#### 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