1. Consider the initial boundary value problem for the temperature $T(x, t)$ in a rod of length $L$ and thermal diffusivity $\kappa$ given by the heat equation
   $\frac{\partial T}{\partial t}=\kappa \frac{\partial^{2} T}{\partial x^{2}} \quad \text { for } \quad 0<x<L, t>0,$
   with the boundary conditions $T_{x}(0, t)=0$ and $T_{x}(L, t)=0$ for $t>0$ and the initial condition $T(x, 0)=T^{*} x(L-x) / L^{2}$ for $0<x<L$, where $T^{*}$ is a positive constant.
   (a) Show that the solution $T(x, t)$ is uniquely determined.
   (b) Use the method of separation of variables, the principle of superposition and the theory of Fourier series to derive the series solution given by
   $T(x, t)=\frac{T^{*}}{6}-\sum_{m=1}^{\infty} \frac{T^{*}}{m^{2} \pi^{2}} \cos \left(\frac{2 m \pi x}{L}\right) \exp \left(-\frac{4 m^{2} \pi^{2} \kappa t}{L^{2}}\right) .$
   (c) What is the behaviour of the temperature $T(x, t)$ in the limit as $t \rightarrow \infty$ ?
   [In part (b) you may assume that the orders of summation and integration may be interchanged as necessary and the identities
   $\int_{0}^{L} \cos \left(\frac{m \pi x}{L}\right) \cos \left(\frac{n \pi x}{L}\right) \mathrm{d} x=\frac{L}{2} \delta_{m n}, \quad \int_{0}^{L} x(L-x) \cos \left(\frac{n \pi x}{L}\right) \mathrm{d} x \mathrm{~d} x=-\frac{L^{3}\left(1+(-1)^{n}\right)}{n^{2} \pi^{2}},$
   where $m$ and $n$ are positive integers and $\delta_{m n}$ is Kronecker's delta.]
   Solution.
   (a)Suppose that $T(x, t)$ and $\tilde T(x, t)$ are solutions and let $W(x, t) = T(x, t) − \tilde T(x,t)$.
   By linearity, $W(x, t)$ satisfies the heat equation $\frac{\partial W}{\partial t}=\frac{\partial T}{\partial t}-\frac{\partial \widetilde{T}}{\partial t}=\kappa \frac{\partial^2T}{\partial x^2}-\kappa \frac{\partial^2\widetilde{T}}{\partial x^2}=\kappa \frac{\partial^2 W}{\partial x^2}$ for $0<x<L, t>0$.
   with the boundary conditions $W(0, t)=T(0, t)-\widetilde{T}(0, t)=0, \quad W(L, t)=T(L, t)-\widetilde{T}(L, t)=0$ for $t>0$ and the initial condition $W(x, 0)=T(x, 0)-\widetilde{T}(x, 0)=0$ for $0<x<L$.
   Analyse the integral $I(t)$ defined by $I(t):=\frac{1}{2} \int_{0}^{L} W(x, t)^{2} \mathrm{~d} x$.
   Evidently $I(t) ≥ 0$ for $t ≥ 0$ and $I(0) = 0$.
   But, for $t>0$\begin{align*}\frac{\mathrm{d} I}{\mathrm{~d} t} &=\int_{0}^{L} W \frac{\partial W}{\partial t} \mathrm{~d} x \\&=\int_0^LW \kappa \frac{\partial^2W}{\partial x^2} \mathrm{~d} x \\&=\left[\kappa W \frac{\partial W}{\partial x}\right]_{x=0}^{x=L}-\kappa \int_0^L\frac{\partial W}{\partial x} \frac{\partial W}{\partial x} \mathrm{~d} x \\&=-\kappa \int_0^L\left(\frac{\partial W}{\partial x}\right)^{2} \mathrm{~d} x \\& \leq 0\end{align*}which means that $I(t)$ cannot increase, so that $I(t) \leq I(0)=0$ for $t \geq 0$.
