General Method of Characteristics Problem

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The discussion focuses on solving the quasi-linear partial differential equation u*ux + uy - u = 0 with the initial condition u(x,0) = x + 10. Participants explore the characteristic equation, compatibility equation, and the value at u(5,10). They discuss the method of characteristics, emphasizing the need for simultaneous solutions of the compatibility and characteristic equations. The conversation highlights the complexity of the problem, noting that multiple methods exist for solving such equations, and concludes with one participant arriving at a potential solution of u = -5.
ZachN
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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.
 
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You're using the method of Lagrange right? If so, you should be solving:

z\phi_x+\phi_y+z\phi_z=0

with characteristic system:

\frac{dy}{dx}=\frac{1}{z},\quad\quad \frac{dz}{dx}=1

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

\phi(x,y,z)=C(z-x,y-\log(z))

using the method of characteristics. Then the solution is:

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

u-x+y-\log(u)=0

So we got the start, the end, and now just need to fill-in.
 
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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 \frac{dy}{dx} = \frac{b}{a}. In this case a = u, b = 1, c = -u.

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

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.
 
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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 \frac{dy}{dx} = \frac{b}{a}. In this case a = u, b = 1, c = -u.

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

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: \frac{dy}{dx}=\frac{b}{a}=\frac{2x}{1}= 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.
 
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