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Let $\Omega$ be an open bounded subset of $\mathbb{R}^n (n\geq 3)$ and let $\Omega_0$ be any domain containing $\Omega$ for which the divergence theorem is true. Let $f$ be bounded and locally Holder continuous (with exponent $\alpha\leq 1$) in $\Omega$ and $f$ is extended to vanish outside $\Omega$.

We also fix a function $\eta\in C^1(\mathbb{R})$ satisfying:

  1. $0\leq\eta\leq 1$,
  2. $0\leq \eta'\leq 2$ and
  3. $\eta (t)=0$ for $t\leq 1$ and $\eta=1$ for $t\geq 2$.

Define for $\varepsilon>0$: \begin{equation} v_{\varepsilon}(x)\equiv \int_{\Omega} D_i\Phi(x-y)\eta_{\varepsilon}f(y) \text{ d}y\quad x\in\Omega, \end{equation} where $\Phi$ is the fundamental solution to Laplace's equation, $\eta_{\varepsilon}=\eta(\vert x-y\vert /\varepsilon)$ and the operator $D_i=\frac{\partial}{\partial x_i}$.

In one of the steps of the proof of lemma 4.2 in Gilbarg-Trudinger's book, Elliptic PDE of Second Order pg 55, they differentiate $v_{\varepsilon}$ to obtain: \begin{align} D_j v_{\varepsilon}(x)&=\int_{\Omega}D_j(D_i\Phi\eta_{\varepsilon})f(y)\text{ d} y\\ &=\int_{\Omega}D_j(D_i\Phi\eta_{\varepsilon})(f(y)-f(x))\text{ d} y+f(x)\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y\\ &=\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})(f(y)-f(x))\text{ d} y-f(x)\int_{\partial\Omega_0}D_i\Phi\nu^j(y)\text{ d} S_y \end{align}

I am trying to justify these steps, particularly the last two equations and I am having a bit of difficulty.

Below is my work.

Since $f=0$ on $\Omega_0\setminus\Omega$ we can write \begin{equation} D_j v_{\varepsilon}(x)=\int_{\Omega}D_j(D_i\Phi\eta_{\varepsilon})f(y)\text{ d} y=\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})(f(y)-f(x))\text{ d} y+f(x)\int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y. \end{equation}

Now I am trying to justify why \begin{equation} \int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y=-\int_{\partial\Omega_0}D_i\Phi\nu^j(y)\text{ d} S_y. \end{equation}

We note the following: \begin{align} \Phi (x-y)&=\frac{\vert x-y\vert^{2-n}}{n\alpha(n)(2-n)}\\ \Phi_{x_i}&=-\Phi_{y_i}\\ \Phi_{y_i x_j}&=-\Phi_{y_iy_j}\\ \frac{\partial\eta_{\varepsilon}}{\partial x_j}&=-\frac{\partial\eta_{\varepsilon}}{\partial y_j} \end{align}

Thus we have (using the above equations and integration by parts formula as appropriate):

\begin{align} \int_{\Omega_0}D_j(D_i\Phi\eta_{\varepsilon})\text{ d} y&=\int_{\Omega_0}\frac{\partial}{\partial x_j}\left[\frac{\partial\Phi}{\partial x_i} \eta_{\varepsilon}\right]\text{ d}y\\ &=-\int_{\Omega_0}\frac{\partial}{\partial x_j}\left[\frac{\partial\Phi}{\partial y_i} \eta_{\varepsilon}\right]\text{ d}y\\ &=\int_{\Omega_0}\frac{\partial^2\Phi}{\partial y_j\partial y_i} \eta_{\varepsilon}\text{ d}y-\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i} \frac{\partial\eta_{\varepsilon}}{\partial x_j}\text{ d}y\\ &=\int_{\partial\Omega_0}\frac{\partial\Phi}{\partial y_i} \eta_{\varepsilon}\nu^{j}(y)\text{ d}S_y+\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i} \frac{\partial\eta_{\varepsilon}}{\partial y_j}\text{ d}y\\ &=-\int_{\partial\Omega_0}\frac{\partial\Phi}{\partial x_i} \nu^{j}(y)\text{ d}S_y+\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i} \frac{\partial\eta_{\varepsilon}}{\partial y_j}\text{ d}y.\\ \end{align} I can't seem to justify why the last integral disappears. In the above $\varepsilon$ is such that $\text{dist}(x, \partial\Omega)>2\varepsilon$ so that $B(x, 2\varepsilon)\subset\Omega$.

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  • $\begingroup$ What do you mean by G-T? $\endgroup$
    – user426277
    May 25, 2017 at 8:53
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    $\begingroup$ @Idonotknow G-T refers to Gilbarg-Trudinger's book on second order elliptic partial differential equations. $\endgroup$
    – Nirav
    Jun 19, 2017 at 14:02

1 Answer 1

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The last integral will not be zero in general. In fact it will not even tend to zero as $\epsilon \rightarrow 0$ in general. The partial derivative of $\eta_{\epsilon}$ is of order $\epsilon^{-1}$ while the partial derivative of $\Phi$ is of order $|x-y|^{-n+1}$ which, since the integral vanishes outside $B_{\epsilon}$, is of order $\epsilon^{-n+1}$ the term in the last integral is therefore of order $\epsilon^{-n}$ and is being integrated over (up to some constant multiple of the radius) $B_{\epsilon}$ which has volume $\epsilon^n$. Therefore the integral will have constant order and will not tend to zero as $\epsilon \rightarrow 0$.

Actually your mistake is a simple one, you have just applied the integration by parts formula incorrectly. The integral

$\displaystyle\int_{\Omega_0} \dfrac{\partial^2 \Phi}{\partial y_j \partial y_i} \eta_{\epsilon}$

does not equal

$\displaystyle\int_{\partial \Omega_0} \dfrac{\partial \Phi}{\partial y_i} \eta_{\epsilon} \nu^j dS$.

There is another term from the integration by parts formula which exactly cancels the last integral term in your last equation.

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  • $\begingroup$ Lets see\begin{equation}\int_{\Omega_0} \frac{\partial\Phi}{\partial y_j\partial y_i}\eta_{\varepsilon}\text{ d}y=-\int_{\Omega_0}\frac{\partial\Phi}{\partial y_i}\frac{\partial\eta_{\varepsilon}}{\partial y_j}\text{ d}y+\int_{\partial\Omega_0}\frac{\partial\Phi}{\partial y_i}\eta_{\varepsilon}\nu^{j}\text{ d}S_y\end{equation}... I can't believe I missed that...thanks James! $\endgroup$
    – Nirav
    Apr 13, 2013 at 10:08

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