Solving a PDE and finding the jump condition (method of characteristics)

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SUMMARY

The discussion focuses on solving a partial differential equation (PDE) using the method of characteristics. The user successfully derived the solution U(x,y) = 0 for x<0 and U(x,y) = Uo(x-1)/(1+Uo*y) for x>0. However, they encountered difficulties in determining the jump curve, x = ξ(y), and initially miscalculated it, obtaining a natural logarithm instead of the correct expression, 1 - sqrt(1+Uo*y), as stated in their reference material. Ultimately, the user resolved their issue independently.

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Here I have my PDE:

http://desmond.imageshack.us/Himg718/scaled.php?server=718&filename=pde.png&res=medium

I have found the solution by using the method of characteristics two times, one for x<0 and the other for x>0.

I have: U(x,y) = o for x<0 and U(x,y) = Uo(x-1)/(1+Uo*y) for x>0

The trouble I am having is finding the jump curve, x = \xi(y),

Im not sure if I am doing this correctly, but I keep getting the natural log of something, while my book says the correct answer is 1 - sqrt(1+Uo*y) for the jump curve...

can anyone help me?

much appreciated
 
Last edited by a moderator:
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nevermind, i have figured out the problem

thanks
 

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