2.1.1 Both ends of string fixed, initial position zero, Fourier series solution

Solve, \(0<x<L\)

\[ \frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}\qquad 0<x<L,t>0 \] Boundary conditions, \(t>0\)\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} & =g\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}\\ u\left ( x,0\right ) & =0 \end{align*}

Let \(L=10,a=1\) for animation.

Solution

Since domain is finite, it is easier to use the series solution for wave equation than D’Alembert solution. The eigenvalue ODE is gives solution for \(\lambda >0\) as\[ X_{n}\left ( x\right ) =c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right ) \] Where \(\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2},n=1,2,3,\cdots \ \)The time solution is \(T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\lambda _{n}}at\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) \). At \(t=0\), this gives \(0=A_{n}\). Therefore \(T_{n}\left ( t\right ) =B_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) \). Hence the complete solution is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }c_{n}T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\sum _{n=1}^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}

To find \(c_{n}\), taking the time derivative of the above gives\[ \frac{\partial }{\partial t}u\left ( x,t\right ) =\sum _{n=1}^{\infty }c_{n}\sqrt{\lambda _{n}}\cos \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \] At \(t=0\) the above becomes\[ g\left ( x\right ) =\sum _{n=1}^{\infty }c_{n}\sqrt{\lambda _{n}}\sin \left ( \sqrt{\lambda _{n}}x\right ) \] Applying orthogonality\begin{align*} \int _{0}^{L}g\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx & =\sqrt{\lambda _{n}}c_{n}\frac{L}{2}\\ c_{n} & =\frac{2}{L\sqrt{\lambda _{n}}}\int _{0}^{L}g\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx \end{align*}

Hence since \(g\left ( x\right ) =\frac{8x\left ( L-x\right ) ^{2}}{L^{3}}\), \(L=10,a=1\) the above becomes\[ c_{n}=\frac{2}{10\left ( \frac{n\pi }{10}\right ) }\int _{0}^{10}\frac{8x\left ( 10-x\right ) ^{2}}{10^{3}}\sin \left ( \frac{n\pi }{10}x\right ) dx \] Integrating the above gives\[ c_{n}=\frac{320\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{4}\pi ^{4}}\] Therefore the solution is\begin{align*} u\left ( x,t\right ) & =\sum _{n=1}^{\infty }\frac{320\left ( 2+\left ( -1\right ) ^{n}\right ) }{n^{4}\pi ^{4}}T_{n}\left ( t\right ) X_{n}\left ( x\right ) \\ & =\frac{320}{\pi ^{4}}\sum _{n=1}^{\infty }\frac{2+\left ( -1\right ) ^{n}}{n^{4}}\sin \left ( \sqrt{\lambda _{n}}at\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \end{align*}

Animation

Animation for 60 seconds

Source code is