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Method of characteristics 
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#1
Jul2111, 01:22 PM

P: 376

1. The problem statement, all variables and given/known data
[itex] x \frac{ \partial u}{ \partial x} + y \frac{ \partial u}{ \partial y}= x^2u^2[/itex] 2. Relevant equations 3. The attempt at a solution characteristics are given by [itex] \frac{ dy}{ dx} = \frac{ y}{ x} [/itex] (a) and [itex] \frac{ du}{ dx} = \frac{x^2u^2 }{ x} [/itex] (b) So i integrate both equations but for (a) do i bring the y across which ends up giving ln(y) = ln(x) + k or leave it where it is and i get y = yln(x) + k ?? 


#2
Jul2111, 05:01 PM

HW Helper
P: 1,583

Hello, still doing characteristics? Why not write the characteristic equations as:
[tex] \dot{x}=x,\quad\dot{y}=y,\quad\dot{u}=x^{2}u^{2} [/tex] You won't be able to write down a complete solution as you have no initial condition to work from. For your equation for characteristics, your first answer was correct, the characteristics are given by [itex]y=kx[/itex] for k constant. 


#3
Jul2211, 12:51 AM

P: 72

To expand on hunt_mat's answer:
Don't we need to separate the variables before integrating? 1/y dy = 1/x dx Hmmm, then again you're using partial deriv. notation, which I've only studied Calc. for, not yet Differential Equations. 


#4
Jul2911, 08:07 AM

P: 376

Method of characteristics
thanks for the replies lads. I guess i never total got to grips with methods of chars.!!!
I also have initial conditions of u(x,1) = x for infinity<x<infinity. y=kx becomes k=y/x then from (b) we have [itex] \frac{ du}{ dx} = \frac{x^2u^2 }{ x}[/itex] which when differentiated gives [itex] u = \frac{ 2}{ x^2} + F(k) [/itex] and transfer in our earlier value of k gives [itex] u = \frac{ 2}{ x^2} + F(\frac{ y}{ x}) [/itex] im getting confused now i think in different methods? 


#5
Jul2911, 01:21 PM

HW Helper
P: 1,583

There are essentially two ways for the method of characteristics (I recognise your name from a number of MOC posts, did I post my notes on the subject?)
From one of your calculations you have: [tex] \frac{du}{dx}=xu^{2} [/tex] Integrating this equation shows that: [tex] \frac{1}{u}=\frac{x^{2}}{2}+F(\xi ) [/tex] We now paramatrise the initial data, so take [itex](\xi ,1)[/itex] as the point which the characteristic passes through, this will give the initial values as [itex]u(\xi ,1)=\xi[/itex], evaluating the characteristic at this point yields [itex]1=k\xi[/itex], giving a value for k which can then be inserted back into the equation for the characteristic. We now evaluate [itex]u[/itex] at the point [itex](\xi ,1)[/itex] to obtain [tex] \frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi ) [/tex] From this you can compute [itex]F(\xi )[/itex] and then from there you can substitute for [itex]\xi[/itex] by using the equation of the characteristic. Now to find u you have the solution 


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