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Riesz operator is injective and, in the finite dimensional setting, automatically surjective as thedual space is of the same dimension as the original space. In order to determine the matrixrepresentation of R, we consider vectors Rej ,

(Re1)(y) = (e1, y) = y1 = (y1 − y2) + y2 = e∗1(y) + e∗2(y)

(Re2)(y) = (e2, y) = y1 + 2y2 = (y1 − y2) + 3y2 = e∗1(y) + 3e∗2(y)

The matrix representation of operator R is thus:(1 11 3

).

• The transpose RT goes from the bidual to the dual space. In the finite dimensional case (in fact,for any Hilbert space), the bidual is identified (canonically isomorphic) with the original space.Due the symmetry of the inner product, transpose RT coincides with R. Indeed,

〈Ru, v〉V ′×V = (u, v)V = (v, u)V = 〈Rv, u〉V ′×V = 〈u,Rv〉V ′′×V ′ .

Consequently, matrix representation of transpose RT coincides with that of R.

Area A-CSEMay 2014

2. A metric space problem. Let X be a set and ρ1(x, y), ρ2(x, y) two metrics on X . Define:

d(x, y) := maxρ1(x, y), ρ2(x, y) . (0.2)

• Is d also a metric on X ? Prove or disprove.

• If the answer to the first question is positive, you have three topologies in X corresponding tothe three metrics. Discuss the relative strength of the corresponding topologies (which one isstronger or weaker than others ?). Hint: Recall the definition of bases of neighborhoods in ametric space.

Answers:

• Yes, it is.

Positive definitness: If d(x, y) = 0 then both ρ1(x, y) = ρ2(x, y) = 0 which implies thatx = y.

Symmetry: We have:ρi(x, y) = ρi(y, x), i = 1, 2 .

Apply maxi=1,2 to both sides.

Triangle inequality: Start with:

ρi(x, y) ≤ ρi(x, z) + ρi(z, y) ≤ maxj=1,2

ρj(x, z) + maxj=1,2

ρj(z, y), j = 1, 2 ,

and take maximum with respect to i on both sides.

• Let Bd(x, ε) and Bρi(x, ε) denote balls corresponding to metrics d and ρi, resp. Inequality

ρi(x, y) ≤ d(x, y), i = 1, 2

implies thatBd(x, ε) ⊂ Bρi(x, ε), i = 1, 2 .

Consequently, if Bd,Bρi denote the bases of neighborhoods in topologies generated by d and ρi,resp., then

Bρi Bd

which demonstrates that metric topogy corresponding to d is stronger than both topologies cor-responding to metrics ρi. We cannot draw any general conclusion about the relative strength ofmetric topologies corresponding to ρi, i = 1, 2.

Area A-CSEMay 2014

3. Contraction Maps. Consider the following Initial-Value Problem (IVP):dq

dt= t ln(q(t)), t > 0

q(0) = 1

• State Banach Contractive Map Theorem.

Answer:Let (X, d) be a complete metric space. Let D ⊂ X (then (D, d) is itself a metric space, too...),and A : D → D is a contraction, i.e.

d(A(f), A(g)) ≤ k d(f, g), ∀f, g ∈ D, k < 1

Then function A has a unique fixed point in set D.

• Use the theorem to prove local existence and uniqueness of solution to the IVP, i.e. that thereexists an interval (0, T ) in which the equation is satisfied. Provide a concrete value of T .

Solution:The problem is equivalent to the solution of the integral equation:

q(t) = 1 +

∫ t

0s ln(q(s)) ds

Consider the Chebyshev space C[0, T ] (with unknown T at this point...) and define the map Ausing the right-hand side of the equation above:

(Aq)(t) = 1 +

∫ t

0s ln(q(s)) ds

First of all, we need to define a set D ⊂ C[0, T ] such that map A sets the set D into itself.Assume that q(t) will vary in the box:

D = q ∈ C[0, T ] : e−1 ≤ q(t) ≤ e, 0 ≤ t ≤ T (0.3)

