Noesis
Level 5 — MasteryPartial Differential Equations

Partial Differential Equations

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Time limit: 90 minutes Total marks: 60 Instructions: Answer all three questions. Full derivations, proofs, and reasoning are required. Where code is requested, pseudocode or Python (NumPy) is acceptable but must be dimensionally and logically consistent.

Question 1 — Classification, Sturm–Liouville, and Separation (20 marks)

Consider the PDE for u(x,t)u(x,t) on 0<x<L0<x<L, t>0t>0:

ut=κ uxx,κ>0,u_t = \kappa\, u_{xx}, \qquad \kappa>0,

with Robin (radiating) boundary conditions

ux(0,t)=0,ux(L,t)+h u(L,t)=0,h>0,u_x(0,t)=0, \qquad u_x(L,t) + h\,u(L,t) = 0, \quad h>0,

and initial data u(x,0)=f(x)u(x,0)=f(x).

(a) Apply the discriminant test to classify the PDE ut=κuxxu_t=\kappa u_{xx} as elliptic, parabolic, or hyperbolic, showing the general second-order form Auxx+2Buxt+Cutt+…Au_{xx}+2Bu_{xt}+Cu_{tt}+\dots and computing B2−ACB^2-AC. (3)

(b) Seek separated solutions u(x,t)=X(x)T(t)u(x,t)=X(x)T(t). Derive the two ODEs and show that the spatial problem is a Sturm–Liouville eigenvalue problem. Write it in the standard S–L form ddx ⁣(p X′)+(q+λw)X=0\frac{d}{dx}\!\left(p\,X'\right)+ (q+\lambda w)X=0, identifying p,q,wp,q,w. (4)

(c) Solve the spatial ODE for eigenvalue λ=μ2>0\lambda=\mu^2>0. Apply both boundary conditions to derive the transcendental eigenvalue equation for μ\mu. (5)

(d) State the orthogonality relation the eigenfunctions XnX_n satisfy, and prove it directly from the S–L problem (integrate by parts, use the boundary conditions). (5)

(e) Write the full series solution u(x,t)u(x,t) and give the integral formula for the coefficients. (3)

Question 2 — Wave Equation: D'Alembert, Energy, and Physics (20 marks)

An infinite string obeys utt=c2uxxu_{tt}=c^2 u_{xx}, −∞<x<∞-\infty<x<\infty, with initial displacement u(x,0)=ϕ(x)u(x,0)=\phi(x) and initial velocity ut(x,0)=ψ(x)u_t(x,0)=\psi(x).

(a) Derive D'Alembert's solution from the characteristic coordinates ξ=x−ct\xi=x-ct, η=x+ct\eta=x+ct. Show the PDE becomes uξη=0u_{\xi\eta}=0 and integrate to obtain the general solution, then apply the initial conditions to get the D'Alembert formula. (6)

(b) Take ϕ(x)=e−x2\phi(x)=e^{-x^2} and ψ(x)=0\psi(x)=0. Write u(x,t)u(x,t) explicitly and describe physically what happens as t→∞t\to\infty (amplitude of each travelling pulse). (3)

(c) Define the total energy E(t)=12∫−∞∞(ut2+c2ux2) dxE(t)=\tfrac12\int_{-\infty}^{\infty}\big(u_t^2 + c^2 u_x^2\big)\,dx. Prove that dEdt=0\dfrac{dE}{dt}=0 for smooth solutions decaying at infinity (conservation of energy). (6)

(d) Coding. Write a finite-difference update scheme (explicit) for utt=c2uxxu_{tt}=c^2u_{xx} on a grid xi=iΔxx_i=i\Delta x, tn=nΔtt_n=n\Delta t. Give the update formula for uin+1u_i^{n+1}, state the CFL stability condition, and explain in one sentence what happens if it is violated. (5)

Question 3 — Fourier Analysis, Parseval, and the Transform (20 marks)

