Question

Substitute (8.25) into (8.22) and use (8.23) and (8.24) to show that (8.25) is a solution of (8.22).

Short answer

It has been proved that (8.25) from the book $u\left(r\right)=\iiint G(\mathbf{r},{\mathbf{r}}^{\text{'}})f\left({\mathbf{r}}^{\text{'}}\right)\text{d}{\tau }^{\text{'}}$is a solution to (8.22):${\nabla }^{2}u=f\left(r\right)$.

Step-by-step solution

Given Information:

It has been given that for reference follow example 1.

Uses of the Green function:

The green function is a function that is used to solve a corresponding partial differential equation in three dimensions, namely Poisson’s equation.

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\\ \mathbf{=}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\end{array}$

Step 3:Take the equation and start solving:

To prove that (8.25) from the book$u\left(r\right)=\iiint G(\mathbf{r},{\mathbf{r}}^{\text{'}})f\left({\mathbf{r}}^{\text{'}}\right)\text{d}{\tau }^{\text{'}}$is a solution to (8.22):

${\nabla }^{2}u=f\left(r\right)$

Use equation (8.23) to write the following equation.

${\nabla }^{2}G(\mathbf{r},{\mathbf{r}}^{\text{'}})=\delta (\mathbf{r}-{\mathbf{r}}^{\text{'}})$

And equation (8.24).

$\iiint f\left({\mathbf{r}}^{\text{'}}\right)\delta (\mathbf{r}-{\mathbf{r}}^{\text{'}})\text{d}{\tau }^{\text{'}}=f\left(r\right)$

Now prove the fact:

Change in (8.25) into the left-hand side of (8.22), and try to calculate the expression.

$\begin{array}{c}{\nabla }^{2}u={\nabla }^2\iiint G(\mathbf{r},{\mathbf{r}}^{\text{'}})f\left({\mathbf{r}}^{\text{'}}\right)\text{d}{\tau }^{\text{'}}\\ =\iiint {\nabla }^{2}G(\mathbf{r},{\mathbf{r}}^{\text{'}})f\left({\mathbf{r}}^{\text{'}}\right)\text{d}{\tau }^{\text{'}}\\ =\iiint \delta (\mathbf{r}-{\mathbf{r}}^{\text{'}})f\left({\mathbf{r}}^{\text{'}}\right)\text{d}{\tau }^{\text{'}}\end{array}$

${\nabla }^{2}u=f\left(r\right)$

Hence it has been proved.