# General Method of Characteristics Problem

• ZachN
In summary, the person is trying to solve a system of linear equations. They have tried using the method of Lagrange, but they keep getting Us or Xs in the result. They are using the method of characteristics to determine the equation of the characteristic passing through (2,4). They find that y = x2 + C1 and that the compatibility equation is valid along that characteristic. They find that u(2,4) = (2)3 + 10 and that u = -5.
ZachN
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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## 1. What is the general method of characteristics problem?

The general method of characteristics problem is a mathematical technique used to solve partial differential equations (PDEs). It involves finding a family of curves, known as characteristics, that satisfy the PDE and then using these curves to construct a solution.

## 2. How does the method of characteristics work?

The method of characteristics works by transforming a PDE into an ordinary differential equation (ODE) along each characteristic curve. The solution to the ODE then provides the functional relationship between the variables involved in the PDE, allowing for the construction of a solution.

## 3. What types of PDEs can be solved using the method of characteristics?

The method of characteristics is most commonly used to solve first order linear PDEs, but it can also be applied to some second order linear PDEs. Additionally, it can be used to solve nonlinear PDEs if they can be reduced to a system of first order equations.

## 4. What are the advantages of using the method of characteristics?

The method of characteristics offers several advantages, including providing an analytical solution to a PDE, allowing for the visualization of the characteristics and their relationship to the solution, and being applicable to a wide range of PDEs.

## 5. Are there any limitations to the method of characteristics?

While the method of characteristics is a powerful tool for solving PDEs, it does have some limitations. It may not be applicable to all types of PDEs, and it can be computationally intensive for complex problems. Additionally, the method may not always yield a unique solution, and it may not provide a solution for all initial or boundary conditions.

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