0 .
Solution: Take Fε as in the hint, and observe that Fε ∈ C1 ∩W 1,∞(R). Therefore, the previousexercise gives that Fε(u) ∈W 1,p(Ω) and that
D(Fε(u)) = F ′ε(u)Du = u+(ε2 + u2)−12Du ,
where the last equality follows by a direct computation. Since u+(ε2 + u2)−12 → χu>0 a.e. in Ω,
by dominated convergence we get that
D(Fε(u))→ χu>0Du
in Lp(Ω;Rn). It follows that u+ belongs to W 1,p(Ω), and that
D(u+) = χu>0Du .
Since u− = u+ − u and |u| = u+ − u− we get that u− , and |u| ∈ W 1,p(Ω), too, and the weakgradients are given by
D(u−) = −χu≤0Du and D(|u|) = χu>0Du− χu≤0Du .
8. Exercises PDE 9-11.01.13
Exercise 1. Prove the following multiplicative form of the trace inequality: if u ∈W 1,2(Rn+), denot-
ing with Tu the trace of u on the hyperplane RN−1 one has
‖Tu‖2L2(RN−1) ≤ C‖u‖L2(RN+ )‖Du‖L2(RN+ ) .
Solution: Being Rn+ an extension domain, for every u ∈ W 1,2(Rn+) there exists a sequence un of
functions in C∞c (Rn) converging to u in W 1,2(Rn+). Now, for every x ∈ RN−1 by the fundamentalTheorem of calculus one has
u2n(x, 0) = −
ˆ +∞
0
∂
∂xNu2n(x, xN ) dxN = −2
ˆ +∞
0
un(x, xN )∂
∂xNun(x, xN ) dxN
for every n ∈ N . This implies by Fubini’s TheoremˆRN−1
u2n(x, 0) dx = −2
ˆRn+un(x)
∂
∂xNun(x) dx .
22 EXERCISES PDE
Since for every n and every x the trace Tun(x) of un on the hyperplane RN−1 is simply equal toun(x, 0), the previous equality and the Cauchy-Schwarz inequality yield
‖Tun‖2L2(RN−1) ≤ 2 ‖un‖L2(Rn+) ‖Dun‖L2(Rn+) .
Since un is converging to u in W 1,2(Rn+), by continuity of the trace operator Tun is converging
to Tu in L2(RN−1), so that we get
‖Tu‖2L2(RN−1) ≤ 2 ‖u‖L2(Rn+) ‖Du‖L2(Rn+) .
by letting n going to +∞ , as required.
Exercise 2. Let Ω be an open set of Rn , and Γ a C1 hypersurface such that Ω = Ω1 ∪Ω2 ∪Γ withΩ1 and Ω2 open connected disjoint.
(a) Let f1 ∈ C1(Ω1) and f2 ∈ C1(Ω2). Under which conditions the function f ∈ L∞(Ω) definedby f = f1 in Ω1 and f = f2 in Ω2 belongs to W 1,∞(Ω)?
(b) Let p ≥ 1, f1 ∈W 1,p(Ω1) and f2 ∈W 1,p(Ω2). Under which conditions the function f ∈ Lp(Ω)defined by f = f1 in Ω1 and f = f2 in Ω2 belongs to W 1,p(Ω)?
Solution: (a) Denote with νΓ the normal vector to the hypersurface Γ pointing from Ω1 to Ω2 . Bythe Divergence Theorem we have, for every ϕ ∈ C∞c (Ω),ˆ
Ω
f(x)divϕ(x) dx =
ˆΩ1
f1(x)divϕ(x) dx+
ˆΩ2
f2(x)divϕ(x) dx =
−ˆ
Ω1
Df1(x) · ϕ(x) dx−ˆ
Ω2
Df2(x) · ϕ(x) dx+
ˆΓ
(f1(ξ)− f2(ξ))ϕ(ξ) · νΓ(ξ) dS(ξ)
therefore f ∈W 1,∞(Ω) if and only if f1 = f2 on Γ. In that case one also has
Df =
Df1 in Ω1
Df2 in Ω2 .
(b) It suffices to repeat the previous reasonings with suitable using of the notion of trace to deducethat the required condition is Tf1 = Tf2 a.e. in Γ, where T denotes the trace operator from W 1,p(Ω1)to Lp(∂Ω1), and from W 1,p(Ω2) to Lp(∂Ω2), respectively.
Exercise 3. Give an example of a connected open set Ω ∈ Rn and of a function u ∈W 1,∞(Ω) suchthat u is not Lipschitz continuous on Ω.
Solution: It simply suffices to modify the example in Exercise 1, 19.12.12. Namely, set Ω : =B(0, 2) \ xN = 0, |x| ≤ 1
2 , which is connected, and let for instance
f(x) :=
(|x|2 − 1
4 )2 if xN > 0, |x| ≤ 12
0 elsewhere .
Denoting with B(0, 1
2
)+the upper hemisphere B
(0, 1
2
)∩ xN > 0 , one has by the Divergence
theorem that for any ϕ ∈ C∞c (Ω;Rn)ˆΩ
f(x)divϕ(x) dx =
ˆB(
0,12
)+ f(x)divϕ(x) dx = −4
ˆB(
0,12
)+(|x|2 − 14 )x · ϕ(x) dx ;
indeed, no boundary term appears since either f or ϕ are 0 on ∂B(0, 1
2
)+. It follows that f ∈
W 1,∞(Ω). On the other hand, for every 0 < ε < 12 , one has
|f(0, . . . , 0, ε)− f(0, . . . , 0,−ε)||(0, . . . , 0, ε)− (0, . . . , 0,−ε)|
=(ε2 − 1
4 )2
2ε
which is unbounded when ε is close to 0. Therefore f cannot be Lipschitz continuous.
