Question

The electrostatic potential u between two concentric spheres of radius $\mathit{r}\mathbf{=}\mathbf{1}$and $\mathit{r}\mathbf{=}\mathbf{4}$is determined from

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{r}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{r}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{u}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{50}\mathbf{,}\mathbf{}\mathit{u}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{.}$

Use the method of this section with $\mathit{n}\mathbf{=}\mathbf{6}$ to approximate the solution of this boundary-value problem.

Short answer

So, the approximate solution is$$\begin{gathered} u_1=72.2222,u_2=83.333,u_3=90 \\ {u}_4=94.444,u_5=97.619 \end{gathered}$$

Step-by-step solution

Definition of finite difference

Finite difference methods are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences.

Solve for finite difference

Given boundary value problem is$\frac{d^{2}u}{d{r}^2}+\frac{2}{r}\frac{du}{dr}=0u\left(1\right)=50,u\left(4\right)=100$

Comparing with (7) we get

$P\left(r\right)=\frac{2}{r},Q\left(r\right)=0,f\left(r\right)=0,a=1,b=4$

Since n=6 we have $h=\frac{4-1}{6}=0.5$

From (8) we get the finite difference equation as

$$\begin{gathered} \left(1+\frac{h}{2}.\frac{2}{r_i}\right)u_{i+1}+(-2+h^2.0)u_i+{\left(1-\frac{h}{2}.\frac{2}{r_i}\right)}_{i-1}=0 \\ \Rightarrow \left(1+\frac{0.5}{r_i}\right)u_{i+1}-2u_i+\left(1-\frac{0.5}{r_i}\right)u_{i-1}=0.......(*) \end{gathered}$$

Solve for interior points

Now the interior points are

$$\begin{gathered} r_1=1+0.5=1.5, \\ {r}_2=2, \\ {r}_3=2.5, \\ {r}_4=3, \\ {r}_5=3.5, \\ {r}_6=4 \end{gathered}$$

Use conditions

For $i=1,2,3,4,5......(*)$yields

$$\begin{gathered} 1.33334u_{i+1}-2u_1+0.66667u_0=0 \\ 1/25u_3-2u_2+0.75u_1=0 \\ 1.2u_4-2u_3+0.8u_2=0 \\ 1.16667u_5-2u_4+0.83334u_3=0 \\ 1.14285u_6-2u_5+0.85714u_4=0 \end{gathered}$$

But from initial conditions $u_0=50$ and $u_6=100$substituting and solving we get

$$\begin{gathered} u_1=72.2222,u_2=83.333,u_3=90 \\ {u}_4=94.444,u_5=97.619 \end{gathered}$$

Thus, the required solution is $$\begin{gathered} u_1=72.2222,u_2=83.333,u_3=90 \\ {u}_4=94.444,u_5=97.619 \end{gathered}$$.