Tutorial 13: Partial Differential Equation II

Tutorial Question

The general solutions of the following PDEs have been found in Tutorial 12, Question 2 –5. Continue to find the particular solution for each case based on the given boundary and initial conditions.

Laplace equation, $u_{xx}+u_{yy}=0$

General solution: $u(x,y)=(c_1+c_2y)(c_3+c_4x)+\left(c_5\cos(\alpha y)+c_6\sin(\alpha y)\right)\left(c_7\cosh(\alpha x)+c_8\sinh(\alpha x)\right)+\left(c_9\cosh(\alpha y)+c_{10}\sinh (\alpha y)\right)\left(c_{11}\cos(\alpha x)+c_{12}\sin(\alpha x)\right)$

Subject to the given boundary conditions:

$\begin{aligned}u(0,y)&=0, &u(a,y)&=0, &0<y<b\\u(x,0)&=0, &u(x,b)&=f(x), &0<x<a\end{aligned}$

Solution
To solve the Laplace equation easily, we use the boundary condition 1 & 2 with zeros first.
For Case 1: $\lambda=0$
BC 1: $u_{1}(0, y)=\left(c_{1}+c_{2} y\right)\left(c_{3}+c_{4}(0)\right)=0$
$\rightarrow$ Assume $c_{1}+c_{2} y \neq 0 ; c_{3}=0$
$\rightarrow u_{1}(x, y)=\left(c_{1}+c_{2} y\right)\left(c_{4} x\right)$
BC 2: $u_{1}(a, y)=\left(c_{1}+c_{2} y\right)\left(c_{4} a\right)=0$
$\rightarrow\left(c_{1}+c_{2} y\right) \neq 0$ & $a \neq 0$ ; hence $c_{4}=0$
$\rightarrow u_{1}(x, y)=\left(c_{1}+c_{2} y\right)(0)=0$
$\therefore u_{1}=0$ ; No solution for case 1
For Case 2: $\lambda=-\alpha^{2}, \alpha>0$
BC 1: $u_{2}(0, y)=\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right)\left(c_{7} \cosh (0)+c_{8} \sinh (0)\right)=0$
$\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right)\left(c_{7}\right)=0$
$\rightarrow\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right) \neq 0 ;$ hence $c_{7}=0$
$\rightarrow u_{2}(x, y)=\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right)\left(c_{8} \sinh (\alpha x)\right)$
BC 2: $u_{2}(a, y)=\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right)\left(c_{8} \sinh (\alpha a)\right)=0$
$\rightarrow\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right) \neq 0, \sinh (\alpha a) \neq 0$ ; hence $c_{8}=0$
$\rightarrow u_{2}(x, y)=\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right)(0)=0$
$\therefore u_{2}=0$ ; No solution for case 2
For Case 3: $\lambda=+\alpha^{2}, \alpha>0$
BC 1: $u_{3}(0, y)=\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right)\left(c_{11} \cos (0)+c_{12} \sin (0)\right)=0$
$\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right)\left(c_{11}\right)=0$
$\rightarrow\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right) \neq 0 ;$ hence $c_{11}=0$
$\rightarrow u_{3}(x, y)=\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right)\left(c_{12} \sin (\alpha x)\right)$
BC 2: $u_{3}(a, y)=\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right)\left(c_{12} \sin (\alpha a)\right)=0$
$\rightarrow\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right), c_{12} \neq 0$ ; hence $\sin (\alpha a)=0$ when $\alpha \mathrm{b}=n \pi, \alpha=\frac{n \pi}{b}$ where $n=1,2,3, \ldots$
$\begin{aligned}\therefore u_{3, n}&=\left(c_{9, n} \cosh \left(\frac{n \pi}{a} y\right)+c_{10, n} \sinh \left(\frac{n \pi}{a} y\right)\right)\left(c_{12, n} \sin (\alpha x)\right) \\\quad&=\left(A_{3, n} \cosh \left(\frac{n \pi}{a} y\right)+B_{3, n} \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right) \text { where } n=1,2,3, \ldots\end{aligned}$
Eigenvalue, $\lambda_{n}=+\alpha_{n}{ }^{2}=\left(\frac{n \pi}{a}\right)^{2}$
Eigenfunction, $\mathrm{u}_{3, \mathrm{n}}=\left(A_{3, \mathrm{n}} \cosh \left(\frac{n \pi}{a} y\right)+B_{3, \mathrm{n}} \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$
$\therefore u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(A_{3, \mathrm{n}} \cosh \left(\frac{n \pi}{a} y\right)+B_{3, \mathrm{n}} \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$
BC 3:
$\begin{aligned}u_{\text {total }}(x, 0)&=\sum_{n=1}^{\infty}\left(A_{3, n} \cosh \left(\frac{n \pi}{a}(0)\right)+B_{3, n} \sinh \left(\frac{n \pi}{a}(0)\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)=0\\&=\sum_{n=1}^{\infty}\left(A_{3, n}\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)=0\end{aligned}$
Recall Half-Range Fourier Sine Series Expansion:
$A_{3, \mathrm{n}}=\frac{2}{L} \int_{0}^{\tau} f(x) \sin \left(\frac{n \pi}{a} x\right) d x=0$
$u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(B_{3, n} \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$
BC 4: $u_{\text {total }}(x, b)=\sum_{n=1}^{\infty}\left(B_{3, n} \sinh \left(\frac{n \pi}{a} b\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)=f(x)$
Half-Range Fourier Sine Series Expansion:
$f(x)=\sum_{n=1}^{\infty}\left(B_{3, n} \sinh \left(\frac{n \pi}{a} b\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$
where $B_{3, n} \sinh \left(\frac{n \pi}{a} b\right)=\frac{2}{L} \int_{0}^{\tau} f(x) \sin n \omega x d x, \omega=\frac{\pi}{a}, p=\frac{2 \pi}{\omega}=2 a, L=\frac{p}{2}=a, \tau=a$
$B_{3, n} \sinh \left(\frac{n \pi}{a} b\right)=\frac{2}{a} \int_{0}^{a} f(x) \sin n \frac{\pi}{a} x d x$
$B_{3, n}=\frac{2}{\operatorname{asinh}\left(\frac{n \pi}{a} b\right)} \int_{0}^{a} f(x) \sin n \frac{\pi}{a} x d x$
$\therefore u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(\frac{2}{\operatorname{asinh}\left(\frac{n \pi}{a} b\right)} \int_{0}^{a} f(x) \sin n \frac{\pi}{a} x d x \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$
$\begin{aligned}u(0,y)&=0, &u(1,y)&=1-y, &0<y<1\\ \left.\frac{\partial u}{\partial y}\right|_{y=0}&=0, &\left.\frac{\partial u}{\partial y}\right|_{y=1}&=0, &0<x<1\end{aligned}$

