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 \begin{center} { \bf Formulas, PDE comprehensive exam} \end{center}
 \begin{itemize}
 \item The { \bf characteristic equations} for the non--linear {\bf first order} equation $\dis F(x,y,z,p,q)=0$, $z=u$, $p=u_x$, $q=u_y$, are given by
\begin{eqnarray*}
     dx/dt = F_p \qquad dy/dt = F_q \qquad dz/dt = pF_p + qF_q \qquad dp/dt = -F_x-F_zp \qquad dq/dt = -F_y-F_zq
 \end{eqnarray*}

 \item {\bf Green's identities:}
 \begin{eqnarray*}
  \int_\Omega (g\Delta f-f\Delta g) \, dx &=& \int_{\partial\Omega} (g\partial_n f- f\partial_n g)\, dS \\
   \int_\Omega (g\Delta f+\nabla g\nabla f) \, dx &=& \int_{\partial\Omega} g\partial_n f \, dS \\
   \int_\Omega \Delta f \, dx &=& \int_{\partial\Omega} \partial_n f \, dS
 \end{eqnarray*}
where $\partial_n$ is the (outward) normal derivative.

 \item The {\bf fundamental solution of the Laplace operator} $\Delta$ in $\mathbb{R}^n$ is given by the potential 
\[ K(x) = \left\{ \begin{array}{ll}
          (2\pi)^{-1} \log \|x\| & \mbox{if $n=2$}\\
          - (4\pi \|x\|)^{-1} & \mbox{if $n=3$} 
          \end{array}
           \right. \]

  \item The {\bf Poisson integral formula} is $\dis u(\xi) = \int_{\partial \Omega} H(x,\xi) u(x) dS_{x}$, where $H(x,\xi)$ is the {\bf Poisson kernel}. The Poisson kernel in the upper half-space in $\mathbb{R}^n$ (that is, $\dis \xi_n>0$) is 
   \[ H(x',\xi) = \frac{2\xi_n}{\omega_n |x'-\xi|^n} \hspace*{1in} x'=(x_1,\ldots,x_{n-1}) \] 
The Poisson kernel for the unit ball in $\mathbb{R}^n$ is 
   \[ H(x,\xi) = \frac{1-|\xi|^2}{\omega_n |x-\xi|^n}  \hspace*{1in} \|x\|=1\] 

  \item {\bf Kirchoff's formula} gives the solution to the pure initial value problem for the three dimensional {\bf wave equation} $\dis u_{tt} = c^2 \Delta u$ with initial data $\dis u(x,0) = g(x)$, $\dis u_t(x,0) = h(x)$.
 \[ u(x,t) =  (4\pi)^{-1} \frac{\partial}{\partial t} \left(t \int_{\|\xi\| = 1} g(x+ct\xi)\, dS_{\xi} \right)
                     + (4\pi)^{-1} t \int_{\|\xi\| = 1} h(x+ct\xi)\, dS_{\xi}\]

 \item The solution to the pure initial value problem for the {\bf heat equation} $u_t=\Delta u$ with initial condition $\dis u(x,0) = g(x)$ is given by the convolution $\dis u(x,t) = \int_{\mathbb{R}^n} K(x-y,t) g(y) \, dy$ of the heat kernel $\dis K(x,t)$ with the initial data.  The heat kernel for $n=1$ is given by 
   \[ K(x,t) = (4\pi t)^{-1/2} \exp(-x^2/4t) \]
 
 \item The {\bf Fourier transform} $\dis \mathcal{F} g$ and the inverse Fourier transform $\dis \mathcal{F}^{-1} h$ are
  %\begin{eqnarray*}
  \[ \mathcal{F} g(\xi) = \int_{\mathbb{R}^n}  \exp(-i x\cdot\xi) g(x)\, dx, \qquad 
          \mathcal{F}^{-1} h (x) = (2\pi)^{-n} \int_{\mathbb{R}^n}  \exp(-i x\cdot\xi) h(\xi)\, d\xi \]
   %\end{eqnarray*}
 Fourier inversion formula: $\mathcal{F}^{-1} (\mathcal{F} g)= g$.
Basic formula: 
 $\dis  \mathcal{F}(\partial_k g)(\xi) = i\xi_k \mathcal{F}g(\xi) $.
 \end{itemize}
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