Deflection of String

Question

I am trying to determine u(x,t) for a string of length L=1 and c^2=1 when the initial velocity is 0 and initial deflection with small k(.01) is as follows:

https://www.desmos.com/calculator/i1xdrpoepa

Answer 1

The displacement $u(x,t)$ of the string is described by the wave equation for $t \ge 0$ and $0 \le x \le 1$: $$ u_{tt} = c^{2}u_{xx},\\ u(x,0)=f(x),\;\;u_{t}(x,0)=0,\\ u(0,t)=u(1,t)=0. $$ The constant $c > 0$ is a physical constant associated with the properties of the string.

You start by looking for separated solutions $T(t)X(x)$ with separation parameter $\lambda$: $$ \frac{T''(t)}{c^{2}T(t)} =\lambda = \frac{X''(x)}{X(x)},\\ T'(0) =0,\;\;\; X(0)=X(1)=0. $$ The parameter $\lambda$ is determined by the $X$ equation because of $X(0)=X(1)=0$. You can't get non-trivial combinations of the hyperbolic functions to vanish at two endpoints, and $\lambda=0$ doesn't work either. So $\lambda < 0$ is required, and you can just about spot the solutions in $x$: $$ X_{n}(x) = \sin(n\pi x),\;\;\; n=1,2,3,\cdots. $$ That means $\lambda = -n^{2}\pi^{2}$ and the corresponding $T$ solutions are $$ T_{n}(t) = \cos(nc\pi t). $$ (Any $\sin(nc\pi t)$ terms are eliminated by $T_{n}'(0)=0$.)

So the proposed solution is $$ u(x,t)=\sum_{n=1}^{\infty}A_{n}\cos(nc\pi t)\sin(n\pi x) $$ The final condition to be matched is $u(x,0)=f(x)$, where $$ f(x) = \left\{ \begin{array}{ll} x, & 0 \le x \le 1/4 \\ 1/4, & 1/4 \le x \le 3/4 \\ 1-x, & 3/4 \le x \le 1 \end{array}\right. . $$ This gives you a Fourier series problem: $$ f(x) = u(x,0) = \sum_{n=1}^{\infty}A_{n}\sin(n\pi x) $$ The $\sin(n\pi x)$ are mutually orthogonal. So, if you multiply both sides by $\sin(m\pi x)$ and integrate over $[0,1]$, all of the integrals involving $\sin(n\pi x)\sin(m\pi x)$ vanish except for $n=m$. Thus, $$ \int_{0}^{1}f(x)\sin(m\pi x)dx = A_{m}\int_{0}^{1}\sin^{2}(m\pi x)dx = \frac{1}{2}A_{m} $$ Once you have the constants $A_{m}$, you're done: $$ \begin{align} A_{m} & = 2\int_{0}^{1}f(x)\sin(m\pi x)dx \\ & = 2\int_{0}^{1/4}x\sin(m\pi x)dx \\ & + 2\int_{1/4}^{3/4}\sin(m\pi x)dx \\ & + 2\int_{3/4}^{1}(1-x)\sin(m\pi x)dx \end{align} $$