Solution
$\begin{aligned} u(0,y)&=0,\rightarrow \text{BC3} &u(1,y)&=1-y,\rightarrow \text{BC4} &0<y<1\\ \left.\frac{\partial u}{\partial y}\right|_{y=0}&=0,\rightarrow \text{BC1} &\left.\frac{\partial u}{\partial y}\right|_{y=1}&=0,\rightarrow \text{BC2} &0<x<1 \end{aligned}$
For Case 1: $\lambda=0$
$u_{1}=\left(c_{1}+c_{2} y\right)\left(c_{3}+c_{4} x\right)$
$\frac{\partial u_{1}}{\partial y}=\left(c_{2}\right)\left(c_{3}+c_{4} x\right)$
BC 1: $\frac{\partial u_{1}(x, 0)}{\partial y}=\left(c_{2}\right)\left(c_{3}+c_{4} x\right)=0$
$\rightarrow$ let $c_{3}+c_{4} x=0 \rightarrow x=-\frac{c_{3}}{c_{4}}$ (Not true since $0<x<1$ hence, $c_{3}+c_{4} x \neq 0, c_{2}=0$
$\rightarrow u_{1}=\left(c_{1}\right)\left(c_{3}+c_{4} x\right)=\left(A_{1}+B_{1} x\right)$
BC 2: $\frac{\partial u_{1}(x, 1)}{\partial y}=\left(c_{2}\right)\left(c_{3}+c_{4} x\right)=0$
$\rightarrow\left(c_{3}+c_{4} x\right) \neq 0$ ; hence $c_{2}=0 \rightarrow u_{1}=\left(c_{1}\right)\left(c_{3}+c_{4} x\right)=\left(A_{1}+B_{1} x\right)$
$\therefore u_{1}(x, y)=\left(A_{1}+B_{1} x\right)$
For Case 2:
$u_{2}=\left(c_{5} \cos (\alpha y)+c_{6} \sin (\alpha y)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)$
$\frac{\partial u_{2}}{\partial y}=\left(-c_{5} \alpha \sin (\alpha y)+c_{6} \alpha \cos (\alpha y)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)$
BC 1: $\frac{\partial u_{2}(x, 0)}{\partial y}=\left(-c_{5} \alpha \sin (0)+c_{6} \alpha \cos (0)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)=0$
$\left(c_{6} \alpha\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)=0$
$\rightarrow\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) \neq 0 \& \alpha \neq 0$ ; hence, $c_{6}=0$
$\rightarrow \frac{\partial u_{2}}{\partial y}=\left(-c_{5} \alpha \sin (\alpha y)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)$
BC 2: $\frac{\partial u_{2}(x, 1)}{\partial y}=\left(-c_{5} \alpha \sin (\alpha)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)=0$
$\rightarrow\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) \neq 0 \& \alpha \neq 0$
$\rightarrow c_{11} \neq 0$ when $\sin (\alpha)=0$ for $\alpha=n \pi$ where $n=1,2,3, \ldots$
$\therefore u_{2, \mathrm{n}}(x, y)=\left(c_{5} \cos (\alpha y)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)$
$=(\cos (n \pi y))\left(A_{2, n} \cosh (n \pi x)+B_{2, n} \sinh (n \pi x)\right)$ where $n=1,2,3, \ldots$
For Case 3:
$u_{3}=\left(c_{9} \cosh (\alpha y)+c_{10} \sinh (\alpha y)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)$
$\frac{\partial u_{3}}{\partial y}=\left(c_{9} \alpha \sinh (\alpha y)+c_{10} \alpha \cosh (\alpha y)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)$
BC 1: $\frac{\partial u_{3}(x, 0)}{\partial y}=\left(c_{9} \alpha \sinh (0)+c_{10} \alpha \cosh (0)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\left(c_{10} \alpha\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\rightarrow\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right) \neq 0 \& \alpha \neq 0$ ; hence, $c_{10}=0$
$\rightarrow \frac{\partial u_{3}}{\partial y}=\left(c_{9} \alpha \sinh (\alpha y)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)$
BC 2: $\frac{\partial u_{3}(x, 1)}{\partial y}=\left(c_{9} \alpha \sinh (\alpha)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\rightarrow\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right) \neq 0 ; \alpha \neq 0$ & $\sinh (\alpha) \neq 0$ ; hence, $c_{9}=0$
$\rightarrow u_{3}(x, y)=((0) \cosh (\alpha y)+(0) \sinh (\alpha y))\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\therefore u_{3}(x, y)=0$ ; No solution for case 3
$\therefore u_{\text {total }}(x, y)=\underbrace{\left(A_{1}+B_{1} x\right)}_{\text {solution } 1}+\underbrace{\sum_{n=1}^{\infty}\left(A_{2, n} \cosh (n \pi x)+B_{2, n} \sinh (n \pi x)\right)(\cos (n \pi y))}_{\text {solution } 2}$
BC 3:
$\begin{aligned}u_{\text {total }}(0, y)&=A_{1}+\sum_{n=1}^{\infty}\left(A_{2, n} \cosh (0)+B_{2, n} \sinh (0)\right)(\cos (n \pi y))=0\\&=A_{1}+\sum_{n=1}^{\infty}\left(A_{2, \mathrm{n}}\right)(\cos (n \pi y))=0\end{aligned}$
Using half-range Fourier cosine series expansion
$\begin{aligned}A_1=0\\A_{2,n}=0\end{aligned}$
Since RHS function is zero
$u_{\text {total }}(x, y)=B_{1} x+\sum_{n=1}^{\infty}\left(B_{3, \mathrm{n}} \sinh (n \pi x)\right)(\cos (n \pi y))$
BC 4: $u_{\text {total }}(1, y)=B_{1}+\sum_{n=1}^{\infty}\left(B_{3, \mathrm{n}} \sinh (n \pi)\right)(\cos (n \pi y))=1-y$
Half-Range Fourier Cosine Series Expansion:
$\begin{aligned}B_{1}&=\frac{1}{L} \int_{0}^{\tau} f(y) d y, \omega=\pi, p=\frac{2 \pi}{\omega}=2, L=\frac{p}{2}=1, \tau=1 \\B_{1}&=\int_{0}^{1}(1-y) d y=\frac{1}{2}\end{aligned}$
$\begin{aligned}B_{3, \mathrm{n}} \sinh (n \pi)&=\frac{2}{L} \int_{0}^{\tau} f(y) \cos (n \pi y) d y \\B_{3, \mathrm{n}} \sinh (n \pi)&=2 \int_{0}^{1}(1-y) \cos (n \pi y) d y \\B_{3, \mathrm{n}}&=\frac{2\left(1-(-1)^{n}\right)}{n^{2} \pi^{2} \sinh (n \pi)} \end{aligned}$
$\therefore u_{t o t a l}(x, y)=\frac{1}{2} x+\frac{2}{\pi^{2}} \sum_{n=1}^{\infty}\left(\frac{\left(1-(-1)^{n}\right)}{n^{2} \sinh (n \pi)} \sinh (n \pi x)\right)(\cos (n \pi y))$
$\begin{aligned}\left.\frac{\partial u}{\partial x}\right|_{y=x}&=u(0,y), &u(\pi,y)&=1,&0<y<\pi\\u(x,0)&=0,&u(x,\pi)&=0,&0<x<1\end{aligned}$

