Find the Equilibrium temperature distribution of a PDE

[Engineer913]

Homework Statement

1) What is the Equilibrium temperature distributions if α > 0?
2) Assume α > 0, k=1, and L=1, solve the PDE with initial condition u(x,0) = x(1-x)

Homework Equations

du/dt = k(d^2u/dx^2) - (α*u)

The Attempt at a Solution

I got u(x) = [(α*u*x)/2k]*[x-L] for Part#1 but this was told to be wrong

Part #2 I got α*u = -2, Also wrong.

Any insight how to correct this?

 

[AdkinsJr]

With the equilibrium condition you should just have a ODE.

$$\frac{d^2u}{dx^2}-\frac{a}{k}u=0$$

 

[HallsofIvy]

Homework Statement

1) What is the Equilibrium temperature distributions if α > 0?
2) Assume α > 0, k=1, and L=1, solve the PDE with initial condition u(x,0) = x(1-x)

Homework Equations

du/dt = k(d^2u/dx^2) - (α*u)

The Attempt at a Solution

I got u(x) = [(α*u*x)/2k]*[x-L] for Part#1 but this was told to be wrong

Yes, it certainly is! How did you get that function?

Part #2 I got α*u = -2, Also wrong.

Any insight how to correct this?

An "equilibrium" solution is one that no longer changes: du/dt= 0 so the partial differential equation becomes k d^2u/dx^2- au= 0, a linear, homogeneous, second order, ordinary differential equations. If you are taking a course in "partial differential equations", you certainly should know how to solve such an equation.