Question

Temperature in a Sphere :

Consider two concentric spheres of radius$\mathit{r}\mathbf{=}\mathit{a}$and $\mathit{r}\mathbf{=}\mathit{b}\mathbf{,}\mathit{a}\mathbf{<}\mathit{b}\mathbf{.}$. See Figure 5.2.10. The temperature in the region between the sphere is determined from the boundary – value problem

$\mathit{r}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{r}}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{u}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{=}{\mathit{u}}_{\mathbf{0}}\mathbf{,}\mathit{u}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{=}{\mathit{u}}_{\mathbf{1}}\mathbf{,}$

Where ${\mathit{u}}_{\mathbf{0}}$and are constants. Solve for .

Short answer

Finally we get the solution as

$u\left(r\right)=-ab\left(\frac{u_1-u_0}{b-a}\right)r^{-1}+\frac{u_{1}b-u_{0}a}{b-a}$

Step-by-step solution

Given Information

The given value is:

$r\frac{d^{2}u}{d{r}^2}+2\frac{du}{dr}=0,u\left(a\right)=u_0,u\left(b\right)=u_1,$

The given differential equation has an auxiliary equation

The auxiliary equation of the given differential equation is

$m(m-1)+2m=m(m+1)=0$

Whose roots are 0 and -1

$u\left(r\right)=c_1{r}^{-1}+C_2$

We get the boundary conditions as

$u\left(a\right)=u_0$

And$u\left(b\right)=u_1$

Which yields the equation

$c_1{a}^{-1}+c_2=u_0$and$c_1{b}^{-1}+c_2=u_1$

Solving we get

localid="1663938589611" $c_2=\frac{u_{1}b-u_{0}a}{b-a}$

Thus We get

$u\left(r\right)=-ab\left(\frac{u_1-u_0}{b-a}\right)r^{-1}+\frac{u_{1}b-u_{0}a}{b-a}$

Finally we get the solution as

$u\left(r\right)=-ab\left(\frac{u_1-u_0}{b-a}\right)r^{-1}+\frac{u_{1}b-u_{0}a}{b-a}$