Solution
$\begin{aligned} \left.\frac{\partial u}{\partial x}\right|_{y=x}&=u(0,y),\rightarrow\text{BC3} &u(\pi,y)&=1,\rightarrow\text{BC4} &0<y<\pi\\u(x,0)&=0,\rightarrow\text{BC1} &u(x,\pi)&=0,\rightarrow\text{BC2} &0<x<1 \end{aligned}$
For Case 1: $\lambda=0$
$u_{1}=\left(c_{1}+c_{2} y\right)\left(c_{3}+c_{4} x\right)$
BC 1: $u_{1}(x, 0)=\left(c_{1}+c_{2}(0)\right)\left(c_{3}+c_{4} x\right)=0$
$\rightarrow$ Assume $c_{3}+c_{4} x \neq 0 ; c_{1}=0$
$\rightarrow u_{1}(x, y)=\left(c_{2} y\right)\left(c_{3}+c_{4} x\right)$
BC 2: $u_{1}(x, \pi)=\left(c_{2} \pi\right)\left(c_{3}+c_{4} x\right)=0$
$\rightarrow\left(c_{3}+c_{4} x\right) \neq 0$ , hence $c_{2}=0$
$\rightarrow u_{1}(x, y)=(0)\left(c_{3}+c_{4} x\right)=0$
$\therefore u_{1}=0$ ; No solution for case 1
For Case 2: $\lambda=-\alpha^{2}, \alpha<0$
BC 1:
$\begin{aligned}u_{2}(x, 0)=\left(c_{5} \cos (0)+c_{6} \sin (0)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)&=0\\\left(c_{5}\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)&=0\end{aligned}$
$\rightarrow\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) \neq 0 ;$ hence $c_{5}=0$
$\rightarrow u_{2}(x, y)=\left(c_{6} \sin (\alpha y)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)$
BC 2: $u_{2}(x, \pi)=\left(c_{6} \sin (\alpha \pi)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)=0$
$\rightarrow\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) \neq 0, c_{6} \neq 0$ ; hence $\sin (\alpha \pi)=0$ when $\alpha \pi=n \pi, \alpha=n$ where $\mathrm{n}=1,2,3, \ldots$
$u_{2, n}=\left(c_{6, n} \sin (n y)\right)\left(c_{7, \mathrm{n}} \cosh (n x)+c_{8, \mathrm{n}} \sinh (n x)\right)$
$=(\sin (n y))\left(A_{2, \mathrm{n}} \cosh (n x)+B_{2, \mathrm{n}} \sinh (n x)\right)$ where $n=1,2,3, \ldots$
For Case 3: $\lambda=+\alpha^{2}, \alpha>0$
BC 1: $u_{3}(x, 0)=\left(c_{9} \cosh (0)+c_{10} \sinh (0)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\left(c_{9}\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\rightarrow\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right) \neq 0$ ; hence $c_{9}=0$
$\rightarrow u_{3}(x, y)=\left(c_{10} \sinh (\alpha y)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)$
BC 2: $u_{3}(x, \pi)=\left(c_{10} \sinh (\alpha \pi)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\rightarrow\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right) \neq 0, \sinh (\alpha \pi) \neq 0, c_{10}=0$
$\rightarrow u_{3}(x, y)=(0)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\therefore u_{3}=0$ ; No solution for case 3
$u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}(\sin (n y))\left(A_{2, \mathrm{n}} \cosh (n x)+B_{2, \mathrm{n}} \sinh (n x)\right)$
$\frac{\partial u}{\partial x}=\sum_{n=1}^{\infty}(\sin (n y))\left(-A_{2, \mathrm{n}} n \sinh (n x)+B_{2, \mathrm{n}} n \cosh (n x)\right)$
BC 3: $\frac{\partial u(0, y)}{\partial x}=\sum_{n=1}^{\infty}(\sin (n y))\left(-A_{2, \mathrm{n}} n \sinh (0)+B_{2, \mathrm{n}} n \cosh (0)\right)$ $=u(0, y)=\sum_{n=1}^{\infty}(\sin (n y))\left(A_{2, \mathrm{n}} \cosh (0)+B_{2, \mathrm{n}} \sinh (0)\right)$
$\rightarrow B_{2, \mathrm{n}} n=A_{2, \mathrm{n}}$
BC 4:
$\begin{aligned}u_{\text {total }}(\pi, y)&=\sum_{n=1}^{\infty}(\sin (n y))\left(B_{2, \mathrm{n}} n \cosh (n \pi)+B_{2, \mathrm{n}} \sinh (n \pi)\right)\\&=\sum_{n=1}^{\infty} B_{2, n}(n \cosh (n \pi)+\sinh (n \pi))(\sin (n y))=1\end{aligned}$
Half-Range Fourier Sine Series Expansion:
$B_{2, \mathrm{n}}(n \cosh (n \pi)+\sinh (n \pi))=\frac{2}{\pi} \int_{0}^{\pi}(\sin (n y)) d y=\frac{2\left(1-(-1)^{n}\right)}{(n \pi)}$
$\therefore u_{\text {total }}(x, y)=\frac{2}{\pi} \sum_{n=1}^{\infty}\left(\frac{\left(1-(-1)^{n}\right)}{n} \frac{n \cosh (n x)+\sinh (n x)}{n \cosh (n \pi)+\sinh (n \pi)}\right)(\sin (n y))$
$\begin{aligned}u(0,y)&=0, &u(a,y)&=50, &0<y<b \\ u(x,0)&=0, &u(x,b)&=0, &0<x<a\end{aligned}$

