Given the boundary-value problem

$\begin{array}{ll}{\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}},} & {0<x<1, \quad 0<t<1} \\ {u(0, t)=0,} & {u(1, t)=0, \quad 0 \leq t \leq 1}\end{array}$

$u(x, 0)=x(1-x),\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 0 \leq x \leq 1$ use $h=k=\frac{1}{5}$ in equation (6) to compute the values of $u_{i, 1}$ by hand.

Solution

VerifiedThe given problem is

$u_{xx}=u_{tt}$

$u(0,t)=0; u(1,t)=0$

$u(x,0)=x(1-x), u_t(x,0)=0$

Now from equation (6), we have

$u_{i,1}=0.5\cdot (u_{i+1,0}+u_{i-1,0})$

Here $g(x_i)=0$ Now,

$u_{1,1}=0.5\cdot (u_{2,0}+u_{0,0})=0.08$

$u_{2,1}=0.5\cdot (u_{3,0}+u_{1,0})=0.12$

$u_{3,1}=0.5\cdot (u_{4,0}+u_{2,0})=0.20$

$u_{4,1}=0.5\cdot (u_{5,0}+u_{3,0})=0.20$

$\cdots \cdots$ so on.. here we use the initial condition $u(x,0)=x(1-x).$

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