APM346 (Summer 2019), Final Exam 1/22

Last (Family) Name: Carruth First (Given) Name: Nathan Student Number: [purposely omitted]

University of Toronto, Faculty of Arts and Science APM346 H1Y, August 2019 Final Examinations Instructor: Nathan Carruth

Duration: 3 hours. No aids allowed.

Please read the following instructions:

• Please fill out the front of this exam booklet, but do not begin writing the actual exam until the announcements are over and the Exam Facilitator has started the exam.

• No aids of any form are allowed on this exam. Possessing an aid during this exam may result in your being charged with an academic offence.

• All cell phones, smart watches, electronic devices, toasters, etc., must be turned off and placed in your bag under your desk. All study materials must also be placed in your bag under your desk. Having such items on your person after the exam has started may be an academic offence.

• [Your instructor does not recommend carrying toasters in your pockets anyway.]

• When you are done with your exam, please raise your hand and wait for someone to come and collect it. Do not collect your bag and jacket while still in possession of the exam paper.

• If you are feeling ill and unable to finish your exam, please let an Exam Facilitator know this prior to leaving the exam hall so it can be properly noted.

• In the event of a fire alarm, do not check your cell phone when escorted outside. Special instructions:

• You must use the definition of the Fourier transform given in class. Use of a different definition (including that given in the textbook) may result in lost marks.

• Use of an incorrect orthogonal set on a problem may result in a very low score for the entire problem. Please check the sets you use. Sets which were derived in class, in the notes, or in the homework solutions on the course webpage may be used without derivation. Other sets, if needed, may be stated without derivation, but then no partial credit will be given for a partially correct set.

• This exam has eight questions, for a total of 125 marks. The weighting is indicated on each question. Note that weighting may not directly correspond either to difficulty or to amount of writing required, and that the ordering of the problems may not be the best order in which to write the exam. You must show all of your work for credit.

• You may use the back sides of the pages, as well as the last four pages, to continue your solutions, as long as this is clearly indicated.

• Unless otherwise stated, you must write out the full form of the final answer for full marks.

APM346 (Summer 2019), Final Exam 2/22 1. [8 marks] Solve the following boundary-value problem on the unit cube

Q = {(x, y, z)|x, y, z ∈ [0, 1]}:

∇^{2}u = 0, u|_{∂Q} = \{ \begin{matrix}1, & z = 1 \\ 0, & otherwise \end{matrix} .

We have from class that the general solution to ∇^{2}u = 0 on Q with u|_{x=0} = u|_{x=1} = u|_{y=0} = u|_{y=1} = 0 is

u = ∑_{ℓ=1}^{∞} ∑_{m=1}^{∞} sin ℓπxsin mπy (a_{ℓm}sinh π\sqrt{ℓ^{2} + m^{2}}z + b_{ℓm}cosh π\sqrt{ℓ^{2} + m^{2}}z) .  [2 marks]

The boundary conditions then give

u|_{z=0} = 0 = ∑_{ℓ,m=1}^{∞} sin ℓπxsin mπy (b_{ℓm}) , [1 mark]  so  b_{ℓm} = 0 [1 mark]

u|_{z=1} = 1 = ∑_{ℓ,m=1}^{∞} sin ℓπxsin mπy (a_{ℓm}sinh π\sqrt{ℓ^{2} + m^{2}}) [1 mark]

a_{ℓm}sinh π\sqrt{ℓ^{2} + m^{2}} = 4 ∫_{0}^{1} ∫_{0}^{1} sin ℓπxsin mπy dy dx = 4 (∫_{0}^{1} sin ℓπx dx) (∫_{0}^{1} sin mπy dy) [1 mark]

= r (−\frac{1}{ℓπ} cos ℓπx|_{0}^{1}) (−\frac{1}{mπ} cos mπy|_{0}^{1})

= \frac{4}{π^{2}ℓm} (1 − (−1)^{ℓ}) (1 − (−1)^{m}) , [1 mark]

so the solution is

u = ∑_{ℓ=1,ℓ odd}^{∞} ∑_{m=1,m odd}^{∞} \frac{16}{π^{2}ℓmsinh π\sqrt{ℓ^{2} + m^{2}}} sin ℓπxsin mπysinh π\sqrt{ℓ^{2} + m^{2}}z.  [1 mark]

NOTES. 1 mark was given if the form for the expansion was not quite correct. Writing out a sum over only ℓ and m odd (as done here) was not required. Taking the initial value of ℓ and m to be 0 instead of 1 should typically result in a deduction of 0.5 marks, since in this case the final expression is meaningless.

APM346 (Summer 2019), Final Exam 4/22 APM346 (Summer 2019), Final Exam 5/22 2. [22 marks] Solve the following boundary-value problem on the spherical shell {(r, θ, ϕ)|1 < r < 2}:

∇^{2}u = 0, u|_{r=1} = \{ \begin{matrix}0, & 0 ≤ θ < \frac{π}{2} \\ sin2ϕ, & \frac{π}{2} < θ ≤ π \end{matrix} , u|_{r=2} = \{ \begin{matrix}sin2ϕ, & 0 ≤ θ < \frac{π}{2} \\ 0, & \frac{π}{2} < θ ≤ π \end{matrix} .

