18.2.1.5 Solution stanexl-b3

Question:

Consider the Laplace equation within a unit circle, but now in polar coordinates:

$$u_{rr} + \frac 1r u_r + \frac 1{r^2} u_{\theta\theta} = 0 \qquad\mbox{for}\qquad r<1$$

The boundary condition on the perimeter of the circle is

$$u(1,\theta) = f(\theta)$$

where $f$ is a given function.

The solution is the Poisson integral formula

$$u(r,\theta) = \frac{1-r^2}{2\pi} \oint\frac{f(\bar\theta){ \rm d}\bar\theta}{1-2r\cos(\bar\theta -\theta)+r^2}$$

Now suppose that function $f(\theta)$ is increased slightly, by an amount $\delta{f}$, and only in a very small interval $\theta_1$ $\raisebox{.3pt}{$<$}$ $\theta$ $\raisebox{.3pt}{$<$}$ $\theta_2$.

Does the solution $u$ change everywhere in the circle, or only in the immediate vicinity of the interval on the boundary at which $f$ was changed. What is the sign of the change in $u$ if $\delta{f}$ is positive?

Answer:

To simplify this, assume that the new solution is $u_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $u_1+\delta{u}$ where $u_1$ is the old solution. So $\delta{u}$ is the change in the solution.

You now have for the original solution

$$u(r,\theta) = \frac{1-r^2}{2\pi} \oint\frac{f(\bar\theta){ \rm d}\bar\theta}{1-2r\cos(\bar\theta -\theta)+r^2}$$

and for the changed solution

$$u(r,\theta)+\delta u(r,\theta) = \frac{1-r^2}{2\pi} \oint\fr... ...theta)]{ \rm d}\bar\theta} {1-2r\cos(\bar\theta -\theta)+r^2}$$

Subtract the two to get

$$\delta u(r,\theta) = \frac{1-r^2}{2\pi} \oint\frac{\delta f(\bar\theta){ \rm d}\bar\theta}{1-2r\cos(\bar\theta -\theta)+r^2}$$

Now $\delta{f}(\bar\theta)$ is only nonzero in the interval from $\theta_1$ to $\theta_2$, so

$$\delta u(r,\theta) = \frac{1-r^2}{2\pi} \int_{\theta_1}^{\th... ...r\theta){ \rm d}\bar\theta}{1-2r\cos(\bar\theta -\theta)+r^2}$$

Now use a little graph to show that for any point not extremely close to the segment from $\theta_1$ to $\theta_2$ on the boundary, the denominator of the integrand is about constant, so

$$\delta u(r,\theta) = \frac{1-r^2}{2\pi} \frac{\int_{\theta_1... ...12}-\theta)+r^2} \qquad\theta_{12}=\frac{\theta_1+\theta_2}{2}$$

Now show that

$$1-2r\cos(\theta_{12}-\theta)+r^2 \ge(1 -r)^2$$

which is positive in the interior of the unit circle.

Then argue that this means that $\delta{u}$ is positive everwhere inside the circle. So the region of influence of the little segment is the interior of the circle, and the temperature increases everywhere.