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1D heat equation

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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 . 


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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

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
...

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