(notice that the box includes the initial value q = 1). Then −1 ≤ ln q(t) ≤ 1, i.e. | ln q(t)| ≤ 1.Consequently,

|∫ t

0s ln q(s) ds| ≤

∫ t

0s ds =

1

2t2

so,

|(Aq)(t)− 1| ≤ 1

2T 2

This gives two bounds for T . From the right:

(Aq)(t) ≤ 1 +1

2T 2 ≤ e ⇒ T ≤

√2(e− 1) ,

Area A-CSEMay 2014

and from the left:

e−1 ≤ 1− 1

2T 2 ≤ (Aq)(t) ⇒ T ≤

√2(1− e−1) .

Now, map A must be a contraction. With flux F (s, q) = s ln q,

|∂F∂q| = s|1

q| ≤ es

so, with q coming form box (0.3), the flux satisfies the Lipschitz condition:

|F (s, q1)− F (s, q2)| ≤ es|qs − q2| .

This leads to the estimate;

|(Aq1)(t)− (Aq2)(t)| ≤∫ t

0es ds ‖q1 − q2‖C[0,T ] ≤

e

2T 2 ‖q1 − q2‖C[0,T ] .

Consequently, a sufficient condition for a contraction is

T <

√2

e.

In conlusion, the IVP will have a unique solution for

T < min√

2(e− 1),√

2(1− e−1),

√2

e .

Area A-CSEMay 2014

4. Consider the following initial-value problem.x+ x = δ(t− 1), t > 0

x(0) = 1, x(0) = 0 ,

where δ denotes the Dirac’s delta “function”.

(a) Define precisely delta functional and reinterpret its action in terms of appropriate jump condi-tions at t = 1 for the solution x(t) and its derivative x(t).

Dirac’s delta at t = 1 is a functional that assigns to every test function φ its value at t = 1,

D(IR) 3 φ→ φ(1) ∈ IR

Delta is the distributional derivative of Heaviside function. Its presence translates into jumpconditions at t = 1,

[x(1)] = 0, [x(1)] = 1

(b) Utilize the jump conditions and solve the problem using elementary means.

For t ∈ (0, 1),x(t) = A+Be−t

Utilizing IC, we getx(t) = 1

For t ∈ (1,∞),x(t) = A+Be−(t−1)

Utilizing jump conditions at t = 1 and the known value of x and x at t = 1−, we get

x(t) = 2− e−(t−1)

(c) Define the Laplace transform. Apply it to both sides of the equation and find the solution in theLaplace domain.

L(f)(s) = f(s) =

∫ ∞0

f(t)e−st dt∫ ∞0

e−stδ(t− 1) dt = e−s

Recall the formulas resulting from integration by parts,

x = sx(s)− x(0)

x = s2x(s)− sx(0)− x(0)

Area A-CSEMay 2014

Transforming both sides of the equation and accounting for the IC, we get;

(s2 + s)x− s− 1 = e−s

which gives the solution in the Laplace domain,

x(s) =1

s+

e−s

s(s+ 1)

(d) Use the Residue Theorem to compute the inverse Laplace transform of the solution in theLaplace domain and compare it with the solution obtained using the elementary calculus.

The following is just a sketch, look up your lecture notes for a detailed explanation.

First term, case: t < 0. Use contour to the right to conclude that x = 0.

First term, case: t > 0. Simple pole at s = 0. Use contour to the left to conclude that

x = Re0est

s= lim

s→0est = 1

The argument showing that the integral over CR vanishes in the limit, needs use of the LebesgueDominated Convergence Theorem.

Second term, case: t < 1. Use contour to the right to conclude that x = 0.

Second term, case: t > 1. Simple poles at s = 0,−1. Use contour to the left to conclude that

x = Re0(es(t−1)

s(s+ 1)) + Re−1(

es(t−1)

s(s+ 1)) = 2− e−(t−1)

The argument showing that the integral over CR vanishes in the limit, is now easy; the denomi-nator is O(R2).

Summing up, we get the result coinciding with the elementary solution.