EXERCISES PDE 23
9. Exercises PDE 16-18.01.13
Exercise 1. Let Ω be a bounded domain with smooth boundary and u ∈ W 1,1(Ω). Assume thatDu ∈ Lp(Ω) for some 1 < p <∞ . Prove that u ∈W 1,p(Ω).
Hint: Prove initially that u ∈ Lploc(Ω), to understand the exercise. Taking a closer look to theproof of Theorem 3, Section 5.3.3 of Evans’ book can help near to the boundary.
Solution: First step: let us prove that u ∈ Lploc(Ω), which requires no regularity of the boundary.Define un = %n ? u ∈ C∞(Rn), and fix a smooth open subset Ω′ ⊂⊂ Ω. For n large, depending onΩ′ , it holds Dun = %n ?Du in Ω′ (the formula Dun = D%n ?u is instead true on the whole Rn : whythis difference?). We therefore deduce by Jensen’s inequality (how?) that
‖Dun‖Lp(Ω′) ≤ ‖Du‖Lp(Ω) . (9.1)
By Poincare-Wirtinger and (9.1), if un,Ω′ denotes the integral mean of un on Ω′ we get that thereexists a constant CΩ′ depending on Ω′ such thatˆ
Ω′|un(x)− un,Ω′ |p dx ≤ CΩ′‖Du‖pLp(Ω) .
Since un is converging to u in L1(Ω′), by Fatou’s Lemma, denoting with uΩ′ the integral mean of uon Ω′ , we arrive at ˆ
Ω′|u(x)− uΩ′ |p dx ≤ CΩ′‖Du‖pLp(Ω)
and this implies the claim.Second step: suitably changing coordinates like in Theorem 3, Section 5.3.3 of Evans’ book, for
every x0 ∈ ∂Ω one can find a neighborhood V := B(x0, r) ∩ Ω and a positive number λ > 0 suchthat the ball B(x + λ
neN ,1n ) ⊂ Ω for all x ∈ V . Therefore, if one defines un(x) = u(x + λ
neN ) one
has un ∈W 1,1(V ) and since we only did a translation of the argument
‖Dun‖Lp(V ) ≤ ‖Du‖Lp(Ω) . (9.2)
Now, defining vn = %n ? un ∈ C∞(Rn), by condition B(x + λneN ,
1n ) ⊂ Ω it holds Dvn = %n ? Dun .
Notice also that un and thus vn converge to u in L1(V ). Therefore, the same argument used in thefirst step plus (9.2) proves that u in Lp(V ).
Third step: Ω can be written as a finite union Ω = ∪mi=1Vi where V0 ⊂⊂ Ω and each Vi , i ≥ 1 issuch that the second step can be applied. Sinceˆ
Ω
|u(x)|p dx ≤m∑i=1
ˆVi
|u(x)|p dx
the finiteness of the right-hand side gives u ∈ Lp(Ω). Since Du ∈ Lp(Ω) by the hypothesis, theexercise is concluded.
Exercise 2. The geometric counterpart of the Sobolev inequality.(a) (optional, if not simply assume the result). Let f ∈ L1(Rn;Rm). Prove thatˆ
Rn|f(x)| dx = sup
ˆRnf(x) · ϕ(x) dx : ϕ ∈ C∞c (Rn;Rm), ‖ϕ‖∞ ≤ 1
. (9.3)
(b) Let C1,N the Sobolev constant in Rn , i.e. the least possible constant such that the Sobolevinequality
‖u‖ NN−1≤ C‖Du‖1
holds for every u ∈W 1,1(Rn). Prove the following isoperimetric inequality: for every smooth boundedopen set Ω ⊂ Rn one has
|Ω|N−1N ≤ C1,NHn−1(∂Ω) , (9.4)
where |Ω| is the Lebesgue measure of Ω and Hn−1(∂Ω) is the (Hausdorff, or surface) measure of theboundary. Hint: approximate the characteristic function χΩ by convolution and use (9.3) and thedivergence theorem to estimate the L1 norm of the gradients.
24 EXERCISES PDE
Solution: (a) Define
f1(x) :=
f(x)|f(x)| if f(x) 6= 0
0 otherwise
Let %n be the usual mollifiers, and, given a sequence ϕn of C∞c (Rn) functions with 0 ≤ ϕn ≤ 1 withϕn(x)→ 1 for every x , set gn := ϕn(%n ?f1). It is clear that gn ∈ C∞c (Rn;Rm) and that ‖gn‖∞ ≤ 1.Furthermore gn converges to f1 a.e. Observing that f(x) · f1(x) = |f(x)| for every x , by dominatedconvergence one hasˆ
Rn|f(x)| dx = lim
n
ˆRnf(x) · gn(x) dx ≤ sup
ˆRnf(x) · ϕ(x) dx : ϕ ∈ C∞c (Rn;Rm), ‖ϕ‖∞ ≤ 1
.
