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Show all of your work. Full credit may not be given for an answer alone. You may use the backs of the pages for scratch work. Do not unstaple or remove pages.

This is a non-calculator exam. One side of one sheet of US letter-size paper on which definitions, theorems, and formulas are handwritten by you may be used. Worked-out examples are not allowed on the formula sheet. You will turn in your formula sheet with the exam.

By taking this exam you are agreeing to abide by these rules and Rutgers University's Academic Integrity Policy.

Problem Possible Points
Number Points Earned
$ \displaystyle\vphantom{\int}$ 1 10  
$ \displaystyle\vphantom{\int}$ 2 10  
$ \displaystyle\vphantom{\int}$ 3 15  
$ \displaystyle\vphantom{\int}$ 4 15  
$ \displaystyle\vphantom{\int}$ 5 15  
$ \displaystyle\vphantom{\int}$ 6 10  
$ \displaystyle\vphantom{\int}$ 7 25  
$ \displaystyle\vphantom{\int}$ Total 100  

  1. (10 points) Show that the function

    $\displaystyle u(x,y) = \arctan\left(\frac y x\right)
$

    is harmonic (i.e., $ \nabla^2 u = 0$) where it is defined.

    Proof. [Solution]This is easy if you remember the derivative of arctan, and I make no apologies for using this function. We have

    $\displaystyle \frac{\partial u}{\partial x}$ $\displaystyle = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \left(\frac{-y}{x^2}\right) = \frac{-y}{x^2 + y^2}.$    
    $\displaystyle \frac{\partial^2 u}{\partial x^2}$ $\displaystyle = \frac{y}{(x^2 + y^2)^2}(2x) = \frac{2xy}{(x^2 + y^2)^2}.$    
    $\displaystyle \frac{\partial u}{\partial y}$ $\displaystyle = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2}.$    
    $\displaystyle \frac{\partial^2 u}{\partial y^2}$ $\displaystyle = \frac{-x}{(x^2 + y^2)^2}(2y) = \frac{-2xy}{(x^2 + y^2)^2}.$    

    Cleary

    $\displaystyle \frac{\partial^2 u}{\partial x^2}
+ \frac{\partial^2 u}{\partial y^2} = 0.
$

    $ \qedsymbol$

  2. (10 points) Formulate a boundary value problem (partial differential equation, boundary conditions, and initial conditions) for the motion of an elastic string of length $ L$, fastened at the left end, and attached with a ring to a pole at the right end. The ring allows the string to slide up and down the pole, so that end is free. The initial position of the string is a function $ f(x)$.

    \includegraphics[height=1.5in]{mt2-2.eps}

    Just set up the problem and stop!

    Proof. [Solution]We expect the string to satisfy the wave equation. So the displacement will be

    $\displaystyle u_{tt}(x,t) = c^2 u_{xx}(x,t)
$

    for some constant $ c$, for all $ 0 < x < L$, $ t > 0$. The fact that the left end is fixed gives us

    $\displaystyle u(0,t) = 0
$

    for all $ t > 0$, and the fact that the right end is free gives us

    $\displaystyle u_x(L,t) = 0
$

    for all $ t > 0$. Initial conditions are

    $\displaystyle u(x,0) = f(x)
$

    for $ 0 \leq x \leq L$. I did not mention initial velocity; in general we should have

    $\displaystyle u_t(x,0) = g(x)
$

    for some function $ g(x)$ and $ 0 \leq x \leq L$. $ \qedsymbol$

  3. (15 points) Solve the following initial value problem for the wave equation in an infinitely long string:

    $\displaystyle u_{tt}$ $\displaystyle = c^2 u_{xx};$ $\displaystyle (-\infty < x < \infty,\ t > 0)$    
    $\displaystyle u(x,0)$ $\displaystyle = \sin(x);$ $\displaystyle (-\infty < x < \infty)$    
    $\displaystyle u_t(x,0)$ $\displaystyle = x^2.$ $\displaystyle (-\infty < x < \infty)$    

    Proof. [Solution]Since we have an infinitely long string, Fourier methods do not apply. Instead, we have to use the d'Alembert methods of Section 16.4. We have

    $\displaystyle u(x,t) = \frac12\left(f(x + ct) + f(x-ct)\right)
+ \frac{1}{2c}\int_{x-ct}^{x+ct} g(\xi) \,d\xi.
$

    Using our data $ f(x) = \sin(x)$, $ g(x) =x^2$, we have

    \begin{displaymath}\begin{split}u(x,t) &= \frac12\left(\sin(x + ct) + \sin(x-ct)\right) + \frac1{2c}\int_{x-ct}^{x+ct}\xi^2 \,d\xi. \\ \end{split}\end{displaymath}    

