#### 2.1.6 Both ends ﬁxed. Initial velocity zero. Dispersion term present.

Solve, $$0<x<L$$$\frac{1}{a^{2}}\frac{\partial ^{2}u}{\partial t^{2}}+\gamma ^{2}u=\frac{\partial ^{2}u}{\partial x^{2}}\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

Dispersion term $$\gamma ^{2}u\,$$ causes the shape 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*} \frac{1}{a^{2}}T^{\prime \prime }X+\gamma ^{2}XT & =X^{\prime \prime }T\\ \frac{1}{a^{2}}\frac{T^{\prime \prime }}{T}+\gamma ^{2} & =\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$T^{\prime \prime }+a^{2}\left ( \gamma ^{2}+\lambda _{n}\right ) T=0$ The solution is$T_{n}\left ( t\right ) =A_{n}\cos \left ( a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }t\right ) +B_{n}\sin \left ( a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }t\right )$ Taking time derivatives gives$T_{n}^{\prime }\left ( t\right ) =-a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }A_{n}\sin \left ( a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }t\right ) +B_{n}a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }\cos \left ( a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }t\right )$ At time $$t=0$$, the above is zero (initial velocity is zero), which gives$0=B_{n}a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }$ Hence $$B_{n}=0$$ and the time ODE solution becomes$T_{n}\left ( t\right ) =A_{n}\cos \left ( a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }t\right )$ Hence the fundamental solution is\begin{align*} u_{n}\left ( x,t\right ) & =T_{n}X_{n}\\ & =c_{n}\cos \left ( a\sqrt{\left ( \gamma ^{2}+\lambda _{n}\right ) }t\right ) \sin \left ( \frac{n\pi }{L}x\right ) \end{align*}

Where constant $$A_{n}$$ merged with $$c_{n}$$. Therefore the solution is$u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\cos \left ( at\sqrt{\left ( \gamma ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\right ) }\right ) \sin \left ( \frac{n\pi }{L}x\right )$ $$c_{n}$$ 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*}

Hence solution is$u\left ( x,t\right ) =\sum _{n=1}^{\infty }\left ( \frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac{n\pi }{L}x\right ) dx\right ) \cos \left ( at\sqrt{\left ( \gamma ^{2}+\left ( \frac{n\pi }{L}\right ) ^{2}\right ) }\right ) \sin \left ( \frac{n\pi }{L}x\right )$

Animation

Let $$L=10$$ cm and do animation for $$\gamma =1/8$$ and$$1/4$$. Let $$a^{2}=1$$, and let initial position $f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}x-4 & & 4\leq x\leq 5\\ 6-x & & 5\leq x\leq 6\\ 0 & & \text{otherwise}\end{array} \right .$ Here is animation for $$\gamma =0.125$$

Here is animation for $$\gamma =0.6$$ (more distortion)

Source code