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Find the temperature distribution $T(x,y)$ in the upper half-plane, given that the temperature along the $x$-axis is at:

$$T(x,0)=T_0, \quad x<-1$$ $$T(x,0)=T_1, \quad x>1$$

And $$\frac{\partial T}{\partial y}(x,0)=0, \quad |x|<1.$$

Assume the temperature follows $$\nabla ^2 T=0.$$

I have only dealt with Dirichlet and Neumann boundary conditions previously. I was unable to find any helpful materials on the internet. I would be grateful for any reference or hints on how to generate a solution. I know that for the Dirichlet conditions one took the Fourier transform of the equation, and thus produced a solution, and in the case of Neumann conditions one expressed the function in terms of the derivative which also satisfied Laplace's.

The only idea I can come up with is to solve for the function $T$ on the segment $|x|<1, y=0$, since we're given the derivative. So there, $T(x,y)$ is really just a function of $x$, and preferably continuous, so on the left its value will be approaching $T_0$ and on the right, $T_1$. But I don't see how to find the rest of the distribution on this segment. I bet one has to use the fact that it satisfies Laplace's equation in the region, but I'm unsure how to proceed. I can't just write that $T(x,y)=f(x)$ on the segment, and $f''(x)=0$, then solve that with $f(-1)=T_0$, $f(1)=T_1$, right?

Spine Feast
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1 Answers1

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First subtracting from $T(x,t)$ a constant $(T_1+T_0)/2$ the problem can be reduced to the case of odd initial data $T_0=-T_1$. The solution will be odd with respect to $x$ too.

Namely, putting $\tilde T(x,t)=T(x,t)-(T_0+T_1)/2\ $ we have $\tilde T(x,0)=\tilde T_1=(T_1-T_0)/2\,$ for $x>1$, $\tilde T(x,0)=-\tilde T_1\,$ for $x<-1$ and $\frac{\partial \tilde T(x,0)}{\partial y}=0\,$ for $x\in(-1,1)\,$.

Function $\frac{\partial^2 T(x,0)}{\partial x\partial y}=0$ on $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$. Hence $\psi(x)=\frac{\partial T(x,0)}{\partial y}$ is piecewise constant and odd: $$ \psi(x)=\begin{cases} -a, & x<-1, \\ 0, & -1<x<1, \\ a, & x>1. \end{cases} $$ So the problem in question reduces to the Neumann problem: $\Delta u=0$, $y>0$, and $\frac{\partial T}{\partial y}\Big|_{y=0}=\psi$. The parameter $a$ to be defined from the condition $T\big|_{y=0}=T_1$ then $x>1$.

Andrew
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  • Thank you for your answer but I do not understand the idea of subtracting the constant to make the function odd. It would be great if you could expand your answer a bit. – Spine Feast Nov 26 '13 at 21:42
  • @DepeHb I edited the answer. – Andrew Nov 27 '13 at 12:39
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    But $\frac{\partial T(x,0)}{\partial x}=0$ does not imply $\frac{\partial^2 T(x,0)}{\partial x\partial y}=0$... e.g. $T=xy$. – Siwel Feb 06 '17 at 17:29