Question

Temperature in a ring:

The temperature u(r) in the circular ring shown in Figure 5.2.11 is determined from the boundary – value proble

Where ${\mathit{u}}_{\mathbf{0}}$and${\mathit{u}}_{\mathbf{1}}$are constants. Show that

$\mathit{u}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{u}}_{\mathbf{0}}\mathbf{}\mathbf{I}\mathbf{n}\mathbf{(}\mathbf{r}\mathbf{/}\mathbf{b}\mathbf{)}\mathbf{-}{\mathbf{u}}_{\mathbf{1}}\mathbf{I}\mathbf{n}\mathbf{(}\mathbf{r}\mathbf{/}\mathbf{b}\mathbf{)}}{\mathbf{I}\mathbf{n}\mathbf{(}\mathbf{a}\mathbf{/}\mathbf{b}\mathbf{)}}$

Short answer

Finally we get the solution as

$u\left(r\right)=\frac{u_0\mathrm{I}n\left(\frac{r}{b}\right)-u_1\mathrm{I}n\left(\frac{r}{a}\right)}{\mathrm{I}n\left(\frac{a}{b}\right)}$

Step-by-step solution

 Step 1: Given Information

The given equation is:

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

Given Information

The auxiliary equation of the given differential equation is

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

Whose roots are 0 and 0

$u\left(r\right)=c_1\mathrm{I}nr+c_2$

We get the boundary conditions as

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

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

Which yields the equation

$c_1\mathrm{I}na+c_2=u_0$and$c_1\mathrm{I}nb+c_2=u_1$

Solve the expression

Solving we get

$c_1=\frac{u_1\mathrm{I}na-u_0\mathrm{I}nb}{\mathrm{I}n\left(\frac{a}{b}\right)}$ and$c_1=\frac{u_0-u_1}{\mathrm{I}n\left(\frac{a}{b}\right)}$

Thus We get

$u\left(r\right)=\frac{u_0\mathrm{I}n\left(\frac{r}{b}\right)-u_1\mathrm{I}n\left(\frac{r}{a}\right)}{\mathrm{I}n\left(\frac{a}{b}\right)}$

Finally we get the solution as

$u\left(r\right)=\frac{u_0\mathrm{I}n\left(\frac{r}{b}\right)-u_1\mathrm{I}n\left(\frac{r}{a}\right)}{\mathrm{I}n\left(\frac{a}{b}\right)}$