Answer to Question #288789 in Differential Equations for Ashok Kumar

Question

Solve the problem of the vibrating string for the following boundary conditions

1. y(0,t)= 0

2. y(l,t)= 0

3. dy/dt(x,0)= v0 sin nπx/l

4. y(x,0)= y0 sin 2πx/l

Answer 1

$\text{The question we are to solve is;}\\ \frac{\partial^2y}{\partial t^2}=c^2\frac{\partial^2y}{\partial x^2}\\ y(0,\text{t})=0, y(L,\text{t})=0,0\leq t \leq \text{L}\\ \text{y(x,0)}=y_0\sin(\frac{2\pi x}{L}), \frac{\partial y(x,0)}{\partial t}=y_0\sin(\frac{n\pi x}{L}), 0\leq x \leq L\\ \text{Let's start off with the product solution;}\\\text{y(x,t)}=\text{X(x)T(t)}\\ \text{Plugging this into the two boundary conditions gives,}\\ \text{X(0)}=0,\text{X(L)=0}\\ \text{Plugging the product solution into the differential equation, separating and}\\ \text{introducing a separation constant gives,}\\ \frac{\partial^2(X(x)T(t))}{\partial t^2}=c^2\frac{\partial^2(X(x)T(t)))}{\partial x^2}\\ \Rightarrow X(x)\frac{d^2 T}{dt^2}=c^2T(t)\frac{d^2X}{dx^2}\\ \Rightarrow \frac{1}{c^2T(t)}\frac{d^2T}{dt^2}=\frac{1}{X(x)}\frac{d^2X}{dx^2}=-\lambda\\ \text{We moved the $c^2$ to the left side for convenience and chose $-\lambda$ for the separation}\\ \text{constant so the differential equation for X would match a known (and solved) case. }\\ \text{The two ordinary differential equations we get from separation of variables are then,}\\ \frac{d^2T}{dt^2}+c^2\lambda T=0 \text{ and } \frac{d^2X}{dx^2}+\lambda X=0 \text{ with the boundary conditions}\\ \text{X(0)}=0 \text{ and } \text{X(L)}=0\\ \text{Now, the eigenvalues and eigenfunctions for this problem are,}\\ \lambda_n=(\frac{n\pi}{L})^2 \text{ and }X_n(x)=\sin(\frac{n\pi x}{L}) \text{ respectively. For n} =1,2,3,\ldots.\\ \text{The first ordinary differential equation is now,}\\ \frac{d^2T}{dt^2}+(\frac{\pi nc}{L})^2 T=0\\ \text{and because the coefficient of the T is clearly positive the solution to this is,}\\ T(t)=c_1\cos(\frac{n\pi ct}{L})+c_1\sin(\frac{n\pi ct}{L})\\ \text{Because there is no reason to think that either of the oefficients above are zero we}\\ \text{ then get two product solutions,}\\ y_n(x,t)=A_n\cos(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L}), \text{and}\\ y_n(x,t)=B_n\sin(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L}) \text{ for n}=1,2,3,\ldots\\ \text{Then the general solution due to the principle of superposition is,}\\ y(x,t)=\sum^\infty_{n=1}[A_n\cos(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L})+B_n\sin(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L})].\\ \text{Now in order to apply the second initial condition we will need to differentiate the}\\ \text{general solution with respect to t. So,} \\ \frac{\partial y}{\partial t}=\sum^\infty_{n=1}[-\frac{n \pi c}{L}A_n\sin(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L})+B_n\cos(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L})].\\ \text{Applying the initial conditions yields,}\\ y(x,0)=y_0\sin(\frac{2\pi x}{L})=\sum^\infty_{n=1}[A_n\sin(\frac{n \pi x}{L})], \text{and}\\ \frac{\partial y(x,0)}{\partial t}=\sum^\infty_{n=1}[\frac{n\pi c}{L}B_n\sin(\frac{n\pi x}{L})].\\ \text{Now, using the formulas from Fourier sine series, we get;}\\ A_n=\frac{2}{L}\int^L_0[(y_0\sin(\frac{2\pi x}{L}))(\sin(\frac{n\pi x}{L}))] \ dx, \text{for n}=1,2,3,\ldots\\ B_n=\frac{2}{n\pi c}\int^L_0[(y_0\sin(\frac{n\pi x}{L}))(\sin(\frac{n\pi x}{L}))] \ dx, \text{for n}=1,2,3,\ldots\\ \text{Thus, upon integrating we have} \\ A_n=0 \text{ if n}\neq2, \text{and A}_2=y_0\\ \text{Also, } B_n=\frac{L y_0}{n\pi c} \text{, for n=}1,2,3,\ldots\\ \text{Hence, y(x,t)}=y_o\cos(\frac{2\pi ct}{L})\sin(\frac{2\pi x}{L})+\frac{Ly_o}{\pi c}\left\{\sum^\infty_{n=1}\left[\frac{1}{n}\sin(\frac{n\pi ct}{L})\sin(\frac{n\pi x}{L})\right]\right\} \\ \text{is the solution of the given initial boundary value problem.}$