# Determine characteristics for the method of characteristics

I'm trying to understand the method of characteristics to solve first-order PDEs. As an example in his course, my professor solve this PDE for $u(x,y)$: $$x\frac{\partial u}{\partial x}-y\frac{\partial u}{\partial y}=R$$

with $u(s,s)=f(s)$ given along the parametrized curve $\Gamma$ defined by $x(s)=s$ and $y(s)=s$.

To determine the characteristics he first writes: $$\frac{dx}{x}=-\frac{dy}{y}$$ And then immediately jumps to: $$\int^x_s \frac{dx'}{x'}=-\int^y_x\frac{dy'}{y'}$$

I guess he simplifies a lot the resolution of the simple ODE but I don't really understand how he does that and how he finds the characteristics that way.

• – Robert Israel Aug 12 '12 at 20:53

## 1 Answer

The basic idea of the method of characteristics is to differentiate the unknown $u$ along a path $(x(t),y(t))$: You will find $$\frac{d}{dt}u(x(t),y(t))=\dot x u_x+\dot y u_y$$ where I use dots for differentiation wrt $t$, and subscripts for partial derivatives.

Now, to get some help from the given PDE, we need the vector $(\dot x,\dot y)$ to be parallel to the vector $(x,-y)$ from the coefficients of the PDE. So long as $x$ and $y$ are nonzero this requirement boils down to $$\frac{\dot x}{x}=-\frac{\dot y}{y}$$ which, if we multiply by $dt$, becomes $$\frac{dx}{x}=-\frac{dy}{y}$$ just as your professor noted. Whether you wish to think of $y$ as a function of $x$, or $x$ as a function of $y$, this is the standard form of a separable ODE, and it is solved by integrating as shown.

• Thanks for the explanation about the method but I still don't understand how to solve the separable ODE with the parameter $s$ (or $t$ as you used) as my professor did (though I have no problem to solve the equation without it). Why does he put $[s,x]$ and $[x,y]$ as domains of integration ? I guess I am mindblocking but could you just show me the steps ? – snickers Aug 12 '12 at 20:55