    Now

    \begin{displaymath}\begin{split}\frac1{2c}\int_{x-ct}^{x+ct}\xi^2 \,d\xi &= \lef...
...t] \\ &= \frac{1}{6c}\left[6x^2ct + 2c^3 t^3\right] \end{split}\end{displaymath}    

    So

    $\displaystyle u(x,t) = \sin(x)\cos(ct) + x^2 t + \tfrac{1}{3}c^2t^3.
$

    $ \qedsymbol$

  4. (15 points) Solve the following boundary-value problem for the wave equation:

    $\displaystyle u_{tt}$ $\displaystyle = c^2 u_{xx};$ $\displaystyle (0 < x < L,\ 0 < y < K \ t > 0)$    
    $\displaystyle u(x,0,t)$ $\displaystyle = u(x,K,t) = 0;$ $\displaystyle (0 \leq x \leq L,\ t > 0)$    
    $\displaystyle u(0,y,t)$ $\displaystyle = u(L,y,t) = 0;$ $\displaystyle (0 \leq y \leq K,\ t > 0)$    
    $\displaystyle u(x,y,0)$ $\displaystyle = x(L-x)y(K-y).$ $\displaystyle (0 \leq x \leq L,\ 0 \leq y \leq K)$    

    Proof. [Solution]Again, I said nothing about initial velocity; most assumed wisely that it was zero. We know that the general solution is

    $\displaystyle u(x,y,t) = \sum_{m,n=1}^\infty
c_{n,m}\sin\left(\frac{\pi n x}{L}...
...m y}{K}\right)
\cos\left(\pi\sqrt{\frac{n^2}{L^2} + \frac{m^2}{K^2}} ct\right)
$

    where

    $\displaystyle c_{n,m} = \frac{4}{LK}\int_0^L\int_0^K
u(x,y,0)\sin\left(\frac{\pi n x}{L}\right)
\sin\left(\frac{\pi m y}{K}\right)\,dy \,dx.
$

    Now as we have done in homework and lecture several times:

    $\displaystyle \frac{2}{L}\int_0^L x(L-x)\sin\left(\frac{\pi n x}{L}\right)
\,dx
= \frac{4L^2}{n^3\pi^3}\left(1 - (-1)^n\right).
$

    Likewise

    $\displaystyle \frac{2}{K}\int_0^K y(K-y)\sin\left(\frac{\pi m y}{K}\right)
\,dy
= \frac{4K^2}{m^3\pi^3}\left(1 - (-1)^m\right).
$

    So

    $\displaystyle c_{n,m} = \frac{16L^2K^2}{n^3m^3\pi^6}
\left(1 - (-1)^n\right)\left(1 - (-1)^m\right).
$

    You could go on, but why? $ \qedsymbol$

  5. (15 points) Solve the following boundary-value problem for the potential equation:

    \begin{displaymath}\begin{array}{cr} u_{xx} + u_{yy} = 0; &(0 < x < \pi,\ 0 < y ...
...) \\ u(0,y) = u(\pi,y) = 0. &(0 \leq y \leq \pi) \\ \end{array}\end{displaymath}    

    Proof. [Solution]Notice that we have zero on all three sides of the rectangle except the top. Thus our general solution is (remember $ L = K = \pi$)

    $\displaystyle u(x,y) = \sum_{n=1}^\infty c_n \sin(nx) \sinh(ny),
$

    where

    $\displaystyle c_n = \frac{2}{\pi \sinh(n\pi)} \int_0^\pi u(x,\pi)\sin(nx)\,dx.
$

    But since $ u(x,\pi) = \sin(2x)$, we have without even doing the integral the result

    $\displaystyle c_n = \begin{cases}
0 & n \neq 2; \\
\frac{1}{\sinh(2\pi)} & n = 2.
\end{cases}$

    So the series has only a single term:

    $\displaystyle u(x,y) = \frac{\sin(2x)\sinh(2y)}{\sinh(2\pi)}.
$

    $ \qedsymbol$

  6. (10 points) Consider heat conduction in a thin rod with insulated ends:

    \begin{displaymath}\begin{array}{cr} u_{t} = k u_{xx}; & (0 < x < L,\ t > 0) \\ u_x(0,t) = u_x(L,t) = 0. & (t > 0) \\ \end{array}\end{displaymath}    

    Remember $ k$ is a positive constant. Show that the integral

    $\displaystyle E(t) = \int_{0}^L u(x,t)^2 \,dx
$

    is a strictly decreasing function of $ t$ as long as $ u_x(x,t) \not\equiv
0$.

