In Stein and Shakarchi's Real Analysis, on page 313, Exercise 6, we have the following problem concerning a special case of Green's second identity.

Suppose $u$ and $v$ are a pair of functions that are in $C^2(\overline{B})$. Then one has $$ \int_B(v\Delta u-u\Delta v)\,dx = \int_{S^{d-1}}\bigg(v(\gamma)\frac{\partial u}{\partial n}(\gamma)-u(\gamma)\frac{\partial v}{\partial n}(\gamma)\bigg)\,d\sigma(\gamma). $$ Here $S^{d-1}$ is the unit sphere with $d\sigma$ the measure defined in Section 3.2 of Chapter 6 of the third volume of Stein and Shakarchi, and $\partial u/\partial n = \nabla u\cdot \gamma, \partial v/\partial n = \nabla v\cdot \gamma$ denote the directional derivatives of $u$ and $v$ (respectively) along the inner normals to $S^{d-1}$.

The claim is that the above can be derived from Lemma 4.5 of the same book in the previous chapter by taking $\eta = \eta_\epsilon^+$ and letting $\epsilon\to0$. The notation we are using here is that for any $x\in\Bbb R^d \setminus\{0\}$, $x = r\gamma$, where $\gamma = x/|x|$ is a point on the unit sphere and $r = |x|$. I can try to elaborate more on the notation if anything is unclear in this or what follows.

My attempted proof is as follows:

Lemma 4.5 states that whenever $u,v,\eta$ are in $C^2(\overline B)$, then we have $$ \int_B(v\Delta u-u\Delta v)\eta\,dx = \int_{B}(v\nabla u-u\nabla v)\cdot \nabla \eta\,dx. $$ Put $\eta = \eta_\epsilon^+$, from the previous section and recall that $\eta = 1$ for $|x| \le 1-\epsilon$, $\eta = 0$ for $|x| \ge 1$, $|\nabla\eta| \le c/\epsilon$ for $1-\epsilon< |x|< 1$, and $|\nabla \eta| = 0$ for all $|x|\notin (1-\epsilon,1)$. Clearly $\eta \to 1$ almost everywhere as $\epsilon\to0$, and the left-hand integrand is dominated, so by dominated convergence, the left-hand side goes to $\int_B(v\Delta u-u\Delta v)\,dx$.

Using the polar coordinate transformation, rewrite the right-hand integral as $$ \int_{B}(v\nabla u-u\nabla v)\cdot \nabla \eta\,dx = \int_{S^{d-1}}\int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \nabla \eta \, r^{d-1}\,dr\,d\sigma. $$ Note that \begin{align*} \bigg|\int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \nabla \eta\, r^{d-1} \,dr\bigg| &= \int_{1-\epsilon}^1 O(1/\epsilon)r^{d-1}\,dr \\ &= O\bigg(\frac{1}{\epsilon}\bigg)[1-(1-\epsilon)^d] \\ &= O\bigg(\frac{1}{\epsilon}\bigg)[\epsilon + O(\epsilon^2)] \\ &= O(1) + O(\epsilon) = O(1)\ \text{as $\epsilon\to0$}. \end{align*} so that $\int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \nabla \eta\, r^{d-1} \,dr$ is dominated by an integrable function of $\gamma\in S^{d-1}$. My problem is that $\eta(x)$ is taken to be $\chi\big(\frac{|x|-1+\epsilon}{\epsilon}\big)$, where $\chi$ is a fixed $C^2$ function that is $1$ for $|x| \le 1/4$, and $0$ for $3/4<|x|$ (you can see an interactive graph of a function like $\chi$ here), so by the Chain Rule, $$ \nabla\eta(r\gamma) = \chi'\big(\frac{r-1+\epsilon}{\epsilon}\big)\cdot\frac{\gamma}{\epsilon}. $$ And it's not clear to me that the limit as $\epsilon\to 0$ of $\int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \nabla \eta\, r^{d-1} \,dr$ is $(v\nabla u-u\nabla v)\cdot \gamma$, which is what we want it to be. Another thought I had was to somehow use the Lebesgue differentiation theorem (LDT) by writing $$ \epsilon\nabla\eta(r\gamma) = \gamma + \gamma\bigg(\chi'\big(\frac{r-1+\epsilon}{\epsilon}\big)-1\bigg), $$ and then rewriting the integral as \begin{align*} \int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \nabla \eta\, r^{d-1} \,dr &= \frac{1}{\epsilon}\int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \epsilon\nabla \eta\, r^{d-1} \,dr \\ &= \frac{1}{\epsilon}\int_{1-\epsilon}^1(v\nabla u-u\nabla v)\cdot \bigg\{\gamma + \gamma\Big(\chi'\big(\frac{r-1+\epsilon}{\epsilon}\big)-1\Big)\bigg\}\, r^{d-1} \,dr, \end{align*} but the $\chi'$ function has $\epsilon$ in its argument as well, so I can't think of how to apply LDT here.

