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\begin{document}



\centerline{\textsc{Advanced Analysis}}
\centerline{\textbf{Math 121, Fall 2004}}
\centerline{\textbf{Solutions, Midterm 2}}


\bigskip\bigskip\noindent
\textbf{1.} [20\%] Use Laplace transforms to find the solution $y(t)$ of the following initial value problem
\beann
&&y^{\prime\prime} + 9 y = f(t),\\
&&y(t) = 0,\quad y^\prime(0) = 0,
\eeann
where $f(t)$ is an arbitrary function. Express your answer as a convolution.


\bigskip\noindent
\textbf{Solution.}
\begin{itemize}
\item Let $Y(p) = \mathcal{L}[y(t)]$ and $F(p) = \mathcal{L}[f(t)]$, where $\mathcal{L}$ denotes the
Laplace transform. Then
\[
p^2 Y + 9 Y = F(p),
\] 
so
\[
Y(p) = \frac{F(p)}{p^2 + 9}.
\]
From the tables,
\[
\mathcal{L}^{-1}\left[\frac{1}{p^2 + 9}\right] = \frac{1}{3}\sin(3t).
\]
Therefore, using the convolution theorem, we have
\[
y(t) = \frac{1}{3} \int_0^t \sin [3(t-s)] f(s)\, ds.
\]
\end{itemize}

\newpage
\noindent
\textbf{2.} [20\%] 
(a) Say what jump conditions the solution
$y(t)$ of the following initial value problem satisfies
at $t=0$, and find the solution directly (do not use Laplace transforms):
\beann
&&y^{\prime} + 2 y = \delta(t),\\
&&y(t) = 0\quad\mbox{for $t<0$}.
\eeann

\smallskip\noindent
(b) Write the solution of the following initial value problem, where $f(t)$ is an arbitrary function,
as a convolution (you don't need to derive your answer):
\beann
&&y^{\prime} + 2 y = f(t),\\
&&y(0) = 0.
\eeann


\bigskip\noindent
\textbf{Solution.}
\begin{itemize}
\item (a) The solution $y(t)$ has a jump discontinuity of size one at $t=0$, so that $y(t)$ for $t>0$ satisfies the
IVP
\beann
&&y^\prime + 2 y = 0,\\
&&y(0) = 1.
\eeann
The solution of the ODE is
\[
y(t) = c e^{-2t}
\]
where $c$ is a constant of integration, and the initial condition implies that $c=1$. Hence,
\[
y(t) = \cases{e^{-2t} & for $t > 0$,\cr
0 & for $t < 0$.}
\]

\item  (b) The solution is given by
\[
y(t) = \int_0^t e^{-2(t-s)}f(s)\,ds.
\]
\end{itemize}


\newpage
\noindent
\textbf{3.} [20\%] Use Fourier series to find the solution $u(x,y)$
of the following boundary value problem
for Laplace's equation in the strip $0<x<1$, $y>0$:
\beann
&&\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0,\\
&&\frac{\partial u}{\partial x} (0,y) = \frac{\partial u}{\partial x}(1,y) = 0,\\
&&u(x,0) = x,\qquad
\mbox{$u(x,y)$ is bounded as $y\to\infty$}.
\eeann
What does the solution $u(x,y)$ approach as $y\to \infty$?


\bigskip\noindent
\textbf{Solution.}
\begin{itemize}
\item The appropriate linear combination of separated solutions of Laplace's equation that
satisfy the boundary conditions at $x=0,1$ and as $y\to\infty$ is
\[
u(x,y) = \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n \cos (n\pi x) e^{-n\pi y},
\]
where the $a_n$ are constants. Imposing the boundary condition at $y=0$, we obtain that
\[
x = \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n \cos (n\pi x),
\]
so
\[
a_n = 2\int_0^1 x\cos(n\pi x) \, dx.
\]
Evaluating this integral, we find that $a_0 = 1$, and for $n\ge 1$
\[
a_n = \cases{-4/(n\pi)^2 & for $n$ odd,\cr
0 & for $n$ even.}
\]
Hence the solution is
\beann
u(x,y) &=& \frac{1}{2} - \frac{4}{\pi^2}\left\{\cos(\pi x) e^{-\pi y} + \frac{1}{3^2} \cos (3\pi x) e^{-3\pi y}\right.\\
&&\qquad\qquad\qquad\left.+ \frac{1}{5^2} \cos (5\pi x) e^{-5\pi y} +\dots\right\}.
\eeann
It follows that $u(x,y) \to 1/2$ as $y\to \infty$. Thus, the temperature approaches a constant value equal to the mean value of the
boundary data at $y=0$.
\end{itemize}

\newpage
\noindent
\textbf{4.} [20\%] Reduce the boundary conditions to homogeneous boundary conditions
and use Fourier series to find the solution
$u(x,t)$ of the following initial-boundary value problem
for the heat equation in $0<x<1$ and $t >0$:
\beann
&&\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2},\\
&&u(0,t) = 0, \qquad u(1,t) = 1,\\
&&u(x,0) = 0.
\eeann
What does the solution $u(x,t)$ approach as $t\to \infty$?


