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PDE by Laplace Transform

  • Thread starter freesnow
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  • #1
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Homework Statement



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

Homework Equations



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
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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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 :wink:
 
  • #3
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Thanks!
 
  • #4
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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
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
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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
 
  • #6
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I finally solved it! Thanks very much!
 

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