$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 5.3: Hyperbolic Equation

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

As an example of a hyperbolic equation we study the wave equation. One of the systems it can describe is a transmission line for high frequency signals, 40m long.

\begin{aligned} \dfrac{\partial^2}{\partial x^2} V &= \underbrace{LC}_{\text{imp}\times \text {capac}}\dfrac{\partial^2}{\partial t^2} V \nonumber\\ \dfrac{\partial}{\partial x} V (0,t) &= \dfrac{\partial}{\partial x} V(40,t) = 0, \nonumber\\ V(x,0) &= f(x),\nonumber\\ \dfrac{\partial}{\partial t} V(x,0) &= 0,\end{aligned} Separate variables, $V(x,t) = X(x) T(t).$ We find $\frac{X''}{X} = LC \frac{T''}{T} = -\lambda .$

Which in turn shows that

\begin{aligned} X'' &=-\lambda X, \nonumber\\ T'' &= -\frac{\lambda}{LC} T .\end{aligned}

We can also separate most of the initial and boundary conditions; we find $X'(0) = X'(40)=0,\;\;T'(0)=0.$ Once again distinguish the three cases $$\lambda>0$$, $$\lambda=0$$, and $$\lambda<0$$:

 $$\lambda>0$$

(almost identical to previous problem) $$\lambda_n = \alpha_n^2$$, $$\alpha_n = \frac{n\pi}{40}$$, $$X_n=\cos(\alpha_n x)$$. We find that

$T_n(t) = D_n\cos \left(\frac{n\pi t}{40\sqrt{LC}}\right) + E_n\sin\left(\frac{n\pi t}{40\sqrt{LC}}\right).$ $$T'(0)=0$$ implies $$E_n=0$$, and taking both together we find (for $$n \geq 1$$) $V_n(x,t) = \cos\left(\frac{n\pi t}{40\sqrt{LC}}\right) \cos\left(\frac{n\pi x}{40}\right).$

 $$\lambda=0$$

$$X(x) = A + B x$$. $$B=0$$ due to the boundary conditions. We find that $$T(t) = D t + E$$, and $$D$$ is 0 due to initial condition. We conclude that $V_0(x,t) = 1.$

 $$\lambda<0$$

No solution.

Taking everything together we find that

$V(x,t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi t}{40\sqrt{LC}}\right) \cos\left(\frac{n\pi x}{40}\right).$ The one remaining initial condition gives

$V(x,0) = f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi x}{40}\right).$

Use the Fourier cosine series (even continuation of $$f$$) to find \begin{aligned} a_0 & = \frac{1}{20} \int_0^{40} f(x) dx, \nonumber\\ a_n & = & \frac{1}{20} \int_0^{40} f(x)\cos\left(\frac{n\pi x}{40}\right) dx.\end{aligned}