Solution
$\begin{aligned} u(0,y)&=0, \rightarrow\text{BC3} &u(a,y)&=50, \rightarrow\text{BC4} &0<y<b \\ u(x,0)&=0, \rightarrow\text{BC1} &u(x,b)&=0, \rightarrow\text{BC2} &0<x<a \end{aligned}$
For Case 1: $\lambda=0$
BC 1: $u_{1}(x, 0)=\left(c_{1}+c_{2}(0)\right)\left(c_{3}+c_{4} x\right)=0$
$\rightarrow$ Assume $c_{3}+c_{4} x \neq 0 ; c_{1}=0$
$\rightarrow u_{1}(x, y)=\left(c_{2} y\right)\left(c_{3}+c_{4} x\right)$
BC 2: u_1(x, b)=\left(c_2 b\right)\left(c_3+c_4 x\right)=0$
$\rightarrow\left(c_{3}+c_{4} x\right) \neq 0 \& b \neq 0$ ; hence $c_{2}=0$
$\rightarrow u_{1}(x, y)=(0)\left(c_{3}+c_{4} x\right)=0$
$\therefore u_{1}=0$ ; No solution for case 1
For Case 2: $\lambda=-\alpha^{2}, \alpha<0$
BC 1:
$\begin{aligned}u_{2}(x, 0)=\left(c_{5} \cos (0)+c_{6} \sin (0)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)&=0 \\ \left(c_{5}\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)&\=0\end{aligned}$
$\rightarrow\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) \neq 0$ ; hence $c_{5}=0$
$\rightarrow u_{2}(x, y)=\left(c_{6} \sin (\alpha y)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)$
BC 2: $u_{2}(x, b)=\left(c_{6} \sin (\alpha b)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right)=0$
$\rightarrow\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) \neq 0, c_{6} \neq 0$ ; hence $\sin (\alpha b)=0$ when $\alpha \mathrm{b}=n \pi, \alpha=\frac{n \pi}{b}$
where $n=1,2,3, \ldots$
$\begin{aligned}u_{2, n}&=\left(c_{6, n} \sin \left(\frac{n \pi}{b} y\right)\right)\left(c_{7, \mathrm{n}} \cosh \left(\frac{n \pi}{b} x\right)+c_{8, \mathrm{n}} \sinh \left(\frac{n \pi}{b} x\right)\right)\\&=\left(\sin \left(\frac{n \pi}{b} y\right)\right)\left(A_{2, \mathrm{n}} \cosh \left(\frac{n \pi}{b} x\right)+B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} x\right)\right) \quad \text { where } n=1,2,3, \ldots\end{aligned}$
For Case 3: $\lambda=+\alpha^{2}, \alpha>0$
BC 1: $u_{3}(x, 0)=\left(c_{9} \cosh (0)+c_{10} \sinh (0)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\left(c_{9}\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\rightarrow\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right) \neq 0$ ; hence $c_{9}=0$
$\rightarrow u_{3}(x, y)=\left(c_{10} \sinh (\alpha y)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)$
BC 2: $u_{3}(x, b)=\left(c_{10} \sinh (\alpha b)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\rightarrow\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right) \neq 0, \sinh (\alpha b) \neq 0, c_{10}=0$
$\rightarrow u_{3}(x, y)=(0)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)=0$
$\therefore u_{3}=0$ ; No solution for case 3
$u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(\sin \left(\frac{n \pi}{b} y\right)\right)\left(A_{2, \mathrm{n}} \cosh \left(\frac{n \pi}{b} x\right)+B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} x\right)\right)$
BC 3:
$\begin{aligned}u_{\text {total }}(0, y)&=\sum_{n=1}^{\infty}\left(\sin \left(\frac{n \pi}{b} y\right)\right)\left(A_{2, \mathrm{n}} \cosh (0)+B_{2, \mathrm{n}} \sinh (0)\right)\\&=\sum_{n=1}^{\infty}\left(A_{2, \mathrm{n}}\right)\left(\sin \left(\frac{n \pi}{b} y\right)\right)=0\end{aligned}$
Recall Half-Range Fourier Sine Series Expansion:
$A_{2, \mathrm{n}}=\frac{2}{L} \int_{0}^{\tau} f(y) \sin \left(\frac{n \pi}{b} y\right) d y=0$
$u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(\sin \left(\frac{n \pi}{b} y\right)\right)\left(B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} x\right)\right)$
BC 4: $u_{\text {total }}(a, y)=\sum_{n=1}^{\infty}\left(\sin \left(\frac{n \pi}{b} y\right)\right)\left(B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} a\right)\right)=50$
Half-Range Fourier Sine Series Expansion:
$f(y)=\sum_{n=1}^{\infty}\left(B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} a\right)\right)\left(\sin \left(\frac{n \pi}{b} y\right)\right)$
where $B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} a\right)=\frac{2}{L} \int_{0}^{\tau} f(y) \sin \left(\frac{n \pi}{b} y\right) d y, \omega=\frac{\pi}{b}, p=\frac{2 \pi}{\omega}=2 a, L=\frac{p}{2}=b, \tau=b$
$B_{2, \mathrm{n}} \sinh \left(\frac{n \pi}{b} a\right)=\frac{2}{b} \int_{0}^{b} 50 \sin n \frac{\pi}{b} y d y$
$\begin{aligned} B_{2, \mathrm{n}}&=\frac{100}{b \sinh \left(\frac{n \pi}{b} a\right)} \int_{0}^{b} \sin n \frac{\pi}{b} y d y \\ &=\frac{100}{\sinh \left(\frac{n \pi}{b} a\right)}\left[\frac{-\cos n \pi-(-1)}{n \pi}\right] \\ &=\frac{100}{n \pi \sinh \left(\frac{n \pi}{b} a\right)}\left[1-(-1)^{n}\right] \\ \end{aligned}$
$\therefore u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(\frac{100}{n \pi \sinh \left(\frac{n \pi}{b} a\right)}\left[1-(-1)^{n}\right] \sinh \left(\frac{n \pi}{b} x\right)\right)\left(\sin \left(\frac{n \pi}{b} y\right)\right)$
$\begin{aligned}u(0,y)&=0, &u(a,y)&=50, &0<y<b \\ u(x,0)&=0, &u(x,b)&=f(x), &0<x<a\end{aligned}$

