Partial Differential Equations

Question 1 [12 marks]

(a) For $A u_{xx}+B u_{xy}+C u_{yy}$, discriminant $= B^2-4AC$. Here $A=x^2$, $B=0$, $C=-y^2$. [1] $B^2-4AC = 0 - 4(x^2)(-y^2) = 4x^2y^2.$ [2] For $x>0,y>0$, $4x^2y^2>0$ ⇒ hyperbolic everywhere in the region. [1]

(b) $A=1$, $B=2$, $C=1-x$. [1] $B^2-4AC = 4 - 4(1-x) = 4x.$ [3]

• Parabolic when $4x=0 \Rightarrow x=0$ (the $y$-axis). [2]
• Hyperbolic when $4x>0 \Rightarrow x>0$. [1]
• Elliptic when $4x<0 \Rightarrow x<0$. [1]

Question 2 [14 marks]

(a) Half-range sine series: $f(x)=\sum b_n \sin(nx)$ with $b_n = \frac{2}{\pi}\int_0^\pi x\sin(nx)\,dx.$ [2] Integrate by parts: $\int_0^\pi x\sin(nx)\,dx = \left[-\frac{x\cos nx}{n}\right]_0^\pi + \frac1n\int_0^\pi \cos nx\,dx = -\frac{\pi\cos n\pi}{n} + 0.$ [2] Since $\cos n\pi=(-1)^n$: $b_n = \frac{2}{\pi}\cdot\left(-\frac{\pi(-1)^n}{n}\right) = \frac{-2(-1)^n}{n} = \frac{2(-1)^{n+1}}{n}.$ [1] $\boxed{\,x = \sum_{n=1}^\infty \frac{2(-1)^{n+1}}{n}\sin(nx)\,}, \quad 0<x<\pi.$ [1]

(b) LHS of Parseval: $\frac{2}{\pi}\int_0^\pi x^2\,dx = \frac{2}{\pi}\cdot\frac{\pi^3}{3} = \frac{2\pi^2}{3}.$ [3] RHS: $\sum_{n=1}^\infty b_n^2 = \sum \frac{4}{n^2}.$ [2] Equate: $\frac{2\pi^2}{3} = 4\sum\frac{1}{n^2} \Rightarrow \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$ [3]

Question 3 [14 marks]

(a) Let $u=X(x)T(t)$. Then $XT'=kX''T \Rightarrow \dfrac{T'}{kT}=\dfrac{X''}{X}=-\lambda$. [2] Spatial: $X''+\lambda X=0$, with $X(0)=X(L)=0$. Nontrivial solutions require $\lambda_n=(n\pi/L)^2$, $X_n=\sin(n\pi x/L)$. [3] Time: $T'=-k\lambda_n T \Rightarrow T_n=e^{-k(n\pi/L)^2 t}$. [2] General solution: $u(x,t)=\sum_{n=1}^\infty b_n \sin\!\frac{n\pi x}{L}\,e^{-k(n\pi/L)^2 t}.$ [1]

(b) Match $u(x,0)=\sum b_n\sin(n\pi x/L) = 3\sin\frac{\pi x}{L}-\sin\frac{3\pi x}{L}$. By orthogonality: $b_1=3$, $b_3=-1$, all others $0$. [2] $\boxed{\,u(x,t)=3\sin\!\frac{\pi x}{L}\,e^{-k(\pi/L)^2 t} - \sin\!\frac{3\pi x}{L}\,e^{-9k(\pi/L)^2 t}\,}.$ [2]

(c) As $t\to\infty$ each exponential $\to 0$, so $u\to 0$. [1] Physically: with both ends at $0$ and no source, heat diffuses out until the rod reaches the steady state $u\equiv 0$. [1]

Question 4 [12 marks]

(a) D'Alembert with $\phi(x)=e^{-x^2}$, $\psi=0$: $u(x,t)=\tfrac12\big[\phi(x-ct)+\phi(x+ct)\big] = \tfrac12\big[e^{-(x-ct)^2}+e^{-(x+ct)^2}\big].$ [4]