Area A-CSEMay 2014

5. (a) Recall the formula for the gradient in polar coordinates,

∇u =∂u

∂rer +

1

r

∂u

∂θeθ

and use integration by parts to derive the corresponding formula for the Laplacian.

Integrate by parts in r, θ,∫Ω

(∂u

∂r

∂v

∂r+

1

r2

∂u

∂θ

∂v

∂θ

)rdr dθ =

∫Ω∇u∇v = −

∫Ω

∆u v + boundary terms

to obtain:

∆u =1

r

∂

∂r(r∂u

∂r) +

1

r2

∂2u

∂θ2

(b) Use separation of variables and whatever means you need, to solve the boundary-value problemsshown in Fig. 1.

Figure 1: Steady-state heat conduction in a wedge domain.

Boundary condition at θ = α,u = 10

prompts for looking first for a particular solution in the form

u = 10Θ(θ)

Substituting into the formula for the Laplacian, we learn that Θ must satisfy the equation

Θ′′ = 0

Along with the BC,Θ(0) = 0, Θ(α) = 1

Area A-CSEMay 2014

this leads to the final form of the particular solution,

u =10

αθ

Now, we look for the ultimate solution in the form,

u(r, θ) =10

αθ + v(r, θ)

where v satisfies homogeneous BCs at θ = 0, α and the nonhomogeneous BC at r = a,

v(a, θ) = −10

αθ

We can use now the standard separation of variables,

v = R(r)Θ(θ)

to arrive atr(rR′)′

R= −Θ′′

Θ= λ

Operator in θ is self-adjoint and positive-definite, so λ = k2, k > 0. We get

Θ = A sin kθ +B cos kθ

BC at θ = 0 implies B = 0, and BC at θ = α implies that

sin kα = 0 =⇒ kα = nπ, n = 1, 2, . . .

Thus,

k = kn =1

αnπ, n = 1, 2, . . .

This leads to Cauchy-Euler eqn in r and the final solution in the form,

v =∞∑n=1

Anrkn sin(knθ)

Coefficients An are computed using L2-orthogonality of sin(knθ),

An = − 10

αakn

∫ α0 θ sin(knθ) dθ∫ α0 sin2(knθ) dθ

Area A-CSEMay 2014

6. Let D be the region shown in the Fig. 2, whose surface S consists of three pieces.

Bottom: S1 : z = f1(x, y), (x, y) ∈ R

Top: S2 : z = f2(x, y), (x, y) ∈ R

Side: S3 : f1(x, y) ≤ z ≤ f2(x, y), (x, y) on C.

where R is the projection of D onto the xy-plane and C is the boundary of R.

State and prove the Divergence Theorem for F = Q(x, y, z)k, where Q is continuous and has contin-uous first partial derivatives on D. Assume that functions f1, f2 and curve C are sufficiently regular,e.g. C1.

Figure 2: A domain in IR3.

Proof: See class notes. This problem is about 1D integration by parts and recalling the definition ofsurface integral.

Area A-CSEMay 2014

Area A-CSEMay 2015

CES Math Qual Exam 2017Tuesday, May 30, 2017, 9:00am - noon.

Solve the following six problems in 3 hours.

1. A linear algebra problem. Consider IR3. LetA be the rotation about the x1 axis by an angle α (positiveα corresponds to the “right-hand rule”).

• Write down an explicit formula for map A : IR3 → IR3, x = (x1, x2, x3)→ y = (y1, y2, y3), i.e.express y in terms of x (and α). Is map A linear ? Explain why ?

• Let ei, i = 1, 2, 3, be the canonical basis for IR3. Write down matrix representation of map A inbasis ei.

• Consider the canonical inner product in IR3,

(x, y) :=3∑i=1

xiyi .

Is the map self-adjoint with respect to the inner product ? Explain, why ?

• Consider vectors a1 = e1, a2 = e1 + e2, a3 = e3. Do they form a basis for IR3 ? Explain, why ?

• Write down the matrix representation of map A in the new basis.

• Determine dual basis a∗j , j = 1, 2, 3.

