Partial Differential Equations

Q1. Discriminant $D = B^2 - 4AC$; elliptic if $D<0$, parabolic if $D=0$, hyperbolic if $D>0$. (1 mark for stated rule, spread across parts)

(a) $A=1,\ B=0,\ C=1 \Rightarrow D = 0 - 4 = -4 < 0$ → elliptic. (2)

(b) $u_{xx} - c^2 u_{tt}=0$: $A=1,\ B=0,\ C=-c^2 \Rightarrow D = 0 - 4(1)(-c^2) = 4c^2 > 0$ → hyperbolic. (2)

(c) $u_t - \alpha u_{xx}=0$: only $u_{xx}$ is second order, so $A=-\alpha,\ B=0,\ C=0 \Rightarrow D = 0$ → parabolic. (2)

Q2. (4 marks)

• IVP: conditions are specified at a single value of the evolution variable (usually time $t=0$); the solution is propagated forward in time. (1) Example: heat equation with $u(x,0)=f(x)$. (1)
• BVP: conditions are specified on the boundary of the spatial domain (at different points in space). (1) Example: Laplace's equation $\nabla^2 u = 0$ with $u$ prescribed on the boundary of a region. (1)

(Many PDE problems are mixed initial-boundary value problems.)

Q3. Dirichlet conditions (over one period): (4 marks, 1 each)

1. $f$ is single-valued and (absolutely) integrable over a period.
2. $f$ has a finite number of maxima and minima in one period.
3. $f$ has a finite number of finite discontinuities (jumps) in one period.
4. Where these hold, the series converges to $f(x)$ at points of continuity and to $\tfrac12[f(x^+)+f(x^-)]$ at a jump.

Q4. For period $2L$: (5 marks) $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos\frac{n\pi x}{L} + b_n\sin\frac{n\pi x}{L}\right)$ (2) $a_0 = \frac{1}{L}\int_{-L}^{L} f(x)\,dx \quad\text{(1)}$ $a_n = \frac{1}{L}\int_{-L}^{L} f(x)\cos\frac{n\pi x}{L}\,dx \quad\text{(1)}$ $b_n = \frac{1}{L}\int_{-L}^{L} f(x)\sin\frac{n\pi x}{L}\,dx \quad\text{(1)}$

Q5. Half-range sine series on $(0,\pi)$, so $L=\pi$: (5 marks) $b_n = \frac{2}{\pi}\int_0^\pi 1\cdot\sin(nx)\,dx = \frac{2}{\pi}\left[-\frac{\cos nx}{n}\right]_0^\pi = \frac{2}{\pi n}(1-\cos n\pi).$ (3) Since $\cos n\pi = (-1)^n$: $b_n = \dfrac{2}{\pi n}\big(1-(-1)^n\big) = \begin{cases}\dfrac{4}{\pi n}, & n \text{ odd}\\[4pt] 0, & n \text{ even}\end{cases}$ (1) $f(x) = \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin\big((2k+1)x\big)}{2k+1}.$ (1)

Q6. Parseval's theorem: (3 marks) $\frac{1}{L}\int_{-L}^{L} [f(x)]^2\,dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}\big(a_n^2 + b_n^2\big).$ (2 marks for correct RHS structure, 1 for the $a_0^2/2$ term / correct LHS factor.)

Q7. (4 marks)

• Heat equation: $u_t = \alpha u_{xx}$ (or $\kappa u_{xx}$). Models temperature $u(x,t)$ in a rod; $\alpha>0$ is the thermal diffusivity. (2)
• Wave equation: $u_{tt} = c^2 u_{xx}$. Models transverse displacement $u(x,t)$ of a string; $c$ is the wave (propagation) speed. (2)

Q8. D'Alembert's solution: (4 marks) $u(x,t) = \frac{1}{2}\big[\phi(x+ct) + \phi(x-ct)\big] + \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds.$ (2 marks for the $\phi$ average term, 2 marks for the correct integral of $\psi$ with $1/2c$ and limits.)

Q9. Let $u=X(x)T(t)$. Then $u_t = X T'$, $u_{xx}=X'' T$. (1) Substitute into $u_t=\alpha u_{xx}$: $X T' = \alpha X'' T$. (1) Divide by $\alpha X T$: $\frac{T'}{\alpha T} = \frac{X''}{X} = -\lambda \quad(\text{constant}).$ (2) Hence the two ODEs: $X'' + \lambda X = 0, \qquad T' + \alpha\lambda T = 0.$ (1) (The negative separation constant $-\lambda$ is chosen so that decaying, non-trivial bounded solutions with typical BCs exist.)

[
  {"claim":"Q1(b) discriminant is positive (hyperbolic): B^2-4AC with A=1,B=0,C=-c^2",
   "code":"c=symbols('c',positive=True); A,B,C=1,0,-c**2; D=B**2-4*A*C; result=simplify(D-4*c**2)==0 and D.subs(c,1)>0"},
  {"claim":"Q1(a) discriminant negative (elliptic)",
   "code":"A,B,C=1,0,1; D=B**2-4*A*C; result=(D==-4) and (D<0)"},
  {"claim":"Q5 sine coefficient b_n = 2/(pi n)*(1-(-1)**n)",
   "code":"n=symbols('n',positive=True,integer=True); x=symbols('x'); bn=simplify(Rational(2,1)/pi*integrate(sin(n*x),(x,0,pi))); result=simplify(bn-2*(1-(-1)**n)/(pi*n))==0"},
  {"claim":"Q5 b_3 = 4/(3 pi), b_2 = 0",
   "code":"x=symbols('x'); b3=simplify(Rational(2,1)/pi*integrate(sin(3*x),(x,0,pi))); b2=simplify(Rational(2,1)/pi*integrate(sin(2*x),(x,0,pi))); result=(simplify(b3-4/(3*pi))==0) and (b2==0)"}
]