# 1D heat equation

see attached

Consider the following problem

ut = 4uxx 0<x<Pi, t>0

u(0,t)=a(t), u(Pi,t)=b(t) t>0

u(x,0)=f(x) 0<x<Pi

(a) Show that the solution (which exists and is unique for reasonably nice functions f,a,b) u(x,t) is of the form

U(x,t) = v(x,t)+(1-x/Pi) a(t)+x/Pi b(t)

where v solves a heat equation of the form vt = 4vxx + q(x, t) with homogeneous boundary conditions:

v(0,t) = v(π,t) = 0 for t > 0. Determine q(x,t).

. (b) Assume a(t) ≡ a0,b(t) ≡ b0 are constant. Determine the steady state solution uE. How does this
solution depend on the initial value f(x)?

. (c) Show that for large t one has u(x, t) ≈ uE (t) + C e−4t sin x, for some constant C . Determine C .

https://brainmass.com/math/fourier-analysis/1d-heat-equation-624657

#### Solution Preview

The equation is:

(1.1)

The boundary conditions are

(1.2)

And the initial condition is

(1.3)

If we set

(1.4)

We see that

(1.5)

And:

(1.6)

Plugging (1.4) into (1.1) we obtain:

(1.7)

Then we define

(1.8)

Therefore

(1.9)

Is a solution to the original equation if

(1.10)

And

(1.11)

Now the boundary functions are constants

(1.12)

For the steady state we require and the boundary conditions still hold

...

#### Solution Summary

The solution shows how to convert an equation with time-dependent boundary values to a simple non-homogeneous equation, and in a special case how to get the steady state and asymptotic behavior