# Reparametrizing characteristic curves for PDE's

I'm looking for solutions for a PDE that looks like this $$\nabla u(\vec x) \cdot f(\vec x) = k.$$ For some clarification, $u$ is a log-probability. And this arises from a Fokker-Plank-like equation, and I'd like to find the marginals $u(x_1)$ for example. The method of characteristics tells me that if i take characteristic curves parametrized by $$\frac{d \vec x}{dt} = f(\vec x),$$ the variation of $u(\vec x)$ along these curves will be $$\frac{du}{dt}= k.$$ But in my particular case, $f$ determines a dynamic system $\dot{\vec x} = f$ which converges to a fixed point only in infinite time. Since I'd like to be able to have $u(x)$ instead of $u(t;x_0)$ I've come to the idea of parametrizing the characteristic curves with constant velocity, that is, take $$\frac{d \vec x}{ds} = c \frac{f(\vec x)}{\|f(\vec x)\|},$$ which will result in a new parametrization s which goes from 0 to the length of the characteristic curve from the boundary condition to the equilibrium point. This will lead to an equation for $u(s)$ $$\frac{du}{ds}= \frac{ck}{\|f(\vec x)\|}.$$ Clearly this only corresponds to the original equation for a correct choice of the constant $c$, yet it seems to me that the PDE is satisfied for any value of $c$. I'm a little confused. Am I doing something terribly stupid along the way, or am I just missing the place where I should constrain the constant $c$.

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## 1 Answer

I suppose you mean du/ds in the last equation. There is one problem with reparametrizing the characteristic curves: the vector field f is zero at the fixed point, so you are dividing by ||f(x)|| which tends to zero (assuming suitable smoothness of f). So either you consider only a region that doesn't include the fixed point, or you have to be careful with continuity to ensure that all constructions exist.

Apart from that you may reparametrize the characteristic curves as you see fit. Notice that this doesn't solve the original 1st order partial differential equation, because to achieve that you still need to compute all solutions to the dynamical system determined by f.

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Thanks for pointing out the typo! But why doesn't the second set of equations solve the pde? If i have $u(s)$ given by the solution to the last equation, and just take $u(t) = u(t(s))$ where $t(s)$ is the change of variables, then $u(t(s))$ will solve the first equation as far as I could think it through. – alexsuse Feb 8 '12 at 10:14
To obtain the solution u(x) from the integral u(s), you need to know x(s,x_0). To solve the PDE completely, you need to know all the integral curves of f starting from points in the region you want to consider. – Pait Feb 8 '12 at 13:05