Recall Legendre’s equation: (1 − x^{2})P′′_{ℓ} − 2xP′_{ℓ} + ℓ(ℓ + 1)P_{ℓ} = 0. [Can you see a certain P_{ℓm} hiding here?] The following identities may be useful: P′_{ℓ+1} − xP′_{ℓ} = (ℓ + 1)P_{ℓ}, (2ℓ + 1)P_{ℓ} = P′_{ℓ+1} − P′_{ℓ−1}. [Hint: the algebra is probably easiest if you write everything in terms of derivatives of P_{n} for various n before integrating.] Your answer may include P_{n}(0) for values of n for which this is nonzero. You may also use the normalisation integral for P_{ℓm}: ∫_{−1}^{1} P_{ℓm}^{2}(x) dx = \frac{(ℓ+m)!}{(ℓ−m)!} \frac{2}{2ℓ+1}.

We have the general solution

u(r, θ, ϕ) = ∑_{ℓ=0}^{∞} ∑_{m=0}^{ℓ} P_{ℓm}(cos θ) [cos mϕ (α_{ℓm}r^{ℓ} + β_{ℓm}r^{−(ℓ+1)}) + sin mϕ (γ_{ℓm}r^{ℓ} + δ_{ℓm}r^{−(ℓ+1)})] . [1 mark]

The first boundary condition [1 mark] then gives

α_{ℓm} + β_{ℓm} = 0,  all ℓ, m  [1 mark]
γ_{ℓm} + δ_{ℓm} = 0,  m ≠ 2  [0.5 marks]
∑_{ℓ=2}^{∞} P_{ℓ2}(cos θ) (γ_{ℓ2} + δ_{ℓ2}) = \{ \begin{matrix}0, & 0 ≤ θ < \frac{π}{2} \\ 1, & \frac{π}{2} < θ ≤ π \end{matrix}  [1 mark]

Similarly, the second boundary condition [1 mark] gives

2^{ℓ}α_{ℓm} + 2^{−(ℓ+1)}β_{ℓm} = 0,  all ℓ, m  [1 mark]
2^{ℓ}γ_{ℓm} + 2^{−(ℓ+1)}δ_{ℓm} = 0,  m ≠ 2  [0.5 marks]
∑_{ℓ=2}^{∞} P_{ℓ2}(cos θ) (2^{ℓ}γ_{ℓ2} + 2^{−(ℓ+1)}δ_{ℓ2}) = \{ \begin{matrix}1, & 0 ≤ θ < \frac{π}{2} \\ 0, & \frac{π}{2} < θ ≤ π \end{matrix}  [1 mark]

Since the matrix

( \begin{matrix}1 & 1 \\ 2^{ℓ} & 2^{−(ℓ+1)} \end{matrix} ) has inverse \frac{1}{2^{−(ℓ+1)} − 2^{ℓ}} ( \begin{matrix}2^{−(ℓ+1)} & −1 \\ −2^{ℓ} & 1 \end{matrix} ) ,

we see that α_{ℓm} = β_{ℓm} = 0 for all ℓ, m [1 mark], while γ_{ℓm} = δ_{ℓm} = 0 for all m ≠ 2 [1 mark]. We now need to expand the two functions appearing in the remaining two conditions. To do this, we note that

P_{ℓ2}(x) = (1 − x^{2})P′′_{ℓ}[0.5 marks] = 2xP′_{ℓ} − ℓ(ℓ + 1)P_{ℓ}[0.5 marks] = 2 [P′_{ℓ+1} − (ℓ + 1)P_{ℓ}] − ℓ(ℓ + 1)P_{ℓ}

= 2P′_{ℓ+1} − (ℓ + 2)(ℓ + 1)P_{ℓ} = 2P′_{ℓ+1} − \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1} (P′_{ℓ+1} − P′_{ℓ−1})

= \frac{4ℓ + 2 − (ℓ^{2} + 3ℓ + 2)}{2ℓ + 1}P′_{ℓ+1} + \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P′_{ℓ−1} = −\frac{ℓ(ℓ − 1)}{2ℓ + 1}P′_{ℓ+1} + \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P′_{ℓ−1},

[3.5 marks] so

∫ P_{ℓ2}(x) dx = −\frac{ℓ(ℓ − 1)}{2ℓ + 1}P_{ℓ+1} + \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P_{ℓ−1} + C.