    Hint 1   Show $ E'(t) < 0$.

    Proof. [Solution]I should have given the hint not to solve the problem for $ u$! Instead, just differentiate under the integral sign, apply the differential equation, and integrate by parts.

    \begin{displaymath}\begin{split}E'(t) &= \frac{d}{dt}\int_0^Lu(x,t)^2 \,dx \\ &=...
..._x^2 \,dx \right]\\ &= - 2k\int_0^L u_x^2 \,dx < 0. \end{split}\end{displaymath}    

    Still, many of you tried to go the route of solving for $ u$ directly. You get

    $\displaystyle u(x,t) = \sum_{n=0}^\infty c_n \cos\left(\frac{n\pi x}{L}\right)
\exp\left(-\left(\frac{n\pi}{L}\right)^2kt\right),
$

    where the $ c_n$ are some constants determined by initial data. (Yes, we know that they are the Fourier cosine coefficients of the initial position, and $ c_0$ is actually off by a factor of $ 2$, but none of that matters.) Then

    \begin{displaymath}
\begin{split}
E(t) = \int_0^L u(x,t)^2 \,dx
&= \int_0^L
\...
...\exp\left(-2\left(\frac{n\pi}{L}\right)^2kt\right).
\end{split}\end{displaymath}

    Therefore

    \begin{displaymath}
\begin{split}
E'(t) &= \frac{L}{2}\sum_{n=1}^\infty c_n^2
\...
...\exp\left(-2\left(\frac{n\pi}{L}\right)^2kt\right)
\end{split}\end{displaymath}

    $ \qedsymbol$

  7. (25 points) We will solve the following boundary-value problem for the heat equation:

    $\displaystyle u_{t}$ $\displaystyle = k u_{xx};$ $\displaystyle (0 < x < L,\ t > 0)$    
    $\displaystyle u_x(0,t)$ $\displaystyle = 0;$ $\displaystyle (t > 0)$    
    $\displaystyle u(L,t)$ $\displaystyle = T;$ $\displaystyle (t > 0)$    
    $\displaystyle u(x,0)$ $\displaystyle = f(x).$ $\displaystyle (0 \leq x \leq L)$    

    1. (5 points) Transform the problem into $ u(x,t) = U(x,t) + \psi(x)$, where $ U$ satisfies the heat equation with homogeneous boundary conditions:

      $\displaystyle U_{t}$ $\displaystyle = k U_{xx};$ $\displaystyle (0 < x < L,\ t > 0)$    
      $\displaystyle U_x(0,t)$ $\displaystyle = 0;$ $\displaystyle (t > 0)$    
      $\displaystyle U(L,t)$ $\displaystyle = 0.$ $\displaystyle (t > 0)$    

      What differential equation and boundary conditions must $ \psi$ satisfy? What initial conditions must $ U$ satisfy?

      Proof. [Solution]A word to the wise: read the problem and answer what it asks. Many people missed the chance for points because they rushed on to part (c) without reading the directions.

      All we have to do is apply both sets of equations to the identity

      $\displaystyle u(x,t) = U(x,t) + \psi(x).
$

      Since $ u_t = ku_{xx}$ and $ U_t = ku_{xx}$, we have

      $\displaystyle \psi''(x) = 0.
$

      Since $ u_x(0,t) = U_x(0,t) = 0$, we must have

      $\displaystyle \psi'(0) = 0.
$

      Since $ u(L,t) = T$ and $ U(L,t) = 0$, we must have

      $\displaystyle \psi(L) = T.
$

      Since $ u(x,0) = f(x)$, we must have

      $\displaystyle U(x,0) = f(x) -\psi(x).
$

      $ \qedsymbol$

    2. (5 points) Solve for $ \psi(x)$.

      Proof. [Solution]The fact that $ \psi''(x) = 0$ means that

      $\displaystyle \psi(x) = ax + b,
$

      for some constants $ a$ and $ b$. Since $ \psi'(0) = a$ and we need it to be zero, we have $ a = 0$ and thus $ \psi(x) = b$ is a constant. Since $ \psi(L) = b$ and we need it to be $ T$, we must have

      $\displaystyle \psi(x) = T.
$

      $ \qedsymbol$

    3. (10 points) Find the special solutions for $ U(x,t)$. Write down the general solution.