(a) For f(x)=xf(x)=x on (−π,π)(-\pi,\pi) extended 2π2\pi-periodically, derive the full Fourier series. State whether Dirichlet's conditions hold and what the series converges to at x=πx=\pi. (6)

(b) Apply Parseval's theorem to the series in (a) to evaluate ∑n=1∞1n2\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}. Show all steps. (5)

(c) Define the Fourier transform f^(k)=∫−∞∞f(x)e−ikx dx\hat{f}(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,dx. Solve the heat equation ut=κuxxu_t=\kappa u_{xx} on −∞<x<∞-\infty<x<\infty with u(x,0)=f(x)u(x,0)=f(x) by transforming in xx; solve the resulting ODE in tt and invert to express uu as a convolution of ff with the heat kernel. Give the heat kernel explicitly. (9)

Answer keyMark scheme & solutions

Question 1

(a) General form: Auxx+2Buxt+Cutt+⋯=0A u_{xx}+2B u_{xt}+C u_{tt}+\dots=0. Writing ut−κuxx=0u_t-\kappa u_{xx}=0 has no uttu_{tt}, no uxtu_{xt}: A=−κA=-\kappa, B=0B=0, C=0C=0. Discriminant B2−AC=0−(−κ)(0)=0B^2-AC = 0 - (-\kappa)(0) = 0. Since B2−AC=0⇒B^2-AC=0 \Rightarrow parabolic. (1 form, 1 identify coeffs, 1 conclusion) (3)

(b) XT′=κX′′T⇒T′κT=X′′X=−λX T' = \kappa X'' T \Rightarrow \frac{T'}{\kappa T}=\frac{X''}{X}=-\lambda (constant).

  • Time: T′+κλT=0T' + \kappa\lambda T=0.
  • Space: X′′+λX=0X'' + \lambda X = 0, i.e. ddx(1⋅X′)+(0+λ⋅1)X=0\frac{d}{dx}(1\cdot X') + (0+\lambda\cdot1)X=0. So p=1p=1, q=0q=0, w=1w=1. Boundary conditions X′(0)=0X'(0)=0, X′(L)+hX(L)=0X'(L)+hX(L)=0 are separated/self-adjoint — genuine S–L problem. (1 each ODE, 1 S–L form + p,q,w, 1 BC transfer) (4)

(c) With λ=μ2>0\lambda=\mu^2>0: X=Acos⁡μx+Bsin⁡μxX=A\cos\mu x + B\sin\mu x, X′=−Aμsin⁡μx+Bμcos⁡μxX'=-A\mu\sin\mu x + B\mu\cos\mu x. X′(0)=Bμ=0⇒B=0X'(0)=B\mu=0\Rightarrow B=0. So X=Acos⁡μxX=A\cos\mu x, X′=−Aμsin⁡μxX'=-A\mu\sin\mu x. Second BC: X′(L)+hX(L)=0⇒−Aμsin⁡μL+hAcos⁡μL=0X'(L)+hX(L)=0 \Rightarrow -A\mu\sin\mu L + hA\cos\mu L=0. Divide by Acos⁡μLA\cos\mu L: −μtan⁡μL+h=0-\mu\tan\mu L + h=0, i.e. tan⁡(μL)=hμ(equivalently μtan⁡μL=h).\boxed{\tan(\mu L)=\frac{h}{\mu}}\quad(\text{equivalently } \mu\tan\mu L=h). Infinitely many roots μn\mu_n, giving λn=μn2\lambda_n=\mu_n^2. (1 general soln, 1 first BC, 1 second BC, 2 transcendental eqn) (5)