Any suggestions for how to finish this off?


1 Answer 1


There was a typo in my original post. In Lemma 4.5, $v$ and $u$ switch places on either side of the equation. Here is the properly cited Lemma 4.5:

Lemma 4.5. We have the identity $$\int_B (v\Delta u - u \Delta v)\eta\,dx = \int_B u(\nabla v\cdot\nabla \eta) - v(\nabla u\cdot\nabla \eta)\,dx.$$

I also believe there is a slight typo in the exercise. What we can show is that $$ \int_Bv\Delta u-u\Delta v\,dx = \int_{S^{d-1}}v\frac{\partial u}{\partial n}-u\frac{\partial v}{\partial n}\,d\sigma, \tag{1} $$ where $(\partial u/\partial n)(x) = (x/|x|)\cdot \nabla v(x)$ is the directional derivative of $v$ in the outward normal direction, as opposed to the inner normal direction, with an analogous statement holding for $\partial v/\partial n$. Alternatively, $$ \int_Bv\Delta u-u\Delta v\,dx = \int_{S^{d-1}}u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\,d\sigma \tag{1'} $$ where in $(1')$, the derivatives are along the inner normals.

Here is how we can finish the proof of the claim. By the cited Lemma 4.5, for $\eta_\epsilon^+(x) = \chi(\frac{|x|-1+\epsilon}{\epsilon})$ and the chain rule, \begin{align*} \int_B(v\Delta u-u\Delta v)\eta_\epsilon^+\,dx &= \int_B u(\nabla v\cdot\nabla \eta_\epsilon^+) - v(\nabla u\cdot \nabla \eta_\epsilon^+)\,dx \\ &= \frac{1}{\epsilon}\int_B\chi'(\frac{|x|-1+\epsilon}{\epsilon})[ u(\nabla v\cdot \frac{x}{|x|})-v(\nabla u\cdot \frac{x}{|x|})]\,dx \\ &=: \frac{1}{\epsilon}\int_B\chi'(\frac{|x|-1+\epsilon}{\epsilon})[ (u\frac{\partial v}{\partial n})(x)-(v\frac{\partial u}{\partial n})(x)]\,dx\\ &= \frac{1}{\epsilon}\int_0^1\chi'(\frac{r-1+\epsilon}{\epsilon})r^{d-1}\int_{S^{d-1}}(u\frac{\partial v}{\partial n})(r\gamma)-(v\frac{\partial u}{\partial n})(r\gamma)\,d\sigma(\gamma)\,dr. \end{align*} Notice that by its definition, $\chi'(t)$ is nonzero only for $t\in[1/4,3/4]$, so if $t=\frac{r-1+\epsilon}{\epsilon}$, then $\chi'(t)$ is nonzero only for $r\in [1-3\epsilon/4,1-\epsilon/4]$, so the integral over $r$ can be taken over $r\in [1-3\epsilon/4,1-\epsilon/4]$. Now change variables $r = r_\epsilon(s) = \epsilon s + 1 - \epsilon$, so that the last expression above becomes \begin{align*} \int_{1/4}^{3/4}\chi'(s)r_\epsilon(s)^{d-1}\int_{S^{d-1}}(u\frac{\partial v}{\partial n})(r_\epsilon(s)\gamma)-(v\frac{\partial u}{\partial n})(r_\epsilon(s)\gamma)\,d\sigma(\gamma)\,ds:= \int_{1/4}^{3/4}\chi'(s)F_\epsilon(s)\,ds. \end{align*} Notice that \begin{align*} F_\epsilon(s) \xrightarrow{\epsilon\to 0} \int_{S^{d-1}}(u\frac{\partial v}{\partial n})(\gamma) - (v\frac{\partial u}{\partial n})(\gamma)\,d\sigma(\gamma), \end{align*} uniformly in $s\in [1/4,3/4]$. Therefore, \begin{align*} \int_{1/4}^{3/4}\chi'(s)F_\epsilon(s)\,ds\xrightarrow{\epsilon\to0} (\int_{1/4}^{3/4}\chi')(\int_{S^{d-1}}u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\,d\sigma). \end{align*} By the fundamental theorem of calculus, $\int_{1/4}^{3/4}\chi'=\chi(3/4)-\chi(1/4) = -1$. The left-hand side $$ \int_B(v\Delta u - u\Delta v)\eta_\epsilon^+\,dx\xrightarrow{\epsilon\to0} \int_B v\Delta u - u\Delta v\,dx $$ by the dominated convergence theorem. In all, $$ \int_B v\Delta u - u\Delta v\,dx = -(\int_{S^{d-1}}u\frac{\partial v}{\partial n}-v\frac{\partial u}{\partial n}\,d\sigma) = \int_{S^{d-1}}v\frac{\partial u}{\partial n}-u\frac{\partial v}{\partial n}\,d\sigma. $$


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