PDE by Laplace Transform


by freesnow
Tags: laplace, transform
freesnow
freesnow is offline
#1
May22-09, 09:03 AM
P: 11
1. The problem statement, all variables and given/known data

Use the Laplace Transform to solve the PDE for u(x,t) with x>0 and t>0:
x(du/dx) + du/dt = xt
with IC: u(x,0) = 0 and BC: u(0,t) = 0

2. Relevant equations

3. The attempt at a solution

After taking LT of the PDE wrt t, the PDE becomes
x(dU/dx) + sU = x/(s2)

Integrating factor :
I = exp([tex]\int(s/x)dx[/tex]) = xs

ODE becomes
d/dx(Uxs) = xs/s2

Integrating both sides:
U = x/(s3+s) + A(s)/xs

then I don't know how to find A(s), if I use BC, the factor 1/0 will come out...or is there some other way to calculate the PDE with LT?

thanks
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gabbagabbahey
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#2
May22-09, 08:44 PM
HW Helper
gabbagabbahey's Avatar
P: 5,004
Quote Quote by freesnow View Post
Integrating both sides:
U = x/(s3+s) + A(s)/xs

then I don't know how to find A(s), if I use BC, the factor 1/0 will come out...
Unless A(s)=0
freesnow
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#3
May23-09, 12:39 AM
P: 11
Thanks!

freesnow
freesnow is offline
#4
May24-09, 01:00 AM
P: 11

PDE by Laplace Transform


so U(x,s) = x/(s3+s2)

but then I don't know how to do the inverse LT to get u(x,t) such that it fits the PDE...

thanks
gabbagabbahey
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#5
May24-09, 01:25 AM
HW Helper
gabbagabbahey's Avatar
P: 5,004
Quote Quote by freesnow View Post
so U(x,s) = x/(s3+s2)

but then I don't know how to do the inverse LT to get u(x,t) such that it fits the PDE...

thanks
Just use partial fractions:

[tex]\frac{1}{s^3+s}=\frac{1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs}{s^2+1}+\frac{C }{s^2+1}[/tex]

Solve for A,B and C
freesnow
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#6
May24-09, 01:47 AM
P: 11
I finally solved it! Thanks very much!


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