#### 2.1.3 Both ends of string ﬁxed, initial velocity zero, d’Alembert 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} & =0\\ u\left ( x,0\right ) & =f\left ( x\right ) \end{align*}

solution

When using d’Alembert solution, the initial position from $$0<x<L$$ is ﬁrst odd extended to $$-L<x<L$$. The new odd extended function is called $$h\left ( x\right )$$ which now has period $$2L$$. The solution becomes$u\left ( x,t\right ) =\frac{1}{2}\left ( h\left ( x-at\right ) +h\left ( x+at\right ) \right )$ Below is speciﬁc example with animation.

Animations

Solve the above, with $$a=2$$ (speed). $f\left ( x\right ) =\begin{array} [c]{ccc}\frac{x}{10} & & 0<x<10\\ \frac{30-x}{20} & & 10<x<30 \end{array}$ Hence $$L=30$$ and period of $$h\left ( x\right )$$ is $$2L=60$$. The animation basically implements $$u\left ( x,t\right ) =\frac{1}{2}\left ( h\left ( x-at\right ) +h\left ( x+at\right ) \right )$$ for time which is 3 periods. The time to travel one period is found from $$a=\frac{2L}{t}$$. Solving for $$t$$ gives $$t=\frac{2L}{a}=\frac{60}{2}=30$$ seconds. So every $$30$$ seconds, the shape $$f\left ( x\right )$$ repeats.

This is plot of $$h\left ( x\right )$$ (the odd extension of $$f\left ( x\right )$$) from $$-L<x<L$$ at time $$t=0$$

This is plot of one $$h\left ( x\right )$$ moving left and one $$h\left ( x\right )$$ moving right at $$t=4$$ seconds (before adding them)

This is plot of average of the above two waves at $$t=4$$ seconds from $$-L<x<L$$

This is the same plot as above, but for $$0<x<L$$

This is animation for 60 seconds showing it for $$-2L<x<2L$$

This is the same animation for 60 seconds but showing it for $$0<x<L$$ since this is the actual physical range

Source code is