Partial Differential Equations

Question 1 (8 marks)

(a) Discriminant $\Delta = B^2 - 4AC$. If $\Delta>0$ hyperbolic, $\Delta=0$ parabolic, $\Delta<0$ elliptic. (1) Here $A=1,\ B=4,\ C=1$: $\Delta = 16 - 4 = 12 > 0$. (1) → hyperbolic (lower-order $3u_x$ term does not affect classification). (1)

(b) $A=x^2$, $B=0$, $C=-y^2$. $\Delta = 0 - 4(x^2)(-y^2)=4x^2y^2$. (2) For $x,y>0$, $\Delta=4x^2y^2>0$ everywhere ⇒ hyperbolic throughout the open first quadrant. (1) (Parabolic only on axes where $x=0$ or $y=0$.) (1) Physically: hyperbolic equations have real characteristics, so signals/information propagate along them at finite speed (wave-like); initial disturbances travel without instant smoothing. (1)

Question 2 (12 marks)

(a) Assume $f=\frac{a_0}{2}+\sum(a_n\cos nx+b_n\sin nx)$. Use orthogonality on $(-\pi,\pi)$: $\int_{-\pi}^{\pi}\cos mx\cos nx\,dx=\pi\delta_{mn},\quad \int_{-\pi}^{\pi}\sin mx\sin nx\,dx=\pi\delta_{mn},\quad \int_{-\pi}^{\pi}\cos mx\sin nx\,dx=0.$ (2) Multiply by the relevant basis function and integrate: $a_0=\frac1\pi\int_{-\pi}^{\pi}f\,dx,\quad a_n=\frac1\pi\int_{-\pi}^{\pi}f\cos nx\,dx,\quad b_n=\frac1\pi\int_{-\pi}^{\pi}f\sin nx\,dx.$ (2)

(b) $f=x$ is odd ⇒ $a_0=0$, $a_n=0$. (1) $b_n=\frac1\pi\int_{-\pi}^{\pi}x\sin nx\,dx=\frac2\pi\int_0^{\pi}x\sin nx\,dx.$ (1) Integration by parts: $\int_0^\pi x\sin nx\,dx=\left[-\frac{x\cos nx}{n}\right]_0^\pi+\frac1n\int_0^\pi\cos nx\,dx=-\frac{\pi\cos n\pi}{n}=-\frac{\pi(-1)^n}{n}.$ (2) So $b_n=\frac2\pi\cdot\left(-\frac{\pi(-1)^n}{n}\right)=\frac{2(-1)^{n+1}}{n}$. (1) $\boxed{\,x=\sum_{n=1}^{\infty}\frac{2(-1)^{n+1}}{n}\sin nx\,}$

(c) Dirichlet conditions: $f$ periodic, with a finite number of maxima/minima and finite number of finite discontinuities per period, and absolutely integrable over a period. (2) At $x=\pi$ (a jump discontinuity of the periodic extension), the series converges to the midpoint $\frac{f(\pi^-)+f(-\pi^+)}{2}=\frac{\pi+(-\pi)}{2}=0$. (1)

Question 3 (12 marks)

(a) Let $u=X(x)T(t)$. Then $XT'=\alpha^2 X''T$, divide by $\alpha^2 XT$: $\frac{T'}{\alpha^2 T}=\frac{X''}{X}=-\lambda \text{ (const)}.$ (2) Gives $X''+\lambda X=0$ and $T'+\alpha^2\lambda T=0$. (2) If $\lambda\le 0$ the spatial solutions (exponential/linear) cannot satisfy $X(0)=X(L)=0$ nontrivially; only $\lambda>0$ (written $-\lambda<0$ as the constant) yields sinusoidal solutions matching the BCs and a decaying $T$. (2)

(b) $X(0)=X(L)=0\Rightarrow \lambda_n=\left(\frac{n\pi}{L}\right)^2$, $X_n=\sin\frac{n\pi x}{L}$. (1) $T_n=e^{-\alpha^2(n\pi/L)^2 t}$. (1) $u(x,t)=\sum_{n=1}^\infty b_n\sin\frac{n\pi x}{L}e^{-\alpha^2(n\pi/L)^2 t},\quad b_n=\frac2L\int_0^L f(x)\sin\frac{n\pi x}{L}dx.$ (2)

(c) $f=\sin 3x$ matches $n=3$, $L=\pi$, $\alpha^2=1$, so $b_3=1$, others $0$: $u(x,t)=e^{-9t}\sin 3x.$ (2)