The converse inequality is trivial so that we get (9.3).(b) By (9.3) and integrating by parts we get that for every u ∈W 1,1(Rn) one has
‖Du‖1 = sup ˆ
Rnudivϕ : ϕ ∈ C∞c (Rn;Rn), ‖ϕ‖∞ ≤ 1
. (9.5)
Now, if %n are the usual mollifiers, we have for every ϕ ∈ C∞c (Rn;Rn) with ‖ϕ‖∞ ≤ 1 that also‖%n ? ϕ‖∞ ≤ 1. Therefore, by symmetry of the mollifiers, standard properties of the convolution andalso using the divergence theorem, we haveˆ
Rn(%n ? χΩ)(x) divϕ(x) dx =
ˆRnχΩ(x) div (%n ? ϕ)(x) dx =
ˆΩ
div (%n ? ϕ)(x) dx =ˆ∂Ω
(%n ? ϕ)(ξ) · ν(ξ) dHn−1(ξ) ≤ Hn−1(∂Ω)
for every smooth bounded open set Ω ⊂ Rn . It follows from this and (9.5) that for every n ∈ N onehas
‖D(%n ? χΩ)‖1 ≤ Hn−1(∂Ω) (9.6)
for every smooth bounded open set Ω ⊂ Rn .Now, combining (9.6) with the Sobolev inequality gives
‖%n ? χΩ‖ NN−1≤ C1,NHn−1(∂Ω)
for every smooth bounded open set Ω ⊂ Rn and every n ∈ N . Letting n to +∞ we get
‖χΩ‖ NN−1≤ C1,NHn−1(∂Ω)
which is exactly (9.4).
Exercise 3. Let p > 2. Show with a counterexample that the Morrey’s Imbedding Theorem is notverified in the nonsmooth two-dimensional domains
Dα := (x, y) : 0 < x < 1, 0 < y < xα
for α sufficiently large.Hint: prove that there are unbounded funtions in W 1,p(Dα).
Solution: Let α > p− 1. We can choose γ > 0 such that α+1γ+1 > p , which is equivalent to
α− (γ + 1)p > −1 . (9.7)
Now, consider the function
uγ(x, y) := x−γ .
The function uγ is unbounded in Dα so that it cannot belong to a space of Holder functions. On theother hand by Fubini’s theorem one hasˆ
Dα
|uγ(x, y)|p dxdy =
ˆ 1
0
xα−γp dx < +∞
EXERCISES PDE 25
since by (9.7) one gets α− γp > −1. Similarly, being Du(x, y) = (−γx−(1+γ), 0) one getsˆDα
|Duγ(x, y)|p dxdy = γpˆ 1
0
xα−(γ+1)p dx < +∞
again using Fubini’s theorem and (9.7). Therefore uγ ∈ W 1,p(Dα) but it cannot satisfy Morrey’simbedding Theorem.
10. Exercises PDE 23-25.01.13
Exercise 1. Fourier analysis in Sobolev spaces. Let f ∈ L2((0, π)). Define for every N ∈ N
fN (t) :=2
π
N∑n=1
an sin(nt)
where an =´ π
0f(t) sin(nt) dt . It is known that fN converges to f in L2((0, π)) (if you need to
convince yourself that no cosine is needed to have L2 convergence, simply apply the general theoremto the odd extension of f to [−π, π]).
(a) Assume that f ∈ H10 ((0, π)). Show that fN converges to f uniformly in [0, π] . Hint: prove
fN to be a Cauchy sequence in some Sobolev space.(b) If f ∈ H1
0 ((0, π)) ∩H2((0, π)) prove that there exists a constant C such that
‖fN − f‖H1 ≤ C
N‖f‖H2 . (10.1)
Solution: (a) Since f(0) = f(π) = 0 integrating by parts we obtain
nan = −ˆ π
0
f ′(t) cos(nt) dt (10.2)
for every n ∈ N . Since √
2π cos(nt) : n ∈ N is an orthonormal system of L2(0, π), Bessel’s inequality
givesN∑n=1
n2a2n =
N∑n=1
( ˆ π
0
f ′(t) cos(nt) dt)2
≤ π
2‖f ′‖22
for every N ∈ N . Being a convergent series,∑Nn=1 n
2a2n is a Cauchy sequence.
On the other hand, since √
2π cos(nt) : n ∈ N is an orthonormal system of L2(0, π), for every
N ∈ N and every M ∈ N with M ≥ N , one has
‖f ′M − f ′N‖22 = ‖M∑n=N
nan cos(n·)‖22 =
M∑n=N
n2a2n‖ cos(n·)‖22 =
M∑n=N
n2a2n . (10.3)
It follows that fN is a Cauchy sequence in L2((0, π)). Since fN was converging to f in L2((0, π)),combining the two gives that fN is converging to f in H1
0 ((0, π)), and therefore uniformly by Morrey’simbedding Theorem.
(b) If f ∈ H2((0, π)) we can once more integrate by parts in (10.2), obtainingˆ π
0
f ′(t) cos(nt) dt = − 1
n
ˆ π
0
f ′′(t) sin(nt) dt
so that
n2an =
ˆ π
0
f ′′(t) sin(nt) dt
and, again using Bessel’s inequality,N∑n=1
n4a2n ≤
π
2‖f ′′‖22
26 EXERCISES PDE
for every N ∈ N . Combining this with (10.3) for every N ∈ N and every M ∈ N with M ≥ N , onehas
‖f ′M − f ′N‖22 =
M∑n=N
n2a2n ≤
1
N2
M∑n=N
n4a2n ≤
π
2N2‖f ′′‖22 .
When M goes to +∞ we therefore obtain
‖f ′ − f ′N‖2 ≤√π
2
1
N‖f ′′‖2
Since by Poincare’s inequality
‖f − fN‖2 ≤ π‖f ′ − f ′N‖2we get that
‖f − fN‖H10 ((0,π)) ≤
√π
2(1 + π)
1
N‖f ′′‖2
as required.
