Week 13

Tutorial 13: Heat & Wave Equations

1. Solve the heat equation
$$\frac{\partial^2 u}{\partial t^2}-\frac{\partial^{2} u}{\partial x^{2}}=0, \quad 0<x<2, t>0$$
subject to
$$\begin{array}{ll} u(0, t)=u(1, t)=0 & t>0 \\ u(x, 0)=\pi x & 0<x<2 \end{array}$$
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Solution
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2u}{\partial t^2}$$
Suppose $u(x,t)=XT$
$$\begin{alignat*}{3}\frac{\partial u}{\partial x}&=X'T\qquad\qquad\frac{\partial u}{\partial t}&&=XT'\\\frac{\partial^2u}{\partial x^2}&=X''T\qquad\qquad\frac{\partial^2 u}{\partial t^2}&&=XT''\end{alignat*}$$
$$\begin{aligned}X''T&=XT''\\\frac{X''}{X}&=\frac{T''}{T}=-\lambda\end{aligned}$$
$$\begin{align}X''+\lambda X&=0\\T'+\lambda T&=0\end{align}$$
Case 1: $\lambda=0$
$$\begin{alignat*}{3}X''&=0\qquad\qquad\quad\qquad\qquad T''&&=0\\m^2&=0\qquad\quad\qquad\qquad\qquad m^2&&=0\\m_{1,2}&=0\qquad\quad\qquad\qquad\qquad m_{1,2}&&=0\\\Rightarrow X(x)&=C_1e^{Cx}+C_2xe^{Cx}\quad \Rightarrow T(t)&&=C_3e^{ct}+C_4e^{Ct}\\&=C_1+C_2x\quad\qquad &&=C_3+C_4t\end{alignat*}\\\therefore u(x,t)=XT=(C_1+C_2x)(C_3+C_4t)$$
Case 2: $\lambda<0$ ($-\alpha^2$)
$$\begin{alignat*}{3}X''-\alpha^2X&=0\qquad\qquad\qquad\qquad\qquad T''-\alpha^2T&&=0\\m^2-\alpha^2&=0\qquad\qquad\qquad\qquad\qquad m^2-\alpha^2&&=0\\m_{1,2}&=\pm\alpha\qquad\qquad\qquad\qquad\quad\qquad m_{1,2}&&=\pm\alpha\\\Rightarrow X(x)&=C_1\cosh\alpha x+C_2\sinh\alpha x\quad\Rightarrow T(t)&&=C_3\cosh\alpha t+C_4\sinh\alpha t\end{alignat*}\\\therefore u(x,t)=XT=(C_1\cosh\alpha x+C_2\sinh\alpha x)(C_3\cosh\alpha t+C_4\sinh\alpha t)$$
Case 3: $\lambda>0$ ($+\alpha^2$)
$$\begin{alignat*}{3}X''+\alpha^2X&=0\qquad\qquad\qquad\qquad T''+\alpha^2T&&=0\\m^2+\alpha^2&=0\qquad\qquad\qquad\qquad m^2+\alpha^2&&=0\\m_{1,2}&=\pm\alpha i\qquad\qquad\qquad\qquad\quad M_{1,2}&&=\pm\alpha i\\\Rightarrow X(x)&=C_1\cos\alpha x+C_2\sin\alpha x\quad\Rightarrow T(t)&&=C_3\cos\alpha t+C_4\sin\alpha t\end{alignat*}\\\therefore u(x,t)=XT=(C_1\cos\alpha x+C_2\sin\alpha x)(X_3\cos\alpha t+C_4\sin\alpha t)$$
Applying the boundary conditions: $u(0,t)=0$, $u(1,t)=0$
Case 1:
$$\begin{aligned}u(0,t)&=(C_1+C_2(0))(C_3+C_4t)\\&=C_1(C_3+C_4 t)=0\Rightarrow C_1=0\\\Rightarrow u(x,t)&=(C_2x)(C_3+C_4t)\\u(1,t)&=C_2(C_3+C_4t)=0\Rightarrow C_2=0\\\end{aligned}\\\therefore u(x,t)=C_3+C_4t$$
Case 2
$$\begin{aligned}u(0,t)&=(C_1\cosh0+C_2\sinh0)(C_3\cosh\alpha t+C_4\sinh\alpha t)=0\Rightarrow C_1=0\\\Rightarrow u(x,t)&=(C_2\sinh\alpha x)(C_3\cosh\alpha t+C_4\sinh\alpha t)\\u(1,t)&=(C_2\sinh\alpha)(C_3\cosh\alpha t+C_4\sinh\alpha t)=0\\C_2\sinh\alpha&=0\Rightarrow C_2=0\end{aligned}\\\therefore u(x,t)=C_3\cosh\alpha t+C_4\sinh\alpha t$$
Case 3
$$\begin{aligned}u(0,t)&=(C_1\cos0+C_2\sin0)(C_3\cos\alpha t+C_4\sin\alpha t)=0\Rightarrow C_1=0\\\Rightarrow u(x,t)&=C_2\sin\alpha x(C_3\cos\alpha t+C_4\sin\alpha t)\\u(1,t)&=C_2\sin\alpha(C_3\cos\alpha t+C_4\sin\alpha t)=0\\C_2\sin\alpha&=0\Rightarrow C_2=0\end{aligned}\\\therefore u(x,t)=C_3\cos\alpha t+C_4\sin\alpha t$$
For non-trivial solutions, let $C_2\neq0$ ⇒ $\sin\alpha=0$ ⇒ $\alpha=n\pi\quad n=1,2,...$
Eigenvalues:
$$\alpha_n=\alpha_n^2=n^2\pi^2,\quad n=1,2,...$$
Eigenvectors:
$$X(x)=C_n\sin\alpha x=C_n\sin n\pi x, \quad n=1,2,...$$
Applying initial conditions: $u(x,0)=\pi x$, $\left.\frac{\partial u}{\partial t}\right|_{t=0}=0$
$$\begin{aligned}u_n(x,t)&=(C_n\sin n\pi x)(C_3\cos\alpha t+C_4\sin\alpha t),\quad n=1,2,...\\&=(A_n\cos n\pi t)+B_n\sin n\pi t)\sin n\pi x\\\Rightarrow u(x,t)&=\sum_{n=1}^\infty (A_n\cos n\pi t+B_n\sin n\pi t)\sin n\pi x\end{aligned}$$
$$\begin{aligned}u(x,0)=\sum_{n=1}^\infty (A_n\cos 0+B_n\sin0)\sin n\pi x&=\sin\pi x\\\sum_{n=1}^\infty A_n\sin n\pi x&=\sin\pi x\end{aligned}\\\Rightarrow A_n=\frac{2}{1}\int_0^1\sin\pi x\sin n\pi x dx$$
$$\begin{aligned}A_n&=2\int_0^1\sin\pi x\sin n\pi xdx\\&=2\int_0^1\frac{1}{2}\left(\cos(\pi x-n\pi x)-\cos(\pi x+n\pi x)\right)dx\\&=\int_0^1\cos(1-n)\pi xdx=\int_0^1\cos(1+n)\pi xdx\\&=\left[\frac{\sin(1-n)\pi x}{(1-n)\pi}-\frac{\sin(1+n)\pi x}{(1+n)\pi}\right]_0^1\\&=\frac{\sin(1-n)\pi}{(1-n)\pi}-\frac{\sin(1+n)\pi}{(1+n)\pi},\quad n=1,2,3,...\end{aligned}$$
For $n=2,3,...$, $A_n=0$
For $n=1$
$$\begin{aligned}A_1&=2\int_0^1\frac{1}{2}(\cos(1-1)\pi x-\cos(1+1)\pi x)dx\\&=\int_0^1(\cos0-\cos 2\pi x)dx\\&=\int_0^11dx-\int_0^1\cos 2\pi xdx\\&=\left.x\right|_0^1-\left.\frac{\sin 2\pi x}{2\pi}\right|_0^1=1\end{aligned}$$
