Conserved Current for a PDE

Question

Let $(x,t) \in \mathbb{R}^2$, $W(x)$ be a (smooth enough) real-valued function 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}{4 m^2} \frac{\partial^4 U}{\partial x^4}+ \frac{W}{m} \frac{\partial^2 U}{\partial x^2} +\frac{W’}{m} \frac{\partial U}{\partial x} + \left( \frac{W’’}{2m} - \frac{W^2}{\hbar^2} \right) U \qquad (I), \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 (I). Let us call the set of solutions $\mathcal{E}$.

Let $D_{x}^k F$ be the set of all partial derivatives of $F$ with respect to $x$ from order 1 to order $k$. I ask whether there exist (smooth enough) real-valued functions $p \geq 0$ and $j$ such that, by setting \begin{equation} P(x,t)=p \left(U(x,t),(D_{x}^k U)(x,t), \frac{\partial U}{\partial t}(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \\ J(x,t)=j \left(U(x,t),(D_{x}^k U)(x,t), \frac{\partial U}{\partial t}(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \end{equation} the following properties hold:

(i) if $U$ is the solution of (I) corresponding to a given function $W(x)$ and given initial conditions $U(x,0)$ and $\frac{\partial U}{\partial t}(x,0)$, and $\tilde{U}$ is the solution of (I) corresponding to the same initial conditions, but to $W(x)+c$, with $c \in \mathbb{R}$, then $P(x,t)$ is the same when computed for $U$ and $\tilde{U}$;

(ii) for every $U \in \mathcal{E}$ the following conservation law holds

\begin{equation} \frac{\partial P}{\partial t} + \frac{\partial J}{\partial x}=0; \end{equation}

(iii) $p$ is not the constant function.

The answer I think should be negative, but I don't how to "prove" this: since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but some heuristic, but convincing argument in this direction.

NOTE (1) 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 $\mathit{Quantum}$ $\mathit{Theory}$ published in 1951. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.

NOTE (2) Bohm's physical discussion is not very clear, so that it can admit different mathematical interpretations. A simpler interpretation of Bohm's original statement is the following. Consider the following equation \begin{equation} \frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 U}{\partial x^4} \qquad{(II)}, \end{equation} and let $\mathcal{F}$ the set of all (smooth enough) solutions of this equation. Do there exists (smooth enough) real-valued functions $p \geq 0$ and $j$ such that, by setting \begin{equation} P(x,t)=p \left(U(x,t),(D_{x}^k U)(x,t), \frac{\partial U}{\partial t}(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \\ J(x,t)=j \left(U(x,t),(D_{x}^k U)(x,t), \frac{\partial U}{\partial t}(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \end{equation} the following properties hold:

(i) for every $U \in \mathcal{F}$ we have \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 $\omega > 0$;

(iii) $p$ is not a constant function?

Maybe this mathematical problem is more easily seen to have a negative answer than the one I formulated above.

Answer 1

Proof: Consider equation (I) with $W(x)=const$ and the following initial conditions: $$U(x,0)=0,$$ $$D_tU(x,0)=A,$$ and periodic boundary conditions: $$U(x,t)=U(x+a,t),$$ where $a$ and $A$ are constants.

Solution of equation (I) depends only in $t$ in this case: $U=U(t)$, and $p=p(t)$, $j=0$. It has the following form: $$U=\frac{A\hbar}{W}\sin{\frac{Wt}{\hbar}}$$

The first derivative is: $$D_tU=A\cos{\frac{Wt}{\hbar}}$$

Function $p=p(U,D_tU)$ has 2 properties:

1 It must not depend on $t$ due to the conservation law.

2 It must not depend on $W$ due to property (i) - $p$ does not change after transformation $W\rightarrow W+c$.

However, such function $p$ is constant: one can invert expressions for $U$ and $D_t U$ to find functions $W(U,D_tU)$ and $t(U,D_tU)$ and use $$\frac{\partial p}{\partial U}=\frac{\partial p}{\partial W}\frac{\partial W}{\partial U}+\frac{\partial p}{\partial t}\frac{\partial t}{\partial U}=0,$$ $$\frac{\partial p}{\partial (D_tU)}=\frac{\partial p}{\partial W}\frac{\partial W}{\partial (D_tU)}+\frac{\partial p}{\partial t}\frac{\partial t}{\partial (D_tU)}=0,$$ because $\frac{\partial p}{\partial W}=0$ and $\frac{\partial p}{\partial t}=0$.

UPDATE: answering the updated question:

Assume that $U=f(x-vt)$, where $v=(\hbar\omega/2m)^{1/2}$ (it follows from (ii) that $f(x)=\cos{((2m\omega/\hbar)^{1/2}x)}$). Assume that there is a function $p$ that depends only on $U$ and its derivatives, and does not depend on $\omega$. In this case, $P(x,t)$ is also a function of $x-vt$: $P(x,t)=F(x-vt)$.

As a consequence: $$\frac{D_t P}{D_x P}=-v=-\sqrt{\frac{\hbar\omega}{2m}},$$ which contradicts the assumption that $P(x,t)$ does not depend on $\omega$.

Answer 2

After some careful thought I must conclude that the mathematical formulation of Bohm's statement I have given in my post above is wrong for the reason that atarasenko identified in his comment. Indeed, if we start from the usual on-dimensional Schrödinger equation for a particle in the case of a potential $W(x)$:

\begin{equation} i \hbar \frac{\partial \psi}{\partial t}(x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x,t) + W(x) \psi(x,t), \end{equation}

and we change $W(x)$ by adding a constant $W_0$, the new solution $\tilde{\psi}$ corresponding to the same initial data $x \mapsto \psi(x,0)$ is $\tilde{\psi}(x,t)=\psi(x,t) \exp(-iW_0 t/ \hbar)$. If $U$ and $V$ are respectively the real and imaginary part of $\psi$, and $\tilde{U}$ and $\tilde{V}$ are respectively the real and imaginary part of $\tilde{\psi}$, then from \begin{equation} \frac{\partial \tilde{\psi}}{\partial t} = \exp(-iW_0t/ \hbar) \left[ \frac{\partial \psi}{\partial t} - \frac{i W_0}{\hbar} \psi \right], \end{equation} we get \begin{equation} \frac{\partial \tilde{U}}{\partial t}(x,0)= \frac{\partial U}{\partial t}(x,0) + \frac{W_0}{\hbar} V(x,0), \end{equation} so $x \mapsto \frac{\partial \tilde{U}}{\partial t}(x,0)$ and $x \mapsto \frac{\partial U}{\partial t}(x,0)$ do not coincide. Since Bohm requires that the usual theory based on the complex-valued wave function $\psi$ and the (eventual) theory based on the real-valued wave function $U$ should be equivalent, we conclude that the mathematical formulation I have given above is incorrect.

As for the second mathematical formulation of Bohm's statement I gave in my NOTE (2) above, I have discussed it at some length in my post on MathOverflow Conserved Positive Charge for a PDE, but no one could give some answer. This problem seems not trivial at all, and maybe it could have a positive answer, in the sense that there exist some functions $p \geq 0$ and $j$ which satisfy properties (i),(ii) and (iii) I listed in NOTE (2) above. For this reason, I suspect that this mathematical formulation is flawed too, this time probably because Bohm had in his mind some kind of invariance argument based on physical considerations (hidden in Condition (3) required for the probability function in Section (4.2) of his book Quantum Theory) which we do not know and which would make the non-existence result transparent.