Question 16.21: Obtain an explicit finite difference scheme for the solution...
Obtain an explicit finite difference scheme for the solution of following variable coefficient problem
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+x \frac{\partial u}{\partial x} \quad 0 \leq x \leq 1, t \geq 0
with following initial and boundary conditions
u(x, 0) = x(2 − x)
u(0,t) = 0, u(1,t) = 1
Replace temporal derivative term with forward difference and spatial derivative terms by central differences. Use this scheme for the solution over a rectangular grid defined by spacing Δx = 0.2 and Δt = 0.005. Solve up to t = 0.02 only.
The spacing is Δx = 0.2 for 0 ≤ x ≤ 1; so, our node points are given by
x_0=0, x_1=0.2, x_2=0.4, x_3=0.6, x_4=0.8, x_5=1
Let u_{i, j}=u\left(x_i, t_j\right) . The initial condition is u(x,0) = x(2 − x), so we have
\begin{aligned} &u_{0,0}=u\left(x_0, t_0\right)=u(0,0)=0(2-0)=0 \\ &u_{1,0}=u\left(x_1, t_0\right)=u(0.2,0)=0.2(2-0.2)=0.36 \\ &u_{2,0}=u\left(x_2, t_0\right)=u(0.4,0)=0.4(2-0.4)=0.64 \\ &u_{3,0}=u\left(x_3, t_0\right)=u(0.6,0)=0.6(2-0.6)=0.84 \\ &u_{4,0}=u\left(x_4, t_0\right)=u(0.8,0)=0.8(2-0.8)=0.96 \end{aligned}
u_{5,0}=u\left(x_5, t_0\right)=u(1,0)=1(2-1)=1 (16.95)
Similarly, boundary conditions are u(0,t) = 0 and u(1,t) =1 . This implies
u_{0, j}=u\left(x_0, t_j\right)=u(0, t)=0
u_{5, j}=u\left(x_5, t_j\right)=u(1, t)=1 ; \quad \text { for } \forall j=0,1,2,3, \cdots (16.96)
The following table shows the initial (16.95) and boundary (16.96) conditions.
| 5(1) | 4(0.8) | 3(0.6) | 2(0.4) | 1(0.2) | 0(0) | j(t) \ i(x) |
| 1 | 0.96 | 0.84 | 0.64 | 0.36 | 0 | 0(0) |
| 1 | 0 | 1(0.005) | ||||
| 1 | 0 | 2(0.01) | ||||
| 1 | 0 | 3(0.015) | ||||
| 1 | 0 | 4(0.02) |
Let the value of u(x, t) at point \left(x_i, t_j\right) \text { be } u_{i, j} \text { i.e. } u\left(x_i, t_j\right)=u_{i, j} Discretizing the given variable coefficient problem at point \left(x_i, t_j\right) , we get
\left.\frac{\partial u}{\partial t}\right|_{\left(x_i, t_j\right)}=\left.\frac{\partial^2 u}{\partial x^2}\right|_{\left(x_i, t_j\right)}+\left.x \frac{\partial u}{\partial x}\right|_{\left(x_i, t_j\right)} (16.97)
The following forward difference (for time derivative term) and central difference (for space derivative terms) formulas can be used to generate the explicit scheme
\begin{aligned} &\frac{\partial u\left(x_i, t_j\right)}{\partial t}=\frac{u_{i, j+1}-u_{i, j}}{\Delta t}+O(\Delta t) \\ &\frac{\partial u\left(x_i, t_j\right)}{\partial x}=\frac{u_{i+1, j}-u_{i-1, j}}{2(\Delta x)}+O\left(\Delta x^2\right) \end{aligned}
\frac{\partial^2 u\left(x_i, t_j\right)}{\partial x^2}=\frac{u_{i+1, j}-2 u_{i, j}+u_{i-1, j}}{(\Delta x)^2}+O\left(\Delta x^2\right) (16.98)
Putting the finite differences (16.98) in Eq. (16.97) and neglecting the error terms in discretization, we have
\frac{u_{i, j+1}-u_{i, j}}{\Delta t}=\frac{u_{i+1, j}-2 u_{i, j}+u_{i-1, j}}{(\Delta x)^2}+x_i \frac{u_{i+1, j}-u_{i-1, j}}{2(\Delta x)}
On rearranging the terms, we have following explicit scheme
u_{i, j+1}=\frac{\Delta t}{2(\Delta x)^2}\left(\left(2+x_i(\Delta x)\right) u_{i+1, j}+\left(2-x_i(\Delta x)\right) u_{i-1, j}\right)+\left\lgroup 1-\frac{2(\Delta t)}{(\Delta x)^2} \right\rgroup u_{i, j}
Using Δx = 0.2 and Δt = 0.005 , we have
u_{i, j+1}=0.0625\left(\left(2+0.2 x_i\right) u_{i+1, j}+\left(2-0.2 x_i\right) u_{i-1, j}\right)+0.75 u_{i, j} (16.99)
For j = 0, we get
u_{i, 1}=0.0625\left(\left(2+0.2 x_i\right) u_{i+1,0}+\left(2-0.2 x_i\right) u_{i-1,0}\right)+0.75 u_{i, 0}
Computing the values for i = 1, 2, 3 and 4, we get
\begin{aligned} &u_{1,1}=0.0625\left(\left(2+0.2 x_1\right) u_{2,0}+\left(2-0.2 x_1\right) u_{0,0}\right)+0.75 u_{1,0}=0.351600 \\ &u_{2,1}=0.0625\left(\left(2+0.2 x_2\right) u_{3,0}+\left(2-0.2 x_2\right) u_{1,0}\right)+0.75 u_{2,0}=0.632400 \\ &u_{3,1}=0.0625\left(\left(2+0.2 x_3\right) u_{4,0}+\left(2-0.2 x_3\right) u_{2,0}\right)+0.75 u_{3,0}=0.832400 \\ &u_{4,1}=0.0625\left(\left(2+0.2 x_4\right) u_{5,0}+\left(2-0.2 x_i\right) u_{3,0}\right)+0.75 u_{4,0}=0.951600 \end{aligned}
These values give the second row of the table.
Proceeding in a similar manner, for j = 1, 2, and 3, we will get different rows of the table.
The results are presented in the following table (only six decimal digits)
| 5(1) | 4(0.8) | 3(0.6) | 2(0.4) | 1(0.2) | 0(0) | j(t) \ i(x) |
| 1 | 0.96 | 0.84 | 0.64 | 0.36 | 0 | 0(0) |
| 1 | 0.951600 | 0.832400 | 0.632400 | 0.351600 | 0 | 1(0.005) |
| 1 | 0.944426 | 0.824694 | 0.624704 | 0.344331 | 0 | 2(0.01) |
| 1 | 0.938159 | 0.817060 | 0.617058 | 0.337898 | 0 | 3(0.015) |
| 1 | 0.932581 | 0.809605 | 0.609559 | 0.332098 | 0 | 4(0.02) |
