# Base Characteristics

1. Apr 8, 2009

### Winzer

1. The problem statement, all variables and given/known data
I am trying to plot base characteristics using mathematica.
The characteristic equations are:

2. Relevant equations
$$\frac{dx}{d\tau}=x u(x,y)$$
$$\frac{dy}{d\tau}=y u(x,y)$$
$$\frac{du}{d\tau}=x^2+y^2$$
u(x,1)=Sqrt(x^2+1) Initial Data

3. The attempt at a solution
How do I solve this with Mathematica. I need to get x(tau,xi),y(tau,xi),u(tau,xi) so I can plot the base characteristics. By the way xi is the other parameter.

2. Apr 8, 2009

### Winzer

i tried for x
Code (Text):

ode1 = D[x[\[Tau]], \[Tau]] == x[\[Tau]] u[x[\[Tau]], y[\[Tau]]]
ode2 = D[y[\[Tau]], \[Tau]] == y[\[Tau]] u[x[\[Tau]], y[\[Tau]]]
ode3 = D[u[\[Tau]], \[Tau]] == x[\[Tau]]^2 + y[\[Tau]]^2
DSolve[{ode1, ode2, ode3, x[0] == \[Xi], y[0] == 1,
u[0] == Sqrt[\[Xi]^2 + 1]}, x[\[Tau]], \[Tau]]

Note: \[Tau]=$$\tau$$ & \[Xi]=$$\xi$$
Its when I copy pasted the code it turned up like this.

3. Apr 8, 2009

### Winzer

Any ideas? I think it's how I express the function in mathematica. Should I show x as a function of tau explicitly everywhere?

4. Apr 8, 2009

### Winzer

What about if i have the intial conditions:
$$x(\xi,0)=\xi$$
$$y(\xi,0)=1$$
$$u(\xi,0)=\sqrt(\xi^2+1)$$

5. Apr 17, 2009

### Winzer

What about uncoupling the equations?

6. Apr 17, 2009

### skyfire

$$\hat{}\downarrow\downarrow$$$$\scshape$$

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