So we now have to assume that
\(\vF\) obeys
\(\frac{\partial F_1}{\partial y}(x,y)
= \frac{\partial F_2}{\partial x}(x,y)\) on
all of
\(\bbbr^2\) and prove that it is conservative. We'll do so using the strategy of Example
2.3.13 to find a function
\(\varphi(x,y)\text{,}\) that obeys
\begin{equation*}
\begin{split}
\frac{\partial \varphi}{\partial x}(x,y) &= F_1(x,y) \\
\frac{\partial \varphi}{\partial y}(x,y) &= F_2(x,y)
\end{split}
\end{equation*}
The partial derivative \(\frac{\partial \ }{\partial x}\) treats \(y\) as a constant. So \(\varphi(x,y)\) obeys the first equation if and only if there is a function \(\psi(y)\) with
\begin{equation*}
\varphi(x,y)
=\int_0^x F_1(X,y)\,\dee{X}
\ +\ \psi(y)
\end{equation*}
This \(\varphi(x,y)\) will also obey the second equation if and only if
\begin{align*}
F_2(x,y)&= \frac{\partial \varphi}{\partial y}(x,y)\\
&=\frac{\partial \ }{\partial y}\Big(\int_0^x F_1(X,y)\,\dee{X}\ +\ \psi(y)\Big)\\
&=\int_0^x \frac{\partial F_1}{\partial y}(X,y)\,\dee{X}\ +\ \psi'(y)
\end{align*}
So we have to find a \(\psi(y)\) that obeys
\begin{equation*}
\psi'(y) = F_2(x,y) -
\int_0^x \frac{\partial F_1}{\partial y}(X,y)\,\dee{X}
\end{equation*}
This looks bad — no matter what
\(\psi(y)\) is, the left hand side is independent of
\(x\text{,}\) while it looks like the right hand side depends on
\(x\text{.}\) Fortunately our screening test hypothesis now rides in to the rescue
7 . (We haven't used it yet, and it has to come in somewhere.)
\begin{align*}
F_2(x,y) - \int_0^x \frac{\partial F_1}{\partial y}(X,y)\,\dee{X}
&=F_2(x,y) - \int_0^x \frac{\partial F_2}{\partial x}(X,y)\,\dee{X}\\
&= F_2(x,y) - F_2(X,y)\Big|_{X=0}^{X=x}\\
&=F_2(0,y)
\end{align*}
In going from the first line to the second line we used the fundamental theorem of calculus. So choosing
\begin{equation*}
\psi(y) = \int_0^y F_2(0,Y)\,\dee{Y} +C
\end{equation*}
for any constant \(C\text{,}\) does the trick.