# Application of Method of Characteristics

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.

$$u_t - yu_x + xu_y = 0$$ $$\frac{dt}{1} = \frac{dx}{-y} = \frac{dy}{x} = \frac{du}{0} \quad \text{is OK.}$$ 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.

• Thank you. One follow up question regarding $F$: I would equate $$u(0, x, y) = F\Big( x^2 + y^2, 0 - \tan^{-1}\Big(\frac{x}{y}\Big)\Big) = x + y.$$ But how does this yield $F$? – Taufi Sep 13 '18 at 18:00
• Did you put $F\left(x^2+y^2\:,\: -\tan^{-1}\left(\frac{y}{x}\right) \right)=x+y$ into polar coordinates ? – JJacquelin Sep 14 '18 at 4:38
• We would arrive at $F(r, -\phi) = r(\cos\phi + \sin\phi)$. But how does this help? – Taufi Sep 16 '18 at 18:28
• So, the function $F$ is determined. We put it into the general solution where there is $(t-\phi)$ instead of $(-\phi)$. $$u=r\left(\cos(\phi-t)+\sin(\phi-t)\right)$$ $$u=r\left(\cos(\phi)\cos(t)+\sin(\phi)\sin(t)+\sin(\phi)\cos(t)-\cos(\phi)\sin(t)\right)$$ $$u(t,x,y)=x\cos(t)+y\sin(t)+y\cos(t)-x\sin(t)$$ – JJacquelin Sep 17 '18 at 6:07
• Thank you, JJacquelin! – Taufi Sep 17 '18 at 20:47