This came up while reading something about the porous medium equation, (see equation 10 of the link if curious) I don't have much background in PDE's so I apologize for my ignorance.
Suppose that $-\nabla^{2} p_{1} = s_{1}$ and $-\nabla^{2} p_{2} =s_{2} $ then does $\int \big( \nabla{p_{1}} \cdot \nabla{p_{2}} \big) = \int s_{1}p_{2}?$
Here the $s_{i}$ and $p_{i}$ are functions on $\mathbb{R}^{N}$ and $\nabla^{2}$ is the Laplacian operator.
This is just integration by parts (sometimes called stokes theorem with multidimensional vectors). For functions vanishing at infinity $$ \int_{\mathbb{R}^d} \nabla p_1\cdot\nabla p_2 = -\int_{\mathbb{R}^d} \mathrm{div}(\nabla p_1)p_2 = -\int_{\mathbb{R}^d} \Delta p_1\,p_2 = \int_{\mathbb{R}^d} s_1\,p_2, $$ where $\Delta$ denotes the Laplacian.
If you are not familiar with multidimensional calculus, just write on a basis and integrate by parts for each coordinate $$ \begin{align*} \int_{\mathbb{R}^d} \nabla p_1\cdot\nabla p_2 &= \sum_j \int_{\mathbb{R}^d} \partial_j p_1\partial_j p_2 \\ &= -\sum_j \int_{\mathbb{R}^d} \partial_j^2p_1\, p_2 = -\int_{\mathbb{R}^d} \Delta p_1\,p_2 = \int_{\mathbb{R}^d} s_1\,p_2, \end{align*} $$