Characteristic curve

Question

Consider the equation $yu_x-xu_y=0$ for $(x,y)\in \mathbb{R} \times (0,\infty)$ with $u(x,0)=x^2$ as the initial condition.

I just need help solving for the characteristic curve. I have that $$x_t=y, y_t=x, u_t=0$$ but I am not sure how to solve for the characteristic curves when there are three independent variables to solve?

score 1 · Accepted Answer · answered Sep 24 '14 at 08:18

1

You have, by the method of characteristics:

$$ \frac{dx}{y} = \frac{dy}{-x} = \frac{du}{0}. $$

If you solve the first equality, you will find:

$$x^2+y^2 = C,$$

where $C$ is a constant. On the other hand, the last fraction tells you that $du = 0$ and therefore $u = K$, being $K$ another constant. Put $K$ as a function of $C$ to have the general solution for $u$, which is finally given by:

$$ u = F(x^2+y^2),$$

where $F$ is an arbitrary function of its argument.

Cheers!

answered Sep 24 '14 at 08:18

Dmoreno

7,517

So, what I do to get $x^2+y^2 = C,$ is the following: $$\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{0} \implies -xdx= ydy=C\implies -\frac{x^2}{2}=\frac{y^2}{y}=C\implies x^2+y^2=C?$$ Will that be correct? – Robben Sep 24 '14 at 14:29
Exactly, except for the typo $y^2/y \to y^2/2$. Cheers! – Dmoreno Sep 24 '14 at 14:33
1

You're welcome @Robben. – Dmoreno Sep 24 '14 at 14:41

Characteristic curve

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