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PS : Indeed, there is a typo in my equation. Thanks to Zachary's observation.

Consider a PDE for $v: [0,1]^2\to (-\infty,0]$ satisfying

$$v_t(t,x) = \frac{v_{xx}(t,x)v(t,x)-v_{x}(t,x)^2}{v(t,x)^2},\quad \forall (t,x)\in (0,1)^2$$

with the terminal condition $v(1,\cdot)=0$ and boundary conditions $v(\cdot,0)=v(\cdot,1)=-1$.

Has the wellposedness of this PDE been studied? Any answer, comments and related references are highly appreciated.

PS : Thanks to Zachary's observation, with the change of variable $-v:=\exp(-u)$, we obtain a new PDE

$$u_t(t,x) = -e^{u(t,x)}u_{xx}(t,x),\quad \forall (t,x)\in (0,1)^2,$$

with the terminal condition $u(1,\cdot)=\infty$ and boundary conditions $u(\cdot,0)=u(\cdot,1)=0$.

PS : Adopting terceira's suggestion by writing $v(t,x)=f(t)g(x)$, one has

$$f'(t)g(x)^3=g''(x)g(x)-g'(x)^2$$

and further

$$f'(t)\quad \equiv\quad \mbox{Constant}\equiv C \quad \equiv \quad \frac{g''(x)g(x)-g'(x)^2}{g(x)^3}.$$

Combining with the terminal condition one obtains $f(t)=C(t-1)$. How to deal with the ODE for $g$?

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    $\begingroup$ It would be nicer if you replaced $v_{xx}$ by $v v_{xx}$. Then your equation would be equivalent to $u_t=e^u \partial_{xx} u$ with $-v(x,t)=e^{-u(x,t)}$. $\endgroup$
    – Dispersion
    Commented Jan 13, 2023 at 1:57
  • $\begingroup$ @Zachary Thanks a lot for the inspiring comment. This seems to be a more friendly parabolic PDE $\endgroup$
    – Fawen90
    Commented Jan 13, 2023 at 6:56
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    $\begingroup$ Hsve you tried separation of variables? This at least provides some explicit solutions in terms of elementary functions which might be helpful. $\endgroup$
    – terceira
    Commented Jan 13, 2023 at 12:00
  • $\begingroup$ @terceira I've tried to the separation of variables $v(t,x)=f(t)g(x)$, while its seems that the boundary conditions $f(t)g(0)=-1=f(t)g(1)$ are not compatible. $\endgroup$
    – Fawen90
    Commented Jan 13, 2023 at 12:37
  • $\begingroup$ The ODE for $g$ can be rewritten as $Ce^h=h''$ with $h=\log g$. $\endgroup$
    – Dispersion
    Commented Jan 13, 2023 at 16:23

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