Showing the non-trivial details only. The map:y1 = x1y2 = cos θx2 − sin θx3y3 = sin θx2 + cos θx3

Matrix representation: 1 0 00 cos θ − sin θ0 sin θ cos θ

The matrix is non-symmetric so this is not a self-adjoint map.

Relating the new basis to the old one:

a1 = e1a2 = e1 + e2a3 = e3

e1 = a1e2 = a2 − a1e3 = a3

Area A-CSEMay 2017with solutions

Computing the representation in the new basis:

Aa1 = Ae1 = e1 = a1Aa2 = Ae1 +Ae2 = e1 + cos θe2 + sin θe3 = (1− cos θ)a1 + cos θa2 + sin θa3Aa3 = Ae3 = sin θe2 + cos θe3 = − sin θa1 + sin θa2 + cos θa3

to obtain: 1 1− cos θ − sin θ0 cos θ sin θ0 sin θ cos θ

Representing x in the new basis:

x = x1e1 + x2e2 + x3e3 = (x1 − x2)a1 + x2a2 + x3a3

so,a∗1(x) = x1 − x2a∗2(x) = x2

a∗3(x) = x3


2. Equivalence of norms on a finite dimensional space. Consider IRN .

• Define the l1 norm ‖x‖1 in IRN and demonstrate that indeed it satisfies the three defining condi-tions for a norm.

Trivial.

• State the Weierstrass Theorem for an arbitrary topological space.

See the book.

• Let ‖x‖ be now any other norm defined on IRn.

(i) Show that there exists a constant C > 0 such that,

‖x‖ ≤ C‖x‖1 ∀x ∈ IRN

(ii) Use (i) to demonstrate that function

IRN 3 x→ ‖x‖ ∈ IR

is continuous in l1-norm.

(iii) Use the Weierstrass Theorem to conclude that there exists a constant D > 0 such that

‖x‖1 ≤ D‖x‖ ∀x ∈ IRN

Conclude that the l1 norm is equivalent to any other norm on IRN . Explain why the result impliesthat any two norms defined on an arbitrary finite-dimensional vector space must be equivalent.

(i) Let ei denote the canonical basis in IRn. Then

‖x‖ = ‖n∑i=1

xiei‖ ≤n∑i=1

|xi| ‖ei‖ ≤ Cn∑i=1

|xi|

whereC = max‖e1‖, . . . , ‖en‖

(ii) This follows immediately from the fact that

|‖x‖ − ‖y‖| ≤ ‖x− y‖

and property (i).

(iii) The l1 unit ball is compact. Consequently, norm ‖ · ‖ attains a minimum on the l1 unit ball,i.e.,

C ≤ ‖ x

‖x‖1‖ ∀x

Positive definitness of the norm implies that C > 0. Multiplying by ‖x‖1/C, we get

‖x‖1 ≤ C−1‖x‖

Take now two arbitrary norms. As each of them is equivalent to norm ‖ · ‖1, they must beequivalent with each other as well.


3. Application of Banach Contractive Map Theorem

• State the Contractive Map Theorem. Make sure to list all assumptions.

• Consider the following initial value problem:dudt = (u− 1)2(t− 1), t ∈ (0, T )

u(0) = 1

where, at this point, T is unknown. Use elementary means to solve the problem. Comment on amaximum T for which the solution exists in interval (0, T ).

• Use the Contractive Map Theorem to prove that the solution exists (kind of after dinner exercisefor this example) and provide a concrete estimate for interval length T . How close can you getto the optimal T determined in the first step ?

Solution:

• See the book.

• If you separate the variables: ∫ u(t)

1

du

(u− 1)2=

∫ t

0(t− 1) dt

and obtain:1

(1− u)|u(t)1 =

t(t− 2)

2|10 =

t(t− 2)

2

then, evaluating the left-hand side at u = 1, you obtain a singular term. So this leads to nowhere.There are two ways out. The first one is to change the integration bounds to:

1

(1− u)|u(t)u(ε) =

t(t− 2)

2|1ε =

(t+ ε− 2)(t− ε)2

to obtain:1− u(t) =

1(t+ε−2)(t−ε)

2 + 11−u(ε)

=2(1− u(ε))

(t+ ε− 2)(t− ε)(1− u(ε)) + 2

Passing with ε→ 0, we see that the right hand side converges to zero. Consequently, u(t) = 1 isthe (constant) solution. The second way out is simply to notice that that u(t) = 1 is the solutionto the problem. Clearly, the solution is determined for any t.

