In Schrödinger's equation well-behaved wave functions at $t=0$ will never change their normalization value no matter what the real valued potential $V(x,t)$ is.
So, I have a counter example for this:
$$\varphi(x,t)=\sqrt{{\Big(\frac{t}{a}\Big)}^2+1}\,\,\frac{e^{\frac{-x^2}{2b^2}}}{\sqrt{\sqrt{\pi}\,\,b}}\exp\left\{i\left\{\frac{-2mt}{\hbar(t^2+a^2)}\int_0^xe^{\frac{l^2}{b^2}}\int_0^le^{\frac{-n^2}{b^2}}\,\,dn\,\,dl\right\}\right\},$$
where $a$ , $b$ , $\hbar$ and $m$ are real values greater than $0$. If we put this complex function which is smooth in the equation
$$i\hbar\frac{\partial \varphi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \varphi}{\partial x^2}+V(x,t)\varphi, $$
we will get a real valued function $V(x,t)$, that is, the imaginary part of $V$ cancels out. Furthermore at $t=0$,
$$\varphi(x)=\frac{e^{\frac{-x^2}{2b^2}}}{\sqrt{\sqrt{\pi}\,\,b}}$$
which I think is a very well behaved wave function, but it’s changing its normalization value $\int_{-\infty}^{\infty}|\varphi|^2\,\,dx$ through time for the real $V(x,t)$ that we get.
Questions
What is happening? Is there a mathematical constraint on what $V(x,t)$ should be for the normalization value to be indepentent of time?
The normalization value will also change no matter what $V(x,t)$ is, for some smooth initial wavefunctions?
Edit: Any one trying to work this one out? When I try to put this into the equation it is checking out. I really am getting a real valued potential.
Computation:
We can write the given wavefunction in the form
$$\varphi(x,t)=A(x,t)e^{i\theta(x,t)},$$ where $$A(x,t)=\sqrt{{\Big(\frac{t}{a}\Big)}^2+1}\,\,\frac{e^{\frac{-x^2}{2b^2}}}{\sqrt{\sqrt{\pi}\,\,b}}$$ and $$\theta(x,t)=\frac{-2mt}{\hbar(t^2+a^2)}\int_0^xe^{\frac{l^2}{b^2}}\int_0^le^{\frac{-n^2}{b^2}}\,\,dn\,\,dl$$
so in the equation, $$i\hbar\frac{\partial Ae^{i\theta}}{\partial t}+\frac{\hbar^2}{2m}\frac{\partial^2 Ae^{i\theta}}{\partial x^2}=VAe^{i\theta}$$
$$i\hbar\left\{iAe^{i\theta}\frac{\partial \theta}{\partial t}+e^{i\theta}\frac{\partial A}{\partial t}\right\}+\frac{\hbar^2}{2m}\frac{\partial}{\partial x}\left\{iAe^{i\theta}\frac{\partial \theta}{\partial x}+e^{i\theta}\frac{\partial A}{\partial x}\right\}=VAe^{i\theta}$$
$$i\hbar\left\{iAe^{i\theta}\frac{\partial \theta}{\partial t}+e^{i\theta}\frac{\partial A}{\partial t}\right\}+\frac{\hbar^2}{2m}\frac{\partial}{\partial x}\left\{e^{i\theta}\left\{iA\frac{\partial \theta}{\partial x}+\frac{\partial A}{\partial x}\right\}\right\}=VAe^{i\theta}$$
$$i\hbar\left\{iAe^{i\theta}\frac{\partial \theta}{\partial t}+e^{i\theta}\frac{\partial A}{\partial t}\right\}+\frac{\hbar^2}{2m}\left\{ie^{i\theta}\frac{\partial \theta}{\partial x}\left\{iA\frac{\partial \theta}{\partial x}+\frac{\partial A}{\partial x}\right\}+e^{i\theta}\left\{i\frac{\partial A}{\partial x}\frac{\partial \theta}{\partial x }+iA\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 A}{\partial x^2}\right\}\right\}=VAe^{i\theta}$$