Thus (making the change of variables x = cos θ, as usual)

[\frac{(ℓ + 2)!}{(ℓ − 2)!} \frac{2}{2ℓ + 1}] [0.5 marks] (2^{ℓ}γ_{ℓ2} + 2^{−(ℓ+1)}δ_{ℓ2}) = −\frac{ℓ(ℓ − 1)}{2ℓ + 1}P_{ℓ+1} + \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P_{ℓ−1}|_{0}^{1} [0.5 marks]

= \frac{ℓ^{2} + 3ℓ + 2 − ℓ^{2} + ℓ}{2ℓ + 1} + \frac{ℓ(ℓ − 1)}{2ℓ + 1}P_{ℓ+1}(0) − \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P_{ℓ−1}(0)

= 2 + \frac{ℓ(ℓ − 1)}{2ℓ + 1}P_{ℓ+1}(0) − \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P_{ℓ−1}(0), [1 mark]

while since P′_{ℓ+1}, P′_{ℓ−1} are even or odd as ℓ is [1 mark],

[\frac{(ℓ + 2)!}{(ℓ − 2)!} \frac{2}{2ℓ + 1}] (γ_{ℓ2} + δ_{ℓ2}) = (−1)^{ℓ} [2 + \frac{ℓ(ℓ − 1)}{2ℓ + 1}P_{ℓ+1}(0) − \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P_{ℓ−1}(0)] .

[0.5 marks] [0.5 marks]

Thus finally

( \begin{matrix}γ_{ℓ2} \\ δ_{ℓ2} \end{matrix} ) = \frac{(ℓ − 2)!}{(ℓ + 2)!} \frac{2ℓ + 1}{2} \frac{1}{2^{−(ℓ+1)} − 2^{ℓ}} ( \begin{matrix}2^{−(ℓ+1)} & −1 \\ −2^{ℓ} & 1 \end{matrix} ) ( \begin{matrix}(−1)^{ℓ} \\ 1 \end{matrix} )

· [2 + \frac{ℓ(ℓ − 1)}{2ℓ + 1}P_{ℓ+1}(0) − \frac{(ℓ + 2)(ℓ + 1)}{2ℓ + 1}P_{ℓ−1}(0)]

= \frac{(−1)^{ℓ}}{2^{−(ℓ+1)} − 2^{ℓ}} [\frac{2ℓ + 1}{(ℓ + 2)(ℓ + 1)ℓ(ℓ − 1)} + \frac{1}{2(ℓ + 2)(ℓ + 1)}P_{ℓ+1}(0) − \frac{1}{2ℓ(ℓ − 1)}P_{ℓ−1}(0)]

· ( \begin{matrix}2^{−(ℓ+1)} − (−1)^{ℓ} \\ −2^{ℓ} + (−1)^{ℓ} \end{matrix} )

[3.5 marks]

and we have the final answer

u = ∑_{ℓ=2}^{∞} P_{ℓ2}(cos θ)sin 2ϕ\frac{2ℓ + 1}{(ℓ + 2)(ℓ + 1)ℓ(ℓ − 1) (2^{−(ℓ+1)} − 2^{ℓ})}
· [r^{ℓ} ((−1)^{ℓ}2^{−(ℓ+1)} − 1) − r^{−(ℓ+1)} ((−1)^{ℓ}2^{ℓ} − 1)]

+ ∑_{ℓ=2,ℓ odd}^{∞} P_{ℓ2}(cos θ)sin 2ϕ\frac{1}{2^{−(ℓ+1)} − 2^{ℓ}} [\frac{1}{2(ℓ + 2)(ℓ + 1)}P_{ℓ+1}(0) − \frac{1}{2ℓ(ℓ − 1)}P_{ℓ−1}(0)]

· [r^{ℓ} ((−1)^{ℓ}2^{−(ℓ+1)} − 1) − r^{−(ℓ+1)} ((−1)^{ℓ}2^{ℓ} − 1)] . [0.5 marks]

NOTES. One can also use the alternative (less general) form for the solution

u(r, θ, ϕ) = ∑_{ℓ=0}^{∞} ∑_{m=0}^{ℓ} P_{ℓm}(cos θ) (a_{ℓm} cos mϕ + b_{ℓm}sin mϕ) (c_{ℓm}r^{ℓ} + d_{ℓm}r^{−(ℓ+1)}) .

However, in either case it is necessary to solve systems for all of the coordinates; and concluding too quickly that (for example) a_{ℓm} = 0 for all ℓ and m led to lost marks. (This is analogous to problem 3 on the midterm.) Additionally, the identity (2ℓ + 1)P_{ℓ} = P′_{ℓ+1} − P′_{ℓ−1} only applies to P_{ℓ}, not to the P_{ℓ2} with which we need to work here: attempting to solve the problem that way probably led to little credit being given.

Beyond the foregoing, most lost marks on this problem were probably due to algebraic errors or simply not finishing.

The alert reader will note that the marks above add up to 22.5, not 22. This was an inadvertant slip on the part of the instructor which was felt not to be serious enough to attempt to correct once it was discovered. Thus this problem had effectively 0.5 bonus marks attached to it.

3. [9 marks] Solve the following boundary-value problem on the cylinder

{(ρ, ϕ, z)|ρ < 1, 0 < z < 2}:

∇^{2}u = 0, u|_{z=0} = u|_{z=2} = 0, u|_{ρ=1} = z cos2ϕ.

