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I am studying for a midterm and confused by this problem from Strauss' Introduction to PDE's:

Solve $u_t = k u_{xx}; u(x,0) = 0; u(0,t) = 1 \;\;\; x \in (0,\infty)$

Now in the proceeding chapter, Strauss explains how to solve the diffusion equation with Dirichlet condition $u(0,t) = 0$ on the half-line, by extending the 'initial data' function $\phi$ to be odd, thereby guaranteeing that $\phi(0) = 0$, and thus that $u(0,t) = 0$.

However, now we have what I believe is called a nonhomogeneous Dirichlet boundary condition at $x=0$. I am a little uneasy about how to proceed. I can easily extend the function $u(x,0) = \phi(x) \equiv 0$ to be either even or odd, but neither will guarantee that $u(0,t) = 1$. Furthermore, because the inital data is zero except at one point, I run into difficulty trying to apply the formula for the solution on the whole line:

$u(x,t) = \int_\infty^\infty S(x-y) \phi(y) dy$

where S is the source function. Either hints or a full solution would be helpful, thanks!

mb7744
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1 Answers1

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You already know how to solve the equation with null boundary condition. Let $u=v+\phi$, where you chose $\phi$ in such a way that $v$ satisfies the same equation and $v(0,t)=0$. Then solve for $v$. There is a very simple choice for $\phi$.

Julián Aguirre
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  • Ah... I'm guessing $\phi \equiv 1$, and then I solve the DE in $v$ with $v(x,0) = -1, v(0,t) = 0$? I am currently working through it. – mb7744 Feb 12 '15 at 15:49
  • Your guess is correct. – Julián Aguirre Feb 12 '15 at 15:50
  • Thank you, I got it. Would it be most appropriate now for me to answer my own question for future readers (leaving yours as the accepted) or just leave this as is? – mb7744 Feb 12 '15 at 15:59
  • I gave a hint and not a full answer to make you think about it and find the answer yourself. I am sure that this has been more helpful than reading a full answer. Because of this, I think that it is better leave things as they are, and let future readers discover the solution by themselves as you have done. – Julián Aguirre Feb 12 '15 at 16:05