Time Derivative of PDE solution

Question

Consider the PDE $$ \frac{\partial \psi}{\partial t}(t,x)=\psi''(t,x)+2\psi\psi'(t,x)-\tilde V'(x)~ on~ [0,D/2]\times(0,\infty)\\ \psi(0,t)=0, \psi(D/2,t)=-k, \\\psi(\cdot,0)=\psi_0 $$ for some $k\in\mathbb{N}$. $\psi_0$ is - except a finite point set- a stationary solution of the corresponding ODE and in such a way that $\psi$ remains under $\psi_0$ at all times, there's also a subsolution, which is a smooth stationary solution and satisfies the same boundary conditions.

I'm trying to prove that $\frac{\partial \psi}{\partial t}\leq 0$. My approach: Differentiating the PDE with respect to $t$ and writing $\eta=\frac{\partial \psi}{\partial t}$ gives $$ \frac{\eta }{\partial t}=\eta''+2\psi\eta'+2\psi'\eta \\ \psi(0,t)=\psi(D/2,t)=0. $$ I'm thinking of applying the strong parabolic maximum principle, since it seems straight forward to me that $\eta$ should be non positiv near $t=0$ ( $\psi$ remains under the supersolution $\psi_0$). However, $\psi$ need not be differentiable with respect to $t$ at $t=0$ and I can't exclude any oscillating behaviour of $\eta$. Does anyone know how to proceed or has another idea? Thanks in advance

Answer 1

The boundary condition $\psi(D/2, -t) = -k$ and under the suggested transformation $\psi(D/2, t) = 0$ seems to be in question.

In the general sense the equation \begin{align} \partial_{t} \psi(x,t) = \partial_{x} \left( \partial_{x} \psi(x,t) - \psi^{2}(x,t) - \tilde{V} \right) \end{align} is of separable form in the sense $\psi(x,t) = \psi_{0}(x) e^{-|E|t}$. Making this substitution then \begin{align} -|E| = \partial_{x} \left( \partial_{x} \psi(x,t) - \psi^{2}(x,t) - \tilde{V} \right) \end{align} which leaves a non-linear pde in $\psi_{0}(x)$. Now, \begin{align} \partial_{t} \psi(x,t) = - |E| \psi(x,t) \leq 0 \end{align} for $\psi(x,t) \geq 0$ over the domain.

Answer 2

I will show a way to transform your equation into a simpler form which might allow for a easier analysis of your problem.

Starting with your equation

$$ \frac{d\psi}{dt} = \frac{d}{dx}\left(\psi' + \psi^2 - V\right)$$

we define $$\zeta(x,t) \equiv \int_0^x\psi(z,t)dz$$ and by integrating your equation over $x$ (we ignore boundary terms; or redefine $V$ to account for them) we find

$$ \frac{d\zeta}{dt} = \zeta'' + (\zeta')^2-V$$

To simplify it further we define $\beta(x,t) \equiv e^{\zeta(x,t)}$ which gives

$$\frac{d\beta}{dt} = \beta'' - V\beta$$

This is the well known Heat equation (with $V$ describing internal heat generation/loss) or if you like: a Schrodinger equation in imaginary time. Note that the original variable is given by

$$\psi(x,t) = \left(\frac{\beta'(x,t)}{\beta(x,t)}\right)$$

Now since the equation for $\beta$ is separable we can, if needed, solve it using separation of variables: $\beta = X(x) T(t)$. For a zero potential (can also be solved analytically for a linear and quadratic potential) we find

$$ \beta(x,t) = A \cos(\omega x)e^{-\omega^2 t}$$

where the second boundary condition imply that $\tan(\omega D/2) = k/\omega$ whose solutions forms a countable set $\{\omega_i\}_{-\infty}^{\infty}$ and the general solution therefore reads

$$\beta(x,t) = \sum_i A_i \cos(\omega_i x)e^{-\omega^2_i t}$$