Answer to Question #187340 in Differential Equations for irfan

Question #187340

Solve IBVP

∂u/∂t (x,t)=2 (∂^2 u)/(∂x^2 )+x+2,0<x<1,t>0

u(x,0)=x+2,0<x<1,u(0,t)=2,u(2,t)=-2


1
Expert's answer
2021-05-07T11:07:02-0400

Givenpde:ut=22ux2+x+2,0<x<1,t>0u(x,0)=x+2,0<x<1u(0,t)=2u(2,t)=2The given pde is non homogeneous heat equation.Given pde : \frac{\partial u}{\partial t} =2 \frac{\partial^2 u}{\partial x^2}+x+2, 0<x<1,t>0\newline u(x,0)=x+2,0<x<1\newline u(0,t)=2\newline u(2,t)=-2\newline \text{The given pde is non homogeneous heat equation.}\newlineNow, let u(x,t)=v(x,t)+Ax+BTherefore, ut=vt, 2ux2=2vx2We get, A=B=2.Then,v(x,0)=x,0<x<1v(0,t)=0v(2,t)=0Therefore, the solution is given byv(x,t)=n=1En(x)sin(nπxl)eαn2π2tl2+n=1sin(nπxl)0teαn2π2(ts)l2Fn(s)dsWhere En(x)=2l0lF(x)sin(nπxl)dx,Fn(s)=2l0tϕ(s,t)sin(nπsl)dsHere, we haveF(x)=x,ϕ(x,t)=x+2,α=2,l=2.Therefore, En(x)=02xsin(nπx2)dx=(1)n+14nπFn(s)=0(s+2)sin(nπs2)dsTherefore, solution isu(x,t)=n=1En(x)sin(nπxl)eαn2π2tl2+n=1sin(nπxl)0teαn2π2(ts)l2Fn(s)ds+2x+2where En(x)=(1)n+14nπFn(s)=0(s+2)sin(nπs2)ds\text{Now, let }u(x,t)=v(x,t)+Ax+B\newline \text{Therefore, } \frac{\partial u}{\partial t} =\frac{\partial v}{\partial t} ,\space \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x^2}\newline % +x+2, 0<x<1,t>0\newline \text{We get, A=B=2.} Then,\newline v(x,0)=x,0<x<1\newline v(0,t)=0\newline v(2,t)=0\newline \text{Therefore, the solution is given by}\newline v(x,t)=\sum_{n=1}^\infty E_{n}(x)sin(\frac{n\pi x}{l}) e^{\frac{- \alpha n ^2 \pi ^2 t }{l^2}}+\sum_{n=1}^\infty sin(\frac{n\pi x}{l}) \int_{0} ^{t} e^{\frac{- \alpha n ^2 \pi ^2 (t-s) }{l^2}}F_{n}(s)ds\newline \text{Where} \space E_{n}(x)=\frac{2}{l} \int_{0} ^{l} F(x)sin(\frac{n\pi x}{l}) dx,\newline F_{n}(s)=\frac{2}{l} \int_{0} ^{t} \phi (s,t)sin(\frac{n\pi s}{l}) ds\newline \text{Here, we have} \newline F(x)=x,\phi (x,t)=x+2, \alpha=2,l=2.\newline \text{Therefore, } E_{n}(x)=\int_{0} ^{2} xsin(\frac{n\pi x}{2}) dx\newline =\frac{(-1)^{n+1}4}{n\pi }\newline F_{n}(s)=\int_{0} ^{\infty} (s+2)sin(\frac{n\pi s}{2}) ds\newline \text{Therefore, solution is} \newline u(x,t)=\sum_{n=1}^\infty E_{n}(x)sin(\frac{n\pi x}{l}) e^{\frac{- \alpha n ^2 \pi ^2 t }{l^2}}+\sum_{n=1}^\infty sin(\frac{n\pi x}{l}) \int_{0} ^{t} e^{\frac{- \alpha n ^2 \pi ^2 (t-s) }{l^2}}F_{n}(s)ds+2x+2\newline where \space E_{n}(x)=\frac{(-1)^{n+1}4}{n\pi }\newline F_{n}(s)=\int_{0} ^{\infty} (s+2)sin(\frac{n\pi s}{2}) ds


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