complex calculation in Schrödinger equation I'm studying a paper with the following Schrödinger equation.
$$i\,y_t+\Delta\,y-F(y)=0$$ subject to Dirichlet boundary conditions where $F$ is supposed to be of the form $F=\displaystyle \frac{\partial V}{\partial \bar{z}}$, $F(0)=0$ where the “potential” $V$ is real valued and satisfies $V(e^{i\,\theta}\,z)=V(z).$
Multiplying the above equation by $\bar{y}_t$, integrating in $\Omega$ and taking the real part we have
$$\frac12\,\frac{d}{dt}\|\nabla y(t)\|_2^2+Re\int_{\Omega}F(y)\,\bar{y}_t\,dx=0\,.$$
The authors claim the following conservation energy:
$$\frac{d}{dt}\left[\|\nabla y(t)\|_2^2+\int_{\Omega} V(y(x,t))\,dx\right]=0 \quad (1)$$
My question: How I show that $\displaystyle Re\int_{\Omega}F(y)\,\bar{y}_t\,dx=\frac{d}{dt}\int_{\Omega} V(y(x,t))\,dx$ or how I arrive in identity (1) ?
I tried to use the relation $\displaystyle\frac{\partial}{\partial\bar{z}}=\frac12\left(\frac{\partial}{\partial x}+i\,\frac{\partial}{\partial y}\right) $, but I didn't get solve it.
Since already, I thank the attention.
 A: \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\int_\Omega V\bigl(y(x,t)\bigr)\,\mathrm{d}x
&=\int_\Omega\left(\Re\bigl(y_t(x,t)\bigr)\frac{\partial V}{\partial x}\bigl(y(x,t)\bigr)+\Im\bigl(y_t(x,t)\bigr)\frac{\partial V}{\partial y}\bigl(y(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\int_\Omega\left(\Re\bigl(y_t(x,t)\bigr)\frac{\partial V}{\partial x}\bigl(y(x,t)\bigr)-\Re\bigl(iy_t(x,t)\bigr)\frac{\partial V}{\partial y}\bigl(y(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\Re\int_\Omega y_t(x,t)\left(\frac{\partial V}{\partial x}\bigl(y(x,t)\bigr)-i\frac{\partial V}{\partial y}\bigl(y(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\Re\int_\Omega\overline{y_t(x,t)\left(\frac{\partial V}{\partial x}\bigl(y(x,t)\bigr)-i\frac{\partial V}{\partial y}\bigl(y(x,t)\bigr)\right)}\,\mathrm{d}x\\
&=\Re\int_\Omega\overline{y_t}(x,t)\left(\frac{\partial V}{\partial x}\bigl(y(x,t)\bigr)+i\frac{\partial V}{\partial y}\bigl(y(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\Re\int_\Omega\overline{y_t}(x,t)F\bigl(y(x,t)\bigr)\,\mathrm{d}x.
\end{align*}
Edit. In a comment you ask to justify the equality:
$$\frac{\mathrm{d}}{\mathrm{d}t}V\bigl(y(x,t)\bigr)=\Re\bigl(y_t(x,y)\bigr)\frac{\partial V}{\partial x}\bigl(y(x,t)\bigr)+\Im\bigl(y_t(x,t)\bigr)\frac{\partial V}{\partial y}\bigl(y(x,t)\bigr).$$
Your problem might comes from a misunderstanding of the notations. The $y$ in $\dfrac{\partial V}{\partial y}$ is not the function $y$ that appears in $y(x,t)$. More precisely: $V$ is a real-valued function defined on $\mathbb{C}$
$$V:\mathbb{C}\longrightarrow\mathbb{R}.$$
We may understand $V$ as being defined on $\mathbb{R}^2$, but let me give it another name, say $V_\mathbb{R}$:
$$\forall (x,y)\in\mathbb{R}^2,\ V_\mathbb{R}(x,y)=V(x+iy).$$
We denote by $\dfrac{\partial}{\partial x}$ and $\dfrac{\partial}{\partial y}$ the partial derivatives of $V_\mathbb{R}$ with respect to its first and second variables. The $x$ and $y$ that appear here are just notations. Then:
$$V\bigl(y(x,t)\bigr)=V_\mathbb{R}\Bigl(\Re\bigl(y(x,t)\bigr),\Im\bigl(y(x,t)\bigr)\Bigr).$$
From the chain rule we obtain:
$$\frac{\mathrm{d}}{\mathrm{d}t}V\bigl(y(x,t)\bigr)=\Re\bigl(y_t(x,t)\bigr)\frac{\partial V_\mathbb{R}}{\partial x}\bigl(y(x,t)\bigr)+\Im\bigl(y_t(x,t)\bigr)\frac{\partial V_\mathbb{R}}{\partial y}\bigl(y(x,t)\bigr).$$
The equality you asked to justify is just this, where we have identified $V$ and $V_\mathbb{R}$.
The notation is, unfortunately, not here to help you. If it's easier for you, don't call $y$ the solution of Schrödinger Equation, give it another name, say $g$. The original computation I gave now becomes:
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\int_\Omega V\bigl(g(x,t)\bigr)\,\mathrm{d}x
&=\int_\Omega\left(\Re\bigl(g_t(x,t)\bigr)\frac{\partial V}{\partial x}\bigl(g(x,t)\bigr)+\Im\bigl(g_t(x,t)\bigr)\frac{\partial V}{\partial y}\bigl(g(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\int_\Omega\left(\Re\bigl(g_t(x,t)\bigr)\frac{\partial V}{\partial x}\bigl(g(x,t)\bigr)-\Re\bigl(ig_t(x,t)\bigr)\frac{\partial V}{\partial y}\bigl(g(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\Re\int_\Omega g_t(x,t)\left(\frac{\partial V}{\partial x}\bigl(g(x,t)\bigr)-i\frac{\partial V}{\partial y}\bigl(g(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\Re\int_\Omega\overline{g_t(x,t)\left(\frac{\partial V}{\partial x}\bigl(g(x,t)\bigr)-i\frac{\partial V}{\partial y}\bigl(g(x,t)\bigr)\right)}\,\mathrm{d}x\\
&=\Re\int_\Omega\overline{g_t}(x,t)\left(\frac{\partial V}{\partial x}\bigl(g(x,t)\bigr)+i\frac{\partial V}{\partial y}\bigl(g(x,t)\bigr)\right)\,\mathrm{d}x\\
&=\Re\int_\Omega\overline{g_t}(x,t)F\bigl(g(x,t)\bigr)\,\mathrm{d}x.
\end{align*}