• Integrate both sides of the ODE and use the initial condition to arrive at the equivalent integralequation:

u(t) = 1 +

∫ t

(u(s)− 1))2 (s− 1) ds︸︷︷︸=:Au


Clearly, u(t) is a solution of the equation above if and only if u(t) is a fixed point on operatorA. Let T be a parameter to be specified later. Consider the usual Chebyshev space C[0, T ] witha subset K ⊂ C[0, T ],

K := u ∈ C[0, T ] : |u(t)− 1| ≤ 1, t ∈ [0, T ] .

We shall make sure first that A maps K into itself, i.e. A : K → K is well defined. Assumeu ∈ K. Clearly, the integrand in the definition of A is bounded and, therefore, Au is Lipschitzcontinuous. Moreover, assuming T ≤ 1, we have,

|(Au)(t)− 1| ≤ |∫ t

0|u(s)− 1|2︸︷︷︸

≤1

|s− 1|ds ≤ −(1− s)2

2|t0 =

1

2− (1− t)2

2≤ 1

2< 1 .

Thus, A is well defined if T ≤ 1. Now, investigate under what conditions A is a contraction.

|(Au)(t)− (Av)(t)| ≤ |∫ t

0

[(u(s)− 1)2 − (v(s)− 1)2

](s− 1) ds|

≤∫ t

0|u(s)− v(s)|︸︷︷︸≤‖u−v‖

|(u(s)− 1) + (v(s)− 1)|︸︷︷︸≤2

|s− 1| ds

≤ (1− (1− T )2)‖u− v‖ .

Consequently, A is a contraction if T < 1. Compare with the analytical solution to see that theestimate is rather conservative.


4. Solve the Neumann problem on the disk of radius 1 centered about the origin:

∆u(r, θ) = 0, 0 ≤ r < 1,−π ≤ θ < π,

∂u

∂r(1, θ) = f(θ), −π ≤ θ < π,

with the necessary condition that ∫ π

−πf(θ)dθ = 0.

Why is this a necessary condition? Justify. Show in your solution process where this condition comesinto play.

Solution:

The condition on f(θ) is necessary by the Divergence Theorem. Applying the Divergence Theoremto the differential equation we find

0 =

∫ π

−π

∂u

∂rdθ =

∫ −ππ

f(θ)dθ.

Using separation of variables we find that the solution should be of the form

u(r, θ) = C +∞∑n=1

anrn cos(nθ) + bnr

n sin(nθ).

Then∂u

∂r(1, θ) = f(θ) =

∞∑n=1

nan cos(nθ) + nbn sin(nθ).

This is a Fourier-type expansion of f(θ). The constant term in this expansion is∫ π

−πf(θ)dθ

and this must be zero for the Fourier expansion to be valid. The coefficients are

an =1

nπ

∫ π

−πf(ξ) cos(nξ)dξ

andbn =

1

nπ

∫ π

−πf(ξ) sin(nξ)dξ

The constant C is undetermined and the solution is unique only up to a constant.


5. Find the inverse Laplace transform (using the Residue Theorem and complex integration) for thefunction

f(s) =1

(s2 − 4)(s− 1)2, s ∈ IC

for t > 0.

Solution:

The inverse Laplace transform is1

2πi

∫ γ+i∞

γ−i∞F (s)estds

for γ sufficiently large. We consider the contour integral 12πi

∫C F (s)estds over a contour C that goes

from γ − iR to γ + iR and closes by a semicircle CR to the left with R large enough so that all polesare inside the contour. The poles of F (s) are at 2,−2 and a pole of strength 2 at s = 1. Therefore wetake γ > 2. Applying the Residue Theorem we get

res2F (s)est =1

4e2t

res−2F (s)est = − 1

36e−2t

res1F (s)est =1

3tet − 2

9et.

