I know that the characteristic function $f(t, u) = \mathbb{E}(e^{iuX_t})$ of some random variable $X_t$ depending on $t \geq 0$ has to solve: $$f_t(t, u) = \left(iu - \frac{u^2}{2}\right) f(t, u) + u f_u(t, u) + \frac{u^2}{2}f_{uu}(t, u),$$ with the following boundary conditions: $f(0, u) = e^{iuy}$, $f(t, 0) = 1$ and $|f(t, u)| \leq 1$, for all $u \in \mathbb{R}$, $t \geq 0$.

I would like to obtain an explicit solution of this equation, if possible. It is not possible to use the separation of variables since the boundary conditions are not homogenous. An other idea is to put $f(t, u) = \exp(g(t, u))$, to obtain: $$g_t(t, u) = \left(iu - \frac{u^2}{2}\right) + u g_u(t, u) + \frac{u^2}{2}\left(g_u(t, u)^2 + g_{uu}(t, u)\right),$$ with the following boundary conditions: $g(0, u) = iuy$, $g(t, 0) = 0$ and $g(t, u) \leq 0$, for all $u \in \mathbb{R}$, $t \geq 0$.

Does anyone have an idea how to solve either equation? Thanks a lot!

  • $\begingroup$ What is it about the initial and boundary conditions that prevents you from using separation of variables? $\endgroup$ – MasterYoda Sep 18 '18 at 1:50
  • $\begingroup$ If we set $f(t, u) = \alpha(t) \beta(u)$ for some functions $\alpha$, $\beta$, the condition $f(t, 0) = 1$, for all $t \geq 0$, would imply $\alpha(t)\beta(0) = 1$, for all $t \geq 0$, or equivalently (that $\beta(0) \neq 0$ and) that $\alpha$ is constant. $\endgroup$ – kantadou Sep 18 '18 at 8:09
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    $\begingroup$ In the condition $f(0,u)=e^{iuy}$ what is $y$ which appears here coming from nowhere ? $\endgroup$ – JJacquelin Sep 23 '18 at 8:55
  • $\begingroup$ Thanks, I forgot to mention that $y \in \mathbb{R}$ is a fixed constant (corresponding to the non-random value of $X_0$). $\endgroup$ – kantadou Sep 24 '18 at 8:21

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