#### 2.1.4 Left end ﬁxed, right end free, initial velocity zero, Fourier series solution

Solve, $$0<x<L$$$\frac{\partial ^{2}u}{\partial t^{2}}=c^{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\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =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

Separation of variables gives the eigenvalue ODE\begin{align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X^{\prime }\left ( L\right ) & =0 \end{align*}

It is know that only $$\lambda >0$$ gives non-trivial solution in this case. Hence solution is$X\left ( x\right ) =A\cos \left ( \sqrt{\lambda }x\right ) +B\sin \left ( \sqrt{\lambda }x\right )$ Since $$X\left ( 0\right ) =0$$ then the above gives $$0=A$$ and the solution becomes$X\left ( x\right ) =B\sin \left ( \sqrt{\lambda }x\right )$ Taking derivatives$X^{\prime }\left ( x\right ) =\sqrt{\lambda }B\cos \left ( \sqrt{\lambda }x\right )$ Since $$X^{\prime }\left ( L\right ) =0$$ the above becomes$0=\sqrt{\lambda }B\cos \left ( \sqrt{\lambda }L\right )$ Which implies $$\sqrt{\lambda }L=\frac{n\pi }{2}$$ for $$n=1,3,5,\cdots$$ or $\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots$ Hence $X_{n}\left ( x\right ) =c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right )$ The time ODE becomes$T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\lambda _{n}}ct\right ) +B_{n}\sin \left ( \sqrt{\lambda _{n}}ct\right )$ Taking derivatives gives$T_{n}^{\prime }\left ( t\right ) =-\sqrt{\lambda _{n}}cA_{n}\sin \left ( \sqrt{\lambda _{n}}ct\right ) +\sqrt{\lambda _{n}}cB_{n}\cos \left ( \sqrt{\lambda _{n}}ct\right )$ Since initial velocity is zero, then the above becomes$0=\sqrt{\lambda _{n}}cB_{n}$ Which means $$B_{n}=0$$ and the time ODE becomes$T_{n}\left ( t\right ) =A_{n}\cos \left ( \sqrt{\lambda _{n}}ct\right )$ Hence the fundamental solution is $$u_{n}=T_{n}X_{n}$$ or by superposition\begin{align*} u\left ( x,t\right ) & =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\cos \left ( \sqrt{\lambda _{n}}ct\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) \\ \lambda _{n} & =\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \end{align*}

$$c_{n}$$ is now found from initial position. At $$t=0$$ the solution becomes$f\left ( x\right ) =\sum _{n=1,3,5,\cdots }^{\infty }c_{n}\sin \left ( \sqrt{\lambda _{n}}x\right )$ By orthogonality$c_{n}=\frac{2}{L}\int _{0}^{L}f\left ( x\right ) \sin \left ( \sqrt{\lambda _{n}}x\right ) dx$ This complete the solution.

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Solve, $$0<x<L$$$\frac{\partial ^{2}u}{\partial t^{2}}=c^{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\\ \left . \frac{\partial u}{\partial x}\right \vert _{x=L} & =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 ) =\left \{ \begin{array} [c]{ccc}\frac{3h}{L}x & & 0<x<\frac{L}{3}\\ h & & \frac{L}{3}<x<L \end{array} \right . \end{align*}

Use $$c=4,h=0.1,L=1.$$

Solution, $$c_{n}=\frac{24h}{\left ( \left ( 2n-1\right ) \pi \right ) ^{2}}\sin \left ( \frac{\left ( 2n-1\right ) }{6}\pi \right ) ,\lambda _{n}=\left ( \frac{\left ( 2n-1\right ) \pi }{2L}\right ) ^{2}$$ with $$n=1,2,3,\cdots$$$u\left ( x,t\right ) =\sum _{n=1}^{\infty }C_{n}\cos \left ( \sqrt{\lambda _{n}}ct\right ) \sin \left ( \sqrt{\lambda _{n}}x\right )$

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