What is the characteristic equation for a first order linear PDE?

In summary, a first order linear PDE is a mathematical equation that involves multiple independent variables and their partial derivatives. Its general form is a(x,y)u<sub>x</sub> + b(x,y)u<sub>y</sub> = c(x,y,u), and it can be used to model various physical phenomena such as heat transfer, wave propagation, and fluid flow. To solve a first order linear PDE, one must use techniques such as separation of variables, method of characteristics, or Green's function. These equations have applications in physics, engineering, and other fields, and are also important in the development of numerical methods for solving more complex PDEs.
  • #1
kingwinner
1,270
0
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.
===========================================

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!
 
Physics news on Phys.org
  • #2
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:

[tex]f_x + f_y\ \frac {dy}{dx} = 0[/tex]

[tex] \frac {dy}{dx} = -\frac {f_x}{f_y}[/tex]

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:

[tex]\frac{dy}{dx} = -\frac {u_x}{u_y}[/tex]

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

[tex]\frac{dy}{dx} = \frac {b(x,y)}{a(x,y)}[/tex]

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
Thank you very much! You are of great help!
 
  • #4
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! :)
 
  • #6
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.

kingwinner said:
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! :)
 
  • #7
kingwinner said:
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.

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
LCKurtz said:
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.

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
Above, you asked:

"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.
 

1. What is a first order linear PDE?

A first order linear PDE (partial differential equation) is a mathematical equation that involves multiple independent variables and their partial derivatives. It is considered linear because the unknown function and its derivatives appear to the first power, and it can be written in the form of a linear combination of the unknown function and its derivatives.

2. What is the general form of a first order linear PDE?

The general form of a first order linear PDE is: a(x,y)ux + b(x,y)uy = c(x,y,u), where u is the unknown function, x and y are the independent variables, and a, b, and c are functions of x, y, and u.

3. What are some examples of first order linear PDEs?

Some common examples of first order linear PDEs include the heat equation, wave equation, and transport equation. These equations are used to model various physical phenomena such as heat transfer, wave propagation, and fluid flow.

4. How do you solve a first order linear PDE?

To solve a first order linear PDE, one must use various techniques such as separation of variables, method of characteristics, or Green's function. The specific method used depends on the type of PDE and its initial/boundary conditions.

5. What are the applications of first order linear PDEs?

First order linear PDEs have many applications in physics, engineering, and other fields. They are used to model and study a wide range of phenomena such as diffusion, vibrations, and fluid dynamics. They are also used in the development of numerical methods for solving more complex PDEs.

Similar threads

  • Differential Equations
Replies
1
Views
1K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
13
Views
2K
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
2
Views
917
  • Differential Equations
Replies
1
Views
1K
  • Differential Equations
Replies
1
Views
1K
Replies
3
Views
740
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
5
Views
608
Back
Top