# Math Help - Understanding method of characteristics for first order equations

1. ## Understanding method of characteristics for first order equations

Hi Folks,
I am new to the method of characteristics for solving 1st order diff eqns. I am looking at an example in the book, which i do not quite follow..here goes

$x^2u_x+yu_y+xyu=1$
which is of the form $a(x,y)u_x+b(x,y)u_y+c(xy)u=f(x,y)$

The characteristic equation is $\frac{dy}{dx}=\frac{b}{a}=\frac{y}{x^2}$
which becomes $ln(y) + 1/x = k = \eta = \varphi(x,y)$

The bit I dont understand is making the transformation with

$\xi = x\text{and}\eta = ln(y) + 1/x$ to the partial differential eqn as above with $\omega(\xi, \eta)=u(x,y)$

$u_x = \omega_\xi\xi+\omega_\eta\eta_x = \omega_\xi+\omega_\eta(\frac{-1}{x^2})$

How is this $u_x$ calculated?

2. Using the chain rule.

$\frac{\partial u}{\partial x}= \frac{\partial u}{\partial\xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial\nu}\frac{\partial\nu}{\partial x}$

3. ## Understanding method of characteristics for first order equations

Originally Posted by HallsofIvy
Using the chain rule.

$\frac{\partial u}{\partial x}= \frac{\partial u}{\partial\xi}\frac{\partial \xi}{\partial x}+ \frac{\partial u}{\partial\nu}\frac{\partial\nu}{\partial x}$
But what is u equal to? All i know its a function of x and y etc, i dont know how omega comes into it?

Thanks

4. ## Understanding method of characteristics for first order equations

To me it appears that
$\frac{\partial u}{\partial\xi} = \omega_\xi$ (not sure why)and
$\frac{\partial \xi}{\partial x} =1$

$\frac{\partial u}{\partial\eta} = \omega_\eta$ (not sure why)and
$\frac{\partial \eta}{\partial x} =\frac{-1}{x^2}$

Thanks

5. Let me see if I can help here.

You have the PDE

$x^2u_x + y u_y = 1 - x y u$

If you introduce a change of coordinates $r = r(x,y), s = s(x,y)$ (I'm using these instead of $\xi$ and $\eta$)
then using the chain rule stated by HallsofIvy

$u_x = u_r r_x + u_s s_x$
$u_y = u_r r_y + u_s s_y$

then your PDE, after regrouping terms becomes

$\left(x^2r_x + y r_y\right) u_r + \left(x^2s_x + y s_y\right) u_s = 1 - x y u$.

The idea is to choose $r$ and $s$ such that one of the two terms vanish. The method of characteristics gives you this. If you choose

$r = x \;(\text{really anything})\; \text{and}\; s = \ln y + \frac{1}{x}$

Then substituting these into your original PDE gives an ODE (since there's no $u_s$, the $s$ is treated as constant).

6. Thank you Danny and Hallsofivy,

That makes it clearer.

Regards