   Since $I(t) \geq 0$ and $I(t) \leq 0$ for $t \geq 0$, we deduce that $I(t)=0$ for $t \geq 0$, and hence that $W(x, t)=0$ for $0 \leq x \leq L, t \geq 0$ (assuming continuity of $W$ there).
   (b) Let $T(x,t)=F(x)G(t)$, then $F(x)G'(t)=κF''(x)G(t)$. Separating the variables by assuming $F(x)G(t)\ne0$ therefore gives ${F(x)\over F''(x)}=\kappa{G(t)\over G'(t)}$. The left-hand side of this expression is independent of $t$, while the right-hand side is independent of $x$. Since the left-hand side is equal to the right-hand side, they must both be independent of $x$ and $t$, and therefore equal to a constant, $−λ ∈\mathbb R$.
   Case (i): $λ = −ω^2(ω > 0)$.
   $F(x) = A\cosh(ωx) + B\sinh(ωx).$ The boundary conditions then require $B=0,A\sin(ωL)=0$, so that $F = 0$.
   Case (ii): $λ = 0$.
   $F(x) = A + Bx.$ The boundary conditions then require $B=0$, so that $F=A$.
   Case (iii): $λ = ω^2(ω>0)$
   $F(x) = A\cos(ωx) + B\sin(ωx).$ The boundary conditions then require $B=0,A\sin(ωL)=0$, for $A\ne0$, we have $ωL = nπ$ for some $n ∈\mathbb N\text{\\}\{0\}$.
   By the principle of superposition, general series solution is given by $\frac{b_0}2+\sum_{n=1}^∞b_n\cos\left(\frac{n \pi x}{L}\right) \exp \left(-\frac{n^{2} \pi^{2} \kappa t}{L^{2}}\right)$.
   To satisfy the initial conditions, $T(x, 0)=\frac{b_0}2+\sum_{n=1}^{\infty} b_{n} \cos \left(\frac{n \pi x}{L}\right)$.
   By the theory of Fourier series, $b_0=\frac2L\int_0^L\frac{T^{*} x(L-x)}{L^2}\mathrm dx=\frac{T^{*}}3$ and $b_{n}=\frac{2}{L} \int_{0}^{L}\frac{T^{*} x(L-x)}{L^2}\cos\left(\frac{n \pi x}{L}\right) \mathrm{d} x=-\frac{2T^*\left(1+(-1)^{n}\right)}{n^{2} \pi^{2}}$$=\begin{cases}-\frac{T^*}{m^2\pi^{2}}&n=2m\\0&n=2m-1\end{cases}(m≥1)$
   Therefore $T(x, t)=\frac{T^{*}}{6}-\sum_{m=1}^{\infty} \frac{T^{*}}{m^{2} \pi^{2}} \cos \left(\frac{2 m \pi x}{L}\right) \exp \left(-\frac{4 m^{2} \pi^{2} \kappa t}{L^{2}}\right)$.
   (c) As $t→∞$, $\exp \left(-\frac{4 m^{2} \pi^{2} \kappa t}{L^{2}}\right)→0$, so $T(x,t)→\frac{T^{*}}{6}$
2. (a) Let $\kappa$ and $\omega$ be positive constants. Show that the heat equation $\frac{\partial T}{\partial t}=\kappa \frac{\partial^{2} T}{\partial x^{2}}$ has complex-valued solutions of the form $F(x) \mathrm{e}^{\mathrm{i} \omega t}$ provided $\kappa F^{\prime \prime}=\mathrm{i} \omega F$.
   Hence find $F$ if $F^{\prime}(x) \rightarrow 0$ as $x \rightarrow \infty$ and $F(0)=T_{1}$, where $T_{1}$ is a positive constant. [You may assume that the roots of $\lambda^{2}=\mathrm{i} \omega / \kappa$ are $\lambda=\pm(1+\mathrm{i}) \sqrt{\omega / 2 \kappa}$.]