Question 4 (10 marks)

(a) With $\xi=x-ct$, $\eta=x+ct$: chain rule gives $u_{tt}=c^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta})$ and $u_{xx}=u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta}$. (2) Substituting into $u_{tt}=c^2u_{xx}$ ⇒ $-4c^2 u_{\xi\eta}=0$ ⇒ $u_{\xi\eta}=0$. (1) Integrate: $u=F(\xi)+G(\eta)=F(x-ct)+G(x+ct)$. (1) Apply ICs: $F(x)+G(x)=\phi(x)$, $-cF'+cG'=\psi(x)$. Integrating the second: $-F+G=\frac1c\int_0^x\psi\,ds+K$. Solve: $u(x,t)=\frac12[\phi(x-ct)+\phi(x+ct)]+\frac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds.$ (2)

(b) $\psi=0$: $u(x,t)=\frac12[e^{-(x-ct)^2}+e^{-(x+ct)^2}].$ (2) Physically: the initial Gaussian bump splits into two half-amplitude copies travelling right and left at speed $c$ without change of shape. (2)

Question 5 (10 marks)

(a) $\widehat{f'}(k)=\int_{-\infty}^\infty f'(x)e^{-ikx}dx$. Integrate by parts: $=[f(x)e^{-ikx}]_{-\infty}^{\infty}+ik\int_{-\infty}^\infty f(x)e^{-ikx}dx.$ (2) Assuming $f(x)\to 0$ as $|x|\to\infty$ (so the boundary term vanishes): $\widehat{f'}(k)=ik\hat f(k)$. (2)

(b) Convolution theorem: $\widehat{f*g}=\hat f\cdot\hat g$ (and inversely, product of transforms ↔ convolution). (2) Transforming the heat equation in $x$: $\hat u_t=-\alpha^2 k^2\hat u$ ⇒ $\hat u(k,t)=\hat f(k)e^{-\alpha^2 k^2 t}$. (2) This is a product of transforms, so by the convolution theorem $u=f * G$ where $G$ is the inverse transform of $e^{-\alpha^2 k^2 t}$ — a Gaussian heat kernel $G(x,t)=\frac{1}{\sqrt{4\pi\alpha^2 t}}e^{-x^2/(4\alpha^2 t)}$. Hence the temperature is the initial data smoothed (convolved) by a spreading Gaussian. (2)

Question 6 (8 marks)

(a) Forward time: $u_t\approx\frac{u_i^{n+1}-u_i^n}{\Delta t}$. Central space: $u_{xx}\approx\frac{u_{i+1}^n-2u_i^n+u_{i-1}^n}{\Delta x^2}$. (2) Equate: $u_i^{n+1}=u_i^n+r(u_{i+1}^n-2u_i^n+u_{i-1}^n)$, with $r=\frac{\alpha^2\Delta t}{\Delta x^2}$. (2)

(b) From memory:

u_new = u.copy()
u_new[1:-1] = u[1:-1] + r*(u[2:] - 2*u[1:-1] + u[:-2])
u_new[0] = uL; u_new[-1] = uR   # Dirichlet ends
u = u_new

(3) Stability (CFL): $r=\dfrac{\alpha^2\Delta t}{\Delta x^2}\le \dfrac12$. (1)

[
  {"claim":"b_n for f(x)=x on (-pi,pi) equals 2(-1)^(n+1)/n",
   "code":"n=symbols('n',positive=True,integer=True); x=symbols('x'); b=Rational(1,pi)*integrate(x*sin(n*x),(x,-pi,pi)); b=simplify(b); expected=2*(-1)**(n+1)/n; result=simplify(b-expected)==0"},
  {"claim":"Fourier series at x=pi converges to 0 (midpoint of jump)",
   "code":"val=(pi+(-pi))/2; result=(val==0)"},
  {"claim":"Heat solution with f=sin(3x),L=pi,alpha=1 is exp(-9t)sin(3x): satisfies u_t=u_xx",
   "code":"x,t=symbols('x t'); u=exp(-9*t)*sin(3*x); result=simplify(diff(u,t)-diff(u,x,2))==0"},
  {"claim":"D'Alembert (psi=0,phi=exp(-x^2)) solves wave eqn u_tt=c^2 u_xx",
   "code":"x,t,c=symbols('x t c'); u=(exp(-(x-c*t)**2)+exp(-(x+c*t)**2))/2; result=simplify(diff(u,t,2)-c**2*diff(u,x,2))==0"}
]