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I am trying to solve the following question in my text book. I've done simple versions of this question, but I fail to get right result using the same approach:

Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:


$$\frac{\partial u }{\partial x}(0,t) =u(0,t)-T$$ $$\frac{\partial u}{\partial x}(L,t)=\alpha$$

My approach: 1) Since $Q=0$, that means my answer should have the structure: $C_1(x)+C_2=u(x)$ 2) Now this is where I get messed up, instead of giving boundaries like $u(0)=0$ and $u(L)=T$, here they give me these differential equations. So what do I attempt? I try to get u(x) back by anti-differential. But this really gives me a weird result with which I don't think I can do anything...Any ideas?

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You haven't included the differential equation they gave you. In case you're referring to the boundary conditions on the partial derivative: That's not a differential equation, it's a single value that you can substitute into the general solution to determine the free parameters, just like you would with a boundary condition on $u$ itself. – joriki Feb 22 '13 at 16:02
They do not include it, this is all they give: u(x, 0) = f (x). However, for easier examples with boundary conditions 0 to L, I just plugged them in and was able to find answer, but now I have these partials, which makes me stuck. :/ – RealityDysfunction Feb 22 '13 at 16:05
I was referring to "here they give me these differential equations" (which was in the singular when I commented). Unless they're giving them without including them, we must have misunderstood each other. – joriki Feb 22 '13 at 16:08
Yeah, I misunderstood it seems, usually I try to get u(0) and u(L), but from these partials how do I get it? – RealityDysfunction Feb 22 '13 at 16:12
As I wrote, you don't need to, you can substitute the boundary conditions on the derivative into the general solution just like you would with the boundary conditions on the function itself. Just differentiate the general solution, substitute, and solve for the free parameters; it's almost exactly the same process, the only difference is that you have to differentiate the general solution first. – joriki Feb 22 '13 at 16:55
up vote 1 down vote accepted

Assuming the context here is the standard heat equation, and $Q=0$ represents the source term, then the associated steady-state problem is given by \begin{align} v''(x)&=0, \quad 0<x<L,\\ v'(0)&=v(0)-T,\\ v'(L)&=\alpha. \end{align}

Thus, $v(x)=Ax+B$. The first boundary condition implies $A=B-T$ while the second implies $A=\alpha$. Hence, $v(x)=\alpha x+\alpha+T$.

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I will attempt using your suggestions. Btw the back of book answer is: u = T + a(x + 1). – RealityDysfunction Feb 22 '13 at 20:43
Oops, when I typed the steady-state problem problem above, I left off the $T$ in the first BC and solved accordingly. I have edited my solution to include that $T$ in the first BC and then the corresponding $v$ solution. – JohnD Feb 22 '13 at 20:49
Great! However the way I am learning it from the book is different, and I am not sure how "The first boundary condition implies A=B−T..." – RealityDysfunction Feb 22 '13 at 21:03
$v(x)=Ax+B\implies v'(x)=A\implies v'(0)=A$ while $v(0)-T=B-T$. Thus, $v'(0)=v(0)-T\implies A=B-T$. – JohnD Feb 22 '13 at 21:05
I understand now, Thank you! – RealityDysfunction Feb 22 '13 at 22:00

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