Register to reply

Method of characteristics

by gtfitzpatrick
Tags: characteristics, method
Share this thread:
Jul21-11, 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)


[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

Phys.Org News Partner Science news on
Apple to unveil 'iWatch' on September 9
NASA deep-space rocket, SLS, to launch in 2018
Study examines 13,000-year-old nanodiamonds from multiple locations across three continents
Jul21-11, 05:01 PM
HW Helper
P: 1,583
Hello, still doing characteristics? Why not write the characteristic equations as:
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.
Jul22-11, 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.

Jul29-11, 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?
Jul29-11, 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:
Integrating this equation shows that:
\frac{1}{u}=\frac{x^{2}}{2}+F(\xi )
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
\frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi )
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

Register to reply

Related Discussions
Method of characteristics Calculus & Beyond Homework 1
Method of characteristics Calculus & Beyond Homework 0
Method of Characteristics Calculus & Beyond Homework 0
Method of characteristics Calculus & Beyond Homework 0
Method of characteristics Differential Equations 0