Exercise 2. Cea’s Lemma: Consider a symmetric, bounded, coercive bilinear form B : H ×H → Ron a Hilbert space H , that is
B(u, v) = B(v, u) , B(u, v) ≤ C1‖u‖H‖v‖H , B(u, u) ≥ c0‖u‖2H .
(a) Fix a linear continuous functional F : H → R . Justify that there exists a unique u ∈ H suchthat
B(u, v) = F (v)
for every v ∈ H .(b) Fix additionally an N -dimensional subspace HN of H . Justify the following fact: there exist
a unique approximate solution uN ∈ HN , that is a unique uN ∈ HN such that
B(uN , vN ) = F (vN )
for every vN ∈ HN .(c) Prove the following equality
B(u− uN , u− uN ) +B(vN − uN , vN − uN ) = B(u− vN , u− vN ) (10.4)
for every vN ∈ HN . Deduce Cea’s estimate:
‖u− uN‖H ≤√
C1
c0inf‖u− vN‖ : vN ∈ HN . (10.5)
(d) Consider now H = H10 ((0, π)) and HN := spant 7→ sin(nt) : n = 1, . . . , N . For a(x) ∈
C1[0, π] with 1 ≤ a(x) ≤ 2 for every x and f ∈ L2((0, π)) set
B(u, v) :=
ˆ π
0
a(t)u′(t) v′(t) dt (10.6)
and F (v) :=´ π
0f(t) v(t) dt . Check that u and uN as in (a) and (b), respectively, exist, and show
that uN → u in H10 ((0, π)) as N goes to +∞ . If additionally u ∈ H2((0, π)) show that there exists
a constant C such that
‖u− uN‖H1 ≤ C
N‖u‖H2 .
Solution: (a) follows from the Lax-Milgram theorem. The same holds for (b) since the restriction ofB to the Hilbert subspace HN ×HN is clearly bilinear, coercive and continuous.
(c) By bilinearity and simmetry of the form B we have
B(u− uN , u− uN ) = B(u, u)− 2B(u, uN ) +B(uN , uN )
B(vN − uN , vN − uN ) = B(vN , vN )− 2B(uN , vN ) +B(uN , uN )
B(u− vN , u− vN ) = B(u, u)− 2B(u, vN ) +B(vN , vN )
EXERCISES PDE 27
therefore (10.4) is equivalent to prove that
2(B(uN , uN )−B(u, uN ) +B(u, vN )−B(uN , vN )) = 0 .
This can be easily proved to be true by observing that by definition of u and uN , for every vN ∈ HN
one has
B(u, vN ) = F (vN ) = B(uN , vN )
and in particular, for vN = uN ,
B(u, uN ) = B(uN , uN ) .
Now, by (10.4), and the coercivity and continuity of B we get for every vN ∈ HN
c0‖u− uN‖2H ≤ B(u− uN , u− uN ) ≤ B(u− uN , u− uN ) +B(vN − uN , vN − uN ) =
B(u− vN , u− vN ) ≤ C1‖u− vN‖2Hwhence
‖u− uN‖H ≤√
C1
c0‖u− vN‖H
for every vN ∈ HN . Taking the infimum in the rigth-hand side gives (10.5).(d) To prove existence of u and uN when B is given by (10.6), it suffices to check that the Lax-
Milgram theorem can be applied. The only relevant point to this end is to prove coercivity of B ,which follows from the Poincare inequality. Namely, one has
B(u, u) ≥ ‖u′‖22 ≥1
1 + π2‖u‖2H1 .
It follows now from the previous exercise that
inf‖u− vN‖H10 ((0,π)) : vN ∈ HN → 0
as N goes to +∞ . In particular, by (10.1), if u ∈ H2 one has for every N ∈ N
inf‖u− vN‖H10 ((0,π)) : vN ∈ HN ≤
C
N‖u‖H2 .
Combining these two facts with (10.5) gives the conclusion.
Exercise 3. Consider a bounded smooth open subset Ω ⊂ Rn , and ΓD ⊂ ∂Ω with Hn−1(ΓD) > 0.Let
H1ΓD (Ω) := u ∈ H1(Ω), Tu = 0 on ΓD ,
with T the trace operator. For λ > 0 define
Bλ(u, v) =
ˆΩ
∇u(x) · ∇v(x) dx− 1
λ
ˆ∂Ω
Tu(ξ)Tv(ξ) dHn−1(ξ)
for u and v ∈ H1ΓD
(Ω). Prove that, when λ is sufficiently large, for every f ∈ L2(Ω) there exists a
unique uf ∈ H1ΓD
(Ω) such that
Bλ(uf , v) =
ˆΩ
f(x) v(x) dx
for every v ∈ H1ΓD
(Ω). Hint: does the Poincare inequality apply in H1ΓD
(Ω)?
Solution: Following the hint, let us first prove that there exists a constant C such that
‖u‖L2(Ω) ≤ C‖∇u‖L2(Ω) (10.7)
for every u ∈ H1ΓD
(Ω). To prove this, it suffices to argue as in the proof of the Poincare-Wirtingerinequality (Theorem 1, Section 5.8.1 in Evans’book): contradicting (10.7) would produce the existenceof a sequence un in H1
ΓD(Ω) with ‖un‖|L2(Ω) = 1 for every n which is converging to a constant c in
H1(Ω). By continuity of the trace operator, H1ΓD
(Ω) is a Hilbert subspace of H1(Ω), so it must be
c = 0, since this one is the only constant in H1ΓD
(Ω). On the other hand, by strong L2 convergence
we must have ‖c‖L2(Ω) = c|Ω| 12 = 1, and this gives a contradiction.
