I am given the PDE, $\ u_t=u_{xx},$ with boundary conditions $u(0,t)=A, \ u(1,t)=B$ and $u(x,0)=f(x)$. I have found the solution of this PDE is $$u(x,t)=A+(B-A)x+\sum_{n=1}^{\infty}B_ne^{n^2\pi^2 t}\sin(n\pi x),$$ where $$B_n=2\int_{0}^{1}\left(f(x)-(B-A)x-A\right)\sin(n\pi x) \ dx.$$ I found this by letting $u(x,t)=U(x)+v(x,t)$. For my solution, $v(x,t)$ solves the PDE in the case where the boundary conditions are homogeneous, e.g. $v(0,t)=0, \ u(1,t)=0$. My question is, why isn't $v(x,0)=0$ necessary to solve for $v(x,t)$? If this were the case, then $$T'-n^2\pi^2T=0\implies T_n(t)=C_1e^{n^2\pi^2 t}\implies C_1=0.$$

  • $\begingroup$ I'm not sure what you're asking? $v(x,0) = 0$ if $f(x) = A + (B-A)x$. If $v(x,0)=0$ then $v(x,y)=0$ $\endgroup$ – Dylan Mar 18 at 14:23
  • $\begingroup$ I think I was trying to treat my initial condition as a boundary condition. If I consider $u(x,0)=f(x)$ after I have constructed a general solution, this works out fine. $\endgroup$ – Betty Mar 18 at 23:15

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