Similarity between Schrodinger and Euler-Bernoulli equations - any possible physical meaning?

This correspondence is a special case of the Madelung equations - a formulation of quantum mechanics that is equivalent to the Schrodinger equation. The Madelung equations are more mathematically complicated than the Schrodinger equation, but some people find them more physically intuitive because they are formally almost identical to the Euler equation for a fluid. They are useful in the de Broglie-Bohm formulation of quantum mechanics.

The free 1D TDSE

$$ i\frac{\partial\Psi}{\partial t}~=~-\frac{\hbar}{2m} \frac{\partial^2\Psi}{\partial x^2}, \qquad \Psi~\equiv~u+iv,$$ $$\Updownarrow$$ $$ \frac{\partial u}{\partial t}~=~- \frac{\hbar}{2m} \frac{\partial^2v}{\partial x^2}, \qquad \frac{\partial v}{\partial t}~=~ \frac{\hbar}{2m} \frac{\partial^2 u}{\partial x^2},\tag{1}$$ is an auto-Bäcklund transformation for the free EBE

$$ \frac{\partial^2\Psi}{\partial t^2}+\frac{\hbar^2}{4m^2} \frac{\partial^4\Psi}{\partial x^4}~=~0, \qquad \Psi~\equiv~u+iv, $$ $$\Updownarrow$$ $$ \frac{\partial^2u}{\partial t^2}+\frac{\hbar^2}{4m^2} \frac{\partial^4 u}{\partial x^4}~=~0, \qquad \frac{\partial^2 v}{\partial t^2}+\frac{\hbar^2}{4m^2} \frac{\partial^4v}{\partial x^4}~=~0.\tag{2}$$

This means in particular that if $u$ is a solution to the EBE (2), then we can solve the TDSE (1) wrt. the imaginary part $v$ [i.e. the EBE (2) for $u$ serves as the integrability condition for the existence of the $v$-solution to the TDSE (1)]. The $v$-solution is unique modulo an affine function $a_2x+b_2$, where $a_2,b_2\in \mathbb{R}$. Then we have generated a partner solution $v(x,t)+a_2x+b_2$ to the EBE (2).

Similarly, note that the TDSE (1) is unaffected if we replace $u(x,t)$ with $u(x,t)+a_1x+b_1$, where $a_1,b_1\in \mathbb{R}$. The two solutions $u(x,t)+a_1x+b_1$ and $v(x,t)+a_2x+b_2$ to the EBE (2) then constitute a "Schrödinger pair".