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\newcommand{\pxx}[2]{\frac{\partial^2 #1}{\partial #2^2}}

\begin{document}

\title{Math 135-2, Homework 8}
\author{Name: \underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}\quad ID: \underline{\qquad\qquad\qquad\qquad}}
\date{}
\maketitle


\begin{problem}{Problem 42.1}
  If $w = F(x,y) = G(r,\theta)$ with $x=r\cos\theta$ and $y=r\sin\theta$, show that
  \begin{align*}
    \pxx{w}{x} + \pxx{w}{y} &= \frac{1}{r} \bk{\px{}{r}\gp{r\px{w}{r}} + \frac{1}{r}\pxx{w}{\theta}} \\
    &= \pxx{w}{r} + \frac{1}{r} \px{w}{r} + \frac{1}{r^2} \pxx{w}{\theta}.
  \end{align*}
  Hint:
  \begin{align*}
    \px{w}{r} = \px{w}{x} \cos \theta + \px{w}{y} \sin \theta \qquad \mbox{and} \qquad \px{w}{\theta} = \px{w}{x} (-r\sin \theta) + \px{w}{y} (r\cos \theta).
  \end{align*}
  Similarly, compute $\displaystyle \px{}{r}\gp{r\px{w}{r}}$ and $\displaystyle \pxx{w}{\theta}$.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem A}
  The Poisson integral in the case of a general disc is (see problem 42.3)
  \begin{align*}
    w(r,\theta) &= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{R^2 - r^2}{R^2 - 2 R r \cos(\theta-\phi) + r^2} f(\phi)\,d\phi.
  \end{align*}
  If $f(\phi)$ is continuous, show that the maximum and minumum values of
  $w(r,\theta)$ are obtained on the boundary.  (It is possible that these values
  may be obtained in the interior and at the boundary.)  How does this statement
  change if the continuity requirement is dropped?  Hint: the substitution
  $\tan(\theta/2) = u$ is useful for integrating rational functions in $\sin
  \theta$ and $\cos \theta$.  This substitution implies $\cos \theta =
  \frac{1-u^2}{1+u^2}$, $\sin \theta = \frac{2u}{1+u^2}$, and $d\theta =
  \frac{2\,du}{1+u^2}$, thus turning the integrand into a rational function.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 43.1}
  The differential equation $P(x) y'' + Q(x) y' + R(x) y = 0$ is called
  \textit{exact} if it can be written in the form $[P(x) y']' + [S(x) y]' = 0$
  for some function $S(x)$.  In this case the second equation can be integrated
  at once to give the first order linear equation $P(x) y' + S(x) y = c_1$,
  which can then be solved by the method of Section 10.  By equating
  coefficients and eliminating $S(x)$, show that a necessary and sufficient
  condition for exactness is $P''(x) - Q'(x) + R(x) = 0$.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 43.3}
  If the equation in Problem 1 is not exact, it can be made exact by multiplying
  by a suitable integrating factor $\mu(x)$.  Thus, $\mu(x)$ must satisfy the
  condition that the equation $\mu(x) P(x) y'' + \mu(x) Q(x) y' + \mu(x) R(x) y
  = 0$ is expressible in the form $[\mu(x) P(x) y']' + [S(x) y]' = 0$ for some
  function $S(x)$.  Show that $\mu(x)$ must be a solution of the \textit{adjoint
    equation}
  \begin{align*}
    P(x) \mu'' + [2 P'(x) - Q(x)] \mu' + [P''(x) - Q'(x) + R(x)] \mu = 0.
  \end{align*}
  In general (but not always), the adjoint equation is just as difficult to
  solve as the original equation.  Find the adjoint equation in each of the
  folliwing cases:
  \qpart{a} Legendre's equation: $(1-x^2) y'' - 2xy' + p(p+1) y = 0$
  \qpart{b} Bessel's equation: $x^2 y'' + x y' + (x^2-p^2) y = 0$
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 43.5}
  Show that the adjoint of the adjoint of the equation $P(x) y'' + Q(x) y' +
  R(x) y = 0$ is the original equation.
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 43.6}
  The equation $P(x) y'' + Q(x) y' + R(x) y = 0$ is called \textit{self-adjoint}
  if its adjoint is the same equation (except for notation).
  \qpart{a} Show that this equation is self-adjoint if and only if $P'(x) =
  Q(x)$.  In this case the equation becomes
  \begin{align*}
    P(x) y'' + P'(x) y' + R(x) y = 0
  \end{align*}
  or
  \begin{align*}
    [P(x) y]' + R(x) y = 0,
  \end{align*}
  which is the standard form of a self-adjoint equation.
  \qpart{b} Which of the equations in Problem 3 are self-adjoint? (Just do (a) and (b).)
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 43.7}
  Show that any equation $P(x) y'' + Q(x) y' + R(x) y = 0$ can be made
  self-adjoint by multipyling through by
  \begin{align*}
    \frac{1}{P} e^{\int Q/P\,dx}.
  \end{align*}
\end{problem}

\begin{solution}
\end{solution}

\begin{problem}{Problem 43.9}
  P
\end{problem}

\begin{solution}
\end{solution}

\end{document}