28 EXERCISES PDE
Now, let C be given by (10.7), and let CT the continuity constant of the trace operator. Assumethat λ > CT (1 + C2). Then for every u ∈ H1
ΓD(Ω) one has, by the continuity of the trace and by
(10.7), that
Bλ(u, u) = ‖∇u‖2L2(Ω) −1
λ‖Tu‖2L2(∂Ω) ≥
1
1 + C2‖u‖2H1(Ω) −
CTλ‖u‖2H1(Ω) = c0‖u‖2H1(Ω)
with c0 := 11+C2 − CT
λ > 0 due to the assumption on λ . Therefore for λ > CT (1 + C2) the bilinear
form Bλ is coercive on H1ΓD
(Ω). For any λ > 0 one has also by the Cauchy-Schwarz inequality andthe continuity of the trace operator that
Bλ(u, v) ≤ (1 +CTλ
)‖u‖H1(Ω)‖v‖H1(Ω)
so that Bλ is continuous. The conclusion follows by the Lax-Milgram theorem.
11. Exercises PDE 30.01.13-01.02.13
Exercise 1. Let H be an Hilbert space, a : H × H → R a bilinear, symmetric, coercive, andcontinuous form, and L : H → R a linear continuous functional. Prove that u ∈ H solves
a(u, v) = L(v) for every v ∈ H (11.1)
if and only if
1
2a(u, u)− L(u) = min
1
2a(v, v)− L(v) : v ∈ H
. (11.2)
Hint: if u is a minimizer, fix v ∈ H and consider the function F (t) := 12a(u+ tv, u+ tv)−L(u+ tv),
t ∈ R .
Solution: “If” part: assume that u ∈ H satisfies (11.2). Fix v ∈ H and consider the functionF (t) := 1
2a(u + tv, u + tv) − L(u + tv), t ∈ R . The function F attains then its minimum value fort = 0. Since by a direct computation
F (t) =1
2a(u, u)− L(u) + t(a(u, v)− L(v)) +
t2
2a(v, v) (11.3)
imposing F ′(0) = 0 we get (11.1).“Only if” part: assume that u ∈ H satisfies (11.1). Fix v ∈ H and, again, consider the function
F (t) := 12a(u+ tv, u+ tv)− L(u+ tv), t ∈ R . By (11.3) and (11.1) we obtain
F (t) =1
2a(u, u)− L(u) +
t2
2a(v, v) .
By coerciveness of a , a(v, v) ≥ 0 which implies that F attains its minimum at 0. In particularF (0) ≤ F (1), therefore
1
2a(u, u)− L(u) ≤ 1
2a(u+ v, u+ v)− L(u+ v) (11.4)
for every v ∈ H . Now, for an arbitrary w ∈ H , applying (11.4) with v = w − u one gets
1
2a(u, u)− L(u) ≤ 1
2a(w,w)− L(w)
that is (11.2).
EXERCISES PDE 29
Exercise 2. Let Ω ⊂ Rn be a bounded open set with smooth boundary, and f ∈ L2(Ω). We saythat u ∈ H1(Ω) is a weak solution of the Neumann problem
−∆u = f in Ω∂u∂ν = 0 on ∂Ω
(11.5)
if ˆΩ
∇u(x) · ∇v(x) dx =
ˆΩ
f(x)v(x) dx (11.6)
for any v ∈ H1(Ω).(a) Justify that actually (11.6) is a weak formulation of the problem (11.5) by showing that if u
solves (11.6) and in addition u ∈ H2(Ω) then −∆u = f in L2(Ω) and ∂u∂ν = 0 in L2(∂Ω) in the sense
of traces.(b) Show that a necessary and sufficient condition for the existence of a solution of (11.6) isˆ
Ω
f(x) dx = 0 . (11.7)
Hint: use Lax-Milgram’s theorem in the Hilbert space
H := u ∈ H1(Ω) :
ˆΩ
u(x) dx = 0 . (11.8)
(c) Let g ∈ L2(∂Ω). Generalise part (a) and (b) to the nonhomogeneuous Neumann problem−∆u = f in Ω∂u∂ν = g on ∂Ω
(11.9)
by finding its weak formulation and a necessary and sufficient condition for existence of a weak solution.
Solution: (a) If u solves (11.6) and in addition u ∈ H2(Ω), for every v ∈ C∞c (Ω) integrating byparts and using (11.6) we have
−ˆ
Ω
∆u(x)v(x) dx =
ˆΩ
∇u(x) · ∇v(x) dx =
ˆΩ
f(x)v(x) dx
which implies −∆u = f in L2(Ω). Using this additional information, integrating by parts and usingagain (11.6) we get for every v ∈ H1(Ω) ˆ
Ω
f(x)v(x) dx =
−ˆ
Ω
∆u(x)v(x) dx =
ˆΩ
∇u(x) · ∇v(x) dx−ˆ∂Ω
∂u
∂ν(ξ)v(ξ) dHn−1(ξ) =ˆ
Ω
f(x)v(x) dx−ˆ∂Ω
∂u
∂ν(ξ)v(ξ) dHn−1(ξ) ,
which gives ∂u∂ν = 0 in L2(∂Ω).