On CR the integral is O(R−3) thus it goes to zero as R → ∞ and the contour integral converges tothe inverse transform. Therefore

f(t) =1

4e2t − 1

36e−2t − 1

3tet − 2

9et


6. Find the Principal Solution (Green’s function) for the Helmholtz operator ∆ + k2I on IR2 up to amultiplicative constant.

Solution:

Switching to polar coordinates and using the fact that the solution should be angle-independent, weget that the solution G should satisfy

Grr +1

rGr + k2G = 0

away from the δ point. Multiplying by r2 and defining r′ = kr we see that this is a Bessel equationof order zero in r′. Therefore

G = AJ0(kr) +BY0(kr)

The function which has the right behavior as r → 0 is the function Y0, therefore we conclude that Gis proportional to Y0(kr). The constant B turns out to be 1/4.


CES Math Qual Exam 2018Thursday, May 31, 2018, 9:00 am - noon.

Solve the following six problems in 3 hours.

1. A linear algebra problem. Let ω = (1, 1, 1) ∈ IR3. Consider the map:

IR3 3 x→ y = Ax := ω × x ∈ IR3 .

• Prove that map A is linear.

• Find the matrix representation of map A in the canonical basis ei.

• Determine inverse map A−1.

• Determine transpose map AT and its matrix representation with respect to (wrt) the dual basise∗j .

• Determine adjoint map A∗ with respect to canonical inner product in IRn.

• Consider a weighted inner product in IR3:

(x,y)w := x1y1 + 2x2y2 + 3x3y3 ,

and determine the adjoint of A wrt to the weighted inner product.

Solution:

• This follows immediately from the fact that the cross product is a bilinear operation. It alsofollows from its explicit representation: y1

y2

y3

=

0 −1 11 0 −1−1 1 0

x1

x2

x3

.

• See above.

• It does not exist. A has a non-trivial null space,

N (A) = αω = (α, α, α) : α ∈ IR .

• Matrix representation of AT in the canonical dual basis equals simply the transpose of matrixrepresentation of A,

AT ==

0 1 −1−1 0 1

1 −1 0

= −A

• Just the transpose, A∗ = AT = −A. A is so-called skew-adjoint.

• This is the only part where you have to do some work.

(Ax,y)w = (x3 − x2)y1 + 2(x1 − x3)y2 + 3(x2 − x1)y3

= x1(2y2 − 3y3) + x2(−y1 + 3y3) + x3(y1 − 2y2)

= x1(2y2 − 3y3) + 2x2(−12y1 + 3

2y3) + 3x3(13y1 − 2

3y2)

= (x, A∗y)w

so the matrix representation of the new adjoint in the canonical basis is: 0 2 −3

−12 0 3

2

13 −2

3 0

.

2. An integration exercise.

• State the Lebesgue Dominated Convergence Theorem.

• Let Ω ⊂ IRn be an arbitrary unbounded open set, and let f ∈ L1(Ω). Prove that∫Ω−B(0,n)

f(x) dx→ 0 as n→∞ .

where B(0, n) denotes the ball centered at 0 with radius n.

Solution:

• See the book.

• Consider

fn(x) :=

0 x ∈ Ω ∩ B(0, n)

f(x) x ∈ Ω−B(0, n)

Obviously, fn(x) → 0 as n → ∞, and |fn(x)| ≤ |f(x)|, so |f(x)| provides a dominatingfunction. By the Lebesque Theorem,∫

Ω−B(0,n)f(x) dx =

∫Ωfn(x) dx→ 0 .

3. Application of Banach Contractive Map Theorem.

• State the Contractive Map Theorem. Make sure to list all assumptions.

• Consider the following initial value problem:dxdt = 1

3x−2(t− 1)−1, t ∈ (0, T )

x(0) = 1

where, at this point, T is unknown. Use elementary means to solve the problem. Comment on amaximum T for which the solution exists in interval (0, T ).

