The equilibrium temperature distribution for the given partial differential equation (PDE) is derived from the equation du/dt = k(d²u/dx²) - (α*u) by setting du/dt to zero, resulting in the ordinary differential equation (ODE) k d²u/dx² - αu = 0. For the specific case where α > 0, k = 1, and L = 1, the correct approach involves solving this linear, homogeneous second-order ODE. The initial condition provided, u(x,0) = x(1-x), is not directly applicable to finding the equilibrium distribution, which requires a different method of solution.
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Yes, it certainly is! How did you get that function?Engineer913 said: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
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.Part #2 I got α*u = -2, Also wrong.
Any insight how to correct this?