General Solution of Quasilinear Equations

**fourierwarrior** · December 9th 2010, 12:59 AM

Hi all,

I understand everything in the following example up until the words "Now, consider..."

Could somebody help me to understand how we get the $\frac{d}{ds}(\frac{x+u}{y})$ part?

**HallsofIvy** · December 9th 2010, 02:12 AM

It's simply the product rule. You are differentiating $\frac{x+ u}{y}= (x+ u)\frac{1}{y}$ with respect to s. Presumably, x and y are functions of s but u, here, is independent of s so we can treat it as a constant. The derivative of $(x+ u)\frac{1}{y}$ is $\frac{x+ u}{dx}\frac{1}{y}+ (x+u)\frac{d\frac{1}{y}}{ds}= \frac{dx}{ds}\frac{1}{y}+ (x+ u)\left(-\frac{1}{y^2}\frac{dy}{ds}\right)$ .

**fourierwarrior** · December 9th 2010, 03:52 AM

Sorry, I should have mentioned that all 3 functions, x, y and u are functions of s here.

I understand the product rule, but I don't have a clue where the (x+u)/y sprung from.

My understanding of this problem is that we're trying to find two functions

$\Phi(x,y,u)$ = constant
$\Psi(x,y,u)$ = constant

as the solution of the equation satisfies $F(\Phi,\Psi)$ = 0

Since $\frac{d}{ds}(\frac{x+u}{y})$ = 0,

(x+u)/y is a constant, and so there's $\Phi(x,y,u)$ .

I just don't understand how we got there.

In the notes I have, there is alternative way of solving these quasilinear equations, which involves integrating the characteristic equations, then parametrising the initial line and setting s = 0. Does this sound familiar to anyone?

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