Solution
By applying superposition principle:
Solution $u=$ Solution $u_{\text {total }, 1}(x, y)$ of Question 4 (d) + Solution $u_{\text {total }, 2}(x, y)$ of Question 4 (a).
$u_{\text {total }, 1}(x, y)=\sum_{n=1}^{\infty}\left(\frac{100}{n \pi \sinh \left(\frac{n \pi}{b} a\right)}\left[1-(-1)^{n}\right] \sinh \left(\frac{n \pi}{b} x\right)\right)\left(\sin \left(\frac{n \pi}{b} y\right)\right)$ where $n=1,2 \ldots$
$u_{\text {total }, 2}(x, y)=\sum_{n=1}^{\infty}\left(\frac{2}{\operatorname{asinh}\left(\frac{n \pi}{a} b\right)} \int_{0}^{a} f(x) \sin n \frac{\pi}{a} x d x \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$ where $n=1,2,\ldots$
$\therefore u_{\text {total }}(x, y)=\sum_{n=1}^{\infty}\left(\frac{100}{n \pi \sinh \left(\frac{n \pi}{b} a\right)}\left[1-(-1)^{n}\right] \sinh \left(\frac{n \pi}{b} x\right)\right)\left(\sin \left(\frac{n \pi}{b} y\right)\right)+\sum_{n=1}^{\infty}\left(\frac{2}{\operatorname{asinh}\left(\frac{n \pi}{a} b\right)} \int_{0}^{a} f(x) \sin n \frac{\pi}{a} x d x \sinh \left(\frac{n \pi}{a} y\right)\right)\left(\sin \left(\frac{n \pi}{a} x\right)\right)$ where $n=1,2 \ldots$
Heat equation, $u_{t}-u_{x x}=0$

General solution: $u(x, t)=A_{1} x+B_{1}+e^{\alpha^{2} t}\left(A_{2} \cosh (\alpha x)+B_{2} \sinh (\alpha x)\right) +e^{-\alpha^{2} t}\left(A_{3} \cos (\alpha x)+B_{3} \sin (\alpha x)\right)$

Subject to the given boundary conditions:

$\begin{aligned}u(0,t)&=0, &u(2,t)&=0, &t>0 \\ u(x,0)&=\sin \frac{\pi}{2}x, &&&0<x<2\end{aligned}$

Solution
For Case 1: $\lambda=0$
BC 1: $u_{1}(0, t)=A_{1}(0)+B_{1}=0 ; B_{1}=0$
$\rightarrow u_{1}(x, t)=A_{1} x$
BC 2: $u_{1}(2, t)=A_{1}(2)=0 ; A_{1}=0$
$\rightarrow u_{1}(x, t)=0$ ; No solution for case 1
For Case 2:
BC 1: $u_{2}(0, t)=e^{\alpha^{2} t}\left(\mathrm{~A}_{2} \cosh (0)+\mathrm{B}_{2} \sinh (0)\right)=0$
Comparing coefficient $e^{\alpha^{2} t}: A_{2} \cosh (0)+B_{2} \sinh (0)=0, A_{2}=0$
Then, we get $u_{2}=e^{\alpha^{2} t}\left(\mathrm{~B}_{2} \sinh (\alpha x)\right)$
BC 2: $u_{2}(2, t)=e^{\alpha^{2} t}\left(\mathrm{~B}_{2} \sinh (2 \alpha)\right)=0$
Comparing coefficient $e^{\alpha^{2} t}:\left(\mathrm{B}_{2} \sinh (2 \alpha)\right)=0$
Since $\sinh (2 \alpha)$ will not be zero for $\alpha>0$ , thus $B_{2}=0$
$\rightarrow u_{2}(x, t)=0$ ; No solution for case 2
For Case 3:
BC 1: $u_{3}(0, t)=e^{-\alpha^{2} t}\left(\mathrm{~A}_{3} \cos (0)+\mathrm{B}_{3} \sin (0)\right)=0$
Comparing coefficient $e^{-\alpha^{2} t}: A_{3} \cos (0)+B_{3} \sin (0)=0 \rightarrow A_{3}=0$
Hence we get $u_{3}=e^{-\alpha^{2} t}\left(\mathrm{~B}_{3} \sin (\alpha x)\right)$
BC 2: $u_{3}(2, t)=e^{-\alpha^{2} t}\left(\mathrm{~B}_{3} \sin (2 \alpha)\right)=0$
Comparing coefficient $e^{-\alpha^{2} t}:\left(\mathrm{B}_{3} \sin (2 \alpha)\right)=0$
$B_3 \neq 0$ when $\sin (2 \alpha)=0$ for $2 \alpha=n \pi$ where $n=1,2,3, \ldots$
$\rightarrow u_{3, \mathrm{n}}=e^{-\left(\frac{n \pi}{2}\right)^{2} t}\left(\mathrm{~B}_{3, n} \sin \left(\frac{n \pi}{2} x\right)\right)$ where $n=1,2,3, \ldots$
$\therefore u_{\text {total }}(x, t)=\sum_{n=1}^{\infty} e^{-\left(\frac{n \pi}{2}\right)^{2} t}\left(\mathrm{~B}_{3, n} \sin \left(\frac{n \pi}{2} x\right)\right)$
BC 3: $u(x, 0)=\sum_{n=1}^{\infty}\left(\mathrm{B}_{3, n} \sin \left(\frac{n \pi}{2} x\right)\right)=\sin \frac{\pi}{2} x$
$B_{3, n}=\int_{0}^{2}\left(\sin \frac{\pi}{2} x\right)\left(\sin \left(\frac{n \pi}{2} x\right)\right) d x=\left\{\begin{array}{ll}1 & n=1 \\ 0 & n \neq 1\end{array} \quad\right.$ *refer to Q4
$\therefore u_{\text {total }}(x, t)=e^{-\frac{\pi^{2}}{4} t}\left(\sin \left(\frac{\pi}{2} x\right)\right)$
Heat equation, $u_{t}-\frac{1}{16} u_{x x}=0$

General solution: $u(x, t)=A_{1} x+B_{1}+e^{\alpha^{2} t}\left(A_{2} \cosh (4 \alpha x)+B_{2} \sinh (4 \alpha x)\right) +e^{-\alpha^{2} t}\left(A_{3} \cos (4 \alpha x)+B_{3} \sin (4 \alpha x)\right)$

Subject to the given boundary conditions:

$\begin{aligned}u(0,t)&=0, &u(1,t)&=0, &t>0 \\ u(x,0)&=2\sin 2\pi x, &&&0<x<1\end{aligned}$

