I'm interested in solving the following PDE via the method of characteristics:

$$\frac{\partial f}{\partial t} - ax\frac{\partial f}{\partial p}+ bp \frac{\partial f}{\partial x} = 0,$$

with $f=f(x,p;t)$ and $a,b>0$ constants, given an arbitratry initial condition


and the relations

$$x=x(t), \quad p=p(t)$$ $$x(0)=x_0 \quad p(0)=p_0$$

I don't really know the method and the examples I found didn't really deal with anything similar...would appreciate any pointers on how to proceed here. I should get


with $x_0(x,p;t)$ and $p_0(x,p;t)$ being sort of inverted from the equations for $x(t)$ and $p(t)$. But I have no idea how to derive this result.

  • $\begingroup$ Let the projected characteristics be parametrized by $s$, then, the eqns are $\dot{t}=1$, $t(0)=0$; $\dot{p}=-ax(s),$ $p(0)=p_0$; $\dot{x}=bp(s)$, $p(0)=p_0$. You get $t(s)=s$, and then you see that the projection on $(p,x)$ plane is some kind of ellipse. Expressing $x_0$ and $p_0$ is kind of messy, you get to divide by $\cos(\sqrt{ab}t)$, but other than that it is straightforward. $\endgroup$ – uvs Oct 29 '14 at 9:43

Let us fix $x_0, p_0$.

The method is to find a curve $$ (x,p,t) = (x(t), p(t), t) := \gamma(t) $$ such as $$ \gamma(0) = (x_0,p_0,0) \\ f\circ\gamma \text{ does not depend on }t $$

The last condition is equivalent (under smoothness assumptions) to $$ 0 = \frac {d(f\circ\gamma)}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial p}\frac{\partial p}{\partial t} + \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} $$ To identify with the PDE, you just have to put $$ \frac{\partial p}{\partial t} = -ax(t); \ \ \frac{\partial x}{\partial t} = bp(t) $$

Now if you can find a solution $(x_0(t), p_0(t))$ of the previous equation with the initial conditions given and if $x(t_1) = x_1, p(t_1) = p_1$, then you must have $$ f(x_1,p_1,t_1) = f\circ\gamma (t_1) = f\circ\gamma(0) = f(x_0(t),p_0(t),0) = f_0(x_0, p_0) $$

  • $\begingroup$ I would like to see a derivation of this result, if possible. I'm not at all sure where it came from. $\endgroup$ – Spine Feast Nov 7 '14 at 15:05
  • $\begingroup$ @DepeHb what don't you get in my message? I don't see what require more precision actually. $\endgroup$ – mookid Nov 7 '14 at 16:26
  • $\begingroup$ The "if..then" part $\endgroup$ – Spine Feast Nov 7 '14 at 18:01
  • 1
    $\begingroup$ this is because $f\circ \gamma $ is a constant, in such a case. $\endgroup$ – mookid Nov 7 '14 at 18:03

first fix the notations: $ab=\lambda^2>0­­$ $$by\frac{\partial f}{\partial x}-ax\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}=0$$ $f$ is a first integral of the vector field $X=by\,\partial_{x}-ax\,\partial_{y}+\partial_{z}$, i.e. $\mathscr{L}_X(f)=0$ or equivalently $f$ is constant on integral curves of $X$.

The integral curves of $X$ are solutions of the differential equation $\gamma'(t)=X(\gamma(t))$: $$x'=by,\ y'=-ax,\ z'=1$$ $x=x_0\cos(\lambda t)+\sqrt{\frac{b}{a}}y_0\sin(\lambda t)$, $y=-\sqrt{\frac{a}{b}}x_0\sin(\lambda t)+y_0\cos(\lambda t)$, $z=t+z_0.$

We can verify that $f_1=ax^2+by^2$ is constant (a first integral). if you write $\frac{x}{y}$ as a fraction of $\tan(\lambda t)$ you can write a relation between $\frac{x}{y}$ and $z$ and deduce a second integral function $f_2$.

The solution of the pde can be written $f(x,y,z)=F(ax^2+by^2,f_2(x,y,z))$ where $F : \mathbb{R}^3\to\mathbb{R}$ is any smooth function.


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