18.2.1.8 Solution stanexl-d

Question:

If for the problem of the previous question, the proposed solution is wrong, then so are the described isotherms.

To get a clue about the correct solution and isotherms, consider the following simpler problem. In this problem the top and right boundaries have been distorted into a quarter circle:

$$\mbox{BC:}\quad u(x,0)=1\quad\quad u(0,y)=0 \quad\quad\frac{\partial u}{\partial n}=0\mbox{ on }x^2+y^2=1$$

Solve this problem. Then neatly draw the $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, 0.25, 0.5, 0.75, and 1 isotherms for this problem.

Also neatly draw $u$ versus the polar angle $\theta$ at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.5. In a separate graph, draw the solution proposed in the previous section, $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for $y$ $\raisebox{.3pt}{$<$}$ $x$ and $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for $y$ $\raisebox{.3pt}{$>$}$ $x$, again against $\theta$ at $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.5.

Now go back to the problem of the previous question and very neatly sketch the correct $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, 0.25, 0.5, 0.75, and 1 isotherms for that problem. Pay particular attention to where the 0.25, 0.5, and 0.75 isotherms meet the boundaries and under what angle.

Answer:

You want to convert the simplified problem into polar coordinates $r$ and $\theta$. Show that the Neumann boundary condition on the quarter circle simplifies to $u_r$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.

Guess that the solution might be a function of $\theta$ only, $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(\theta)$. Show that this satisfies the Neumann boundary condition. Show that it satisfies the two Dirichlet boundary conditions if $f(0)$ and $f(\frac 12\pi)$ have suitable values.

Look up the Laplacian in polar coordinates in a table book. Plug in $u$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(\theta)$. That produces an ordinary differential equation. Solve this equation and plug in the boundary conditions on $f$. That gives the exact solution $f(\theta)$ to the given problem. Draw the lines with the given values of $u$ in the $x,y$-plane.

Note that very close to the origin, there is presumably not much difference between the current simplified problem and the original problem of the previous question. In that vicinity the boundary conditions are the same. So draw the isotherms in the original problem the same way near the origin.

The 0.5 isotherm can be drawn all the way based on antisymmetry of $u-\frac 12$ around the 45$\POW9,{\circ}$ line. Compare with the relevant earlier homework problem. Show that the 0.25 and 0.75 lines cannot hit the left and bottom boundaries. Based on the boundary conditions on the other two boundaries, show that they must hit the boundary normally wherever they leave the square. Then draw them neatly.