Answer to Question #179503 in Differential Equations for ariel

Question #179503

Given the partial differential equation π‘ˆπ‘‘ + π‘ˆπ‘₯π‘₯ = 0, 0 < π‘₯ < πœ‹,𝑑 > 0. Use the method of separation of variables to solve the given equation with the following conditions. π‘ˆπ‘₯ (0,𝑑) = 0,𝑑 > 0

π‘ˆπ‘₯ (πœ‹,𝑑) = 0,𝑑 > 0 π‘ˆ(π‘₯, 0) = π‘₯(πœ‹ βˆ’ π‘₯), 0 < π‘₯ < πœ‹


1
Expert's answer
2021-04-15T07:27:47-0400

Given, "U_t+U_{xx}=0,0<x<\\pi,t>0"


So, "\\dfrac{du}{dt}=-\\dfrac{d^2u}{dx^2}"


Taking fourier sine transform of Above equation-

"\\dfrac{d}{dt}(usinsxdx)=-\\dfrac{d^2}{dx^2}(sinsxdx)"


"\\dfrac{du_s}{dt}=-(-s^2u(0,t)-su_s) \\text{, Where } u_s= \\int_0^{\\infty}usinstdt"


"\\dfrac{du_s}{dt}=s^2(0)+su_s"


"\\dfrac{du_s}{u_s}=-sdt"


Integrating and we get,


"logu_s=-st+logc"

"log\\dfrac{u_s}{c}=-st"


"u_s=ce^{-st}~~~~~~-(1)"



As "u_s= \\int_0^{\\pi}u(x,0)sinsxdx"



"=\\int_0^{\\pi}x(\\pi-x)sinsxdx=1"



From 1 we have, c=1


So "u_s=e^{-st}"


Now taking inverse fourier transform and we get-


"u(x,t)=\\int_0^{\\pi}e^{-st}sinstds"



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