We have the general expansion

u(ρ, ϕ, z) = ∑_{m=0}^{∞} ∑_{n=1}^{∞} I_{m} (\frac{nπ}{2} ρ) [cos mϕβ_{nm}sin \frac{nπ}{2} z + sin mϕδ_{nm}sin \frac{nπ}{2} z];  [1.5 marks]

applying the boundary condition [1 mark] gives

δ_{nm} = 0  for all n, m  [1 mark]
β_{nm} = 0  for all m ≠ 2  [0.5 marks]
∑_{n=1}^{∞} I_{2} (\frac{nπ}{2} ) (β_{n2}sin \frac{nπ}{2} z) = z;  [1 mark]

thus (since ∫_{0}^{2} sin^{2} \frac{nπ}{2} z dz = 1 [0.5 mark])

β_{n2}I_{2} (\frac{nπ}{2} ) = ∫_{0}^{2} zsin \frac{nπ}{2} z dz [0.5 marks] = [−\frac{2}{nπ} z cos \frac{nπ}{2} z|_{0}^{2} + \frac{4}{n^{2}π^{2}} sin \frac{nπ}{2} z|_{0}^{2}] [1 mark]

= \frac{4}{nπ} (−1)^{n+1}, [1 mark]

so β_{n2} = \frac{4(−1)^{n+1}}{nπI_{2}(\frac{nπ}{2} )} [0.5 marks], and the solution is

u(ρ, ϕ, z) = ∑_{n=1}^{∞} I_{2} (\frac{nπ}{2} ρ) cos 2ϕ(−1)^{n+1} \frac{4}{nπI_{2} (\frac{nπ}{2} )}sin \frac{nπ}{2} z.  [0.5 marks]

NOTES. Probably the single most common mistake on this problem was forgetting the factor of \frac{1}{2} in the z separation constant, i.e., using nπ instead of \frac{nπ}{2} in the foregoing. This fails to give a correct answer since {sin nπz} is not a complete set on the interval [0, 2]. This generally resulted in the deduction of 0.5 marks. As with problem 1, beginning the sum for n at 0 instead of 1 should generally result in a deduction of 0.5 marks.

APM346 (Summer 2019), Final Exam 8/22 4. [12 marks] Suppose that n ∈ Z, n > 0. Solve the following problem on

(0, +∞) × R^{3}, using Fourier transforms:

\frac{∂u}{∂t} = ∇^{2}u + (4πt)^{−\frac{3}{2}} e^{−\frac{x^{2}}{4t}} , u|_{t=0} = ( \frac{π}{n^{2}}) ^{−\frac{3}{2}} e^{−n^{2}|x|^{2}}.

Find the limit of the solution as n → ∞. What does the initial data behave

like in this limit?

We have, upon Fourier transforming in space,

\frac{∂û}{∂t} = −4π^{2}|k|^{2}û [1 mark] + (4πt)^{− \frac{3}{2}} ( \frac{π}{\frac{1}{4t}} )^{\frac{3}{2}} e^{−4π^{2}|k|^{2}t}
= −4π^{2}|k|^{2}û + e^{−4π^{2}|k|^{2}t} [1 mark]

û|_{t=0} = (\frac{π}{n^{2}} )^{− \frac{3}{2}} (\frac{π}{n^{2}} )^{\frac{3}{2}} e^{− \frac{π^{2}|k|^{2}}{n^{2}}} = e^{− \frac{π^{2}|k|^{2}}{n^{2}}} [1 mark]

whence, using the integrating factor e^{4π^{2}|k|^{2}t} [1 mark],

\frac{∂}{∂t} (e^{4π^{2}|k|^{2}t}û) = 1,
û = [û(0) [1 mark] + t [1 mark]] e^{−4π^{2}|k|^{2}t}
= te^{−4π^{2}|k|^{2}t} + e^{−|k|^{2}π^{2}(4t+ \frac{1}{n^{2}} )}, [0.5 marks]

whence we obtain upon inverse transforming

u = t (\frac{π}{4π^{2}t})^{\frac{3}{2}} e^{− \frac{|x|^{2}}{4t}} [1 mark] + (\frac{π}{π^{2} (4t + \frac{1}{n^{2}} )})^{\frac{3}{2}} e^{−_{\frac{|x|^{2}}{4t+ \frac{1}{n^{2}}}}} [1 mark]

= \frac{1}{8π^{\frac{3}{2}} t^{\frac{1}{2}}} e^{− \frac{|x|^{2}}{4t}} [0.5 marks] + \frac{1}{(π (4t + \frac{1}{n^{2}} ))^{\frac{3}{2}}} e^{−_{\frac{|x|^{2}}{4t+ \frac{1}{n^{2}}}}} [1 mark].

In the limit as n → ∞, the second term becomes simply \frac{1}{(4πt)^{\frac{3}{2}}} e^{− \frac{|x|^{2}}{4t}}, and the whole solution is

u = \frac{1}{(4πt)^{\frac{3}{2}}} (1 + t)e^{− \frac{|x|^{2}}{4t}} .  [1 mark]

Since

∫_{R_{3}} π^{− \frac{3}{2}} e^{−|x|^{2}} dx = π^{− \frac{3}{2}} (\frac{π}{1} )^{\frac{3}{2}} = 1,

and

(\frac{π}{n^{2}} )^{− \frac{3}{2}} e^{−n^{2}|x|^{2}} = n^{3} [π^{− \frac{3}{2}} e^{−|nx|^{2}}] ,

we see that the initial data is an approximate identity and behaves like the delta function δ(x) in the limit n → ∞.[1 mark] NOTES. Probably the most common mistake here was incorrectly taking the forward or inverse Fourier transform of a Gaussian. I think almost nobody correctly found the indicated limit of the initial data (many people said it was zero, which is true only for x ≠ 0).

