An example solution of Poisson's equation in 1-d

Let us now solve Poisson's equation in one dimension, with mixed boundary conditions, using the finite difference technique discussed above. We seek the solution of

$$\frac{d^2 u(x)}{dx^2} = v(x),$$ (142)

in the region $0\leq x\leq 1$, with $v(x) = 1 - 2\,x^2$. The boundary conditions at $x_l=0$ and $x_h = 1$ take the mixed form specified in Eqs. (132) and (133). Of course, we can solve this problem analytically. In fact,

$$u(x) = g + h\,x + \frac{x^2}{2} - \frac{x^4}{6},$$ (143)

where

$$g = \frac{\gamma_l\,(\alpha_h+\beta_h) - \beta_l\,[\gamma_h-(\alpha_h+\beta_h)/3]} {\alpha_l\,\alpha_h + \alpha_l\,\beta_h-\beta_l\,\alpha_h},$$ (144)

$$h = \frac{ \alpha_l\,[\gamma_h-(\alpha_h+\beta_h)/3]-\gamma_l\,\alpha_h} {\alpha_l\,\alpha_h + \alpha_l\,\beta_h-\beta_l\,\alpha_h}.$$ (145)

Figure 62 shows a comparison between the analytic and finite difference solutions for $N=100$. It can be seen that the finite difference solution mirrors the analytic solution almost exactly.

Figure 62: Solution of Poisson's equation in one dimension with $v=1-2\,x^2$, $\alpha _l = 1$, $\beta _l=-1$, $\gamma _l=1$, $\alpha _h=1$, $\beta _h=1$, and $\gamma _h=1$. The dotted curve (obscured) shows the analytic solution, whereas the open triangles show the finite difference solution for $N=100$.