Observe in addition that via integration by parts also the converse holds: if u ∈ H2(Ω) solves(11.5), then (11.6) holds.
(b) Taking v = 1 in (11.6) gives that (11.7) is necessary. To prove that it is sufficient, defining thesubspace H as in (11.8), we first prove that, if u ∈ H satisfiesˆ
Ω
∇u(x) · ∇v(x) dx =
ˆΩ
f(x)v(x) dx (11.10)
for every v ∈ H , then (11.6) holds. Indeed, when v ∈ H1(Ω), then v = v − 1|Ω|
´Ωv(x) dx belongs to
H ; furthermore, ∇v = ∇v and, by (11.7),ˆΩ
f(x)v(x) dx =
ˆΩ
f(x)v(x) dx .
30 EXERCISES PDE
Therefore (11.10) implies (11.6). Now, existence of a unique solution to (11.10) in H follows by theLax-Milgram theorem once it is checked that the bilinear form B : H ×H → R defined by
B(u, v) :=
ˆΩ
∇u(x) · ∇v(x) dx
is continuous and coercive. Continuity is obvious, while coercivity follows by the Poincare-Wirtingerinequality
‖∇u‖2 ≥ CΩ‖u‖2for every u ∈ H (add the details!).
(c) The solution is very similar to the previous case, so I only sketch it. Integrating by parts, oneeasily gets that a smooth solution of (11.9) satisfiesˆ
Ω
∇u(x) · ∇v(x) dx =
ˆΩ
f(x)v(x) dx+
ˆ∂Ω
g(ξ)v(ξ) dHn−1(ξ) . (11.11)
This is therefore our candidate weak formulation. Testing (11.11) on C∞c functions we get that if usolves it and in addition u ∈ H2(Ω) then −∆u = f in L2(Ω). Using this additional information,integrating by parts and using again (11.11) we get for every v ∈ H1(Ω)ˆ
Ω
f(x)v(x) dx =
−ˆ
Ω
∆u(x)v(x) dx =
ˆΩ
∇u(x) · ∇v(x) dx−ˆ∂Ω
∂u
∂ν(ξ)v(ξ) dHn−1(ξ) =ˆ
Ω
f(x)v(x) dx−ˆ∂Ω
(∂u∂ν
(ξ)− g(ξ))v(ξ) dHn−1(ξ) ,
which gives ∂u∂ν = g in L2(∂Ω).
We then show that (11.11) has a solution if and only ifˆΩ
f(x) dx+
ˆ∂Ω
g(ξ) dHn−1(ξ) = 0 . (11.12)
Indeed, taking v = 1 in (11.11) produces (11.12) as a necessary condition. On the other hand, if(11.12) holds, exactly as in part (b) we can show that, defining the subspace H as in (11.8), if u ∈ Hsatisfies ˆ
Ω
∇u(x) · ∇v(x) dx =
ˆΩ
f(x)v(x) dx+
ˆ∂Ω
g(ξ)v(ξ) dHn−1(ξ) . (11.13)
for every v ∈ H , then (11.11) holds. Existence of a solution to (11.13) in H follows again by theLax-Milgram Theorem; indeed, the bilinear form B : H ×H → R defined by
B(u, v) :=
ˆΩ
∇u(x) · ∇v(x) dx
is continuous and coercive, while the linear functional
L(v) :=
ˆΩ
f(x)v(x) dx+
ˆ∂Ω
g(ξ)v(ξ) dHn−1(ξ)
is continuous from H to R by continuity of the trace operator.
Exercise 3. Let Ω ⊂ Rn be a bounded open set with smooth boundary, with ∂Ω = Γ1 ∪ Γ2 , andassume that Hn−1(Γ1) > 0. Let f ∈ L2(Ω). Find the weak formulation of the mixed boundary-valueproblem
−∆u = f in Ω
u = 0 on Γ1
∂u∂ν = 0 on Γ2
(11.14)
and prove existence of a weak solution by using Lax-Milgram’s theorem in a suitable Hilbert subspaceof H1(Ω).
EXERCISES PDE 31
Solution: The presence of the Dirichlet boundary condition suggests to operate in the Hilbert sub-space
H := u ∈ H1(Ω), u = 0 on Γ1 ,where we have already seen that the Poincare inequality
‖∇u‖2 ≥ CΩ‖u‖2holds (short reminder: the only constant function in H is 0 and ∂Ω is smooth).
The weak formulation is: finding u ∈ H such thatˆΩ
∇u(x) · ∇v(x) dx =
ˆΩ
f(x)v(x) dx (11.15)
for every v ∈ H . Indeed, any smooth solution of (11.14) satisfies (11.15). Conversely, if u ∈ Hsatisfies (11.15) and additionally u ∈ H2(Ω), testing (11.15) on C∞c functions gives −∆u = f inL2(Ω). Using this additional information, integrating by parts and using again (11.15) we get forevery v ∈ H ˆ
Ω
f(x)v(x) dx =
−ˆ
Ω
∆u(x)v(x) dx =
ˆΩ
∇u(x) · ∇v(x) dx−ˆ∂Ω
∂u
∂ν(ξ)v(ξ) dHn−1(ξ) =ˆ
Ω
f(x)v(x) dx−ˆ
Γ2
∂u
∂ν(ξ)v(ξ) dHn−1(ξ) ,
since v = 0 on Γ1 . This gives ∂u∂ν = 0 in L2(Γ2). Since u ∈ H we also have u = 0 a.e. on Γ1 , so
(11.15) is the weak formulation of (11.14).Existence of a unique solution to (11.15) in H follows again by the Poincare inequality and the
Lax-Milgram Theorem.