   Solution.
   From $\frac{∂T}{∂t}=\mathrm iω·T$ and $\frac{\partial^2T}{\partial x^2}=\frac{F''}{F}T=\frac{iω·T}κ$ we have $\frac{\partial T}{\partial t}=\kappa\frac{\partial^2T}{\partial x^2}$.
   The general solution of $\kappa F^{\prime \prime}=\mathrm{i} \omega F$ is $F=A\mathrm e^{(1+\mathrm{i}) \sqrt{\omega / 2 \kappa}\;x}+B\mathrm e^{-(1+\mathrm{i}) \sqrt{\omega / 2 \kappa}\;x}$. To satisfy $F'→0$ as $x \rightarrow \infty$, we have $A=0$. To satisfy $F(0)=T_{1}$, we have $B=T_1$.
   Therefore $F=T_1\mathrm e^{-(1+\mathrm{i}) \sqrt{\omega / 2 \kappa}\;x}$.
   (b) Now let $T(x, t)=T_{0}+\operatorname{Re}\left(F(x) \mathrm{e}^{\mathrm{i} \omega t}\right)$, where $T_{0}$ is a real constant. Verify that $T(x, t)=T_{0}+T_{1} \exp \left(-\sqrt{\frac{\omega}{2 \kappa}} x\right) \cos \left(\omega t-\sqrt{\frac{\omega}{2 \kappa}} x\right),$ and explain why $T(x, t)$ is a solution of the heat equation for which $T_{x}(x, t) \rightarrow 0$ as $x \rightarrow \infty$ and $T(0, t)=T_{0}+T_{1} \cos (\omega t)$.
   Solution.
   $F(x) \mathrm{e}^{\mathrm{i} \omega t}=T_1\mathrm e^{\mathrm iωt-(1+\mathrm{i}) \sqrt{\omega / 2 \kappa}\;x}=T_1\exp \left(-\sqrt{\frac{\omega}{2 \kappa}} x\right)(\cos \left(\omega t-\sqrt{\frac{\omega}{2 \kappa}} x\right)+\mathrm i\sin\left(\omega t-\sqrt{\frac{\omega}{2 \kappa}} x\right))$
   Therefore $T(x, t)=T_{0}+\operatorname{Re}\left(F(x) \mathrm{e}^{\mathrm{i} \omega t}\right)=T_{0}+T_{1} \exp \left(-\sqrt{\frac{\omega}{2 \kappa}} x\right) \cos \left(\omega t-\sqrt{\frac{\omega}{2 \kappa}} x\right)$
   Using part (a), $T_x(x,t)=\operatorname{Re}\left(F'(x) \mathrm{e}^{\mathrm{i} \omega t}\right)=0$ and $T(0, t)=T_0+\operatorname{Re}\left(F(0) \mathrm{e}^{\mathrm{i} \omega t}\right)=T_0+\operatorname{Re}\left(T_1 \mathrm{e}^{\mathrm{i} \omega t}\right)=T_{0}+T_{1} \cos (\omega t)$
   (c) A root cellar is used to store crops, ideally by keeping them as cool as possible in the summer, but as warm as possible in the winter. Consider a root cellar buried in soil of thermal diffusivity $\kappa=10^{-6} \mathrm{~m}^{2} \mathrm{~s}^{-1}$. Use the temperature profile in part (b) to predict
   (i) the shallowest ideal depth of the root cellar;
   (ii) the factor by which the amplitude of the temperature oscillations at ground level are reduced at the shallowest ideal depth.
   Solution.
