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# 8.1: Example

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# Separation of variables in polar coordinates

## Example

Consider a circular plate of radius $$c \text{ m}$$, insulated from above and below. The temperature on the circumference is $$100^\circ~\text{C}$$ on half the circle, and $$0^\circ~\text{C}$$ on the other half.

The differential equation to solve is $\rho^2 \pdrhor{u} + \rho \pdrho{u} + \pdphip{u} = 0,$ with boundary conditions $u(c,\phi) = \begin{cases} 100 & \text{if 0 < \phi < \pi} \\ 0 & \text{if \pi < \phi < 2\pi } \end{cases}\quad .$

### Periodic BC

There is no real boundary in the $$\phi$$ direction, but we introduce one, since we choose to let $$\phi$$ run from $$0$$ to $$2\pi$$ only. So what kind of boundary conditions do we apply? We would like to see “seamless behaviour”, which specifies the periodicity of the solution in $$\phi$$, \begin{aligned} u(\rho,\phi+2\pi)&=&u(\rho,\phi),\\ \pdphi{u}(\rho,\phi+2\pi)&=&\pdphi{u}(\rho,\phi).\end{aligned} If we choose to put the seem at $$\phi=-\pi$$ we have the periodic boundary conditions \begin{aligned} u(\rho,2\pi)&=&u(\rho,0),\\ \pdphi{u}(\rho,2\pi)&=&\pdphi{u}(\rho,0).\end{aligned}

We separate variables, and take, as usual $u(\rho,\phi) = R(\rho) \Phi(\phi).$ This gives the usual differential equations \begin{aligned} \Phi''-\lambda\Phi &=&0,\\ \rho^2 R'' + \rho R' + \lambda R &=& 0.\end{aligned} Our periodic boundary conditions gives a condition on $$\Phi$$, $\Phi(0)=\Phi(2\pi),\;\;\Phi'(0)=\Phi'(2\pi). \label{eq:phiBC}$ The other boundary condition involves both $$R$$ and $$\Phi$$.

## Three cases for $$\lambda$$

As usual we consider the cases $$\lambda>0$$, $$\lambda<0$$ and $$\lambda=0$$ separately. Consider the $$\Phi$$ equation first, since this has the most restrictive explicit boundary conditions ([eq:phiBC]).
We have to solve $\Phi'' = \alpha^2\Phi,$ which has as a solution $\Phi(\phi) = A \cos \alpha \phi + B \sin \alpha \phi.$ Applying the boundary conditions, we get \begin{aligned} A &=& A \cos(2\alpha\pi)+ B\sin(2\alpha\pi), \\ B \alpha &=& -A \alpha\sin(2\alpha\pi)+ B\alpha\cos(2\alpha\pi).\end{aligned} If we eliminate one of the coefficients from the equation, we get $A = A \cos(2\alpha\pi)-A\sin(2\alpha\pi)^2/(1-cos(2\alpha\pi))$ which leads to $\sin(2\alpha\pi)^2=-(1-cos(2\alpha\pi))^2,$ which in turn shows $2\cos(2\alpha\pi)=2,$ and thus we only have a non-zero solution for $$\alpha=n$$, an integer. We have found $\lambda_n=n^2,\;\;\Phi_n(\phi) = A_n \cos n \phi + B_n \sin n \phi.$ We have $\Phi'' = 0 .$ This implies that $\Phi = A\phi + B.$ The boundary conditions are satisfied for $$A=0$$, $\Phi_0(\phi)=B_n.$ The solution (hyperbolic sines and cosines) cannot satisfy the boundary conditions.

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Now let me look at the solution of the $$R$$ equation for each of the two cases (they can be treated as one), $\rho^2 R''(\rho) + \rho R'(\rho) -n^2R(\rho)=0.$ Let us attempt a power-series solution (this method will be discussed in great detail in a future lecture) $R(\rho) =\rho^\alpha.$ We find the equation $\rho^\alpha[\alpha(\alpha-1)+\alpha^2-n^2]= \rho^\alpha[\alpha^2-n^2]=0$ If $$n\neq 0$$ we thus have two independent solutions (as should be) $R_n(\rho) = C\rho^{-n}+D\rho^n$ The term with the negative power of $$\rho$$ diverges as $$\rho$$ goes to zero. This is not acceptable for a physical quantity (like the temperature). We keep the regular solution, $R_n(\rho) = \rho^n.$ For $$n=0$$ we find only one solution, but it is not very hard to show (e.g., by substitution) that the general solution is $R_0(\rho) = C_0 + D_0\ln(\rho).$ We reject the logarithm since it diverges at $$\rho=0$$.

## Putting it all together

In summary, we have $u(\rho,\phi) = \frac{A_0}{2} + \sum_{n=1}^\infty \rho^n \left(A_n\cos n\phi+B_n\sin n \phi \right).$ The one remaining boundary condition can now be used to determine the coefficients $$A_n$$ and $$B_n$$, \begin{aligned} U(c,\phi) &=& \frac{A_0}{2} + \sum_{n=1}^\infty c^n \left(A_n\cos n\phi+B_n\sin n \phi \right)\nonumber\\ &=& \begin{cases} 100 & \text{if 0 < \phi < \pi} \\ 0 & \text{if \pi < \phi < 2\pi} \end{cases}\quad.\end{aligned} We find \begin{aligned} A_0 &=& \frac{1}{\pi} \int_0^\pi 100\, d\phi = 100, \nonumber\\ c^n A_n &=& \frac{1}{\pi}\int_0^\pi 100\cos n\phi\, d\phi = \frac{100}{n\pi} \sin(n\phi)|^\pi_0=0,\nonumber\\ c^n B_n &=& \frac{1}{\pi}\int_0^\pi 100\sin n\phi\, d\phi = -\frac{100}{n\pi} \cos(n\phi)|^\pi_0 \nonumber\\&=& \begin{cases} 200/(n\pi) & \text{if n is odd}\\ 0 & \text{if n is even} \end{cases}\quad.\end{aligned} In summary $u(\rho,\phi) = 50 + \frac{200}{\pi} \sum_{n~\rm odd} \left(\frac{\rho}{c}\right)^n \frac{\sin n \phi}{n}. \label{eq:T0-100}$ We clearly see the dependence of $$u$$ on the pure number $$r/c$$, rather than $$\rho$$. A three dimensional plot of the temperature is given in Fig. [fig:T0-100].