12. Exercises PDE 14.02.13
Exercise 1. Differentiation of convolutions revisited. Recall preliminarly the following form of theJensen inequality: let F ∈ L∞(Rn), F ≥ 0 withˆ
RnF (y) dy = 1 .
Then for every convex function ϕ : R→ R and every g ∈ L1(Rn) it holds
ϕ(ˆ
Rng(y)F (y) dy
)≤ˆRnϕ(g(y))F (y) dy .
In the following %ε is a positive symmetric C∞ mollifier withˆRn%ε(y) dy = 1 (12.1)
and supp%ε ⊂ B(0, ε).(a) Let 1 ≤ p <∞ and f ∈W 1,p(Rn). Prove that D(%ε ? f) = %ε ? Df in Lp(Rn).(b) Let 1 ≤ p <∞ and f ∈W 1,p(Rn). Prove that ‖%ε ? f‖W 1,p(Rn) ≤ ‖f‖W 1,p(Rn) .
(c) Let Ω be a bounded open subset of Rn and f ∈ W 1,10 (Ω). Prove that D(%ε ? f) = %ε ? Df in
L1(Ω). Hint: extend f by setting it equal to 0 outside Ω. Is this extension in W 1,1(Rn)?(d) Let Ω be a bounded open subset of Rn and f ∈W 1,1(Ω). Define
Ωε := x ∈ Ω : dist (x, ∂Ω) ≥ ε .
Let Ω′ ⊂⊂ Ωε . Prove that D(%ε ? f) = %ε ? Df in L1(Ω′). Hint: try first with C∞ functions, thenapproximate.
32 EXERCISES PDE
Solution: (a) Let ϕ ∈ C∞c (Rn;Rn). Using the properties of the convolution with the mollifiers andthe definition of weak gradient we haveˆ
Rn%ε ? f(x) divϕ(x) dx =
ˆRnf(x) %ε ? divϕ(x) dx =ˆ
Rnf(x) div (%ε ? ϕ)(x) dx = −
ˆRnDf(x) · %ε ? ϕ(x) dx = −
ˆRn%ε ? Df(x) · ϕ(x) dx
as required.(b) Applying Jensen’s inequality with F (y) = %ε(x − y) for fixed x and using Fubini’s Theorem
and (12.1), we get ˆRn|%ε ? f(x)|p dx =
ˆRn
∣∣∣ ˆRn%ε(x− y)f(y) dy
∣∣∣p dx ≤ˆRn
ˆRn%ε(x− y)|f(y)|p dy dx =
ˆRn|f(y)|p dy
which proves ‖%ε ? f‖Lp(Rn) ≤ ‖f‖Lp(Rn) . The same argument with Df in place of f gives the claim.
(c) Let f be the extension of f simply obtained by f = 0 out of Ω. Trivially f ∈ W 1,1(Rn).
Indeed, since f ∈W 1,10 (Ω) there is a sequence fn ∈ C∞c (Ω) converging to f in W 1,1 . The extensions
fn simply obtained by fn = 0 out of Ω form now easily a Cauchy sequence in W 1,1(Rn) with limit
f . Step (a) gives now D(%ε ? f) = %ε ?Df . Since Df = Df in Ω and Df = 0 outside, we conclude.(d) Let us only see the proof when f ∈ C∞(Ω) leaving to the reader the easy approximation
argument. Since supp%ε ⊂ B(0, ε) and by symmetry of the convolution
%ε ? f(x) =
ˆRn%ε(x− y)f(y) dy =
ˆRn%ε(y)f(x− y) dy =
ˆB(0,ε)
%ε(y)f(x− y) dy .
Now, when x ∈ Ω′ , the gradient Df(x − y) (in the variable x) exists for every y ∈ B(0, ε), sincein this case x − y ∈ Ω where f is differentiable. We can therefore derive under the sign of integral,getting the required equality.
Exercise 2. Let N = 2 and B(0, 1) ⊂ R2 be the unit ball in the plane.(a) For 0 < r < 1 let ur be the unique solution of the problem
∆ur = 0 in B(0, 1) \B(0, r)
ur(x) = 1 if |x| = 1
ur(x) = 0 if |x| = r .
Compute ur .(b) Define
vr(x) :=
ur(x) in B(0, 1) \B(0, r)
0 if |x| ≤ r .Prove that vr ∈ H1(B(0, 1)) and that
‖vr − 1‖H1 → 0
when r → 0.(c) Let f ∈ C1(B(0, 1)). Construct a sequence fr with suppfr ⊂ B(0, 1)\B(0, r) and ‖fr−f‖H1 →
0.(d) Deduce that H1
0 (B(0, 1)\0) = H10 (B(0, 1)).
(e) Let N = 1. Is that true that H10 ((−1, 1)\0) = H1
0 ((−1, 1))? Hint: Sobolev convergence isparticularly strong in dimension 1.
Solution: (a) The radial symmetry of the domain and of the data suggests a radial solution. Arguingin the same way used to construct the fundamental solution and imposing the boundary conditionsone gets
ur(x) =log(|x|)| log r|
+ 1 .
EXERCISES PDE 33
(b) vr are indeed Lipschitz continuous since obtained by gluing together two C1 functions withthe same trace on the interface (see Exercise 2, 9-11.01.13). Furthermore, using polar coordinatesˆ
B(0,1)
|vr(x)− 1|2 dx = |B(0, r)|+ 2π
log2 r
ˆ 1
r
% log2 % d% ≤ |B(0, r)|+ 2Cπ
log2 r.
