Proving convexity for linear parabolic PDE

[emfert]

I have a parabolic PDE of the form $$a\frac{\partial^2 f}{\partial x^2} - b\frac{\partial f}{\partial x} + \frac{\partial f}{\partial t} = 0$$, where $$(x,t) \in (-\infty, \infty) \times [0, T]$$. In addition, $\lim_{x \to \infty} f(x,t) = 0$, $\lim_{x\to -\infty} = k$ (a known positive constant), and $f(x,T) = \psi(x)$ where $\frac{d\psi}{dx}<0$, $\frac{d^2\psi}{dx^2}>0$, and $\frac{d^3\psi}{dx^3}<0$. Further, $b \geq a > 0$.

I'm pretty convinced that $\frac{\partial^2 f}{\partial x^2} > 0$ and $\frac{\partial^3f}{\partial x^3} < 0$ but I don't know how to prove it. I'd appreciate any ideas.

 

[gtoguy97]

Since the equation is parabolic, it follows that $\frac{\partial^2 f}{\partial x^2} > 0$. This can be shown by considering the sign of the second derivative test. For this to hold, we also need to show that $\frac{\partial f}{\partial x} < 0$. This follows from the boundary conditions given at $x = - \infty$ and $x = \infty$. Since $\lim_{x \to \infty} f(x,t) = 0$ and $\lim_{x \to -\infty} f(x,t) = k$, where $k$ is a known positive constant, it follows that $\frac{\partial f}{\partial x} < 0$. To show that $\frac{\partial^3f}{\partial x^3} < 0$, we can use the given boundary conditions. From the given boundary conditions, it follows that $\frac{d\psi}{dx} < 0$ and $\frac{d^2\psi}{dx^2} > 0$. This implies that $\frac{\partial^3f}{\partial x^3}$ is negative near the boundaries and thus, must be negative throughout.