I found this identity in Lawrence C.Evans' book 'Partial Differential Equations' 2ed edition, page293, where $\phi \in C_{c}^{\infty}(V) \ and\ V\subset \subset U$, then $\int_{V}u(x)\frac{\phi(x+he_i)-\phi(x)}{h}dx=-\int_{V}[\frac{u(x)-u(x-he_i)}{h}]\phi(x)dx$ It says "this is the 'integration-by-parts' formula for difference quotients." I'd like to know is there any proof or reference of this "integration-by-parts" formula for difference quotients?


1 Answer 1


It's just a change of variables,

$$\int_V u(x)\phi(x+he_i)\,dx = \int_{V+he_i} u(x-he_i)\phi(x)\,dx,$$

and since $\phi$ has support in $V$, the latter integral effectively only extends over a subset of $V$ (the support of $\phi$). So you have

$$\begin{align} \int_V u(x) \frac{\phi(x+he_i)-\phi(x)}{h}\,dx &= \frac{1}{h} \left(\int_V u(x)\phi(x+he_i) \,dx - \int_V u(x)\phi(x)\,dx\right)\\ &= \frac{1}{h}\left(\int_V u(x-he_i)\phi(x)\,dx - \int_V u(x)\phi(x)\,dx \right)\\ &= \int_V \frac{u(x-he_i) - u(x)}{h}\phi(x)\,dx\\ &= - \int_V \frac{u(x)-u(x-he_i)}{h}\phi(x)\,dx. \end{align}$$


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