# Conserved Positive Charge for a PDE

Let $$(x,t) \in \mathbb{R}^2$$ and consider the following partial differential equation for the real-valued function $$U(x,t)$$: $$$$\frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 U}{\partial x^4},$$$$ where $$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 $$$$P(x,t)=p \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \\ J(x,t)=j \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right),$$$$ the following properties hold:

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

(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 $$$$U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right),$$$$ $$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$$: $$$$\left. \frac{\partial P}{\partial t}(x,t) \middle/ \frac{\partial P}{\partial x}(x,t) \right. = -v = \sqrt{\frac{\hbar \omega}{2m}},$$$$ 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.

• If you compute the continuity equation at t=0 you get a relation which depends entirely on the initial conditions for U, which are completely arbitrary. I have the feeling that by playing with suitable choices of U(t=0) and U_t(t=0) you can prove P,J to be constant – Piero D'Ancona Nov 6 '18 at 18:13
• Well, if you compute the continuity equation at $t=0$, by using the PDE in order to express $U_{tt}$, you actually get an equation which, given the arbitrariness of $U(t=0)$ and $U_t(t=0)$ and the Borel's Lemma, gives us a PDE for $p$ and $j$. But for sure this PDE has non-constant solutions. Actually, there exist non-constant functions $p \geq 0$ and $j$ satisfying the continuity equation: see the example given by Bohm in my post Nonexistence of a Probability for Real Wave Equations. – Maurizio Barbato Nov 6 '18 at 18:42
• We conclude that in any case property (ii) must play some role in the proof of the non-existence result (if it is true). – Maurizio Barbato Nov 6 '18 at 18:42

This is not an answer, but only a remark too long to be posted as a comment.

We can make the formulation of the problem more explicit in the following way. Let $$y=(y_0,y_1,\dots,y_k) \in \mathbb{R}^{k+1}$$ and $$z=(z_0,z_1,\dots,z_k) \in \mathbb{R}^{k+1}$$. Then to say that the functions $$(y,z) \mapsto p(y,z)$$ and $$(y,z) \mapsto j(y,z)$$ satisfy property (i) means that for every $$U \in \mathcal{E}$$ $$$$\nabla_{y} p \cdot D_{x}^{k} \frac{\partial U}{\partial t} + \nabla_{z} p \cdot D_{x}^{k} \frac{\partial^2 U}{\partial t^2} + \nabla_{y} j \cdot D_{x}^{k} \frac{\partial U}{\partial x} + \nabla_{z} j \cdot D_{x}^{k} \frac{\partial^2 U}{\partial x \partial t}=0,$$$$ which is equivalent to say that $$$$\nabla_{y} p \cdot D_{x}^{k} \frac{\partial U}{\partial t} - \frac{\hbar^2}{4 m^2} \nabla_{z} p \cdot D_{x}^{k} \frac{\partial^4 U}{\partial x^4} + \nabla_{y} j \cdot D_{x}^{k} \frac{\partial U}{\partial x} + \nabla_{z} j \cdot D_{x}^{k} \frac{\partial^2 U}{\partial x \partial t}=0.$$$$ Now Borel Theorem tells us that for every real sequence $$(a_{n})_{n=0}^{\infty}$$ and every fixed $$x_0 \in \mathbb{R}$$ there exists a smooth function $$f:\mathbb{R} \rightarrow \mathbb{R}$$ such that $$$$\frac{d^m f}{d x^m}(x_0)=a_{n} \quad (n=0,1,2,\dots),$$$$ so that from the arbitrariness of the initial data $$x \mapsto U(x,0)$$ and $$x \mapsto \frac{\partial U}{\partial t}(x,0)$$, by replacing $$\frac{\partial^m U}{\partial x^m}$$ by $$y_m \in \mathbb{R}$$ and $$\frac{\partial^{m+1} U}{\partial^m x \partial t}$$ by $$z_m \in \mathbb{R}$$ for every non-negative integer $$m$$, we conclude that $$p$$ and $$j$$ satisfy the following equation:

$$$$\sum_{m=0}^{k}\frac{\partial p}{\partial y_m} z_{m} - \frac{\hbar^2}{4 m^2}\sum_{m=0}^{k} \frac{\partial p}{\partial z_m} y_{m+4} + \sum_{m=0}^{k} \frac{\partial j}{\partial y_m} y_{m+1} + \sum_{m=0}^{k} \frac{\partial j}{\partial z_m} z_{m+1} =0 \qquad (I).$$$$ From this equation we immediately deduce that $$p$$ does not depend on $$z_{k-2}, z_{k-1}, z_k$$ and that $$j$$ does not depend on $$z_k$$. Moreoever, we also get that $$$$- \frac{\hbar^2}{4m^2} \frac{\partial p}{\partial z_{k-3}}+ \frac{\partial j}{\partial y_k} = 0,$$$$ and since $$p$$ and $$j$$ do not depend on $$z_k$$ it also holds $$$$\frac{\partial p}{\partial y_k} + \frac{\partial j}{\partial z_{k-1}}=0.$$$$

There are no other simple consequences of Equation (I). Now, the issue in the post amounts to see whether there exist solutions $$p \geq 0$$ and $$j$$ of Equation (I), with $$p$$ non-constant and satisfying property (ii). This latter request makes all the matter quite involved.

Let us explicitly note there are simple non-constant functions $$p(y,z) \geq 0$$ which satisfy (ii). Consider e.g. $$p(y,z)=(y_1 z_1 - z_0 y_2)^2$$. It is easy to see that in this case there is no $$k$$ and no function $$j$$ such that $$p$$ and $$j$$ satisfy equation (I). Indeed, equation (I) would imply in our case that $$j$$ depends only on $$y_0,\dots, y_4, z_0,z_1$$. Moreover it would also imply that

$$$$\frac{\partial j}{\partial y_4}= \frac{\hbar^2}{2 m^2}(y_1 z_1 - z_0 y_2) y_1, \\ \frac{\partial j}{\partial z_1}= 2(y_1 z_1 - z_0 y_2) z_0.$$$$

From the last equation we would get for some function $$g$$: $$$$j=2(y_1 z_1 - z_0 y_2) z_0 z_1 + g(y_0,\dots,y_4,z_0),$$$$ so that we should have $$$$\frac{\partial g}{\partial y_4} = \frac{\partial j}{\partial y_4} = \frac{\hbar^2}{2 m^2}(y_1 z_1 - z_0 y_2) y_1,$$$$

which is a contradiction since $$g$$ does not depend on $$z_1$$.

The difficulty of the problem is to show in general that conditions (i) and (ii) are not compatible, and I have no idea for now of how to tackle the problem.