(b) At $x=0$: $u(0,t)=\tfrac12\big[e^{-c^2t^2}+e^{-c^2t^2}\big]=e^{-c^2t^2}.$ [3] As $t\to\infty$, $u(0,t)\to 0$: the two travelling half-pulses separate, leaving the origin. [1]

(c) With $\phi=0$: $u(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds.$ [2] Boundedness requires the antiderivative $G(x)=\int_0^x g$ to be bounded, i.e. $\int_{x-ct}^{x+ct}g\,ds = G(x+ct)-G(x-ct)$ stays bounded. Precise sufficient condition: $G(\pm\infty)$ finite, i.e. $\int_{-\infty}^{\infty} g\,ds$ converges (integral of $g$, not $|g|$, need be finite for both limits). Equivalently $g$ integrable in the improper sense over $\mathbb R$. [2]

Question 5 [8 marks]

Separate $u=X(x)Y(y)$: $X''/X=-Y''/Y=-\lambda$. [1] $X(0)=X(a)=0 \Rightarrow \lambda_n=(n\pi/a)^2$, $X_n=\sin(n\pi x/a)$. [2] $Y''-(n\pi/a)^2 Y=0$ with $Y(0)=0 \Rightarrow Y_n=\sinh(n\pi y/a)$. [2] $u(x,y)=\sum_{n=1}^\infty c_n \sin\!\frac{n\pi x}{a}\sinh\!\frac{n\pi y}{a}.$ [1] Apply $u(x,b)=f(x)$: $c_n\sinh\!\frac{n\pi b}{a} = \frac{2}{a}\int_0^a f(x)\sin\!\frac{n\pi x}{a}\,dx,$ $\boxed{\,c_n=\frac{2}{a\,\sinh(n\pi b/a)}\int_0^a f(x)\sin\!\frac{n\pi x}{a}\,dx\,}.$ [2]

[
  {"claim":"Q1a discriminant of x^2 u_xx - y^2 u_yy is 4x^2y^2 >0 (hyperbolic)","code":"x,y=symbols('x y',positive=True); A=x**2; B=0; C=-y**2; disc=B**2-4*A*C; result=simplify(disc-4*x**2*y**2)==0 and disc.subs({x:1,y:1})>0"},
  {"claim":"Q1b discriminant 4 - 4(1-x) = 4x","code":"x=symbols('x'); disc=2**2-4*1*(1-x); result=simplify(disc-4*x)==0"},
  {"claim":"Q2a sine coefficient b_n = 2(-1)^(n+1)/n","code":"import sympy as sp; x=sp.symbols('x'); n=sp.symbols('n',positive=True,integer=True); bn=sp.integrate(x*sp.sin(n*x),(x,0,sp.pi))*2/sp.pi; result=sp.simplify(bn - 2*(-1)**(n+1)/n)==0"},
  {"claim":"Q2b sum 1/n^2 = pi^2/6 via Parseval","code":"lhs=sp.Rational(2,1)*sp.pi**2/3; s=sp.summation((2*(-1)**(sp.Symbol('m',integer=True,positive=True)+1)/sp.Symbol('m',integer=True,positive=True))**2,(sp.Symbol('m',integer=True,positive=True),1,sp.oo)); result=sp.simplify(lhs-4*sp.pi**2/6)==0 and sp.simplify(sp.summation(1/sp.Symbol('k',integer=True,positive=True)**2,(sp.Symbol('k',integer=True,positive=True),1,sp.oo))-sp.pi**2/6)==0"},
  {"claim":"Q4b u(0,t)=exp(-c^2 t^2)","code":"c,t=symbols('c t'); u=(sp.exp(-(0-c*t)**2)+sp.exp(-(0+c*t)**2))/2; result=sp.simplify(u-sp.exp(-c**2*t**2))==0"}
]