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Consider the following PDE problem for $u = u(t, x), \ 0 \leq x \leq 4, \ t > 0$

\begin{align} u_{t} &= u_{xx} \\ u_{x}(t,0) &= u_{x}(t,4) \\ &= -2 \\ u(0,x) &= \begin{cases} 0 & 0 \leq x \leq 2 \\ 2x-4 & 2 \leq x \leq 4 \end{cases} \end{align}

I am trying to find the steady state solution to this problem. This means solving $u_{xx}=0$ which has solution $u(x)=c_1 + c_2x$ for arbitrary constants $c_1$ and $c_2$. Now we appeal to the boundary conditions which quickly gives us that $c_2=-2$. Now my question is how do we find the value of the constant $c_1$?

The question comes with a hint to choose the constant so that the total heat energy as $t \rightarrow \infty$ is the same as when $t=0$. This hint just confuses me more, because $u(x)$ doesn't depend on $t$ so I'm not sure what to do with this hint.

Any help is appreciated, thanks!

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2 Answers 2

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Heat is conserved for $u$ because it is entering on the left and exiting on the right at the same constant rate of $2$. The initial heat is $\int_{0}^{4}u(x,0)dx=x^2-4x|_{2}^{4}=4$. And the equilibrium solution would have to be $C-2x$ where $\int_{0}^{4}(C-2x)dx=4$ which means $4C-16=4$ or $C=5$.

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I think the hint means you should consider $$\frac{d}{dt}\int_0^4 u(t,x)dx=\int_0^4 u_t(t,x)dx = \int_0^4 u_{xx}(t,x)dx=\dots =0.$$ Hence $\int_0^4 u(0,x)dx=\int_0^4 u(x)dx$.

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