Method of Characteristics for Solving Partial Differential Equations

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SUMMARY

The discussion focuses on solving the partial differential equation (PDE) ∂u/∂x + (xy)(∂u/∂y) + 2x²zLog[y](∂u/∂z) = 0 using the method of characteristics. The user attempts to parameterize the dependent variables as x=x(r), y=y(r), z=z(r) and derives a system of autonomous equations: dx/dr = 1, dy/dr = xy, dz/dr = 2x²Log[y]. The main confusion arises in transforming the dz/dr equation into the form 2x²zLog[y] = -r⁴zLog[y₀], which is referenced in a handout. The user seeks clarification on the steps to achieve this transformation.

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Homework Statement


I am trying to solve the following equation using the method of characteristics:

∂u/∂x + (xy)(∂u/∂y) + 2x2zLog[y](∂u/∂z) = 0

I'm really just trying to follow along the solution provided here:
http://www.ucl.ac.uk/~ucahhwi/LTCC/sectionA-firstorder.pdf

on page 9.

The Attempt at a Solution



I get that we have to parameterize the dependent variables so x=x(r), y=y(r), z=z(r). Then demand that du/dr=0, from which we get a system of three autonomous equations:

dx/dr = 1; dy/dr = xy; dz/dr = 2x2Log[y].

Then we have to solve these to get the form of u. My trouble is with solving the z equation. According to the handout I linked to above the right hand side of the dz/dr equation becomes:

2x2zLog[y] = -r4zLog[y0], where r=x and y=y0Exp[r^2/2] (these expressions for x and y come from solving the other two equations).

I don't understand how we can make this step. When I plug in for y and x to the left side of the above I come up with y=2zr^2Log[y0] + r^4/2 ...

Can somebody tell me what I'm doing wrong and how to get the answer given in the handout?

Thank you!
 
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Sturk200 said:
I come up with y=2zr2Log[y0] + r4/2...
I don't see how you get that. Please post your working.
 

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