9 03/20 F

  1. See whether any terms must be changed in expression (5) in the notes on elliptic equations in the three dimensional case. Then determine how (6) differs from the two-dimensional case.

  2. Assume that you choose

    \begin{displaymath}
u_{\rm out}(r,\vartheta,\varphi) =
A \frac{1}{r} \bar u(\b...
...a,\varphi)
\qquad \mbox{with}\quad \bar r \equiv \frac{1}{r}
\end{displaymath}

    where $A$ is a constant still to be chosen and function

    \begin{displaymath}
\bar u(\bar r,\vartheta,\varphi)
\quad\mbox{and}\quad u(r,\vartheta,\varphi)
\end{displaymath}

    are the same function using different names. Show that then on the surface of the unit sphere,

    \begin{displaymath}
u_{\rm out}(1,\vartheta,\varphi) = A \bar u(1,\vartheta,\varphi)
= A u(1,\vartheta,\varphi)
\end{displaymath}

    Show next that

    \begin{displaymath}
u_{{\rm out},r} =
- A \left(\frac{\bar u}{r^2} + \frac{1}{r^3} \bar u_{\bar r}\right)
\end{displaymath}

    where subscripts indicate partial derivatives with respect to that variable. Conclude that on the surface on the unit sphere

    \begin{displaymath}
u_{{\rm out},r}(1,\vartheta,\varphi) =
- A \bar u(1,\varth...
...i) =
- A u(1,\vartheta,\varphi) - A u_r(1,\vartheta,\varphi)
\end{displaymath}

    Now look up the Laplacian in spherical coordinates and evaluate $\nabla^2u_{\rm {out}}$ in terms of $\bar
u(\bar{r},\vartheta,\varphi)$ using the chain rule of differentiation. Show that

    \begin{displaymath}
\nabla^2 u_{\rm out} = \frac{A}{r^5}
\left(
\frac{1}{\bar...
...c{1}{\bar r^2\sin^2\vartheta}\bar u_{\varphi\varphi}
\right)
\end{displaymath}

    Hint: in the right hand side, differentiate out the first product before comparing with what you got from differentiating out $\nabla^2u_{\rm {out}}$. Conclude that the Laplacian of $u_{\rm {out}}$ is zero.

  3. In view of the formulae derived in the previous question, show that $A$ must be $A=-1$ in order for the surface integrals in (6) only to involve the function

    \begin{displaymath}
u(1,\vartheta,\varphi)=g(\vartheta,\varphi)
\end{displaymath}

    given in a Dirichlet problem, and not the then unknown function $u_r(1,\vartheta,\varphi)$. Also show that it would not be possible to choose $A$ so that $u$ drops out from both integrals, so that the Neumann problem cannot be solved this way. For the Dirichlet problem, use spherical coordinates for the point $\vec{x}$ at which $u$ is to be evaluated and for the generic integration point $\vec{\xi}$ as in

    \begin{displaymath}
\vec x = r \hat\imath_r \quad
\hat\imath_r =
\left(
\beg...
...ta\sin\bar\varphi \\
\cos\bar\vartheta
\end{array} \right)
\end{displaymath}

    In those terms, show that the integrals to be evaluated reduce to

    \begin{displaymath}
\int_S g(\bar\vartheta,\bar\varphi)
\left(2 \frac{\partial...
...o^2 \sin\bar\vartheta {\rm d}\bar\vartheta{\rm d}\bar\varphi
\end{displaymath}

    Show that the parenthetical expression may be simplified to

    \begin{displaymath}
2 \frac{\partial G}{\partial \rho} + G = \frac{1}{4\pi d}
...
...^2\rho}-1\right)
\qquad d \equiv \vert\vec x - \vec \xi\vert
\end{displaymath}

    Use this plus the made assumption that $\rho=R=1$ on the boundary to derive the Poisson integral (9).

  4. 3.38. This does not require solution of the problem using the Poisson integral formula. You can just examine what symmetry properties the solution $u(x,y)$ should have to figure out the value at the origin. However, feel free to check your result against the Poisson integral.

  5. 3.39. Again this does not require solution of the problem using the Poisson integral formula. You should be able to find the complete solution $u(x,y)$ by mere inspection. However, feel free to check your result against the Poisson integral; in that case, first write the boundary values in the form

    \begin{displaymath}
f(\phi) = A + B \cos(\phi-\theta) + C \sin(\phi-\theta)
\end{displaymath}

    The integrals for $A$ and $B$ can be found in a table of definite integrals.

  6. 3.40. This is about the solution to the Dirichlet problem in a circle, which, as derived in class, is given by the Poisson integral formula (also listed in 3.37.) Because the problem for $u$ is linear, the change in $u$ due to the change in boundary condition is the Poisson integral solution when $f$ is zero everywhere except in the interval $(\theta_1,\theta_2)$, where it equals the change in boundary condition.