# Equilibrium solution

#### Markov

##### Member
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
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
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
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
Shouldn't you have a constant of integration?
Oh yes, yes, so then I just put the initial conditions and that's it?