Indeed, the function %→ % log2 % is uniformly bounded in (0, 1] since it has a limit (namely 0) when%→ 0. Now, the right-hand side vanishes when r → 0. About the first derivatives, again by a directcomputation and using polar coordinatesˆ
B(0,1)
|D(vr(x)− 1)|2 dx =
ˆB(0,1)\B(0,r)
|Dur(x)|2 dx =2π
log2 r
ˆ 1
r
1
%d% =
2π
| log r|which again vanishes when r → 0. Thus, the claim is proved.
(c) Set fr := vrf . Clearly suppfr ⊂ B(0, 1) \ B(0, r). Furthermore, since f is C1 up to theboundary, we easily have
‖fr − f‖H1 ≤ ‖f‖C1‖vr − 1‖H1
which gives the claim.(d) We need to use a double approximation argument. First, of all, observe that clearly
C∞c (B(0, 1\0) ⊂ H10 (B(0, 1)) ;
taking the closure at the left-hand side, the inclusion H10 (B(0, 1)\0) ⊂ H1
0 (B(0, 1)) is trivial. Now,let f ∈ H1
0 (B(0, 1)) and let fn ∈ C∞c (B(0, 1)) a sequence H1 -converging to f . By step (c) applied toeach fn , for every n we can construct gn ∈ H1
0 (B(0, 1)\0) such that ‖gn−fn‖H1 ≤ 1n (construct a
sequence approximating fn and then choose a sufficiently close element of the sequence). Now, easilygn converges to f in H1 ! Since H1
0 (B(0, 1)\0) is a Banach space, f ∈ H10 (B(0, 1)\0), and this
proves the reverse inclusion.(e) No. If the equality holds, any f ∈ H1
0 ((−1, 1)) could be approximated by a sequence fn ∈C∞c ((−1, 1)\0) in the H1 topology, and in particular, by Morrey’s imbedding in dimension 1,uniformly. This would imply f(0) = 0 which has no general reason to be.
Exercise 3. Let Ω ⊂ Rn be a smooth bounded connected open set, and u ∈ C∞(Ω) a solution of the
elliptic equation Lu = −∑Ni,j=1 aij(x) ∂2
∂xi∂xju = 0 in Ω. Assume that the coefficients aij ∈ C1(Ω).
Show that for λ sufficiently large not depending on u the function v := |∇u|2 + λu2 satisfies Lv ≤ 0in Ω. Deduce that
‖∇u‖L∞(Ω) ≤ C(‖∇u‖L∞(∂Ω) + ‖u‖L∞(∂Ω)
). (12.2)
Solution: We called A(x) the matrix whose coefficients are aij(x) and we denote with 〈·, ·〉 thescalar product between vectors in Rn . The norm we consider on matrices is the Frobenius norm.
We start by observing that by a direct computation, for any w ∈ C∞(Ω) one has
Lw2(x) = −N∑
i,j=1
aij(x)∂2
∂xi∂xjw2(x) = −〈A(x)∇w(x),∇w(x)〉+ 2w(x)Lw(x) .
For w =√λu , since u is a solution and by ellipticity, we get
L(λu2(x)) ≤ −λα|∇u(x)|2 (12.3)
with α the coercivity constant of A . For k = 1, . . . , N and w = ∂u∂xk
we have
L(( ∂u
∂xk
)2
(x))≤ −α
∣∣∣∇(∂u
∂xk)(x)
∣∣∣2 + 2∂u
∂xk(x)L
( ∂u∂xk
(x)).
By a direct computation and since Lu = 0 we have
L( ∂u∂xk
(x))
=∂
∂xkLu(x) +
N∑i,j=1
∂
∂xkaij(x)
∂2
∂xi∂xju(x) =
N∑i,j=1
∂
∂xkaij(x)
∂2
∂xi∂xju(x) .
34 EXERCISES PDE
Since aij ∈ C1(Ω) we get easily
2∂u
∂xk(x)L
( ∂u∂xk
(x))≤ CN |∇u(x)||D2u(x)|
with D2u the Hessian matrix and CN a constant depending on the C1 norm of the aij ’s and on thedimension N . Summing over k from 1 to N we have
L(|∇u|2(x)) ≤ −α|D2u(x)|2 + CN |∇u(x)||D2u(x)| . (12.4)
We now recall Young’s inequality in the following form: for every a, b ∈ R and ε > 0 one has
ab ≤ ε2
2a2 +
1
2ε2b2 . (12.5)
Using now (12.3), (12.4) and (12.5) with a = |D2u(x)| , b = |∇u(x)| and ε =√
α2CN
we obtain
L(|∇u|2(x) + λu2(x)) ≤ −λα|∇u(x)|2 +4C2
N
α|∇u(x)|2 .
For λ ≥(
2CNα
)2the right-hand side is nonpositive so that setting v := |∇u|2 + λu2 one has Lv ≤ 0
in Ω. By the weak maximum principle
‖v‖L∞(Ω) ≤ ‖v‖L∞(∂Ω) .
Now, trivially ‖∇u‖2L∞(Ω) ≤ ‖v‖L∞(Ω) . It is also easy to se that there is a constant C depending on
N and λ , thus on N , CN and α such that ‖v‖L∞(∂Ω) ≤ C(‖∇u‖L∞(∂Ω) + ‖u‖L∞(∂Ω)
)2
. From this,
(12.2) follows.