APM346 (Summer 2019), Final Exam 10/22 APM346 (Summer 2019), Final Exam 11/22 5. (a) [19 marks] Solve the following problem on (0, +∞) × B, where B is the unit ball {(r, θ, ϕ)|r < 1}:

\frac{∂u}{∂t} = ∇^{2}u, u|_{t=0} = r^{2}sin^{2}θsin2ϕ, u|_{∂B} = 0.

[If you wish to use quantities like κ_{ℓn}, you must define them explicitly.] Find the limit of the solution as t → +∞.

(b) [4 marks] Suppose that the condition u|_{∂B} = 0 were replaced by the condition u|_{∂B} = cosθ. Explain how you would solve the problem in this case (you need not actually calculate anything). What would you expect the limit of the solution to be in this case as t → +∞? [You need not give an explicit formula, but your answer must be a definite function, not just a description in words.]

(a) The Laplacian on B with Dirichlet boundary conditions has eigenfunctions

j_{ℓ} (κ_{ℓn}r) P_{ℓm}(cos θ) \{ \begin{matrix}cos mϕ \\ sin mϕ \end{matrix}  [1 mark]

(where κ_{ℓn}, n = 1, 2, …, is the nth positive root of j_{ℓ} [0.5 marks]) with corresponding eigenvalues λ_{ℓnm} = −κ_{ℓn}^{2} [1 mark]. Suppose that we expand u in this basis as

u = ∑_{ℓ=0}^{∞} ∑_{m=0}^{ℓ} ∑_{n=1}^{∞} j_{ℓ} (κ_{ℓn}r) P_{ℓm}(cos θ) (a_{ℓnm} cos mϕ + b_{ℓnm}sin mϕ) .  [1 mark]

Then substituting into the equation gives

∑_{ℓ=0}^{∞} ∑_{m=0}^{ℓ} ∑_{n=1}^{∞} j_{ℓ} (κ_{ℓn}r) P_{ℓm}(cos θ) (a′_{ℓnm} cos mϕ + b′_{ℓnm}sin mϕ) [1 mark]

= ∑_{ℓ=0}^{∞} ∑_{m=0}^{ℓ} ∑_{n=1}^{∞} −κ_{ℓn}^{2}j_{ℓ} (κ_{ℓn}r) P_{ℓm}(cos θ) (a_{ℓnm} cos mϕ + b_{ℓnm}sin mϕ) ,

[1 mark]

so

a′_{ℓnm} = −κ_{ℓn}^{2}a_{ℓnm}, [1 mark] b′_{ℓnm} = −κ_{ℓn}^{2}b_{ℓnm}, [1 mark]

and

a_{ℓnm}(t) = a_{ℓnm}(0)e^{−κ_{ℓn}^{2}t}, [1 mark] b_{ℓnm}(t) = b_{ℓnm}(0)e^{−κ_{ℓn}^{2}t}.[1.5 marks]

The initial values can be obtained from u|_{t=0}:

∑_{ℓ=0}^{∞} ∑_{m=0}^{ℓ} ∑_{n=1}^{∞} j_{ℓ} (κ_{ℓn}r) P_{ℓm}(cos θ) (a_{ℓnm}(0) cos mϕ + b_{ℓnm}(0)sin mϕ) = r^{2}sin^{2}θsin 2ϕ.  [1 mark]

Since P_{22}(cos θ) = 3sin^{2}θ, we see that b_{ℓnm}(0) = 0 unless ℓ = m = 2 [0.5 marks], and a_{ℓnm} = 0 for all ℓ, n, m [1 mark]; finally

∑_{n=1}^{∞} b_{2n2}(0)j_{2} (κ_{2n}r) = \frac{r^{2}}{3} ,  [1 mark]

b_{2n2}(0) = \frac{2}{j_{3}^{2} (κ_{2n})}[1 mark] ∫_{0}^{1} \frac{r^{4}}{3} j_{2} (κ_{2n}r) dr[1 mark] = \frac{2}{3j_{3}^{2} (κ_{2n})}\sqrt{\frac{π}{2κ_{2n}}} ∫_{0}^{1} r^{\frac{7}{2}} J_{\frac{5}{2}} (λ_{\frac{5}{2} ,n}r) [0.5 marks] dr

= \frac{2}{3j_{3}^{2} (κ_{2n})}\sqrt{\frac{π}{2}} \frac{1}{κ2n^{\frac{3}{2}}} J_{\frac{7}{2}} (κ_{2n}) [1 mark] = \frac{2}{3j_{3}^{2} (κ_{2n}) κ_{2n}} j_{3}(κ_{2n})

