Answer to Question #283371 in Differential Equations for Rickey

Question

A string of iength L is stretched and fastened to two fix points. Find the solution of

the r.{ave equatiorl (vibrating string) ytt = a^2.yxx, when initial displacernent

y(x,0) = f (x) = b sin (pi.x / t).

also find the Fourier cosine transformation of exp(-x^2)

Answer 1

$\displaystyle \text{The given question is incomplete thus we assume a complete form below;}\\ \frac{\partial^2y}{\partial t^2}=a^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)}=b\sin(\frac{\pi x}{L}),\ \frac{\partial y(x,0)}{\partial t}=\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}=a^2\frac{\partial^2(X(x)T(t)))}{\partial x^2}\\ \Rightarrow X(x)\frac{d^2 T}{dt^2}=a^2T(t)\frac{d^2X}{dx^2}\\ \Rightarrow \frac{1}{a^2T(t)}\frac{d^2T}{dt^2}=\frac{1}{X(x)}\frac{d^2X}{dx^2}=-\lambda\\ \text{We moved the $a^2$ 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}+a^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 na}{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 at}{L})+c_2\sin(\frac{n\pi at}{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 at}{L})\sin(\frac{n\pi x}{L}), \text{and}\\ y_n(x,t)=B_n\sin(\frac{n\pi at}{L})\sin(\frac{n\pi x}{L}) \text{ for n}=1,2,3,\ldots\\ \quad\\ \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 at}{L})\sin(\frac{n\pi x}{L})+B_n\sin(\frac{n\pi at}{L})\sin(\frac{n\pi x}{L})].\\ \qquad\\ \qquad\\ \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 and since we have even functions, we get;}\\ A_n=\frac{2}{L}\int^L_0[(b\sin(\frac{\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[(b\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}\neq1, \text{and A}_2=b\\ \text{Also, } B_n=\frac{L b}{n\pi a} \text{, for n=}1,2,3,\ldots\\ \text{Hence, y(x,t)}=b\cos(\frac{2\pi ct}{L})\sin(\frac{2\pi x}{L})+\frac{Lb}{\pi a}\left\{\sum^\infty_{n=1}\left[\frac{1}{n}\sin(\frac{n\pi at}{L})\sin(\frac{n\pi x}{L})\right]\right\} \\ \text{is the solution of the given initial boundary value problem.}$

2] $\displaystyle f(x)=e^{-x^2}\\$

$\displaystyle F_c(e^{-x^2})=\int^\infty_0e^{-x^2}\cos(sx)\ dx=I\\ \Rightarrow \frac{dI}{ds}=-\int^\infty_0xe^{-x^2}\sin(sx)\ dx=\frac{1}{2}\int^\infty_0(-2xe^{-x^2})\sin(sx)\ dx\\ \text{Using integration by parts, we have}\\ \Rightarrow \frac{dI}{ds}=-\int^\infty_0xe^{-x^2}\sin(sx)\ dx=\frac{1}{2}\int^\infty_0(-2xe^{-x^2})\sin(sx)\ dx\\ \qquad\quad=\frac{1}{2}\left\{\left[\sin(sx)\left(\int-2xe^{-x^2}\ dx\right)\right]^\infty_0-s\int^\infty_0\cos(sx)\left(\int-2xe^{-x^2\ }dx\right)\ dx\right\}\\ \qquad\quad=\frac{1}{2}\left\{\left[\sin(sx)\left(e^{-x^2}\right)\right]^\infty_0-s\int^\infty_0\cos(sx)\left(e^{-x^2}\right)\ dx\right\}\\ \text{ since, }\int(-2xe^{-x^2})\ dx=-2\int(xe^{-x^2})\ dx=\int e^p\ dp=e^p=e^{-x^2}, \text{ if }p=-x^2\\ \qquad\quad=0-\frac{s}{2}\int^\infty_0e^{-x^2}\cos (sx)\ dx\\ \Rightarrow \frac{dI}{ds}=-\frac{s}{2}\int^\infty_0e^{-x^2}\cos(sx)\ dx=-\frac{s}{2}I\\ \Rightarrow \frac{dI}{I}=-\frac{s}{2}ds\\$

Integrating both sides yields;

$\displaystyle \ln (I)=\int\left(-\frac{s}{2}\right)\ ds+\ln(a), \text{ where ln(a) is an arbitrary constant.}\\ \ \qquad=-\frac{s^2}{4}+\ln(a)\\ \Rightarrow \ln(I)-\ln(a)=-\frac{s^2}{4}\\ \Rightarrow\ln\left(\frac{I}{a}\right)=-\frac{s^2}{4}\\ \Rightarrow \frac{I}{a}=e^{-\frac{s^2}{4}}\\ \Rightarrow I=ae^{-\frac{s^2}{4}}\\$

Thus,

$\displaystyle F_c(e^{-x^2})=\int^\infty_0e^{-x^2}\cos(sx)\ dx=I\\ \Rightarrow \int^\infty_0e^{-x^2}\cos(sx)\ dx=ae^{-\frac{s^2}{4}}$

and, $\displaystyle s = 0\Rightarrow a=\int^\infty_0e^{-x^2}\ dx=\frac{\sqrt{\pi}}{2}$

Hence,

$\displaystyle F_c(e^{-x^2})=\int^\infty_0e^{-x^2}\cos(sx)\ dx=I=ae^{-\frac{s^2}{4}}=\frac{\sqrt{\pi}}{2}e^{-\frac{s^2}{4}}$