I want to solve the following transport equation with the method of characteristics.
$$u_t – yu_x + xu_y = 0$$
$$u(0, x, y) = x + y$$
I read several questions here and found various sources for the method itself. Hence, I started with the Lagrange-Charpit equation
$$\frac{dt}{1} = \frac{dx}{-y} = \frac{dy}{x} = \frac{du}{0} $$
and got first that $u = const$.
Therefore, I have the ordinary differential equations
$$ -y \, dt = dx $$
$$ x \, dt = dy $$
$$ x \, dx = -y\, dy $$
From this I would conclude that
$$x(t) = -yt $$
$$ y(t) = xt $$
$$ x^2 = -y^2 $$
Now I should somehow insert these results for $x$ and $y$ into an arbitrary function $F$ which is the solution. However, there is the initial condition so I have to choose this function. But what I don't see is how to insert it? There are three results but only two variables for which I could substitute.
The problems I got with this exercise stem from the yet insufficient understanding of the method. I thought that we want to substitute the coordinates $x, y$ with functions $\Phi(s), \xi(s)$ that depend on only one new variable $s$ in order to reduce the partial differential equation to an ordinary one as we could find $\Phi(s), \xi(s)$ in a way that stay constant with respect to $s$. I don't yet see how this being constant with regard to $s$ is necessary for $\Phi(s), \xi(s)$. In sum I don't see what to insert where and why exactly.
Best Answer
$$u_t - yu_x + xu_y = 0$$ $$\frac{dt}{1} = \frac{dx}{-y} = \frac{dy}{x} = \frac{du}{0} \quad \text{is OK.}$$ A first family of characteristic equations comes from $\frac{du}{0}\neq 0$ which implies $$u=c_1$$ A second family of characteristic equations comes from $\frac{dx}{-y} = \frac{dy}{x}$ which solution is : $$x^2+y^2=c_2$$ A third family of characteristic equations comes from $\frac{dt}{1} = \frac{dx}{-y}$ thus
$\frac{dt}{1} = \frac{dx}{-\sqrt{c_2-x^2}}$ which solution is $t=\int\frac{dx}{-\sqrt{c_2-x^2}}=-\tan^{-1}\left(\frac{x}{y}\right)+$constant.
Since $\tan^{-1}\left(\frac{x}{y}\right)=\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right)$ and with a change of constant : $$t-\tan^{-1}\left(\frac{y}{x}\right)=c_3$$
The general solution of the PDE $u_t - yu_x + xu_y = 0$ expressed on the form of implicit equation is : $$\Phi\left(c_1\:,\: c_2\:,\: c_3 \right)=\Phi\left(u\:,\: x^2+y^2\:,\: t-\tan^{-1}\left(\frac{y}{x}\right) \right)=0$$ where $\Phi$ is an arbitrary function of three variables.
Or, on explicit form : $$u(t,x,y)=F\left(x^2+y^2\:,\: t-\tan^{-1}\left(\frac{y}{x}\right) \right)$$ where $F$ is an arbitrary function of two variables.
Finally, $F$ has to be determined according to the boundary condition $u(0,x,y)=x+y$.
HINT : The calculus for $F$ is easier in polar coordinates.