$$\frac{\partial u}{\partial t}=\sum_{n=1}^\infty \left(-A_nn\pi\sin n\pi t+B_nn\pi\cos n\pi t\right)\sin\pi x\\\left.\frac{\partial u}{\partial t}\right|_{t=0}=\sum_{n=1}^\infty(-A_nn\pi\sin0+B_nn\pi\cos0)\sin\pi x=0\\\sum_{n=1}^\infty(B_nn\pi)\sin\pi x=0\Rightarrow B_n=0$$
Solution exists only when $n=1$. Thus,
$$u(x,t)=\cos\pi t\sin\pi x$$

2. Solve the heat equation
$$\frac{\partial u}{\partial t}-\frac{1}{16} \frac{\partial^{2} u}{\partial x^{2}}=0, \quad 0<x<1, t>0$$
subject to
$$\begin{array}{ll} u(0, t)&=u(1, t)=0 & t>0 \\ u(x, 0)&=2 \sin 2 \pi x & 0<x<1 \end{array}$$
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Solution
Suppose $u(x,t)=X(x)T(t)$
$$\begin{alignat*}{3}\frac{\partial u}{\partial x}&=X'T\qquad\qquad\frac{\partial u}{\partial t}&&=XT'\\\frac{\partial^2u}{\partial x^2}&=X''T\end{alignat*}$$
$$\begin{aligned}XT'-\frac{1}{16}X''T&=0\\XT'&=\frac{1}{16}X''T\\\frac{X''}{16X}&=\frac{T'}{T}=-\lambda\end{aligned}$$
$$\begin{align}X''+16\lambda X&=0\\T'+\lambda T&=0\end{align}$$
Case 1: $\lambda=0$
$$X''=0\quad\Rightarrow\quad X(x)=C_1x+C_2\\u(0,t)=0,\quad X(0)=0=C_2\\u(1,t)=0,\quad X(1)=C_1=0$$
⇒ This is a trivial solution.
Case 2: $\lambda=-\alpha^2<0$
$$X''-16\alpha^2X=0\\X(x)=C_1\cosh(4\alpha x)+C_2\sinh(4\alpha x)\\\begin{aligned}u(0,t)=0,\quad &C_1=0\\u(1,t)=0,\quad &C_2\sinh4\alpha=0\end{aligned}$$
Since $\sinh4\alpha\neq0,\quad C_2=0$
⇒ This is a trivial solution
Case 3:
$$X''+16\alpha^2X=0\\X(x)=C_1\cos(4\alpha x)+C_2\sin(4\alpha x)\\\begin{aligned}u(0,t)=0,\quad&C_1=0\\u(1,t)=0,\quad&C_2=\sin4\alpha=0\end{aligned}$$
To obtain non-trivial solution, $\sin4\alpha=0$
$$\begin{aligned}4\alpha=n\pi,\quad&n=1,2,3,...\\\alpha=\frac{n\pi}{4},\quad&n=1,2,3,...\end{aligned}$$
Eigenvalue:
$$\lambda=\alpha^2=\frac{n^2\pi^2}{16},\quad n=1,2,3,...$$
Eigenfunction:
$$X_n(x)=C_2\sin\left(\left(\frac{n\pi}{4}\right)x\right)=C_2\sin(n\pi x)$$
$$T'+\alpha^2T=0\\\begin{aligned}T(t)&=C_3e^{-\alpha^2t}\\T_n(t)&=C_3e^{-\frac{n^2\pi^2}{16}t}\end{aligned}$$
$$u_n(x,t)=X_nT_n=B_ne^{-\frac{n^2\pi^2}{16}t}\sin(n\pi x),\quad n=1,2,3,...$$
By superposition principle,
$$u(x,t)=\sum_{n=1}^\infty B_ne^{-\frac{n^2\pi^2}{16}t}\sin(n\pi x)$$
Given $u(x,0)=2\sin(2\pi x)$