= \frac{2}{3j_{3} (κ_{2n}) κ_{2n}} , [0.5 marks]

and the final solution is

u = ∑_{n=1}^{∞} j_{2} (κ_{2n}r) P_{22}(cos θ)sin 2ϕ\frac{2}{3j_{3} (κ_{2n}) κ_{2n}} e^{−κ_{2n}^{2}t}.  [0.5 marks]

Since κ_{2n} > 0 for all n, we see that u → 0 as t → +∞. [1 mark]

(b) In this case we would first solve the problem on B

∇^{2}U_{1} = 0, U_{1}|_{∂B} = cos θ,  [1 mark]

and then solve on (0, +∞) × B

\frac{∂u_{2}}{∂t} = ∇^{2}u_{2}, u_{2}|_{t=0} = r^{2}sin^{2}θsin 2ϕ − U_{1}, u_{2}|_{∂B} = 0;  [1 mark]

the full solution would be u = U_{1} + u_{2} [1 mark]. We expect lim_{t→+∞} u = U_{1} [1 mark] in this case.

NOTES. Probably the biggest single reason for deducted marks in (a) was not deriving the equations satisfied by the coefficients, but rather assuming the solutions from the outset. For (b), the single biggest quantitative error was probably taking u_{2}|_{t=0} = r^{2}sin^{2}θsin 2ϕ − cos θ, or even dropping the subtracted term altogether.

Starting the n sum at 0 instead of 1 should not result in lost marks (since n is just a counter, which can just as well be started at 0 as at 1, though in class we always started it at 1).

The curious asymmetry in marking the expressions for a_{ℓnm}(t) and b_{ℓnm}(t) was not intended to create any asymmetry in practice, in that if only one appeared, it would be given the higher mark. (I probably had some reason in mind when I wrote 1 mark for a and 1.5 marks for b, but I have long since forgotten what it was.)

APM346 (Summer 2019), Final Exam 13/22 6. [24 marks] Solve the following problem on the unit disk D = {(ρ, ϕ)|ρ < 1}:

\frac{∂^{2}u}{∂t^{2}} = ∇^{2}u, u|_{∂D} = 0, u|_{t=0} = 0, \frac{∂u}{∂t}|_{t=0} = ρ^{2}sin2ϕ.

[As in problem 5, if you wish to use quantities like λ_{mi}, you must define them explicitly.] What is the lowest frequency occurring? [A symbolic answer is sufficient.]

In this case we have the eigenfunctions J_{m} (λ_{mi}ρ) \{ \begin{matrix}cos mϕ \\ sin mϕ \end{matrix} [1 mark] (where λ_{mi} is the ith positive root of J_{ℓm}(x) [0.5 marks]) with eigenvalues −λ_{mi}^{2} [1 mark]. Expanding u as

u = ∑_{m=0}^{∞} ∑_{i=1}^{∞} J_{m} (λ_{mi}ρ) (a_{mi} cos mϕ + b_{mi}sin mϕ) ,  [1 mark]

we have, upon substituting into the equation,

∑_{n=0}^{∞} ∑_{i=1}^{∞} J_{m} (λ_{mi}ρ) (a′′_{mi} cos mϕ + b′′_{mi}sin mϕ) [1 mark]

= ∑_{m=0}^{∞} ∑_{i=1}^{∞} J_{m} (λ_{mi}ρ) (−λ_{mi}^{2}) (a_{mi} cos mϕ + b_{mi}sin mϕ) , [1 mark]

so that the a_{mi} and b_{mi} satisfy

a′′_{mi} = −λ_{mi}^{2}a_{mi}, [1 mark] b′′_{mi} = −λ_{mi}^{2}b_{m}i, [1 mark]

so

a_{mi}(t) = α_{mi} cos λ_{mi}t + β_{mi}sin λ_{mi}t, b_{mi}(t) = γ_{mi} cos λ_{mi}t + δ_{mi}sin λ_{mi}t. [1 mark]

Now we see that

a_{mi}(0) = α_{mi}, a′_{mi}(0) = λ_{mi}β_{mi}, b_{mi}(0) = γ_{mi}, b′_{mi}(0) = λ_{mi}δ_{mi};

and these initial values can be determined from the initial conditions for u:

0 = u|_{t=0} = ∑_{m=0}^{∞} ∑_{i=1}^{∞} J_{m} (λ_{mi}ρ) (a_{mi}(0) cos mϕ + b_{mi}(0)sin mϕ) [1 mark]
so α_{mi} = γ_{mi} = 0 for all m, i [1 mark];
ρ^{2}sin 2ϕ = u_{t}|_{t=0} = ∑_{m=0}^{∞} ∑_{i=1}^{∞} J_{m} (λ_{mi}ρ) (a′_{mi}(0) cos mϕ + b′_{mi}(0)sin mϕ) , [1 mark]

so a′_{mi}(0) = 0 for all m, i [1 mark], which gives β_{mi} = 0 and a_{mi}(t) = 0 for all t, all m, i [2 marks], while b′_{mi}(0) = 0 for all m ≠ 2 [0.5 marks] , which gives δ_{mi} = 0, hence b_{mi}(t) = 0 for all t [0.5 marks], for m ≠ 2 [1 mark]; finally,

