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A stochastic control problem has led me to the following PDE:

State space: $t \in [0,T]$ and $x \in [-1,1]$.

$$4 \frac{\partial f}{\partial t} \frac{\partial^2 f}{\partial x^2} = 1 \quad \forall (t,x) \in [0,T) \times (-1,1)$$ and the boundary conditions are: $$ \begin{split} f(T,x) &= 0\quad \forall x\\ &\text{and} \\ f(t,\pm 1) &= T-t \end{split}$$ Is there an argument that can let me say that a solution exists to this PDE? I am not after a closed form expression obviously.

From the properties of the problem I can argue that $f(t,x) = f(t,-x)$ and $\arg \max_x f(t,x) = 0$. Don't know if it is of any use though.

Thanks.

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    $\begingroup$ By inspection, $f(t,x) = (T-t)(9-x^2)/8$ is a solution. $\endgroup$
    – user126920
    Commented Jan 29, 2019 at 20:29
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    $\begingroup$ Not sure I see it. The right hand side of the PDE is 1, not $f$. If we evaluate the left hand side with your expression we have $-(T-t)(9-x^2)=-8f(t,x)$. Am I missing something? $\endgroup$
    – avk255
    Commented Jan 30, 2019 at 6:45
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    $\begingroup$ @StanleySnelson: the initial condition $f(T,x)=0$ for all $x$ implies that, if $f$ is $C^2$ near $t=T$, then $f_{xx}=0$. But the equation gives $4f_tff_{xx}=1$, a contradiction. So there is no $C^2$ solution, with the given boundary conditions. $\endgroup$
    – Ben McKay
    Commented Jan 30, 2019 at 8:17
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    $\begingroup$ Not sure I agree entirely. I follow that $f_{xx} \to 0$ as $t \to T$. However, we could have $f_t \to \infty$ at the same time to have $f$ being $C^{1,2}$ on $(0,T) \times (-1,1)$, isn't it? $\endgroup$
    – avk255
    Commented Jan 30, 2019 at 8:49
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    $\begingroup$ At least formally, taking the Legendre transform $u$ of $f$ in the $x$ variable, we have that $w := u(x, T-t)$ solves the heat equation. Taking initial data $w(x,0) = |x|$ and transforming back gives a nontrivial solution with $0$ data on the sides. It seems feasible that playing with the initial data for $w$ could give a solution with the desired data (which corresponds to $w_{xx} = 1$ where $w_x = \pm 1$). $\endgroup$
    – Connor Mooney
    Commented Jan 30, 2019 at 17:24

1 Answer 1

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Putting $T=1$, I obtain with Maple:

sol := pdsolve(4*(diff(f(t, x), t))*(diff(f(t, x), x, x)) = 1,
 {f(1, x) = 0, f(t, -1) = 1-t, f(t, 1) = 1-t}, numeric):sol:-plot3d(t = 0 .. 1, x = -1 .. 1);

solution becomes undefined, problem may be ill posed or method may be ill suited to solution

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  • $\begingroup$ Mathematica reports "NDSolve::ndnum: Encountered non-numerical value for a derivative at t == 1." $\endgroup$
    – user64494
    Commented Jan 30, 2019 at 10:16
  • $\begingroup$ My sense is that, in light of what Ben pointed out above, we must have a blowup for $f_t$ near $T$. That maybe the reason why numerical solutions are showing non existence? $\endgroup$
    – avk255
    Commented Jan 30, 2019 at 18:01

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