Answer to Question #113691 in Quantitative Methods for miyuki

Question

Approximate the solution to the following partial differential equation using the Backward-Difference method. ∂u /∂t − ∂2u/ ∂x2 =0, 0 < x < 2, 0 < t; u(0,t) =u(2,t) =0, 0 < t, u(x,0) =sin( π /2) x,0 ≤x ≤2. Use m = 4, T = 0.1, and N = 2, and compare your results to the actual solution u(x,t) = e−((π2/4)t)* sin( π)

Solve this without MATLAB and need Step by step

Answer 1

$u_{t}-u_{xx}=0$

$u(0,t)=u(2,t)=0$

$u(x,0)=sin(\dfrac{\pi x}{2})$

We know that:

$f'(x)=\lim\limits_{\Delta x\rightarrow0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}$

So, we instead use:

$\dfrac{u_{i,j+1}-u_{i,j}}{\Delta t}=\dfrac{u_{i-1,j}-2u_{i.j}+u_{i+1,j}}{\Delta x^2}$

Multiply both side by $\Delta t:$

$u_{i,j+1}-u_{i,j}=\dfrac{\Delta t \times (u_{i-1,j}-2u_{i.j}+u_{i+1,j})}{\Delta x^2}$

Writing that $r=\dfrac{\Delta t}{\Delta x^2}$ , we can write:

$ru_{i-1,j}+(1-2r)u_{i,j}+ru_{i+1,j}=u_{i,j+1}$