Answer to Question #283371 in Differential Equations for Rickey

Question #283371

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)


1
Expert's answer
2022-02-04T06:23:47-0500

"\\displaystyle\n\\text{The given question is incomplete thus we assume a complete form below;}\\\\\n\\frac{\\partial^2y}{\\partial t^2}=a^2\\frac{\\partial^2y}{\\partial x^2}\\\\\ny(0,\\text{t})=0,\ny(L,\\text{t})=0,0\\leq t \\leq \\text{L}\\\\\n\\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\\\\\n\\text{Let's start off with the product solution;}\\\\\\text{y(x,t)}=\\text{X(x)T(t)}\\\\\n\\text{Plugging this into the two boundary conditions gives,}\\\\\n\\text{X(0)}=0,\\text{X(L)=0}\\\\\n\\text{Plugging the product solution into the differential equation, separating and}\\\\\n\\text{introducing a separation constant gives,}\\\\\n\\frac{\\partial^2(X(x)T(t))}{\\partial t^2}=a^2\\frac{\\partial^2(X(x)T(t)))}{\\partial x^2}\\\\\n\\Rightarrow X(x)\\frac{d^2 T}{dt^2}=a^2T(t)\\frac{d^2X}{dx^2}\\\\\n\\Rightarrow \\frac{1}{a^2T(t)}\\frac{d^2T}{dt^2}=\\frac{1}{X(x)}\\frac{d^2X}{dx^2}=-\\lambda\\\\\n\\text{We moved the $a^2$ to the left side for convenience and chose $-\\lambda$ for the separation}\\\\\n\\text{constant so the differential equation for X would match a known (and solved) case. }\\\\\n\\text{The two ordinary differential equations we get from separation of variables are then,}\\\\\n\\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}\\\\\n\\text{X(0)}=0 \\text{ and } \\text{X(L)}=0\\\\\n\\text{Now, the eigenvalues and eigenfunctions for this problem are,}\\\\\n\\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.\\\\\n\\text{The first ordinary differential equation is now,}\\\\\n\\frac{d^2T}{dt^2}+(\\frac{\\pi na}{L})^2 T=0\\\\\n\\text{and because the coefficient of the T is clearly positive the solution to this is,}\\\\\nT(t)=c_1\\cos(\\frac{n\\pi at}{L})+c_2\\sin(\\frac{n\\pi at}{L})\\\\\n\\text{Because there is no reason to think that either of the oefficients above are zero we}\\\\\n\\text{ then get two product solutions,}\\\\\ny_n(x,t)=A_n\\cos(\\frac{n\\pi at}{L})\\sin(\\frac{n\\pi x}{L}), \\text{and}\\\\\ny_n(x,t)=B_n\\sin(\\frac{n\\pi at}{L})\\sin(\\frac{n\\pi x}{L}) \\text{ for n}=1,2,3,\\ldots\\\\\n\\quad\\\\\n\\text{Then the general solution due to the principle of superposition is,}\\\\\ny(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})].\\\\\n\\qquad\\\\\n\\qquad\\\\\n\\text{Now in order to apply the second initial condition we will need to differentiate the}\\\\\n\\text{general solution with respect to t. So,} \\\\\n\\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})].\\\\\n\\text{Applying the initial conditions yields,}\\\\\ny(x,0)=y_0\\sin(\\frac{2\\pi x}{L})=\\sum^\\infty_{n=1}[A_n\\sin(\\frac{n \\pi x}{L})], \\text{and}\\\\\n\\frac{\\partial y(x,0)}{\\partial t}=\\sum^\\infty_{n=1}[\\frac{n\\pi c}{L}B_n\\sin(\\frac{n\\pi x}{L})].\\\\\n\\text{Now, using the formulas from Fourier sine series and since we have even functions, we get;}\\\\\nA_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\\\\\nB_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\\\\\n\\text{Thus, upon integrating we have} \\\\\nA_n=0 \\text{ if n}\\neq1, \\text{and A}_2=b\\\\\n\\text{Also, } B_n=\\frac{L b}{n\\pi a} \\text{, for n=}1,2,3,\\ldots\\\\\n\\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\\} \\\\\n\\text{is the solution of the given initial boundary value problem.}"


2] "\\displaystyle\nf(x)=e^{-x^2}\\\\"

"\\displaystyle\nF_c(e^{-x^2})=\\int^\\infty_0e^{-x^2}\\cos(sx)\\ dx=I\\\\\n\\Rightarrow \\frac{dI}{ds}=-\\int^\\infty_0xe^{-x^2}\\sin(sx)\\ dx=\\frac{1}{2}\\int^\\infty_0(-2xe^{-x^2})\\sin(sx)\\ dx\\\\\n\\text{Using integration by parts, we have}\\\\\n\\Rightarrow \\frac{dI}{ds}=-\\int^\\infty_0xe^{-x^2}\\sin(sx)\\ dx=\\frac{1}{2}\\int^\\infty_0(-2xe^{-x^2})\\sin(sx)\\ dx\\\\\n\\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\\}\\\\\n\\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\\}\\\\\n\\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\\\\\n\\qquad\\quad=0-\\frac{s}{2}\\int^\\infty_0e^{-x^2}\\cos (sx)\\ dx\\\\\n\n\\Rightarrow \\frac{dI}{ds}=-\\frac{s}{2}\\int^\\infty_0e^{-x^2}\\cos(sx)\\ dx=-\\frac{s}{2}I\\\\\n\\Rightarrow \\frac{dI}{I}=-\\frac{s}{2}ds\\\\"

Integrating both sides yields;

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


Thus,

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

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


Hence,

"\\displaystyle\nF_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}}"

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