Differential equation Steady state solution Hi I am looking for the steady state solution (stationnary solution independent of t when t grow to infinity) of the following equation:
$$ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$$
Initial condition: $u(x,0) = 0 $ and $u(0,t) = 0 $  and $u(1,t) = 1 $ 
I tried to separate the function $u$ like this $u(x,t) = f(x)g(t)$ to find a basis of solutions.
$$ \frac{g'}{g}(t) = \frac{f''}{f}(x)$$
gives me that both ratio are equal to the same constant $\lambda$ for all $x$ and $t$
$g(t)=ae^{\lambda t}$ and $f(x)=be^{x \sqrt \lambda } + ce^{-x \sqrt \lambda }$
$u(x,t)=ae^{\lambda t}(be^{x \sqrt \lambda } + ce^{-x \sqrt \lambda })$
$u(0,t) = 0 $ gives $b = -c$
From that I can not find any solution but the null function.
Could you help me find the steady state solution? Thank you
 A: You are correct, solutions are of the form
$$
\frac{g'(x)}{g(x)} = \frac{f''(x)}{f(x)} = \lambda
$$
However, the value of $\lambda$ could take different signs.

$\lambda = \kappa^2 > 0$, 

In this case
$$
g(x) = a e^{\kappa^2 t} ~~~\mbox{and}~~~ f(x) = b e^{\kappa t} + c e^{-\kappa t}
$$
so that
$$
u(x, t) = ae^{\kappa^2 t}(b e^{\kappa t} + c e^{-\kappa t})
$$
This solution will explode in the limit $t\to \infty$, unless $b=c = 0$

$\lambda = -\kappa^2 < 0$

I will leave this case for you, but the idea is the same, this solution goes to zero in the long term

$\lambda = 0$

For this case
$$
g(x) = a ~~~~\mbox{and}~~~ f(x) = bx + c
$$
The solution then has the form
$$
u(x, t) = Ax + B
$$
Now set the conditions
$$
u(0, t) = 0 = B ~~~\mbox{and}~~~ u(1, t) = 1 = A
$$
So the stationary solution is
$$
u(x, t) = x
$$

EDIT
There's a simpler solution to this: in the stationary case the solution does not depend on $t$, so we have the problem
$$
\frac{\partial^2 }{\partial x^2}u_{\rm stationary}(x) = 0
$$
whose solution is just
$$
u_{\rm stationary}(x) = A x + B
$$
with the boundary conditions we get
$$
u_{\rm stationary}(x) = x
$$