Solution
For Case 1: $\lambda=0$
BC 1: $u_{1}(0, t)=A_{1}(0)+B_{1}=0 ; B_{1}=0$
$\rightarrow u_{1}(x, t)=A_{1} x$
BC 2: $u_{1}(1, t)=A_{1}(1)=0 ; A_{1}=0$
$\rightarrow u_{1}(x, t)=0$ ; No solution for case 1
For Case 2:
BC 1: $u_{2}(0, t)=e^{\alpha^{2} t}\left(\mathrm{~A}_{2} \cosh (0)+\mathrm{B}_{2} \sinh (0)\right)=0$
Comparing coefficient $e^{\alpha^{2} t}: A_{2} \cosh (0)+B_{2} \sinh (0)=0, A_{2}=0$
Then, we get $u_{2}=e^{\alpha^{2} t}\left(\mathrm{~B}_{2} \sinh (4 \alpha x)\right)$
BC 2: $u_{2}(1, t)=e^{\alpha^{2} t}\left(\mathrm{~B}_{2} \sinh (4 \alpha)\right)=0$
Comparing coefficient $e^{\alpha^{2} t}:\left(\mathrm{B}_{2} \sinh (4 \alpha)\right)=0$
Since $\sinh (4 \alpha)$ will not be zero for $\alpha>0$ , thus $B_{2}=0$
$\rightarrow u_{2}(x, t)=0$ ; No solution for case 2
For Case 3:
BC 1: $u_{3}(0, t)=e^{-\alpha^{2} t}\left(\mathrm{~A}_{3} \cos (0)+\mathrm{B}_{3} \sin (0)\right)=0$
Comparing coefficient $e^{-\alpha^{2} t}: A_{3} \cos (0)+B_{3} \sin (0)=0 \rightarrow A_{3}=0$
Hence we get $u_{3}=e^{-\alpha^{2} t}\left(\mathrm{~B}_{3} \sin (4 \alpha x)\right)$
BC 2: $u_{3}(1, t)=e^{-\alpha^{2} t}\left(\mathrm{~B}_{3} \sin (4 \alpha)\right)=0$
Comparing coefficient $e^{-\alpha^{2} t}:\left(\mathrm{B}_{3} \sin (4 \alpha)\right)=0$
$B_{3} \neq 0$ when $\sin (4 \alpha)=0$ for $4 \alpha=n \pi$ where $n=1,2,3, \ldots$
$\rightarrow u_{3, \mathrm{n}}=e^{-\left(\frac{n \pi}{4}\right)^{2} t}\left(\mathrm{~B}_{3, n} \sin (n \pi x)\right)$ where $n=1,2,3, \ldots$
$\therefore u_{\text {total }}(x, t)=\sum_{n=1}^{\infty} e^{-\left(\frac{n \pi}{4}\right)^{2} t}\left(\mathrm{~B}_{3, n} \sin (n \pi x)\right)$
BC 3: $u(x, 0)=\sum_{n=1}^{\infty}\left(\mathrm{B}_{3, n} \sin (n \pi x)\right)=2 \sin 2 \pi x$
$\mathrm{B}_{3, n}=2 \int_{0}^{2}(2 \sin 2 \pi x)(\sin (n \pi x)) d x= \begin{cases}2 & n=2 \\ 0 & n \neq 2\end{cases}$
$\therefore u_{\text {total }}(x, t)=2 e^{-\frac{\pi^{2}}{4} t}(\sin (2 \pi x))$
Wave equation, $u_{t t}-u_{x x}=0$

General solution: $u(x, t)=\left(c_{1}+c_{2} t\right)\left(c_{3}+c_{4} x\right) +\left(c_{5} \cosh (\alpha t)+c_{6} \sinh (\alpha t)\right)\left(c_{7} \cosh (\alpha x)+c_{8} \sinh (\alpha x)\right) +\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right)\left(c_{11} \cos (\alpha x)+c_{12} \sin (\alpha x)\right)$

Subject to the given boundary conditions:

$\begin{aligned}u(0,t)&=0, &u(1,t)&=0, &t>0 \\ u(x,0)&=\sin\pi x, &\frac{\partial u}{\partial t}(x,0)&=0, &0<x<1\end{aligned}$

