# Solve PDE Method of Characteristics help

1. Sep 20, 2011

### lackrange

The problem: Solve for u(x,y,z) such that

$$xu_x+2yu_y+u_z=3u\; \;\;\;\;u(x,y,0)=g(x,y)$$

So I write

$\frac{du}{ds}=3u \implies \frac{dx}{ds}=x,\; \frac{dy}{ds}=2y\;\frac{dz}{ds}=1 .$

Thus $$u=u_0e^{3s},\;\;x=x_0e^{s}\;\;y=y_0e^{2s}\;\;z=s+z_0$$
but from here I can't figure out what to do, there are several ways I can write s....I have only done the method of characteristics before with two variables, and those are pretty much the only examples I can find. Can someone help please?

2. Sep 20, 2011

### HallsofIvy

Staff Emeritus
If you have done the method of characteristiced with two variables it pretty much generalizes to three variables directly- write out the characteristics in terms of x, y, and z and use those to define new variables.

If $x= x_0e^s$ and $y= y_0e^{2s}$, you have $s= ln(x/x_0)= (1/2)ln(y/y_0)$ so that ln(x/x_0)- ln(y^{1/2}/y_0^{1/2})= ln((y_0^{1/2}/x_0)(x/y^{1/2}= 0 which then gives $x/y^{1/2}= constant$. Each curve with different constant is a characteristic. With $z= s+ z_0$ that gives also $s= z-z_0= ln(x/x_0)$, have $x/x_0= e^{z-z_0}= e^{-z_0}e^z$ so $x= (constant)e^z$ or $xe^{-z}= constant$. Putting $s= z-z_0= ln(y^{1/2}/y_0^{1/2})$, we get $y^{1/2}e^{-z}= constant$.

We want to use those "characteristics" or "characteristic curves" as axes: let $u= x/y^{1/2}$, $v= xe^{-z}$, and $w= y^{1/2}e^{z}$.