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Equilibrium solution

Markov

Member
Feb 1, 2012
149
Well I don't know if is the correct term for this for here goes:

Let

$\begin{align}
& {{u}_{t}}=K{{u}_{xx}}+\gamma ,\text{ }0<x<L,\text{ }t>0, \\
& u(0,t)=\alpha ,\text{ }u(L,t)=\beta ,\text{ }t>0, \\
& u(x,0)=0,
\end{align}
$

where $\alpha,\beta,\gamma$ are constant, then find the equilibrium solution. I don't know what I need to do.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Well I don't know if is the correct term for this for here goes:

Let

$\begin{align}
& {{u}_{t}}=K{{u}_{xx}}+\gamma ,\text{ }0<x<L,\text{ }t>0, \\
& u(0,t)=\alpha ,\text{ }u(L,t)=\beta ,\text{ }t>0, \\
& u(x,0)=0,
\end{align}
$

where $\alpha,\beta,\gamma$ are constant, then find the equilibrium solution. I don't know what I need to do.
To find the equilibrium solution $U(x)$, set

$0=Ku_{xx}+\gamma$
$U(0)=\alpha$
$U(L)=\beta$
 

Markov

Member
Feb 1, 2012
149
So I get $u(x)=-\dfrac\gamma{2K} x^2,$ does this make sense? Why the boundary conditions aren't something like $u(0,t)$ ?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Why the boundary conditions aren't something like $u(0,t)$ ?
We removed the t parameter.

---------- Post added at 02:45 PM ---------- Previous post was at 02:43 PM ----------

So I get $u(x)=-\dfrac\gamma{2K} x^2,$ does this make sense?
Shouldn't you have a constant of integration?
 

Markov

Member
Feb 1, 2012
149