Derivation of Green's function in Evans' PDE book. In the book of Evans, on page 34 equation $(25)$ isn't the RHS should be minus what is written there, I mean he uses the fact that $\Delta \Phi(y-x) = \delta(y-x)$ on $U$, and he moves the second term in eq. $(24)$ to the RHS.
If this is not the case then how did he derive equation $(25)$?
The book has a preview on pages 33-34.
https://books.google.co.il/books?id=Xnu0o_EJrCQC&printsec=frontcover#v=onepage&q&f=false
 A: The following is Theorem 1 at page 23 in Evans' book

Let $u = \Phi * f$, then $u \in C^2$ and $\color{red}{-}\Delta u = f.$

Then one adopts the notation $$-\Delta \Phi = \delta_0,$$ thanks to which we can formally compute $$-\Delta u = (-\Delta \Phi) * f = \int \delta(x - y)f(y) = f(x).$$
This should fix your sign problem.
A: Recently I also faced the problem. Here is my way of thinking:
Begining with Eq. (\ref{eq-24}) in p. 33 of Evans' book:
\begin{align}
&\int_{V_\varepsilon} u(y)\Delta\Phi(y-x)-\Phi(y-x)\Delta u(y)\,dy\\
&\qquad=\int_{\partial V_\varepsilon} u(y)\frac{\partial\Phi}{\partial\nu}(y-x)-\Phi(y-x)\frac{\partial u}{\partial\nu}(y)\,dS(y)\tag{24}\label{eq-24}\\
&\qquad=\int_{\partial(U-B(x,\varepsilon))} u(y)\frac{\partial\Phi}{\partial\nu}(y-x)-\Phi(y-x)\frac{\partial u}{\partial\nu}(y)\,dS(y)\\
&\qquad=\int_{\partial U} u(y)\frac{\partial\Phi}{\partial\nu}(y-x)-\Phi(y-x)\frac{\partial u}{\partial\nu}(y)\,dS(y)\\
&\qquad\qquad+\int_{\partial B(x,\varepsilon)} u(y)\frac{\partial\Phi}{\partial\nu^\ast}(y-x)-\Phi(y-x)\frac{\partial u}{\partial\nu^\ast}(y)\,dS(y)
\end{align}
where $\nu^\ast$ denotes the inward pointing unit normal along $\partial B(x,\varepsilon)$, which has the same property as $\nu$ in p. 24 of the book. Therefore, as what p. 24 calculates thereafter, we may obtain the equation in the last of p. 33, implying Eq. (25).
