Let a string with linear density $\rho$ and tension $k$. Find $u(x,t)$

Question

Let a string with linear density $\rho$ and tension $k$ Its left and right hand ends $ [-\pi,\pi]$ are held fixed at height zero (Maybe not at $t=0$, but it are for $t>0$). Initial velocity $v_0(x)\equiv 0$ and $u_0(x)=2\sin x$. Find $u(x,t)$

What I did

$\left\{\begin{array}{cccc}\dfrac{1}{c^2}u_{tt}&=&u_{xx}\hspace{0.25cm}x\in [-\pi,\pi]&\\ u_x(-\pi,t)&=&u_x(\pi,t)=0\\ u(x,0)&=&2\sin x\\ u_t(x,0)&=&0 \end{array}\right.$

$u(x,t)=\Psi(x)\theta(t)\\ u_{tt}=\Psi(x)\ddot{\theta}(t)\\ u_{xx}=\theta(t)\ddot{\Psi}(x)$

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$\Psi(x)=e^{rx}\\ \Psi'(x)=re^{rx}\\ \Psi''(x)=r^2e^{rx}$

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$r^2e^{rx}=-\lambda \Psi(x)\Rightarrow r=\sqrt{\lambda}i\Rightarrow \Psi(x)=e^{\sqrt{\lambda}i}\Rightarrow \Psi(x)=c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)\\ \Psi(x)'=-c_1\sqrt{\lambda}\sin(\sqrt{\lambda}x)+c_2\sqrt{\lambda}\cos(\sqrt{\lambda}x)\\ \Psi(\pi)'=-c_1\sqrt{\lambda}\sin(\sqrt{\lambda}\pi)+c_2\sqrt{\lambda}\cos(\sqrt{\lambda}\pi)=0$

My questions

Am I doing it right? Now I don't know how to continue. Is there any other way?

Answer 1

All solutions to the $1$D wave equation are of the form

$$u(x,t) = f(x-ct) + g(x+ct)$$

where $f,g:\Bbb{R}\to\Bbb{R}$ are at least twice differentiable. Plugging in our boundary conditions we have that

$$u(x,0) = 2\sin x = f(x) + g(x)$$

$$u_t(x,0) = 0 = -cf'(x) + cg'(x)$$

Trivially, the solution is given by $f(x) = g(x) = \sin x$. We can check that this satisfies the last boundary condition as well, meaning that this problem had too many boundary conditions in the first place. Thus the final answer is

$$u(x,t) = \sin\left(x-t\sqrt{\frac{k}{\rho}}\right)+\sin\left(x+t\sqrt{\frac{k}{\rho}}\right)$$

Uniqueness means that any solution we guess that works has to be the only solution.

Answer 2

Using the Laplace transform

$$ s^2U(s,x)+c^2U_{xx}(s,x)= s u_0(x) + v_0(x) = 2 s \sin x $$

solving for $x$ we have

$$ U(s,x) = \phi_1(s)e^{\frac{sx}{c}}+\phi_2(s)e^{-\frac{sx}{c}}+\frac{2s\sin x}{c^2+s^2} $$

with the boundary conditions $U(s,-\pi) = U(s,\pi) = 0$ we have

$$ U(s,x) = \frac{2s\sin x}{c^2+s^2} $$

and after anti-transforming

$$ u(t,x) = 2\sin x\cos(c t) $$