ρ^{2} = ∑_{i=1}^{∞} J_{2} (λ_{2i}ρ) b′_{2i}(0),  [1 mark]

so

b′_{2i}(0) = \frac{2}{J_{3}^{2} (λ_{2i})} ∫_{0}^{1} ρ^{3}J_{2} (λ_{2i}ρ) dρ = \frac{2}{J_{3}^{2} (λ_{2i})} \frac{1}{λ_{2i}} J_{3} (λ_{2i}) = \frac{2}{λ_{2i}J_{3} (λ_{2i})},  [3.5 marks]

whence

δ_{2i} = \frac{2}{λ_{2i}^{2}J_{3} (λ_{2i})}  [1 mark]

and we have finally for u

u(t, ρ, ϕ) = ∑_{i=1}^{∞} \frac{2}{λ_{2i}^{2}J_{3} (λ_{2i})}J_{2} (λ_{2i}ρ) sin 2ϕsin λ_{2i}t.  [1 mark]

The lowest frequency is thus \frac{λ_{21}}{2π}. [1 mark] NOTES. As with problem 5, probably the biggest reason for lost marks was starting directly with the solutions for the coefficients rather than deriving them as here. For the last part of the question, an answer λ_{21} was also acceptable (missing the factor of 2π did not result in lost marks): while technically only \frac{λ_{21}}{2π} is the frequency, λ_{21} is the so-called angular frequency, and since we didn’t spend much time on this point in class I didn’t see a point in deducting marks for missing the 2π.

As with problem 5, starting the i sum at 0 should not result in lost marks.

APM346 (Summer 2019), Final Exam 15/22 7. [18 marks] Solve the following problem on the unit cube Q (defined in problem 1):

∇^{2}u = sin4πxsin2πy cosπz, \frac{∂u}{∂n}|_{∂Q} = 0, u(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = 0.

(Here \frac{∂}{∂n} denotes the derivative in the normal direction to the surface ∂Q.)

We have the eigenfunctions cos ℓπx cos mπy cos nπz [2 marks], with eigenvalues −π^{2} (ℓ^{2} + m^{2} + n^{2}) [1 mark]. Expanding u as

u(x, y, z) = ∑_{ℓ=0}^{∞} ∑_{m=0}^{∞} ∑_{n=0}^{∞} a_{ℓmn} cos ℓπx cos mπy cos nπz,  [1 mark]

we see that the equation gives

∑_{ℓ,m,n=0}^{∞} −π^{2} (ℓ^{2} + m^{2} + n^{2}) a_{ℓmn} cos ℓπx cos mπy cos nπz = sin 4πxsin 2πy cos πz,  [1 mark]

whence we see that

−π^{2} (ℓ^{2} + m^{2} + n^{2}) a_{ℓmn} = n_{ℓ}n_{m}n_{n} ∫_{Q} sin 4πxsin 2πy cos πz cos ℓπx cos mπy cos nπz dV,  (1)

where n_{ℓ} = \{ \begin{matrix}2, & ℓ ≠ 0 \\ 1, & ℓ = 0 \end{matrix} is the appropriate normalisation constant. Now we see that the integral above vanishes for n ≠ 1, while

∫_{0}^{1} sin 2kπx cos ℓπx dx[0.5 marks] = \frac{1}{2} ∫_{0}^{1} sin [(2kπ + ℓπ) x] + sin [(2kπ − ℓπ) x] dx[1 mark]
[ = 0 if ℓ = 2k][1 mark]

= −\frac{1}{2} [ \frac{1}{(2k + ℓ)π} cos [(2k + ℓ) πx]|_{0}^{1} + \frac{1}{(2k − ℓ)π} cos [(2k − ℓ) πx]|_{0}^{1}] ,

[0.5 marks] ℓ ≠ 2k

= \frac{1}{2π} (1 − (−1)^{ℓ}) (\frac{1}{2k + ℓ} + \frac{1}{2k − ℓ}) = \frac{2k}{π} (1 − (−1)^{ℓ}) \frac{1}{4k^{2} − ℓ^{2}} ,

[1 mark]

so for (ℓ, m, n) ≠ (0, 0, 0) we have

a_{ℓmn} = \{ \begin{matrix}0, & n ≠ 1, or m = 2,[0.5 marks] or ℓ = 4 [0.5 marks] \\ \begin{matrix}n_{ℓ}n_{m} \frac{8}{π^{2}} (1−(−1)^{ℓ})(1−(−1)^{m}) \frac{1}{16−ℓ^{2}} \\ · \frac{1}{4−m^{2}} [1 mark]· \frac{−1}{π^{2}(ℓ^{2}+m^{2}+1)} \end{matrix} , & otherwise \end{matrix}