$$\begin{aligned}2\sin(2\pi x)&=\sum_{n=1}^\infty B_n\sin (n\pi x)\\B_n&=2\int_0^12\sin(2\pi x)\sin(n\pi x)dx\end{aligned}$$
$$\int_0^12\sin(2\pi x)\sin(n\pi x)dx=\begin{cases}1,&n=2\\0,&\text{otherwise}\end{cases}$$
$$B_n=\begin{cases}2,&n=2\\0,&\text{otherwise}\end{cases}$$
$$\begin{aligned}\therefore u(x,t)&=B_2e^{-\frac{\pi^2}{4}t}\sin(2\pi x)\\&=2e^{-\frac{\pi^2}{4}t}\sin(2\pi x)\end{aligned}$$

3. Solve the wave equation
$$\frac{\partial^{2} u}{\partial t^{2}}-\frac{\partial^{2} u}{\partial x^{2}}=0, \quad 0<x<1, t>0$$
subject to the given conditions
$$\begin{aligned} u(0, t)&=u(1, t)=0 & t>0 \\ u(x, 0)&=\sin \pi x, \quad \frac{\partial u}{\partial t}(x, 0)=0 &0<x<1 \end{aligned}$$
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Solution
Suppose $u(x,t)=X(x)T(t)$
$$\begin{aligned}XT'-X''T&=0\\\frac{1}{X}X''&=\frac{1}{T}T'=-\lambda\end{aligned}\\\therefore X''=-\lambda X\qquad T'=-\lambda T$$
$$u(0,t)=0\quad,\quad\therefore X(0)=0\\u(2,t)=0\quad,\quad\therefore X(2)=0$$
For $X^4+\lambda X=0$, $X(0)=0$ and $X(2)=0$
Case 1: $\lambda>0$
$$r_{1,2}=\pm\sqrt{\lambda}i\\\begin{aligned}X(x)=&C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x)\\&\Rightarrow C_1=0\quad,\quad\sin(2\sqrt{\lambda})=0\\\end{aligned}\\\\\lambda_n=\left(\frac{n\pi}{2}\right)^2,\quad n=1,2,3,...$$
Case 2: $\lambda=0$
$$X''=0\\X(c)=C_1x+C_2\\\Rightarrow C_1=0,\quad C_2=0$$
⇒ Only trivial solution
Case 3: $\lambda<0$
$$r_{1,2}=\pm\sqrt{\lambda}\\X(c)=C_1\cosh(\sqrt{\lambda}x)+C_2\sinh(\sqrt{\lambda}x)\\\Rightarrow C_1=0,\quad C_2=0$$
⇒ Only trivial solution
For $T’=-\lambda T$
From (1), $\lambda=\lambda_n=\left(\frac{n\pi}{2}\right)^2$
$$\therefore T(t)=C_1e^{-\lambda_n t}\quad,\quad n=1,2,3,...$$
Hence,
$$\begin{aligned}U_n(x,t)&=X_n(x)T_n(t)\\&=B_n\sin\left(\frac{n\pi x}{2}\right)e^{-\left(\frac{n\pi}{2}\right)^2t}\quad,\quad n=1,2,3,...\\\\U(x,t)&=\sin\frac{\pi x}{2}=B_n\sin\left(\frac{n\pi x}{2}\right)\end{aligned}$$
Thus,
$$\begin{aligned}B_n&=\int_0^2\sin\frac{\pi x}{2}\cdot\sin\left(\frac{n\pi x}{2}\right)dx\\&=\begin{cases}1,&\text{for }n=1\\0,&\text{for all }n\neq1\end{cases}\end{aligned}$$
$$\therefore U(x,t)=\sin\left(\frac{\pi x}{2}\right)e^{-\frac{\pi^2}{4}t}$$