General Method of Characteristics Problem

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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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  • #2
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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.
 
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  • #3
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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 [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.
 
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  • #4
1,796
53
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.
 
  • #5
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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.
 
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