I am trying to solve the heat equation $$ u_t=u_{xx} $$ with $$u(0,t)=0, u(1,t)=100, u(x,0)=\sin(\pi\,x), 0<x<1$$

obviously one solution is $100\,x$ for $u(0,t)=0, u(1,t)=100$, and I am supposed to use superposition to come up with a series solution.

my initial reaction is to consider this like a neumann problem letting $$u_x(1,t)=0, u_x(0,t)=0$$ and using seperation of variables to get $$u_n(x,t)=\cos(n\,\pi\,x)e^{n^2 \pi^2 t}$$ Then utilizing super position I can rerwrite my solution as

$$ u(x,t) = a_0/2 + \sum_{n=1}^\infty a_n\cos(n\,\pi\,x)e^{n^2 \pi^2 t} +100x $$

keeping in mind that for $x=1$ the terms other then $100x$ go to $0$ as the $a_n$ and $a_0$ are picked in a way that for $u(x,0)$ we get $\sin(n \pi \,x)e^{n^2 \pi^2 t} + 100x$ which is $0+100x$. thus satisfying the initial condition I still feel as if I messed up as it does not satisfy $u(x,0)=\sin(\pi\,x)$

maybe I pick my $a$ values differently? Or maybe pick $100x$ to be $100 \alpha$ where $\alpha$ is a picecewise function that is 0 everywhere except x=1?


Why Neumann boundary conditions? Let $u=v(x,t)+100\,x$. Then $v$ satisfies the heat equation $v_t-v_{xx}=0$ with boundary and initial conditions $$ v(0,t)=0,\quad v(1,t)=0,\quad v(x,0)=\sin(\pi\,x)-100\,x. $$ The solution will be of the form $$ v(x,t)=\sum_{k=1}^\infty a_n\,e^{n^2\pi^2t}\sin(n\,\pi\,x). $$

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  • $\begingroup$ Was thinking neumann, as if I take the derivative of my boundary conditions, I can make them homogeneous which I know how to solve $\endgroup$ – yipper Aug 8 '18 at 10:12
  • $\begingroup$ You cannot take the derivative of the boundary conditions: there is no $x$ in $u(0,t)=0$. $\endgroup$ – Julián Aguirre Aug 8 '18 at 10:15
  • $\begingroup$ I see, thankyou $\endgroup$ – yipper Aug 8 '18 at 10:15
  • $\begingroup$ if you dont mind, could you explain to me why I cannot take the derivative of $u(0,t)=0$ by using separation of variables I have $X(0)T(t)=u(0,t)$ so wouldn't we assume any derivative of this would be 0? (sorry I dont have any experience taking derivatives of boundary conditions I only saw it as a potential way to get the problem into a familiar form) $\endgroup$ – yipper Aug 8 '18 at 10:36
  • $\begingroup$ $$\frac{\partial u}{\partial x}(0,t)\ne\frac{\partial}{\partial x}\bigl(u(0,t)\bigr)$$ $\endgroup$ – Julián Aguirre Aug 8 '18 at 10:39

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