Solution
For Case 1: $\lambda=0$
BC 1: $u_{1}(0, t)=\left(c_{1}+c_{2} t\right)\left(c_{3}+c_{4}(0)\right)=0$
$\rightarrow c_{1}+c_{2} t \neq 0$ because the vibration is changed with time, hence $T(t) \neq 0 ; c_{3}=0$
$\rightarrow u_{1}(x, t)=\left(c_{1}+c_{2} t\right)\left(c_{4} x\right)$
BC 2: $u_{1}(1, t)=\left(c_{1}+c_{2} t\right)\left(c_{4}\right)=0$
$\rightarrow\left(c_{1}+c_{2} t\right) \neq 0$ ; hence $c_{4}=0$
$\rightarrow u_{1}(x, y)=\left(c_{1}+c_{2} t\right)(0)=0$
$\therefore u_{1}=0$ ; No solution for case 1
For Case 2: $\lambda=-\alpha^{2}, \alpha>0$
BC 1: $u_{2}(0, t)=\left(c_{5} \cos h(\alpha t)+c_{6} \sin h(\alpha t)\right)\left(c_{7} \cos h(0)+c_{8} \sin h(0)\right)=0$
$\left(c_{5} \cosh (\alpha t)+c_{6} \sinh (\alpha t)\right)\left(c_{7}\right)=0$
$\rightarrow\left(c_{5} \cos h(\alpha t)+c_{6} \sin h(\alpha t)\right) \neq 0$ ; hence $c_{7}=0$
$\rightarrow u_{2}(x, t)=\left(c_{5} \cosh (\alpha t)+c_{6} \sinh (\alpha t)\right)\left(c_{8} \sinh (\alpha x)\right)$
BC 2: $u_{2}(1, t)=\left(c_{5} \cos h(\alpha t)+c_{6} \sinh (\alpha t)\right)\left(c_{8} \sinh (\alpha)\right)=0$
$\rightarrow\left(c_{5} \cos h(\alpha t)+c_{6} \sinh (\alpha t)\right) \neq 0, \sinh (\alpha) \neq 0$ ; hence $c_{8}=0$
$\rightarrow u_{2}(x, t)=\left(c_{5} \cos h(\alpha t)+c_{6} \sinh (\alpha t)\right)(0)=0$
$\therefore u_{2}=0$ ; No solution for case 2
For Case 3: $\lambda=+\alpha^{2}, \alpha>0$
BC 1: $u_{3}(0, t)=\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right)\left(c_{11} \cos (0)+c_{12} \sin (0)\right)=0$
$\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right)\left(c_{11}\right)=0$
$\rightarrow\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right) \neq 0$ ; hence $c_{11}=0$
$\rightarrow u_{3}(x, y)=\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right)\left(c_{12} \sin (\alpha x)\right)$
BC 2: $u_{3}(1, t)=\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right)\left(c_{12} \sin (\alpha)\right)=0$
$\rightarrow\left(c_{9} \cos (\alpha t)+c_{10} \sin (\alpha t)\right) \neq 0, c_{12} \neq 0$ ; hence $\sin (\alpha)=0$ when $\alpha=n \pi$ , where $n=1,2,3, \ldot$
$\begin{aligned} \therefore u_{3, n}=\left(c_{9} \cos (n \pi t)+c_{10} \sin (n \pi t)\right)\left(c_{12} \sin (n \pi)\right) \\ &=\left(A_{3, n} \cos (n \pi t)+B_{3, n} \sin (n \pi t)\right)(\sin (n \pi x)) \text { where } n=1,2,3, \ldots \\ \end{aligned}$
$\therefore u_{\text {total }}(x, t)=\sum_{n=1}^{\infty}\left(A_{3, n} \cos (n \pi t)+B_{3, n} \sin (n \pi t)\right)(\sin (n \pi x))$
BC 3:
$\begin{aligned}u_{\text {total }}(x, 0)&=\sum_{n=1}^{\infty}\left(A_{3, n} \cos (0)+B_{3, n} \sin (0)\right)(\sin (n \pi x))=\sin \pi x\\&=\sum_{n=1}^{\infty}(A_{3, n})(\sin(n\pi x))=\sin\pi x\end{aligned}$
Half-Range Fourier Sine Series Expansion:
$\begin{aligned} A_{3, \mathrm{n}}&=2 \int_0^1(\sin \pi x)(\sin (n \pi x)) d x \\ &=2 \int_0^1 \frac{1}{2}(\cos (\pi x-n \pi x))-(\cos (\pi x+n \pi x)) d x \\ &=\int_0^1(\cos (\pi x)(1-n))-(\cos (\pi x)(1+n)) d x \\ &=\frac{\sin (1-n)(\pi)}{(1-n) \pi}-\frac{\sin (1+n)(\pi)}{(1+n) \pi} \text { where } n \neq 1 \\ \end{aligned}$
$\begin{aligned} A_{3,1}&=2 \int_0^1 \frac{1}{2}(\cos ((1-1) \pi x))-(\cos (1+1) \pi x)) d x \\ &=\int_0^1(\cos (0))-(\cos (2 \pi x)) d x \\ &=\int_0^1 1 d x-\int_0^1(\cos (2 \pi x)) d x \\ &=1 \\ \end{aligned}$
$A_{3, n}= \begin{cases}1 & n=1 \\0 & n \neq 1\end{cases}$
$\frac{\partial u(x, t)}{\partial t}=\sum_{n=1}^{\infty}\left(-A_{3, n} n \pi \sin (n \pi t)+B_{3, n} n \pi \cos (n \pi t)\right)(\sin (\pi x))$
BC 4:
$\begin{aligned} \frac{\partial u(x, 0)}{\partial t}=\sum_{n=1}^{\infty}\left(-A_{3, n} n \pi \sin (0)+B_{3, n} n \pi \cos (0)\right)(\sin (\pi x)&)=0 \\ & \sum_{n=1}^{\infty}\left(B_{3, n} n \pi\right)(\sin (\pi x))=0, B_{3, n}&=0 \\ \end{aligned}$
$\therefore u_{\text {total }}(x, t)=(\cos (\pi t))(\sin (\pi x)) \text { only when } \mathrm{n}=1$
Continue to estimate the solution of Q2 at $(0.9,0.1)$ by using 2 summation terms. Assume the actual answer is 0.74045 and the allowable percentage of error is 1%. Please comment if the estimation is acceptable or not. If no, please suggest how to improve the solution.

Solution
$(0.9,0.1)=(x,y)$
$\begin{aligned} u_{\text {total }}(0.9,0.1) &\approx \frac{1}{2}(0.9)+\frac{2}{\pi^2} \sum_{n=1}^2\left(\frac{\left(1-(-1)^n\right)}{n^2 \sinh (n \pi)} \sinh (0.9 n \pi)\right)(\cos (0.1 n \pi)) \\ & \approx \frac{1}{2}(0.9)+\frac{2}{\pi^2} {\left[\left(\frac{\left(1-(-1)^1\right)}{1^2 \sinh (1 \pi)} \sinh (0.9(1) \pi)(\cos (0.1(1) \pi))\right.\right.} \left.\left.\quad+\frac{\left(1-(-1)^2\right)}{2^2 \sinh (2 \pi)} \sinh (0.9(2) \pi)(\cos (0.1(2) \pi))\right)\right] & \approx 0.45+0.28107+0\\ & \approx 0.73107\\ \end{aligned}$
Actual answer is 0.74045
Percentage of error $=\frac{0.74045-0.73107}{0.74045} x 100 \%=1.27 \%$
The estimation is not acceptable, therefore higher number of summation terms is needed for improving the estimation.
Identify the eigenvalue and eigenvector of PDE solutions that has been solved in Q2.

Solution
For case 1 , eigenvalue, $\lambda=0$ ; eigenfunction of PDE: $u_1=\left(A_1+B_1 x\right)$
For case 2, eigenvalue, $\lambda=-(n \pi)^2$ ; eigenfunction of PDE:
$u_{2, \mathrm{n}}(x, y)=(\cos (n \pi y))\left(A_{2, n} \cosh (n \pi x)+B_{2, n} \sinh (n \pi x)\right) \text { where } n=1,2,3, \ldots$
No eigenvalue & eigenfunction for case 3.