\documentclass{article}
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\begin{document}


\begin{center}
Wave Equation
\end{center}

\begin{equation*}
\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}=0\qquad \text{c constant}
\end{equation*}

Note: We can also write this as a first order system. Let $u$ be the basic
unknown, e.g. the height of the string and define $\varphi =c\frac{\partial u%
}{\partial x}$ and $v=\frac{\partial u}{\partial t}$. Then%
\begin{eqnarray*}
\frac{\partial \varphi }{\partial t} &=&c\frac{\partial \psi }{\partial x}%
\qquad \text{by definition} \\
\frac{\partial \psi }{\partial t} &=&c\frac{\partial \varphi }{\partial x}%
\qquad \text{by wave equation}
\end{eqnarray*}

or%
\begin{equation*}
\frac{\partial }{\partial t}%
\begin{pmatrix}
\varphi \\ 
\psi%
\end{pmatrix}%
=%
\begin{pmatrix}
0 & c \\ 
c & 0%
\end{pmatrix}%
\frac{\partial }{\partial x}%
\begin{pmatrix}
\varphi \\ 
\psi%
\end{pmatrix}%
\end{equation*}

\textit{Method I:}

\bigskip

\begin{align*}
\left( \frac{\partial }{\partial t}-c\frac{\partial }{\partial x}\right) & 
\underbrace{\left( \frac{\partial u}{\partial t}+c\frac{\partial u}{\partial
x}\right) }=0 \\
& \qquad \quad v
\end{align*}

Then 
\begin{equation*}
\frac{\partial v}{\partial t}-c\frac{\partial v}{\partial x}
\end{equation*}

so \ \ \ $v=h(x+ct)$ and%
\begin{equation*}
\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x}=v=h(x+ct)
\end{equation*}

The general solution is a particular solution by the general solution to the
homogenous equation. To find a particular soltion let us guess \ \ \ $%
u=f(x+ct)$. Then $\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x%
}=cf^{\prime }+cf^{\prime }=h$. So $f^{\prime }=\frac{h}{2c}$.

\noindent The general solution to the homogenous equation is $u=g(x-ct)$ .

\noindent So the general solution of the wave equations is%
\begin{equation*}
u(x,t)=f(x+ct)+g(x-ct)
\end{equation*}

\noindent i.e. a wave going to the right plus a wave moving to the left both
at speed $c$ .

\bigskip

\textit{Method II: }change of variables

\begin{align*}
\xi &= x+ct\qquad \eta =x-ct \\
0 &= u_{tt}-c^{2}u_{xx}=-4c^{2}u_{\xi \eta }
\end{align*}

So 
\begin{equation*}
u(x,t)=f(\xi )+g(\eta )=f(x+ct)+g(x-ct)
\end{equation*}

\noindent For now we have two arbitraty functions in $C^{2}$.

\noindent \textbf{Note}: If there are discontinuouities in the initial data
they travel along the characteristics $x+ct=$ constant \quad and \quad $%
x-ct= $ constant . Across the characteristics the solution is smooth.

\bigskip

\begin{center}
Initial data - D'Alembart
\end{center}

\bigskip

Consider the following initial value problem%
\begin{align*}
\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}& =0\qquad \qquad -\infty <x<\infty \\
u(x,0)& =\varphi (x)\qquad \frac{\partial u}{\partial t}(x,0)=\psi (x)
\end{align*}

\noindent From the above we know

\begin{align*}
u(x,t) &= f(x+ct)+g(x-ct) \\
\varphi (x) &= u(x,0)=f(x)+g(x) \\
\frac{\partial u}{\partial t}(x,t) &= cf^{\prime}(x+ct)-cg^{\prime}(x-ct) \\
\psi (x) &= \frac{\partial u}{\partial t}(x,0)=c\left[ f^{\prime}
(x)-g^{\prime}(x)\right]
\end{align*}

\noindent Solving

\begin{align*}
f^{\prime}(x) &= \frac{\varphi^{\prime}(x) +\frac{1}{c}\psi (x)}{2} \\
g^{\prime}(x) &= \frac{\varphi^{\prime}(x)- \frac{1}{c}\psi^{\prime}(x)}{2}
\end{align*}

\noindent Integrating we get%
\begin{align*}
f(s) &= \frac{\varphi (s)}{2}+\frac{1}{2c}\int^{s}\psi (z)dz+A \\
g(s) &= \frac{\varphi (s)}{2}-\frac{1}{2c}\int^{s}\psi (z)dz+B
\end{align*}

\noindent D'Alembart's formula

\begin{equation*}
u(x,t)=\frac{\varphi (x+ct)+\varphi (x-ct)}{2}+\frac{1}{2c}%
\int_{x-ct}^{x+ct}\psi (z)dz
\end{equation*}

\newpage

\textbf{Properties}:

\begin{itemize}
\item the solution exists and is unique

\item no part of the wave goes faster than the speed $c$. In particular it
takes a finite time for propagation of signals (causality).

\begin{itemize}
\item In three dimensions signals move only at speed $c$ (Huygen's principle)

\item In two dimensions the entire domain between the characteristics is
filled (flatland). So echoes stay forever !!
\end{itemize}

\item Domain of influence: Initial conditions within a region can affect the
solution only for points/time within the characteristics

\item Domain of Dependence: The solution at point/time (x,t) is influence
only by initial data at (x-ct, x+ct)

\item Cauchy data on noncharacteristic surface determines the solution

\item The solution depends continuously on the initial and boundary data.