(d) Orthogonality: ∫0LXn(x)Xm(x) dx=0\displaystyle\int_0^L X_n(x)X_m(x)\,dx=0 for n≠mn\neq m (weight w=1w=1). Proof: Xn′′=−λnXnX_n''=-\lambda_n X_n, Xm′′=−λmXmX_m''=-\lambda_m X_m. Then (λm−λn)∫0LXnXm dx=∫0L(XnXm′′−XmXn′′) dx=[XnXm′−XmXn′]0L.(\lambda_m-\lambda_n)\int_0^L X_nX_m\,dx=\int_0^L (X_n X_m'' - X_m X_n'')\,dx = \big[X_n X_m' - X_m X_n'\big]_0^L. At x=0x=0: Xn′(0)=Xm′(0)=0X_n'(0)=X_m'(0)=0 → boundary term =0=0. At x=Lx=L: Xm′(L)=−hXm(L)X_m'(L)=-hX_m(L), Xn′(L)=−hXn(L)X_n'(L)=-hX_n(L), so Xn(L)Xm′(L)−Xm(L)Xn′(L)=−hXnXm+hXmXn=0X_n(L)X_m'(L)-X_m(L)X_n'(L)=-hX_nX_m + hX_mX_n=0. Hence RHS =0=0. Since λm≠λn\lambda_m\neq\lambda_n, ∫0LXnXm dx=0\int_0^L X_nX_m\,dx=0. (5) (1 statement, 2 the identity via IBP, 2 boundary evaluation)

(e)