\bigskip\noindent
\textbf{Solution.}
\begin{itemize}
\item The steady-state solution $u=x$ satisfies the non-homogeneous boundary conditions and the PDE. We write
\[
u(x,t) = x + v(x,t).
\]
Then $v(x,t)$ satisfies
\beann
&&\frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial x^2},\\
&&v(0,t) = 0, \qquad v(1,t) = 0,\\
&&v(x,0) = -x.
\eeann
We look for a solution for $v$ of the form
\[
v(x,t) = \sum_{n=1}^\infty b_n \sin(n\pi x) e^{-n^2\pi^2 t}.
\]
This function satisfies the PDE and the boundary conditions at $x=0,1$ for any choice of the constants $b_n$.
Imposing the initial condition at $t=0$ we get that
\[
- x = \sum_{n=1}^\infty b_n \sin(n\pi x).
\]
Using the expression for the Fourier sine coefficients, we get
\beann
b_n &=& - 2\int_0^1 x \sin (n\pi x)\, dx\\
&=& \frac{2(-1)^n}{n\pi}.
\eeann
The solution of the original problem is therefore
\[
u(x,t) = x + \frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin (n\pi x) e^{-n^2\pi^2 t}.
\]
As $t\to\infty$, we have $u(x,t) \to x$, so the temperature approaches the steady-state temperature
distribution associated with the given non-homogeneous boundary conditions.
\end{itemize}

\newpage
\noindent
\textbf{5.} [20\%] (a) Consider the following initial value problem
for $u(x,t)$
\beann
&&\frac{\partial u}{\partial t} = \frac{\partial^3 u}{\partial x^3},\\
&&u(x,0) = f(x),
\eeann
where  $-\infty < x < \infty$, $t >0$. Solve for the Fourier transform of $u(x,t)$ with respect to $x$, and
write the solution for $u(x,t)$ as a Fourier integral, but do not attempt to invert the transform explicitly.


\smallskip
\noindent
(b) State Parseval's theorem for the Fourier transform. Use your solution from (a) to
show that
\[
\int_{-\infty}^\infty u^2(x,t)\, dx = \int_{-\infty}^\infty f^2(x)\, dx
\]
for all times $t$. (This result expresses `conservation of energy'.)


\bigskip\noindent
\textbf{Solution.}
\begin{itemize}
\item (a) Let
\[
\widehat{u}(k,t) = \frac{1}{2\pi} \int_{-\infty}^\infty u(x,t) e^{-ikx}\, dx
\]
be the Fourier transform of $u(x,t)$ with respect to $x$. Then, Fourier tranforming the PDE, we find that
\beann
&&\frac{\partial \widehat{u}}{\partial t} = (i k)^3\widehat{u},\\
&&\widehat{u}(k,0) = \widehat{f}(k),
\eeann
where $\widehat{f}$ is the Fourier transform of $f$. Solving this ODE, we get
\[
\widehat{u}(k,t) =  \widehat{f}(k) e^{(ik)^3 t}.
\]
Using this expression in the Fourier inversion formula, 
\[
u(x,t) = \int_{-\infty}^\infty \widehat{u}(k,t) e^{ikx}\, dk,
\]
and writing $(ik)^3 = -ik^3$, we obtain the solution
\[
u(x,t) = \int_{-\infty}^\infty \widehat{f}(k)e^{ikx - i k^3 t}\, dk.
\]

\item (b) Parseval's theorem states that
\[
\frac{1}{2\pi} \int_{-\infty}^\infty |f(x)|^2\, dx = \int_{-\infty}^\infty \left|\widehat{f}(k)\right|^2\, dk.
\]
Since $|e^{i\theta}| = 1$ for any real number $\theta$, we have that
\[
\left|e^{-ik^3 t}\right| = 1.
\]
Hence, for any $t >0$, we have
\[
\left|\widehat{u}(k,t)\right| = \left|\widehat{f}(k) e^{-ik^3 t}\right| = \left|\widehat{f}(k)\right|.
\]
It then follows from Parseval's theorem and this equation that
\beann
\int_{-\infty}^\infty u^2(x,t)\, dx &=&  2\pi \int_{-\infty}^\infty \left|\widehat{u}(k,t)\right|^2\, dk\\
 &=&  2\pi \int_{-\infty}^\infty \left|\widehat{f}(k)\right|^2\, dk\\
 &=&  \int_{-\infty}^\infty f^2(x)\, dx.
\eeann
\end{itemize}

\end{document}