So the equation is well posed

\item Discontinuities in second derivative propagate along characteristics

\item D'Alembert's solution makes sense even if the initial data are such
that $f$ and $g$ are only piecewise differentiable. Hence, we define a
generalized solution as a limit of "classical solutions". So we have
solutions that don't have two continuous derivatives everywhere.
\end{itemize}

\medskip

\noindent Discuss vibrating string and drum

\medskip

\begin{align*}
\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}+r\frac{\partial u}{\partial t}& =0\qquad \text{resistance} \\
\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}+ku& =0\qquad \text{coiled spring}
\end{align*}

\newpage

\begin{center}
Energy
\end{center}

\noindent We start with the wave equation and multiply by $u$ and integrate.
Let $c^{2}=\frac{T}{\rho }$

\noindent Define: \ \ \ \ \ \ \ \ Kinetic energy (KE) = $\frac{1}{2}\int
\rho \left( \frac{\partial u}{\partial t}\right) ^{2}dx$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ Potential energy \ (PE)\ \ \ = $\frac{1}{2}\int
T\left( \frac{\partial u}{\partial x}\right)^2 dx$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ Total energy T = KE + PE = $\frac{1}{2}%
\int_{-\infty }^{\infty }\left[ \rho \left( \frac{\partial u}{\partial t}%
\right) ^{2}+T\left( \frac{\partial u}{\partial x}\right) ^{2}\right] dx$

\bigskip

\noindent Then 
\begin{equation*}
\frac{d}{dt}(KE)=\int \rho \frac{\partial u}{\partial t}\frac{\partial ^{2}u%
}{\partial t^{2}}dx\overset{\text{wave eq}}{=}\int T\frac{\partial u}{%
\partial t}\frac{\partial ^{2}u}{\partial x^{2}}dx\overset{\text{integrate
by parts}}{=}-\int T\frac{\partial ^{2}u}{\partial x\partial t}\frac{%
\partial u}{\partial x}dx
\end{equation*}

\begin{equation*}
\frac{d}{dt}(PE)=\int T\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{%
\partial x\partial t}dx
\end{equation*}

\noindent So we have conservation of energy

\begin{equation*}
\frac{dE}{dt}=0
\end{equation*}

\newpage

\begin{center}
Telegraph Equation
\end{center}

\bigskip 
\begin{equation*}
Lu=\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}+2\beta \frac{\partial u}{\partial t}=0\qquad \beta ,c\text{ constants}
\end{equation*}

\bigskip

\begin{align*}
\frac{d}{dt}(KE)& =\int \rho \frac{\partial u}{\partial t}\frac{\partial
^{2}u}{\partial t^{2}}dx\overset{\text{telegraph eq}}{=}\int T\frac{\partial
u}{\partial t}\left( \frac{\partial ^{2}u}{\partial x^{2}}-2\frac{\beta }{T}%
\frac{\partial u}{\partial t}\right) dx\overset{\text{integrate by parts}}{=}
\\
& -\int T\frac{\partial ^{2}u}{\partial x\partial t}\frac{\partial u}{%
\partial x}dx-2\int \beta \left( \frac{\partial u}{\partial t}\right) ^{2}dx
\end{align*}

\begin{equation*}
\frac{d}{dt}(PE)=\int T\frac{\partial u}{\partial x}\frac{\partial ^{2}u}{%
\partial x\partial t}dx
\end{equation*}

\noindent

So we have loss of energy

\begin{equation*}
\frac{dE}{dt}=-2\int \beta \left( \frac{\partial u}{\partial t}\right)
^{2}dx<0
\end{equation*}

\bigskip 

Define a new variable by $v(x,t)=e^{\beta t}u(x,t)$ or $u(x,t)=e^{-\beta
t}v(x,t)$ . Then

\bigskip

\begin{eqnarray*}
\frac{\partial u}{\partial t} &=&e^{-\beta t}(\frac{\partial v}{\partial t}%
-\beta v) \\
\frac{\partial ^{2}u}{\partial t^{2}} &=&e^{-\beta t}(\frac{\partial ^{2}v}{%
\partial t^{2}}-2\beta \frac{\partial v}{\partial t}+\beta ^{2}v)
\end{eqnarray*}

\begin{center}
\bigskip 

So the telegraph equation in terms of $v$ becomes%
\begin{eqnarray*}
e^{-\beta t}\left[ \frac{\partial ^{2}v}{\partial t^{2}}-2\beta \frac{%
\partial v}{\partial t}+\beta ^{2}v-c^{2}\frac{\partial ^{2}v}{\partial x^{2}%
}+2\beta \left( \frac{\partial v}{\partial t}-\beta v\right) \right]  &=&0 \\
\frac{\partial ^{2}v}{\partial t^{2}}-c^{2}\frac{\partial ^{2}v}{\partial
x^{2}}-\beta ^{2}v &=&0
\end{eqnarray*}

Klein-Gordon Equation
\end{center}

\bigskip 
\begin{equation*}
\frac{\partial ^{2}u}{\partial t^{2}}-c^{2}\frac{\partial ^{2}u}{\partial
x^{2}}+m^{2}u=0\qquad 
\end{equation*}

Similar to equation for $v$ but opposite sign of lower order term

\bigskip

\end{document}
