2.1.7 Both ends fixed. Initial velocity zero. Telegraphy PDE (dispersion and decay)

Solve, \(0<x<L\)\[ u_{tt}+2hu_{t}=c^{2}u_{xx}\qquad 0<x<L,t>0 \] Boundary conditions, \(t>0\) (both ends fixed)\begin{align*} u\left ( 0,t\right ) & =0\\ u\left ( L,t\right ) & =0 \end{align*}

Initial conditions, \(t=0\)\begin{align*} \frac{\partial u\left ( x,0\right ) }{\partial t} & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

Solution

\(hu_{t}\,\) causes dispersion and decay (damping) of the original wave to distort with time. Using separation of variables, Let \(u=X\left ( x\right ) T\left ( t\right ) .\) Substituting this back in the PDE gives\begin{align*} T^{\prime \prime }X+2hT^{\prime }X & =c^{2}X^{\prime \prime }T\\ \frac{T^{\prime \prime }}{c^{2}T}+\frac{2h}{c^{2}}\frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}

The eigenvalue ODE is\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end{align*}

The eigenvalues are \(\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \) and the eigenfunctions are \(X_{n}\left ( x\right ) =c_{n}\sin \left ( \frac{n\pi }{L}x\right ) \). The time ODE becomes\[ T^{\prime \prime }+2hT^{\prime }+c^{2}\lambda T=0 \] The solution is (with the condition that \(\sqrt{\lambda }c>h\))\[ T_{n}\left ( t\right ) =e^{-ht}\left ( A_{n}\cos \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) \right ) \] Taking time derivatives gives\begin{align*} T_{n}^{\prime }\left ( t\right ) & =-he^{-ht}\left ( A_{n}\cos \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) \right ) \\ & +e^{-ht}\left ( -\sqrt{\lambda _{n}c^{2}-h^{2}}A_{n}\sin \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) +\sqrt{\lambda _{n}c^{2}-h^{2}}B_{n}\cos \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) \right ) \end{align*}

At time \(t=0\), the above is zero (initial velocity is zero), which gives\begin{align*} 0 & =-hA_{n}+\sqrt{\lambda _{n}c^{2}-h^{2}}B_{n}\\ B_{n} & =\frac{h}{\sqrt{\lambda _{n}c^{2}-h^{2}}}A_{n} \end{align*}

And the time ODE solution becomes\[ T_{n}\left ( t\right ) =A_{n}e^{-ht}\left ( \cos \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) +\frac{h}{\sqrt{\lambda _{n}c^{2}-h^{2}}}\sin \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) \right ) \] Hence the fundamental solution is\[ u_{n}\left ( x,t\right ) =T_{n}X_{n}\] Therefore the solution is\[ u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}e^{-ht}\left ( \cos \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) +\frac{h}{\sqrt{\lambda _{n}c^{2}-h^{2}}}\sin \left ( \sqrt{\lambda _{n}c^{2}-h^{2}}t\right ) \right ) \sin \left ( \frac{n\pi }{L}x\right ) \] Where constant \(A_{n}\) merged with \(c_{n}\). \(\ c_{n}\) is found from initial position. At \(t=0\) the above becomes\[ f\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sin \left ( \frac{n\pi }{L}x\right ) \] Applying orthogonality gives\begin{align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx & =c_{n}\frac{L}{2}\\ c_{n} & =\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx \end{align*}

Animation

Let \(L=\pi \), \(c=1\) and let initial position \[ f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}0 & & 0\leq x\leq \frac{L}{4}\\ 4\left ( \frac{x}{L}-\frac{1}{4}\right ) & & \frac{L}{4}\leq x\leq \frac{L}{2}\\ 4\left ( \frac{3}{4}-\frac{x}{L}\right ) & & \frac{L}{2}\leq x\leq \frac{3L}{4}\\ 0 & & \frac{3L}{4}\leq x\leq L \end{array} \right . \] Here is animation for \(h=0.5\). This is large damping. Animation is run for 8 seconds.

Here is animation for \(h=0.1\) for 12 seconds

Source code