Solving PDE Linear 1st Order: Help with Colton's Example

[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 , \\<br /> <br /> \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

 

[saltydog]

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

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.

 

[robousy]

Thanks saltydog.

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

 

[saltydog]

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

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.