now we can take $e^{i\theta}$ common outside and cancel both sides,
$$i\hbar\left\{iA\frac{\partial \theta}{\partial t}+\frac{\partial A}{\partial t}\right\}+\frac{\hbar^2}{2m}\left\{i\frac{\partial \theta}{\partial x}\left\{iA\frac{\partial \theta}{\partial x}+\frac{\partial A}{\partial x}\right\}+\left\{i\frac{\partial A}{\partial x}\frac{\partial \theta}{\partial x }+iA\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 A}{\partial x^2}\right\}\right\}=VA$$
$$-\hbar A\frac{\partial \theta}{\partial t}+i\hbar\frac{\partial A}{\partial t}+\frac{\hbar^2}{2m}\left\{-A{\Big(\frac{\partial \theta}{\partial x}\Big)}^2+i\frac{\partial \theta}{\partial x}\frac{\partial A}{\partial x}+i\frac{\partial A}{\partial x}\frac{\partial \theta}{\partial x }+iA\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 A}{\partial x^2}\right\}=VA$$
$$-\hbar A\frac{\partial \theta}{\partial t}+i\hbar\frac{\partial A}{\partial t}+\frac{\hbar^2}{2m}\left\{-A{\Big(\frac{\partial \theta}{\partial x}\Big)}^2+2i\frac{\partial \theta}{\partial x}\frac{\partial A}{\partial x}+iA\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 A}{\partial x^2}\right\}=VA$$
$$\frac{1}{A}\left\{i\hbar\frac{\partial A}{\partial t}+\frac{\hbar^2}{2m}\left\{2i\frac{\partial A}{\partial x}\frac{\partial \theta}{\partial x}+iA\frac{\partial^2 \theta}{\partial x^2}\right\}-\hbar A\frac{\partial \theta}{\partial t}+\frac{\hbar^2}{2m}\left\{-A{\Big(\frac{\partial \theta}{\partial x}\Big)}^2+\frac{\partial^2 A}{\partial x^2}\right\}\right\}=V$$
we can see that imaginary part of V is,
$$\frac{1}{A}\left\{i\hbar\frac{\partial A}{\partial t}+\frac{\hbar^2}{2m}\left\{2i\frac{\partial A}{\partial x}\frac{\partial \theta}{\partial x}+iA\frac{\partial^2 \theta}{\partial x^2}\right\}\right\}$$
we can take i common and multiply and divide by 2A so we get,
$$\frac{i}{2A^2}\left\{\hbar(2A)\frac{\partial A}{\partial t}+\frac{\hbar^2}{m}\left\{2A\frac{\partial A}{\partial x}\frac{\partial \theta}{\partial x}+A^2\frac{\partial^2 \theta}{\partial x^2}\right\}\right\}$$
so we can write is as,
$$\frac{i}{2A^2}\left\{\hbar\frac{\partial A^2}{\partial t}+\frac{\hbar^2}{m}\left\{\frac{\partial A^2}{\partial x}\frac{\partial \theta}{\partial x}+A^2\frac{\partial^2 \theta}{\partial x^2}\right\}\right\}$$
we can see that in the second term its a multiplication rule so,
$$\frac{i\hbar}{2A^2}\left\{\frac{\partial A^2}{\partial t}+\frac{\hbar}{m}\frac{\partial A^2\frac{\partial \theta}{\partial x}}{\partial x}\right\}$$
so this is the imaginary part, now compute $A^2\frac{\partial \theta}{\partial x}$, $$A^2\frac{\partial \theta}{\partial x}=\left\{{\Big(\frac{t}{a}\Big)}^2+1\right\}\,\,\frac{e^{\frac{-x^2}{b^2}}}{\sqrt{\pi}\,\,b}\frac{\partial}{\partial x}\left\{\frac{-2mt}{\hbar(t^2+a^2)}\int_0^xe^{\frac{l^2}{b^2}}\int_0^le^{\frac{-n^2}{b^2}}\,\,dn\,\,dl\right\}$$
