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.