# How to solve 3-variable transport equation $u_t + a u_x + b u_y =0$ by the method of characteristics?

3-variable transport equation:

\begin{align*} u_t + a u_x + b u_y =0 \qquad \text{on} \, [-3,3]^2 \times [0,+\infty) \end{align*}

where $a = -y(1-x^2-y^2)$ and $b = x(1-x^2-y^2)$ for $x^2+y^2 < 1$, otherwise $a = 0$ and $b = 0$. Thus the solution is time-independent on the boundary.

Initial condition:

\begin{align*} u(x,y,0) = u_0(x,y) = \sqrt{(x-1)^2 + (y-1)^2} - 1 \qquad \text{on} \, [-3,3]^2 \end{align*}

To solve this PDE I think the method of characteristics should be applied. But I don't know the particular implementation for the 3-variable case. Could you please show me how to find the analytical solution of this PDE problem step by step? (This is NOT homework)

Just like in the 2-dimensional case you get $$\frac{dt}{ds}=1\\ \frac{dx}{ds}=a\\ \frac{dy}{ds}=b$$ which especially implies $\frac{d}{ds}(x^2+y^2)=0$ so that the factor $(1-x^2-y^2)$ is constant along the characteristics.
• Right, but the difficulty is on solving the nonlinear system of odes, i.e., what a set of $x$ and $y$ will satisfy $x'=-y(1-x^2-y^2)$ and $y'=x(1-x^2-y^2)$? – user460398 Aug 10 '17 at 9:18
• Not difficult, as I said, it reduces to $x'=-ay$, $y'=ax$ with the constant $a=1-x^2-y^2=1-x_0^2-y_0^2$ which has a well-known solution. – LutzL Aug 10 '17 at 9:37
• Thanks for your answer! I've got the solution $u=u_0(x\cos ct+y\sin ct,y\cos ct-x\sin ct)$ where $c=1-x^2-y^2$. Now if we change the parameters $a,b$ to be more complicated, $a=-y(1-x^2-y^2)$ and $b=x(1-x^2-y^2)$ for $t<3$ but inversely $a=y(1-x^2-y^2)$ and $b=-x(1-x^2-y^2)$ for $t\ge3$, I think the solution will not keep the same, because it follows different characteristics, and how to deal with that case (when $t\ge3$)? – user460398 Aug 10 '17 at 14:30