you can take the term dependent upon time outside, $$A^2\frac{\partial \theta}{\partial x}=\left\{{\Big(\frac{t}{a}\Big)}^2+1\right\}\,\,\frac{e^{\frac{-x^2}{b^2}}}{\sqrt{\pi}\,\,b}\Big(\frac{-2mt}{\hbar(t^2+a^2)}\Big)\frac{\partial}{\partial x}\left\{\int_0^xe^{\frac{l^2}{b^2}}\int_0^le^{\frac{-n^2}{b^2}}\,\,dn\,\,dl\right\}$$
you can easily differentiate it now, $$A^2\frac{\partial \theta}{\partial x}=\left\{{\Big(\frac{t}{a}\Big)}^2+1\right\}\,\,\frac{e^{\frac{-x^2}{b^2}}}{\sqrt{\pi}\,\,b}\Big(\frac{-2mt}{\hbar(t^2+a^2)}\Big)e^{\frac{x^2}{b^2}}\int_0^xe^{\frac{-n^2}{b^2}}\,\,dn$$
$$A^2\frac{\partial \theta}{\partial x}=\frac{(t^2+a^2)}{a^2}\,\,\frac{e^{\frac{-x^2}{b^2}}}{\sqrt{\pi}\,\,b}\Big(\frac{-2mt}{\hbar(t^2+a^2)}\Big)e^{\frac{x^2}{b^2}}\int_0^xe^{\frac{-n^2}{b^2}}\,\,dn$$
$$A^2\frac{\partial \theta}{\partial x}=\frac{-2mt}{\hbar a^2b\sqrt{\pi}}\int_0^xe^{\frac{-n^2}{b^2}}\,\,dn$$
now if we differentiate with respect to x again, $$\frac{\partial A^2\frac{\partial \theta}{\partial x}}{\partial x}=\frac{-2mt}{\hbar a^2b\sqrt{\pi}}\frac{\partial \int_0^xe^{\frac{-n^2}{b^2}}\,\,dn}{\partial x}$$
$$\frac{\partial A^2\frac{\partial \theta}{\partial x}}{\partial x}=\frac{-2mt}{\hbar a^2b\sqrt{\pi}}e^{\frac{-x^2}{b^2}}$$
so , $$\frac{\hbar}{m}\frac{\partial A^2\frac{\partial \theta}{\partial x}}{\partial x}=\frac{-2t}{a^2b\sqrt{\pi}}e^{\frac{-x^2}{b^2}},$$
now try to compute $\frac{\partial A^2}{\partial t}$ we get,
$$\frac{\partial A^2}{\partial t}=\frac{\partial}{\partial t}\left\{\frac{(t^2+a^2)e^{\frac{-x^2}{b^2}}}{a^2b\sqrt{\pi}}\right\}$$
this is,
$$\frac{\partial A^2}{\partial t}=\frac{2te^{\frac{-x^2}{b^2}}}{a^2b\sqrt{\pi}}$$
we can see that $\frac{\partial A^2}{\partial t}$ and $\frac{\hbar}{m}\frac{\partial A^2\frac{\partial \theta}{\partial x}}{\partial x}$ both cancel eachother,
and therefore making the imaginary part $0$.
The potential being real valued is not enough to preserv the normalization, you also need some condition on $\pm \infty$. Notice that your solution oscillates wildly as $x\rightarrow \pm \infty$. The proof typicaly relies on the formula $$ \partial_t(\lvert \phi \rvert^2) = \frac{i\hbar}{2m}(\phi^*\phi_{xx} - \phi \phi^*_{xx}),$$ and after integrating in $\mathbb R$ and using integration by parts everything cancels and the norm is preserv, but this is true only if the boundary terms vanish, namely
$$\frac{d}{dt} \int_{\Bbb R } \lvert \phi \rvert^2 \, dx = -\frac{i\hbar}{2m} \int_{\Bbb R} \phi^*_x \phi_x - \phi_x \phi^*_x + \frac{i\hbar}{2m} \left \{ \lim_{x\to \infty}\phi^*(x) \phi_x(x) - \lim_{x\to -\infty}\phi^*(x) \phi_x(x) - \lim_{x\to \infty}\phi(x)\phi^*_x(x)+\lim_{x\to \infty}\phi(x) \phi^*_x(x) \right \}$$ And for your solution those limits probably don't vanish.