Question 16.20: Solve the wave equation, ∂²u/∂t² = 4∂²u/∂x²; 0 ≤ x ≤ 1, t ≥ ...
Solve the wave equation, \frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2} ; 0 ≤ x ≤ 1, t ≥ 0 with
initial conditions u(x, 0)=\sin (\pi x) \text { and }\left.\frac{\partial u}{\partial t}\right|_{t=0}=0
boundary conditions u(0, t) = u(1, t) = 0 t ≥ 0
Take the step size for time t is 1/6, and step size for x is 1/3. Use implicit scheme to compute the solution up to time t = 1/3. Use central difference formula for the derivative term in the initial condition.
The spacing h = 1/3 and k = 1/6 for variables x and t provide the following grid points x = 0,1/3,2/3,1 and t = 0,1/6,1/3.
The initial and boundary conditions are summarize in the following table
| 3(1) | 2(2/3) | 1(1/3) | 0(0) | j(t) \ i(x) |
| 0 | 0.8660 | 0.8660 | 0 | 0(0) |
| 0 | 0 | 1 (1/6) | ||
| 0 | 0 | 2 (1/3) |
The implicit scheme for the solution of wave equation is as follows
-r u_{i-1, j+1}+2(1+r) u_{i, j+1}-r u_{i+1, j+1}=r u_{i-1, j-1}-2(1+r) u_{i, j-1}+r u_{i+1, j-1}+4 u_{i, j}
Using r=\frac{c k^2}{h^2}=\frac{(1 / 6)^2}{(1 / 3)^2}=1 / 4 , we get
-u_{i-1, j+1}+10 u_{i, j+1}-u_{i+1, j+1}=u_{i-1, j-1}-10 u_{i, j-1}+u_{i+1, j-1}+16 u_{i, j} (16.93)
This implicit scheme will be used to compute solution for different j.
j = 1.
The initial condition is \left.\frac{\partial u}{\partial t}\right|_{t=0}=0 . Using central difference formula, we get
\left.\frac{\partial u}{\partial t}\right|_{t=0}=\left.\frac{u_{i, j+1}-u_{i, j-1}}{2 k}\right|_{j=0}=\left.\frac{u_{i, 1}-u_{i,-1}}{2 k}\right|_{j=0}=0
\Rightarrow u_{i, 1}=u_{i,-1} (16.94)
Implicit scheme (16.93) for j = 0 is given by
-u_{i-1,1}+10 u_{i, 1}-u_{i+1,1}=u_{i-1,-1}-10 u_{i,-1}+u_{i+1,-1}+16 u_{i, 0}
Using Eq. (16.94), we have
-u_{i-1,1}+10 u_{i, 1}-u_{i+1,1}=8 u_{i, 0}
For i = 1, 2, we get following two equations respectively
\begin{aligned} &-u_{0,1}+10 u_{1,1}-u_{2,1}=8 u_{1,0} \\ &-u_{1,1}+10 u_{2,1}-u_{3,1}=8 u_{2,0} \end{aligned}
Using the values u_{0,1}=u_{3,1}=0 \text { and } u_{1,0}=u_{2,0}=0.8660 , and on solving resulting two equations, we get
u_{1,1}=u_{2,1}=0.7698
So, we have
| 3(1) | 2(2/3) | 1(1/3) | 0(0) | j(t) \ i(x) |
| 0 | 0.8660 | 0.8660 | 0 | 0(0) |
| 0 | 0.7698 | 0.7698 | 0 | 1 (1/6) |
| 0 | 0 | 2 (1/3) |
j = 2.
Implicit scheme (16.93) for j = 1 is given by
-u_{i-1,2}+10 u_{i, 2}-u_{i+1,2}=u_{i-1,0}-10 u_{i, 0}+u_{i+1,0}+16 u_{i, 1}
The following equations are obtained for i = 1, 2
\begin{aligned} &-u_{0,2}+10 u_{1,2}-u_{2,2}=u_{0,0}-10 u_{1,0}+u_{2,0}+16 u_{1,1} \\ &-u_{1,2}+10 u_{2,2}-u_{3,2}=u_{1,0}-10 u_{2,0}+u_{3,0}+16 u_{2,1} \end{aligned}
Use u_{0,0}=u_{0,2}=u_{3,2}=u_{3,0}=0, u_{1,1}=u_{2,1}=0.7698 \text { and } u_{1,0}=u_{2,0}=0.8660 in these two equations, then solution of the equations are given by
u_{1,2}=u_{2,2}=0.5025
So, we have
| 3(1) | 2(2/3) | 1(1/3) | 0(0) | j(t) \ i(x) |
| 0 | 0.8660 | 0.8660 | 0 | 0(0) |
| 0 | 0.7698 | 0.7698 | 0 | 1 (1/6) |
| 0 | 0.5025 | 0.5025 | 0 | 2 (1/3) |