• Use the Contractive Map Theorem to prove that the solution exists (kind of after dinner exercisefor this example) and provide a concrete estimate for interval length T .

Solution:

• See the book.

• Use separation of variables to obtain:

3x2dx =dt

x− 1

to obtain:x3|x(t)

1 = ln |t− 1||t0 ,

and, finally,x(t) = [ln |t− 1|+ 1]

13 .

Verify that the solution indeed satisfies the ODE and IC. The solution has a blow up at t = 1.

• The IVP is equivalent to the integral equation:

x(t) = 1 +1

3

∫ t

0x−2(s) (s− 1)−1 ds =: (Ax)(t)

and x(t) is a solution to the integral eqn iff it is a fixed point of operator A. More precisely, wewill identify a set D ⊂ C([0, T ]) (with T to be determined !) such that a) A is well defined, i.e.it maps D into itself, and b) A is a contraction on D.

The flux f(x, s) = 13x−2(t− 1)−1 is undefined for x = 0 and t = 1. This motivates use to seek

T < 1 and define,

D := x ∈ C([0, T ]) : x(t) ≥ 1

2 .

The 12 in the definition above is somehow arbitrary, could have used any positive constant. Def-

inition of D is consistent with the IC and all functions from D are bounded pointwise by 2.Condition a) requires that

1 +1

3

∫ t

0x−2(s)(s− 1)−1 ds ≥ 1

2for t < T .

Equivalently,

|∫ t

0x−2(s)(s− 1)−1 ds| ≤ 3

2

Working on the sufficient side, we get

|∫ t

0x−2(s)(s− 1)−1 ds| ≤

∫ t

0|x−2(s)|︸︷︷︸≤4

|(s− 1)−1| ds ≤ 4

∫ t

0

ds

1− s= −4 ln(1− t) ≤ 3

2.

or, equivalently,

− ln(1− t) ≤ 3

8.

Condition b) requires that

|x1(t)− x2(t)| ≤∫ t

0|x−2

1 (s)− x−22 (s)|(1− s)−1 ds ≤ C‖x1 − x2‖C([0,T ])

with C < 1. By the Mean-Value Theorem, for f(x) = x−2,

f(x2)− f(x1) = f ′(ξ)(x2 − x1) where ξ ∈ (x1, x2) ,

with f ′(x) = −2x−3 and | − 2x−3| ≤ 16 for x ∈ D. Consequently, the integral above isestimated by,

16

3

∫ t

0

ds

1− s︸︷︷︸=− ln(1−t)

maxt∈[0,T ]

|x1(t)− x2(t)|︸︷︷︸=‖x1−x2‖C([0,T ])

so the contraction condition is satisfied if

− ln(1− t) < 3

16.

Of the two conditions, the second one is more restrictive and it leads to the final condition for T ,

T < 1− e−316 .

4. Let u and w be two scalar fields in R2.

• Consider a rectangular domain Ω = (a1, a2)×(b1, b2) ⊂ R2, i.e, if (x, y) ∈ Ω then a1 ≤ x ≤ a2

and b1 ≤ y ≤ b2. First prove that∫Ω

∂u

∂yw dΩ = −

∫Ωu∂w

∂ydΩ +

∫∂Ωuw ny ds,

where ny is the y-component of the unit outward normal vector n on the boundary ∂Ω.

• Using the above (and/or similar) result to prove the following First Green Identity∫Ω∇ · Fw dΩ = −

∫ΩF · ∇w dΩ +

∫∂Ω

F · nw ds,

where F is a vector field in R2.

• Derive the Gauss divergence theorem∫Ω∇ · F dΩ =

∫∂Ω

F · n ds,

Solution:

• We have ∫Ω

∂u

∂yw dΩ =

∫ a2

a1

(∫ b2

b1

∂u

∂yw dy

)dx

=

∫ a2

a1

(−∫ b2

b1

u∂w

∂ydy + u (x, b2)w (x, b2)− u (x, b1)w (x, b1)

)dx

= −∫

Ωu∂w

∂ydΩ +

∫∂Ωuw ny ds,

where we have used the fact that ny = −1, 0, 1 on the lower y-boundary, x-boundaries, andupper y-boundary, respectively. Here, we also use the fact that dx = −nyds on these boundaries.