   (i)To get antiphase of $T(0,t)=T_0+T_1\cos(\omega t)$, let $\cos(\omega t+\sqrt{\omega\over2\kappa}x)=-\cos(\omega t)$, so $\sqrt{\omega\over2\kappa}x=(2k+1)\pi,k\in\Bbb N$, the smallest $x$(shallowest)is when $k=0$, $x=\sqrt{2\kappa\over\omega}\pi$, substituting $\kappa=10^{-6}{\rm m^2s^{-1}}$ and $\omega=\frac{2\pi}{365·25·24·60·60}{\rm s}^{-1}$, we get $x\approx9.95{\rm m}$
   (ii)the factor by which the amplitude of the temperature oscillations at ground level are reduced  is $\exp(-\sqrt{\omega\over2\kappa}\sqrt{2\kappa\over w}·\pi)=e^{-π}≈4\%$.
3. Consider the initial boundary value problem for the temperature $T(x, t)$ in a rod of length $L$ given by the inhomogeneous heat equation
   $\rho c \frac{\partial T}{\partial t}=k \frac{\partial^{2} T}{\partial x^{2}}+Q(x, t)$     for    $0<x<L, t>0,$
   with the boundary conditions $T(0, t)=\phi(t)$ and $T(L, t)=\psi(t)$ for $t>0$ and the initial condition $T(x,0)=f(x)$ for $0<x<L$, where $\rho, c$ and $k$ are positive constants and the functions $Q(x,t),\phi(t),\psi(t)$ and $f(x)$ are given.
   (a) Let $T(x, t)=\phi(t)\left(1-\frac{x}{L}\right)+\psi(t) \frac{x}{L}+U(x, t)$. Determine the functions $\widetilde{Q}$ and $\widetilde{f}$ for which $U$ satisfies the initial boundary value problem given by $\rho c \frac{\partial U}{\partial t}=k \frac{\partial^{2} U}{\partial x^{2}}+\widetilde{Q}(x, t)$ for $0<x<L, t>0,$ with $U(0, t)=U(L, t)=0$ for $t>0$ and $U(x, 0)=\widetilde{f}(x)$ for $0<x<L$.
   Solution.
   Substituting ${\partial T\over\partial t}=\phi'(t)\left(1-\frac{x}{L}\right)+\psi'(t) \frac{x}{L}+{\partial U\over\partial t}$ and ${\partial^2 T\over\partial x^2}={\partial^2 U\over\partial x^2}$ into $\rho c \frac{\partial T}{\partial t}=k \frac{\partial^{2} T}{\partial x^{2}}+Q(x, t)$ and $T(x, 0)=f(x)$, we have $\tilde Q(x,t)=-\rho c\left(\phi'(t)\left(1-\frac{x}{L}\right)+\psi'(t) \frac{x}{L}\right)$.
   From $T(x, t)=\phi(t)\left(1-\frac{x}{L}\right)+\psi(t) \frac{x}{L}+U(x, t)$ and $T(x,0)=f(x)$ we have $f(x)=\phi(0)+U(x,0)$, so $\tilde f(x)=f(x)-\phi(0)$.
   (b) By considering your answer to question 3 of sheet 4 , write down the solution for $U(x,t)$ in the special case in which $\widetilde{Q}(x, t)=0$ for $0<x<L,t>0$.
   Solution.
   $U(x, t)=\sum_{m=0}^∞\frac{8 T^*}{(2 m+1)^3\pi^3} \sin \left(\frac{(2 m+1)\pi x}{L}\right) \exp \left(-\frac{(2 m+1)^{2} \pi^{2} k t}{L^{2}ρc}\right) .$
   (c) Consider now the case in which $\widetilde{Q}$ is not identically zero. Suppose that $U(x, t)$ and $\widetilde{Q}(x, t)$ may be expanded as the Fourier sine series
   $U(x, t)=\sum_{n=1}^{\infty} U_{n}(t) \sin \left(\frac{n \pi x}{L}\right), \quad \widetilde{Q}(x, t)=\sum_{n=1}^{\infty} \widetilde{Q}_{n}(t) \sin \left(\frac{n \pi x}{L}\right),$
   where the Fourier coefficients are given by
   $U_n(t)=\frac{2}{L} \int_{0}^{L} U(x, t) \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x, \quad \widetilde{Q}_{n}(t)=\frac{2}{L} \int_{0}^{L} \widetilde{Q}(x, t) \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x$.
