General Method of Characteristics Problem

  • Context: Graduate 
  • Thread starter Thread starter ZachN
  • Start date Start date
  • Tags Tags
    General Method
Click For Summary

Discussion Overview

The discussion revolves around solving a quasi-linear hyperbolic partial differential equation using the method of characteristics. Participants are exploring the characteristic and compatibility equations, as well as the initial conditions provided. The scope includes theoretical aspects of the method and its application to specific problems.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to solve the equation u*ux + uy - u = 0 with initial conditions and expresses uncertainty about their results.
  • Another participant suggests using the method of Lagrange and provides a characteristic system, indicating a potential solution involving the Lambert W function.
  • A participant mentions learning from a specific textbook and describes a method that involves finding the characteristic from dy/dx = b/a, leading to a compatibility equation.
  • One participant raises a question about the necessity of multiple characteristic equations for quasi-linear equations and expresses unfamiliarity with the term "compatibility equation."
  • Another participant explains the need for compatibility equations in the context of governing relations for flow-fields and discusses the simultaneous solution of characteristic and compatibility equations.
  • A participant shares an example from a textbook, detailing the steps to find the characteristic and compatibility equations, and expresses uncertainty about their own results, suggesting a possible numerical method may be required.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the methods to be used or the results obtained. Multiple competing views on the approach to solving the problem remain, and uncertainty persists regarding the correct application of the method of characteristics.

Contextual Notes

Participants express limitations in their understanding of the compatibility equation and its role in the solution process. There is also mention of potential dependencies on specific methods or assumptions that may not have been fully explored.

ZachN
Messages
29
Reaction score
0
Hello!

I am trying to solve...

u*ux + uy -u = 0 with i.c. u(x,0) = x +10

Determine: a.) characteristic equation
b.) compatibility equation
c.) value at u(5,10)

I've tried the general method of determining the characteristic equation dy/dx=b/a and attempted a parametrization but I don't think I am coming-up with the correct answer. I keep getting Us or Xs in the result - I am assuming there is a clean answer but I could be wrong.
 
Last edited:
Physics news on Phys.org
You're using the method of Lagrange right? If so, you should be solving:

[tex]z\phi_x+\phi_y+z\phi_z=0[/tex]

with characteristic system:

[tex]\frac{dy}{dx}=\frac{1}{z},\quad\quad \frac{dz}{dx}=1[/tex]

Now, I haven't worked it out manually yet, but we should get:

[tex]\phi(x,y,z)=C(z-x,y-\log(z))[/tex]

using the method of characteristics. Then the solution is:

[tex]C(u-x,y-\log(u))=0[/tex]. And an interesting particular case involving the Lambert W function is:

[tex]u-x+y-\log(u)=0[/tex]

So we got the start, the end, and now just need to fill-in.
 
Last edited:
Well, I'm learning this from a section in Gas Dynamics by Zucrow. It only presents one method...

It says to find the characteristic from [tex]\frac{dy}{dx}[/tex] = [tex]\frac{b}{a}[/tex]. In this case a = u, b = 1, c = -u.

So, [tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{u}[/tex], integrating yields uy=x+c1 or y=[tex]\frac{x+c_{1}}{u}[/tex].

The compatibility equation would be adu +cdx = 0.

But maybe I am supposed to use one of the other methods as you suggested, it just didn't go over any of that. I didn't know there were multiple methods.
 
Last edited:
ZachN said:
Well, I'm learning this from a section in Gas Dynamics by Zucrow. It only presents one method...

It says to find the characteristic from [tex]\frac{dy}{dx}[/tex] = [tex]\frac{b}{a}[/tex]. In this case a = u, b = 1, c = -u.

So, [tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{u}[/tex], integrating yields uy=x+c1 or y=[tex]\frac{x+c_{1}}{u}[/tex].

The compatibility equation would be adu +cdx = 0.

But maybe I am supposed to use one of the other methods as you suggested, it just didn't go over any of that. I didn't know there were multiple methods.

I'm not familiar with that approach. Also, it's quasi-linear so seems to me, need two characteristic equations but I'm not sure. Never seen "compatibility equation" also. The method I use, which ultimately yields the equation C(r,s)=r-10 for you initial conditions, is summarized in "Basic Partial Differential Equations" by Bleecker and Csordas.
 
Yeah, you're correct, quasi-linear hyperbolic PDE.

The reason the "compatibility equation" is needed is that the governing relations for the particular flow-field are PDEs, they need to be reduced to total differential equations. The characteristics are the paths of a physical disturbance, in this case, Mach waves. The total differential equation (i.e., compatibility equation) is valid only along the particular characteristic specified by dy/dx = b/a.

It notes that the compatibility and characteristic equations must, in general, be solved simultaneously. This may be why I am getting 'u's in the results... but I wouldn't imagine that it would have given me such as a problem in the book, at least without noting that it required a numerical method.

IOW, I need to know 'u' in order to solve for 'u'.

Here is the book example...

ux + 2xuy - 3x2 = 0

i.c., u(o,y) = 5y +10

Determine: a) the equation of the characteristic passing through (2,4)
b) compatibility equation valid along that characteristic
c) the value u(2,4)

so, coefficients are a = 1, b = 2x, and c = - 3x2

Characteristic Equation: [tex]\frac{dy}{dx}[/tex]=[tex]\frac{b}{a}[/tex]=[tex]\frac{2x}{1}[/tex]= 2x

Integrating: y = x2 +C1

Characteristic through point (2,4): C1 = 0, y = x2

Comaptibility Equation is obtained from adu + cdx = 0,

substituting and integrating gives: u = x3 + C2

The value of C2 valid along the characteristic that we found above is determined from the i.c. And the i.c. requires that y at x = 0 be determined on the characteristic.

so, y = (0)2 = 0

Hence the characteristic passing through (2,4) crosses the initial-value line at (0,0).

u(0,0) = 5(0) +10 = 10 (Value of u along the characteristic at (0,0))

substituting...

10 - (0)3 = C2, C2 = 10

giving, u = x3 + 10 (Compatibility equation valid along the characteristic passing through (2,4))

so, u(2,4) = (2)3 + 10 = 18

I don't know if that helps or not.

Edit:

I just reworked everything and got u = -5, so that might be correct.
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K