Let $(x,t) \in \mathbb{R}^2$ and consider the following partial differential equation for the real-valued function $U(x,t)$: \begin{equation} \frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 U}{\partial x^4}, \end{equation} where $m$ and $\hbar$ are positive constants. In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $\frac{\partial U}{\partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to this partial differential equation. Let us call the set of solutions $\mathcal{E}$.

Let us define for ease of notation $D_{x}^k F=\left( F, \frac{\partial F}{\partial x}, \dots, \frac{\partial^k F}{\partial x^k} \right)$ for every non-negative integer $k$ and for every smooth enough function $F(x,t)$. I ask whether there exist some non non-negative integer $k$ and some (smooth enough) real-valued functions $p \geq 0$ and $j$, with $p$ non-constant, such that, by setting \begin{equation} P(x,t)=p \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \\ J(x,t)=j \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \end{equation} the following properties hold:

(i) for every $U \in \mathcal{E}$ the following continuity equation holds \begin{equation} \frac{\partial P}{\partial t} + \frac{\partial J}{\partial x}=0; \end{equation}

(ii) for the special solution $U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right)$, we have that $P(x,t)$ is independent of $\omega > 0$.

The answer should be negative, but I have no idea of a possible proof. Obviously, since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but only some heuristic, convincing argument in this direction.

NOTE (1) Let us explicitly note a trivial consequence of property (ii). Since for the special solution \begin{equation} U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right), \end{equation} $P(x,t)$ is independent of $\omega > 0$ for every $(x,t) \in \mathbb{R}^2$, we have that $(x,t) \mapsto P(x,t)$ is a constant function. Indeed, for this special solution, we have that $P(x,t)=F(x-vt)$ for some function $F$, where $v=\sqrt{\frac{\hbar \omega}{2m}}$. If we had $F'(\xi)\neq 0$ for some $\xi \in \mathbb{R}$, we would get for $x-vt=\xi$: \begin{equation} \left. \frac{\partial P}{\partial t}(x,t) \middle/ \frac{\partial P}{\partial x}(x,t) \right. = -v = \sqrt{\frac{\hbar \omega}{2m}}, \end{equation} a contradiction.

NOTE (2) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise *Quantum Theory* published in 1951. Bohm explicitly states that the problem has a **negative** answer, without giving any proof or heuristic argument. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.