# Homework Help: Help with PDE please

1. Jan 10, 2006

### robousy

Hi,
I'm working through 'Partial Differential Equations, an introduction' by Colton and am not finding it as clear as I hoped to.
I'm working through an example on how to solve a linear 1st order PDE.
I'll post Colton's example and Italic my questions:
Find the GS of

$$xu_x-yu_y+u=x$$

The characteristic equation is

$$\frac{dy}{dx}=-\frac{y}{x}$$

Integrating gives:

$$logy=-logx+c$$

or

$$xy=\gamma$$

Hence setting:

$$\zeta=x , \\ \eta=xy$$

in our first order PDE yields:

$$\frac{\partial w}{\partial \zeta}+\frac{1}{\zeta}w=1$$

Ok, my first question, why set $$\zeta=x , \eta=xy$$

I don't really see how this relates to the original equation, and also eta doesn't seem to 'do' anything, why not set it to y, or xy^2 etx...i don't follow the logic

ctd...whose solution is:

$$w(\zeta,\eta)=\frac{\zeta}{2}+\frac{1}{\zeta}d(\eta)$$

next question, this only works for $$d(\eta)=0$$ so what is the purpose of introducing it?

If anyone can help I'd be very grateful.
Rich

Last edited: Jan 10, 2006
2. Jan 10, 2006

### saltydog

Your solution contains an arbitrary function of xy. That's the d(xy) part or to avoid confusion with the d standing for derivatives, just call it K(xy) so that the solution in terms of x and y is:

$$u(x,y)=\frac{x}{2}+\frac{K(xy)}{x}$$

Try back-substituting this into the PDE and use for example the cases:

$$K(xy)=xy$$

$$K(xy)=Sin[xy]$$

$$K(xy)=xy+(xy)^2+e^{xy}$$

Should work.

3. Jan 10, 2006

### robousy

Thanks saltydog.

How about the first part to my question - the (in my 'non-lucid' mind) arbitrary assignment of eta and zeta?

4. Jan 11, 2006

### saltydog

Hey Robousy, I'm not clear on the tangent vector explanation of this so I'll explain it this way:

When we have the first-order PDE:

$$a(x,y)u_x+b(x,y)u_y+u=f$$

we try to make a change of variables:

$$w=h(x,y)\quad z=g(x,y)$$

to eliminate one of the partials so that we end up with an equation:

$$v_z+v=f(w,z)\quad\text{with}\; v(w,z)\equiv u(x,y)$$

or some variant thereof. This then can be treated as an ordinary ODE considering the variable w as a constant which we can then "partially integrate" with respect to z remembering when such partial integrations are done, the arbitrary constants become arbitrary functions of the variables kept constant. So what's:

$$\int f(x,y,z)\partial x$$

but I digress.

In order to find suitable change of variables, we substitute v(w,z) into the PDE above and obtain (via the chain rule):

$$a(v_w w_x+v_z z_x)+b(v_w w_y+v_z z_y)+\text{the rest of it}$$

or:

$$v_w(aw_x+bw_y)+(az_x+bz_y)v_z$$

Thus, in order to convert the PDE into one in which only one derivative is present, we'd like:

$$aw_x+bw_y=0$$

right?

Well, it turns out that this can be done if we let w equal to the implicit solution of the characteristic equation:

$$\frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}$$

That is, suppose the solution is in the form:

$$h(x,y)=c$$

Then we let:

$$w=h(x,y)$$

z can be any function of x and y as this will not affect the part we wish to collapse but we of course choose the simplest one.

Last edited: Jan 11, 2006