• Easy: ∫Ω

∂F1

∂xw dΩ = −

∫ΩF1∂w

∂xdΩ +

∫∂ΩF1wnx ds,

∫Ω

∂F2

∂yw dΩ = −

∫ΩF2∂w

∂ydΩ +

∫∂ΩF2wny ds,

• take w = const

5. Consider a flexible string with mass density (mass per unit length) ρ(x) tied between x = 0 and x = `.The string is assumed to be under a constant tension τ at any point along the string at any time. It canbe shown that the potential energy and the kinetic energy of the string are given by

V =τ

2

∫ `

0y2x dx, and T =

1

2

∫ `

0ρ (x) y2

t dx,

where y is the vertical displacement of the string, yx and yt are partial derivative of y with respect tox and t.

• According the Hamilton’s principle, the equation of motion of the string is given by the Euler-Lagrange equation of the following functional∫ T

0(T − V ) dt.

Derive in details the equation of motion for the string.

• Now assume the density ρ is constant. Solve in details for the displacement of the string giventhe initial displacement

y0 (x) = sin

(2π

`x

),

and zero initial velocity yt (x) = 0.

Solution:

• ForI =

∫t

∫xf (x, t, y, yx, yt) dx dt,

the Euler-Lagrange equation for fixed end points of y is given as

∂f

∂y− ∂

∂x

(∂f

∂yx

)− ∂

∂t

(∂f

∂yt

)= 0.

Applying this result for our case yields

τyxx − ρytt = 0,

which is the wave equation.

• Solve the wave equation using separation of variables

y = v(x)w(t)

to get tov′′

v=

1

α2

w′′

w= −k2,

where α2 = τρ . We then have

vn = sin(nπ`x), wn = cos

(αnπ`t),

and the general solution is then

y(x, t) =∑n

en sin(nπ`x)

cos(αnπ

`t).

For

y0 (x) = sin

(2π

`x

),

only one term survives and hence the solution is

y(x, t) = sin

(2π

`x

)cos

(α2π

`t

).

6. Let Ω = (0, 1)× (0, 1).

• Solve in details the following eigenvalue problem

−∆w = λw in Ω

with the homogeneous boundary condition w (x, y) = 0 on ∂Ω. Moreover, show that the oper-ator −∆ with homogeneous boundary condition is a self-adjoint operator. As a result, argue indetails that “any function” v (x, y) can be expressed as

v (x, y) =

∞,∞∑m=1,n=1

vmnφmn (x, y) ,

where φmn (x, y) are eigenfunctions of −∆ and vmn are coefficients in the expansion.

• Denote (φmn (x, y) , λmn) as eigenpairs of the previous question. Consider the following PDE

−∆w = f (x, y) in Ω,

and boundary condition w (x, y) = 0 on ∂Ω. Solve in details for the solution of this equationfor a general f (x, y) and then deduce the solution for f (x, y) = 2sin (πx) sin (2πy).

HINT: You may want to use the following

f (x, y) =

∞,∞∑m=1,n=1

fmnφmn (x, y) ,

and

−∆v (x, y) = −∞,∞∑

m=1,n=1

vmn∆φmn (x, y) =

∞,∞∑m=1,n=1

vmnλmnφmn (x, y) .

Solution:

• By separation of variables w = X(x)Y (x), we conclude that

λmn = π2(m2 + n2

), φmn = sin (mπx) sin (nπy) .

The self-adjointness is clear by integration by parts two times

• Using the hint and the orthogonality of the eigenfunctions, we conclude that

wmn =fmnλmn

,

and for f (x, y) = 2sin (πx) sin (2πy), we see that

w =f12

λ12sin (πx) sin (2πy) =

2

5π2sin (πx) sin (2πy)

area a-cse july 2003 - ices.utexas.edu · may 2009. area a-cse may 2009. area a-cse may 2009. area...

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