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

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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

$2.19