   (i) By differentiating $U_n(t)$ under the integral sign, using the heat equation and integrating by parts, derive an ordinary differential equation for $U_n(t)$. What is the initial condition?
   (ii) Explain without any further calculations how to determine the temperature $T(x, t)$ given the functions $Q(x, t), \phi(t), \psi(t)$ and $f(x)$.
   (iii) What are the advantages of expanding $U$ as a Fourier sine series rather than $T$ ?
   Solution.
   (i)Differentiate $U_n(t)$ with respect to $t$ to obtain$$\rho c\frac{\mathrm{d} U_{n}}{\mathrm{~d} t}=\frac{2}{L} \int_{0}^{L} \rho c \frac{\partial U}{\partial t} \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x=\frac{2}{L} \int_{0}^{L}\left(k \frac{\partial^{2} U}{\partial x^{2}}+\widetilde{Q}\right) \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x$$where we used Leibniz's Integral Rule in the first equality and the heat equation $\rho c \frac{\partial U}{\partial t}=k \frac{\partial^{2} U}{\partial x^{2}}+\widetilde{Q}$ in the second equality. Integrating by parts using the identity$$\int_{0}^{L} u v^{\prime \prime}-u^{\prime \prime} v \mathrm{~d} x=\int_{0}^{L}\left(u v^{\prime}-u^{\prime} v\right)^{\prime} \mathrm{d} x=\left[u v^{\prime}-u^{\prime} v\right]_{0}^{L}$$with $u=U$ and $v=\sin(n \pi x / L)$ gives$$\int_{0}^{L} U\left(-\frac{n^{2} \pi^{2}}{L^{2}} \sin \left(\frac{n \pi x}{L}\right)\right)-U_{x x} \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x=\left[U\frac{n \pi}{L}\cos \left(\frac{n \pi x}{L}\right)-U_{x} \sin \left(\frac{n \pi x}{L}\right)\right]_{0}^{L}=0$$by the boundary conditions, so that$$\frac{2}{L} \int_{0}^{L} U_{x x} \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x=-\frac{n^{2} \pi^{2}}{L^{2}} \frac{2}{L} \int_{0}^{L} U \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x=-\frac{n^{2} \pi^{2}}{L^{2}} U_{n}$$Hence, we find that $U_{n}(t)$ is governed by the ODE$$\rho c \frac{\mathrm{d} U_{n}}{\mathrm{~d} t}+\frac{k n^{2} \pi^{2}}{L^{2}} U_{n}=\widetilde{Q}_{n}(t) \quad \text { for } \quad t>0$$Since $U(x, 0)=\widetilde{f}(x)$, the initial condition for $U_n$ is $U_n(0)=\frac{2}{L} \int_{0}^{L}\widetilde{f}(x) \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x$.
   (ii) Using an integrating factor, we find that the solution of $U_n$ is $U_n(t)=\left(\frac{1}{\rho c_{v}} \int_{0}^{t} \widetilde{Q}_{n}(s) \mathrm{e}^{\kappa_{n} s} \mathrm{~d} s+U_{n}(0)\right) \mathrm{e}^{-\kappa_{n} t}$ where $\kappa_{n}=n^{2} \pi^{2} \kappa / L^{2}$ in terms of the thermal diffusivity $κ=k/(ρc)$.
   $T(x, t)=\phi(t)\left(1-\frac{x}{L}\right)+\psi(t) \frac{x}{L}+U(x, t)$.
   (iii) In general the method of shifting the data (to render homogeneous the boundary conditions) results in a solution that converges more rapidly, especially if Gibb’s phenomenon can be avoided by doing so.