c_n=\frac{\int_0^L f(x)\cos(\mu_n x)\,dx}{\int_0^L \cos^2(\mu_n x)\,dx}.$$ *(2 series form, 1 coefficient integral)* **(3)** --- ## Question 2 **(a)** $\xi=x-ct$, $\eta=x+ct$. Chain rule: $u_x=u_\xi+u_\eta$, $u_t=c(-u_\xi+u_\eta)$. $u_{xx}=u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta}$; $u_{tt}=c^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta})$. $u_{tt}-c^2u_{xx}=c^2(-4u_{\xi\eta})=0\Rightarrow u_{\xi\eta}=0$. Integrate: $u=F(\xi)+G(\eta)=F(x-ct)+G(x+ct)$. ICs: $u(x,0)=F+G=\phi$; $u_t(x,0)=-cF'+cG'=\psi \Rightarrow -F+G=\frac1c\int_0^x\psi+K$. Solving: $$u(x,t)=\tfrac12[\phi(x-ct)+\phi(x+ct)]+\frac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds.$$ *(2 change of variable, 1 reduce to $u_{\xi\eta}=0$, 1 integrate, 2 apply ICs)* **(6)** **(b)** $\psi=0$: $u(x,t)=\tfrac12 e^{-(x-ct)^2}+\tfrac12 e^{-(x+ct)^2}$. Two Gaussian pulses each of amplitude $\tfrac12$, travelling right and left at speed $c$; they separate. As $t\to\infty$ each pulse retains amplitude $\tfrac12$ (no dispersion, no decay), just moving apart. **(3)** **(c)** $E(t)=\tfrac12\int(u_t^2+c^2u_x^2)dx$. $$\frac{dE}{dt}=\int (u_t u_{tt}+c^2 u_x u_{xt})\,dx.$$ Use $u_{tt}=c^2u_{xx}$: first term $=c^2\int u_t u_{xx}$. Integrate second term by parts: $$c^2\int u_x u_{xt}\,dx = c^2[u_x u_t]_{-\infty}^{\infty}-c^2\int u_{xx}u_t\,dx = -c^2\int u_{xx}u_t\,dx$$ (boundary term vanishes by decay). Adding: $$\frac{dE}{dt}=c^2\int u_t u_{xx}\,dx - c^2\int u_{xx}u_t\,dx=0.$$ Hence $E$ constant. *(2 differentiate, 2 substitute PDE + IBP, 2 cancellation & conclusion)* **(6)** **(d)** Central differences: $u_{tt}\approx\frac{u_i^{n+1}-2u_i^n+u_i^{n-1}}{\Delta t^2}$, $u_{xx}\approx\frac{u_{i+1}^n-2u_i^n+u_{i-1}^n}{\Delta x^2}$. Let $r=\frac{c\Delta t}{\Delta x}$: $$u_i^{n+1}=2u_i^n-u_i^{n-1}+r^2\big(u_{i+1}^n-2u_i^n+u_{i-1}^n\big).$$ CFL stability: $r=\dfrac{c\,\Delta t}{\Delta x}\le 1$. If violated ($r>1$), the numerical domain of dependence excludes the true one → solution blows up / unstable oscillations. *(2 update, 2 CFL, 1 explanation)* **(5)** --- ## Question 3 **(a)** $f(x)=x$ is odd → $a_n=0$, $a_0=0$. $b_n=\frac1\pi\int_{-\pi}^\pi x\sin nx\,dx=\frac2\pi\int_0^\pi x\sin nx\,dx$. $\int_0^\pi x\sin nx\,dx=\left[-\frac{x\cos nx}{n}\right]_0^\pi+\frac1n\int_0^\pi\cos nx\,dx=-\frac{\pi\cos n\pi}{n}=-\frac{\pi(-1)^n}{n}$. So $b_n=\frac2\pi\cdot\left(-\frac{\pi(-1)^n}{n}\right)=\frac{-2(-1)^n}{n}=\frac{2(-1)^{n+1}}{n}$. $$x=\sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n}\sin nx.$$ Dirichlet conditions hold ($f$ piecewise smooth, bounded). At $x=\pi$ (a jump of the periodic extension between $\pi$ and $-\pi$), the series converges to the average $\tfrac12(\pi+(-\pi))=0$. *(1 odd, 3 compute $b_n$, 1 series, 1 Dirichlet/convergence)* **(6)** **(b)** Parseval (for $f$ on $(-\pi,\pi)$ with only sine terms): $$\frac1\pi\int_{-\pi}^\pi f^2\,dx=\sum_{n=1}^\infty b_n^2.$$ LHS: $\frac1\pi\int_{-\pi}^\pi x^2 dx=\frac1\pi\cdot\frac{2\pi^3}{3}=\frac{2\pi^2}{3}$. RHS: $\sum \left(\frac{2(-1)^{n+1}}{n}\right)^2=\sum\frac{4}{n^2}$. So $\frac{2\pi^2}{3}=4\sum\frac1{n^2}\Rightarrow \sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}.$ *(1 Parseval form, 2 LHS, 1 RHS, 1 result)* **(5)** **(c)** Transform in $x$: let $\hat u(k,t)=\int u e^{-ikx}dx$. Since $\widehat{u_{xx}}=(ik)^2\hat u=-k^2\hat u$: $$\hat u_t=-\kappa k^2\hat u \Rightarrow \hat u(k,t)=\hat f(k)\,e^{-\kappa k^2 t}.$$ Invert: $u=\mathcal F^{-1}[\hat f\cdot e^{-\kappa k^2 t}]$. By the convolution theorem, product of transforms = transform of convolution: $$u(x,t)=f * G_t,\quad G_t(x)=\mathcal F^{-1}[e^{-\kappa k^2 t}].$$ Compute the inverse of the Gaussian: $\mathcal F^{-1}[e^{-\kappa k^2 t}]=\frac1{2\pi}\int e^{-\kappa t k^2}e^{ikx}dk=\frac{1}{\sqrt{4\pi\kappa t}}e^{-x^2/(4\kappa t)}$. Hence heat kernel $$G_t(x)=\frac{1}{\sqrt{4\pi\kappa t}}\,e^{-x^2/(4\kappa t)},$$ and $$u(x,t)=\frac{1}{\sqrt{4\pi\kappa t}}\int_{-\infty}^\infty f(y)\,e^{-(x-y)^2/(4\kappa t)}\,dy.$$ *(2 transform PDE, 2 solve ODE, 2 convolution theorem, 2 kernel via Gaussian inversion, 1 final formula)* **(9)** ```verify [ {"claim":"Heat u_t=kappa u_xx is parabolic: B^2-AC=0 with A=-kappa,B=0,C=0", "code":"A,B,C=-1,0,0\nresult=(B*B-A*C==0)"}, {"claim":"Fourier coeff of x on (-pi,pi): b_n = 2(-1)^(n+1)/n, verify for n=1,2,3", "code":"x,