# First order linear PDE

1. Sep 18, 2009

### kingwinner

Suppose we have a first order linear PDE of the form:
a(x,y) ux + b(x,y) uy = 0

Then dy/dx = b(x,y) / a(x,y) [assumption: a(x,y) is not zero]

The characteristic equation for the PDE is
b(x,y) dx - a(x,y) dy=0
d[F(x,y)]=0
"F(x,y)=constant" are characteristic curves
Therefore, the general solution to the PDE is u(x,y)=f[F(x,y)] where f is an arbitrary function.
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I don't understand the parts in red.

1) Why would dy/dx = b(x,y) / a(x,y) ? This doesn't seem obvious to me at all...how can we derive (or prove) it?

2) Also, what is the meaning of the equation d[F(x,y)]=0?

Thanks for explaining!

2. Sep 19, 2009

### LCKurtz

I'm not a PDE expert, but here's my take:

1) Why would dy/dx = b(x,y) / a(x,y) ?

Suppose you have any function f satisfying f(x,y) = C. This defines y implicitly as a function of x and you can differentiate both sides by the chain rule:

$$f_x + f_y\ \frac {dy}{dx} = 0$$

$$\frac {dy}{dx} = -\frac {f_x}{f_y}$$

Now if u(x,y) is a solution to your DE, the characteristic curves are the level curves of u, that is, the curves u(x,y) = C. By the analysis above these level curves define y as a function of x and you would have:

$$\frac{dy}{dx} = -\frac {u_x}{u_y}$$

If you solve your equation a(x,y) ux + b(x,y) uy = 0 for this expression you get:

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

This gives an ordinary differential equation for the characteristic curves. Then if F(x,y) = C is the solution to this DE giving the characteristic curves, you proceed to get your solution

u(x,y)=f[F(x,y)] where f is an arbitrary function.

3. Sep 19, 2009

### kingwinner

Thank you very much! You are of great help!

4. Sep 19, 2009

### kingwinner

What if we were to consider the more general first order linear PDE of the form
a(x,y) ux + b(x,y) uy + c(x,y) u= 0 ?

Consider the ODE dy/dx = b(x,y) / a(x,y).
Suppose that the general solution to the above ODE is the curves given by F(x,y)=c where c is an arbitrary constant.
A) In this case, does it imply that u(x,y) along these curves are constant along each curve?
B) Does it also imply that the general solution is u(x,y) = f(F(x,y)), where f is an arbitrary differentiable funciton of one variable? Why or why not?

Thanks for explaining! :)

5. Sep 20, 2009

### LCKurtz

As I said before, PDE isn't really my field of expertise. The method can be applied to such equations, even when the u term isn't linear. You can find a pretty readable exposition of this topic at:

http://www.stanford.edu/class/math220a/handouts/firstorder.pdf

Hope that helps.

6. Sep 20, 2009

### kingwinner

I browsed the above link, I can see the cases of quasilinear, and linear without the "u" term, but it doesn't seem to be dealing with the case of first order linear PDE involving a "u" term anywhere, so I can't find my answer there...

Hopefully, someone else may answer these questions, and I would really appreciate.

7. Sep 21, 2009

### LCKurtz

Look at section 2.2, the semi-linear case. It includes the case where c(x,y,u) doesn't have any x or y in it: c(x,y,u) = u.

8. Sep 21, 2009

### kingwinner

Now but that's actually non-linear, and also they aren't solving dy/dx = b(x,y) / a(x,y).
Thanks, I know you're trying to help, but unfortunately I can't find my answer there.

9. Sep 21, 2009

### LCKurtz

"What if we were to consider the more general first order linear PDE of the form
a(x,y) ux + b(x,y) uy + c(x,y) u= 0 ?"

Then I posted:
"Look at section 2.2, the semi-linear case. It includes the case where c(x,y,u) doesn't have any x or y in it: c(x,y,u) = u."

The semi-linear equation to which they refer is

a(x,y)ux + b(x,y)uy= c(x,y,u)

and if you let c(x,y,u) = -c(x,y)u you get exactly the equation you asked about, and yes, it is linear. The difference in their approach and what you are doing is they are expressing the characteristic curves in parametric form instead of implicit form. If you don't like the parametric approach, surely the book you are studying must have an appropriate explanation.