      Proof. [Solution]We seek special solutions of the form

      $\displaystyle U(x,t) = X(x) T(t).
$

      Then if we are to satisfy the heat equation we must have

      $\displaystyle XT'$ $\displaystyle = k X''T$    
      $\displaystyle \frac{X''}{X}$ $\displaystyle = \frac{T'}{T} =$   constant$\displaystyle = -\lambda.$    

      So we have a problem for $ X$ as well as one for $ T$. The $ X$ problem is

      $\displaystyle X'' + \lambda X$ $\displaystyle = 0;$    
      $\displaystyle X'(0)$ $\displaystyle = 0;$    
      $\displaystyle X(L)$ $\displaystyle = 0.$    

      There is a solution to the differential equation for each $ \lambda$, but that solution may or may not fit the boundary conditions. For example, if $ \lambda = 0$, then $ X(x)$ is a straight line. Since $ X'(0) = 0$, $ X$ is a constant, and since $ X(L) = 0$, that constant is zero. So the only linear solution is the trivial solution.

      If $ \lambda < 0$, then $ \lambda = -\alpha^2$ for some $ \alpha > 0$, and

      $\displaystyle X(x) = a \cosh(\alpha x) + b \sinh(\alpha x)
$

      for some constants $ a$ and $ b$. But

      $\displaystyle 0 = X'(0) = b\alpha,
$

      so $ b = 0$, and

      $\displaystyle 0 = X(L) = a \cosh(\alpha L)
$

      so $ a = 0$. Thus the only hyperbolic solution is the trivial solution.

      If $ \lambda > 0$, then $ \lambda = \alpha^2$ for some $ \alpha > 0$, and then

      $\displaystyle X(x) = a \cos(\alpha x) + b \sin(\alpha x)
$

      for some constants $ a$ and $ b$. But

      $\displaystyle 0 = X'(0) = b\alpha,
$

      so $ b = 0$, and

      $\displaystyle 0 = X(L) = a \cos(\alpha L)
$

      so we must have that $ \alpha L$ is an odd multiple of $ \frac{\pi}{2}$ (not just any multiple, because the set of multiples of $ \frac{\pi}{2}$ includes the set of multiples of $ \pi$, which we definitely do not want). Thus for each $ n$ we have a special solution

      $\displaystyle X_n(x) = \cos\left(\frac{(2n-1)\pi x}{2L}\right).
$

      The $ T$ problem is easier. Since $ T' + k\lambda T = 0$, we have

      $\displaystyle T_n(t) = \exp\left(-\left(\frac{(2n-1)\pi}{2L}\right)^2kt\right).
$

      So our special solution is

      $\displaystyle U_n(x,t) = \cos\left(\frac{(2n-1)\pi x}{2L}\right)
\exp\left(-\left(\frac{(2n-1)\pi}{2L}\right)^2kt\right).
$

      The general solution is therefore

      $\displaystyle U(x,t) = \sum_{n=1}^\infty c_n
\cos\left(\frac{(2n-1)\pi x}{2L}\right)
\exp\left(-\left(\frac{(2n-1)\pi}{2L}\right)^2kt\right).
$

      $ \qedsymbol$

    4. (5 points) How would one find the coefficients in the general solution? You may find the following fact useful: If $ n$ and $ m$ are integers, then

      $\displaystyle \int_0^L \cos\left(\frac{(2n-1)\pi x}{2L}\right)
\cos\left(\frac{...
...2L}\right)\,dx
= \begin{cases}0 & n \neq m; \\ \frac{L}{2} & n = m.\end{cases}$

      Proof. [Solution]If we take the general solution and plug in $ t = 0$, we have

      $\displaystyle f(x) - \psi(x) = U(x,0) = \sum_{n=1}^\infty c_n
\cos\left(\frac{(2n-1)\pi x}{2L}\right).
$

      Now pick a positive integer $ m$, multiply by $ \cos\left(\frac{(2m-1)\pi
x}{2L}\right)$, and integrate between 0 and $ L$. We have

      \begin{multline*}
\int_0^L \left(f(x) - \psi(x)\right)\cos\left(\frac{(2m-1)\pi ...
...)\pi x}{2L}\right)
\cos\left(\frac{(2m-1)\pi x}{2L}\right)\,dx.
\end{multline*}

      Using the fact, we have

      $\displaystyle \int_0^L \left(f(x) - \psi(x)\right)\,dx
= \frac{L}{2} c_m.
$

      Thus for any $ m$,

      $\displaystyle c_m = \frac{2}{L} \int_0^L \left(f(x) - T\right)
\cos\left(\frac{(2m-1)\pi x}{2L}\right)\,dx.
$

      $ \qedsymbol$



Matthew Leingang 2002-11-25