1
$\begingroup$

I wish to solve the following PDE:

\begin{equation} H(v) \frac{\partial P(u,v)}{\partial v} = c H(u) \frac{\partial P(u,v)}{\partial u} \end{equation}

where $c$ is some constant in $(- \infty, 0)$ and $P(u, v)$ is defined on $[0, 1]^2$, satisfying the boundary conditions:

\begin{align} P(0, v) &= 0 \\ P(u, 0) &= 0 \\ P(u, 1) &= u \\ P(1, v) &= v \end{align}

I have already ruled out the following solutions:

  1. $P(u, v) = A(u) B(v)$. This admits a solution to the PDE but fails to satisfy the boundary constraints.

  2. $P(u, v) = A(B(u, v))$ where $B$ is symmetric, i.e., $B(u, v) = B(v, u)$.

Background:

If $F$ is the cumulative distribution function of some random variable $X$ and $H$ is defined by

\begin{equation} H(x) = F^{-1}(x) f(F^{-1}(x)) \end{equation}

where $f$ is the density of $X$ (i.e. $\frac{dF(x)}{dx}$ if it exists) and $F^{-1}$ is the inverse function of the cumulative distribution function $F$, then the provided PDE is satisfied by the copulas of bivariate random variables where both marginals are distributed according to $F$ and the correlation between the two is zero.

$\endgroup$
  • $\begingroup$ Use \partial for the partial derivative notation $\partial$. Also, is this not just a linear, first order PDE which can be solved using the method of characteristics? You may not be able to find the characteristics explicitly unless you know the form of $H$, but I think you can solve this. $\endgroup$ – mattos Sep 18 '16 at 13:55
  • $\begingroup$ Thanks for the edit and pointing out characteristics. Can't believe I forgot about those; I'll give it a try and report back. $\endgroup$ – R.G. Sep 18 '16 at 14:09
  • $\begingroup$ @Mattos It seems that for given functions $H$ characteristics seem to admit solutions (at least in very easy cases). Thanks again for the heads-up! $\endgroup$ – R.G. Sep 19 '16 at 12:36
  • $\begingroup$ All good mate, hope it helps. $\endgroup$ – mattos Sep 19 '16 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.