Question

Find a solution u(x,t) to the problem 𝜕𝑢 𝜕𝑡 = 1.71 𝜕^2𝑢/𝜕𝑥^2 , 𝑢(𝑥, 0) = 𝑠𝑖𝑛 ( 𝜋𝑥/2 ) + 3 𝑠𝑖𝑛 ( 5𝜋𝑥/2 ) , 0 < 𝑥 < 2

Answer 1

$\begin{aligned} &\frac{\partial u}{\partial t}=1.71 \frac{\partial^{2} u}{\partial x^{2}} ...(1) \\ & u(x, 0)=\sin \left(\frac{\pi x}{2}\right)+3 \sin \left(\frac{5 \pi x}{2}\right) ,0<x<2 \end{aligned}$

By using variable separation method

$\begin{aligned} & u(x, t)=X(x) \cdot T(t)...(2) \\ \& & \frac{\partial^{2} u}{\partial x^{2}}=X^{\prime } \cdot T, \quad \frac{\partial u}{\partial t}=X T^{\prime} \\ & \therefore E q^{n}(1) \text { becomes, } \\ & X T^{\prime}=1.71X''T \\ & \frac{T^{\prime}}{1.71 . T}=\frac{X^{\prime \prime}}{X}=K \\ \Rightarrow & \frac{T^{\prime}}{1.71 T}=K , \frac{X^{\prime \prime}}{X}=K \quad(\text { say }) \end{aligned}$

$\begin{aligned} &\text { Solving, } \frac{T^{\prime}}{1.71 . T}=k\\ &\Rightarrow T^{\prime}=1.71 \mathrm{KT}\\ &\Rightarrow T^{\prime}-1.71 \mathrm{KT}=0\\ &\text { Put } T^{\prime}=m \quad \& \quad D(T)=0\\ &\Rightarrow m-1.71 k=0 \text {. }\\ &\Rightarrow m=1.71 \mathrm{k}\\ &\therefore T(t)=A e^{1.71 k t}...(3)\\ &\text { \& solving } \frac{X^{\prime \prime}}{X}=k\\ &\Rightarrow \quad X^{\prime \prime}-k X=0\\ &\text { Put } X^{\prime}=m \text {. }\\ &\Rightarrow m^{2}-k=0\\ &\Rightarrow m^{2}=k\\ &\Rightarrow m=\pm k\\ &\therefore X(x)=B e^{k x}+C e^{-k x}...(4) \end{aligned}$

Put eqn (3) \& (4) in eqn (2), we get

$\begin{aligned} &u(x, t)=A \cdot e^{1,71 k t} \cdot\left(B e^{k x}+c e^{-k x}\right)...(5) \\ &\& B y \text { condition. } \\ &u(x, 0)=\sin \left(\frac{\pi x}{2}\right)+3 \sin \left(\frac{5 \pi x}{2}\right) \quad 0<x<z \\ &\quad \text { i.e. } \quad x=1 \\ &\Rightarrow u(x, 0)=\sin \left(\frac{\pi}{2}\right)+3 \sin \left(\frac{5 \pi}{2}\right) \\ &u(x, 0)=1+3(1) \\ &u(x, 0)=1+3=4 \\ &u(x, 0)=4 \end{aligned}$

$\begin{aligned} \therefore u(x, 0) &=A e^{1.71 k(0)}\left(B e^{k x}+c e^{-k x}\right)=4 \\ \Rightarrow A(1)\left(B e^{k x}+C e^{-k x}\right) &=4 \\ \Rightarrow A &=\frac{4}{B e^{k x}+c e^{-k x}} \end{aligned}$

Eqn (5) becomes

 

$u(x, t)=\frac{4 e^{1.71 k t}}{B e^{k x}+c e^{k x}} \cdot\left(B e^{k x}+c e^{-k x}\right)$

 $u(x, t)=4 e^{1.71 k t}$