= \{ \begin{matrix}0[0.5 marks], & n ≠ 1, or m or ℓ even[0.5 marks] \\ −\frac{128}{π^{2}} \frac{1}{16−ℓ^{2}} \frac{1}{4−m^{2}} [0.5 marks]\frac{1}{π^{2}(ℓ^{2}+m^{2}+1)}[1 mark], & otherwise. \end{matrix}

Now if ℓ = m = n = 0, then the integral in (1) is zero, as is the left-hand side. Thus this equation is consistent but tells us nothing about a_{000} [1 mark]. However, since our series for u only has nonzero coefficients for ℓ, m, n all odd, and

cos \frac{ℓπ}{2} cos \frac{mπ}{2} cos \frac{nπ}{2} = 0  [1 mark]

in such a case, the final condition gives a_{000} = 0 [1 mark]. Thus

u = ∑_{ℓ,m=1,ℓ,m odd}^{∞} −\frac{128}{π^{2}} \frac{1}{16 − ℓ^{2}} \frac{1}{4 − m^{2}} \frac{1}{π^{2} (ℓ^{2} + m^{2} + 1)} cos ℓπx cos mπy cos πz.  [1.5 marks]

NOTES. Again, some marks were lost by simply assuming the general form of the solution to Poisson’s equation rather than deriving it as here (though this is less of an issue than with 5 and especially 6). Marks were also lost for being insufficiently careful with the term a_{000}.

8. [9 marks] Solve the following problem on R^{3} (here x is the first coordinate

of x = (x, y, z)):

\frac{∂u}{∂t} = ∇^{2}u + \frac{∂u}{∂x}, u|_{t=0} = e^{−|x|^{2}}.

We have the Fourier transform:

\frac{∂û}{∂t} = −4π^{2}|k|^{2}û[1 mark] + 2πik_{1}û[1 mark], û|_{t=0} = π^{\frac{3}{2}} e^{−π^{2}|k|^{2}}.[1 mark]

The equation gives

\frac{∂}{∂t} (e^{(4π^{2}|k|^{2}−2πik_{1})t}û[1 mark]) = 0,
û = û(0)e^{−(4π^{2}|k|^{2}−2πik_{1})t}
= π^{\frac{3}{2}} e^{−π^{2}|k|^{2}(4t+1)}e^{2πik_{1}t}.[1 mark]

Since

F^{−1} [π^{\frac{3}{2}} e^{−π^{2}|k|^{2}(4t+1)}] = π^{\frac{3}{2}} (\frac{1}{π(4t + 1)})^{\frac{3}{2}} e^{− \frac{|x|^{2}}{4t+1}} = \frac{1}{(4t + 1)^{\frac{3}{2}}} e^{− \frac{|x|^{2}}{4t+1}} ,  [2 marks]

we see by properties of Fourier transforms that

u = \frac{1}{(4t + 1)^{\frac{3}{2}}} e^{− \frac{1}{4t+1}(y^{2}+z^{2}+(x+t)^{2})}.  [2 marks]

NOTES. Again, probably the biggest issue with this problem was the mishandling of the relevant Fourier transforms. Another issue which came up was failure to use the property

F[f(x − x_{0})](k) = e^{−2πik·x_{0}} f̂(k).

Some solutions wrote effectively F [\frac{∂u}{∂x}] = \frac{∂û}{∂x}, probably by analogy with a similar (though correct) formula for \frac{∂u}{∂t}: unfortunately this formula is not only wrong in actuality but meaningless even in principle, since û is a function of k and t and hence does not depend on x. The point behind the analogous formula for \frac{∂u}{∂t} is that we are taking a function of (t, x) and transforming only in x, meaning that t is essentially a parameter with respect to which we can differentiate either before or after transforming (assuming, as always, that our functions are sufficiently well-behaved that we are allowed to take the derivative inside the integral representing the Fourier transform). x, however, is one of the variables with respect to which we are transforming; i.e., it will be one of the variables over which we integrate, and hence it does not appear in the transformed function and it makes no sense to speak of the derivative of the transform with respect to it. More explicitly:

F [\frac{∂u}{∂t}] = ∫_{R_{3}} \frac{∂u}{∂t}e^{−2πik·x} dx = \frac{∂}{∂t} ∫_{R_{3}} ue^{−2πik·x} dx

= \frac{∂}{∂t}F[u] = \frac{∂û}{∂t},

while attempting to do the same thing with \frac{∂u}{∂x} leads to

F [\frac{∂u}{∂x}] = ∫_{R_{3}} \frac{∂u}{∂x}e^{−2πik·x} dx,

and now there is no way to take the derivative outside of the integral since the integral over x includes an integral over x: one needs instead to do an integration by parts, which leads to the formula

F [\frac{∂u}{∂x}] = 2πik_{1}û

used here, as derived in class. (Here k_{1} represents the component of k corresponding to x.)

APM346 (Summer 2019), Final Exam 18/22 APM346 (Summer 2019), Final Exam 19/22

Scratch paper APM346 (Summer 2019), Final Exam 20/22

Scratch paper APM346 (Summer 2019), Final Exam 21/22

Scratch paper APM346 (Summer 2019), Final Exam 22/22

Scratch paper

– End of exam booklet –