# Ngô Quốc Anh

## July 24, 2010

### Regularity theory for integral equations

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 20:54

My purpose is to derive some regularity result concerning the following integral equation $\displaystyle u(x) = \int_\Omega {\frac{{u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy}$

where $\Omega \subset \mathbb R^n$ is open and bounded and $0<\alpha. To this purpose, in this entry we first consider the equation $\displaystyle u(x) = \int_\Omega {\frac{{f(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy}$

for a suitable choice of $f$.

The case $f \in L^\infty(\Omega)$. We will prove that $u \in C^{1,\beta}(\Omega)$ for any $\beta\in (0,1)$. Indeed, up to a constant factor, the first derivative of $u$ are given by $\displaystyle {D_i}u(x) = \int_\Omega {\frac{{{x_i} - {y_i}}}{{{{\left| {x - y} \right|}^{n + 2 - \alpha }}}}f(y)dy}$.

From this formula, $\displaystyle\left| {{D_i}u({x^1}) - {D_i}u({x^2})} \right| = \mathop {\sup }\limits_\Omega |f|\int_\Omega {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy}$.

By the intermediate value theorem, on the line from $x^1$ to $x^2$, there exists some $x^3$ with $\displaystyle\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right| \leqslant \frac{C}{{{{\left| {{x^3} - y} \right|}^{n + 2 - \alpha }}}}\left| {{x^1} - {x^2}} \right|$.

We put $\displaystyle \delta = 2|x^1-x^2|$.

Since $\Omega$ is bounded, we can find $R>\delta$ with $\Omega \subset B_R(x^3)$ and replace the integral on $\Omega$ by the integral on $B_R(x^3)$. We now write $\displaystyle\begin{gathered} \int_{{B_R}({x^3})} {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} \hfill \\ = \int_{{B_\delta }({x^3})} {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} + \int_{{B_R}({x^3})\backslash {B_\delta }({x^3})} {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} \hfill \\ = {I_1} + {I_2} \hfill \\ \end{gathered}$

Clearly, by co-area formula $\displaystyle\begin{gathered} \int_{{B_\delta }({x^3})} {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} \hfill \\ \qquad\leqslant \int_{B({x^1},2\delta )} {\left| {\frac{{x_1^i - {y^i}}}{{|{x^1} - y{|^{n + 2 - \alpha }}}}} \right|dy} + \int_{B({x^2},2\delta )} {\left| {\frac{{x_2^i - {y^i}}}{{|{x^2} - y{|^{n + 2 - \alpha }}}}} \right|dy} \hfill \\ \qquad\leqslant \int_{B({x^1},2\delta )} {\frac{1}{{|{x^1} - y{|^{n + 1 - \alpha }}}}dy} + \int_{B({x^2},2\delta )} {\frac{1}{{|{x^2} - y{|^{n + 1 - \alpha }}}}dy} \hfill \\ \qquad= C\delta \int_0^{2\delta } {\left[ {\int_{\partial {B_t}({x^1})} {\frac{{d{\sigma _y}}}{{{{\left| {{x^1} - y} \right|}^{n + 1 - \alpha }}}}} + \int_{\partial {B_t}({x^2})} {\frac{{d{\sigma _y}}}{{{{\left| {{x^2} - y} \right|}^{n + 1 - \alpha }}}}} } \right]} dt \hfill \\ \qquad= C\delta {\omega _n}\int_0^{2\delta } {\frac{{dt}}{{{t^{1 - \alpha }}}}} \hfill \\ \qquad= C{\delta ^{\alpha + 1}}. \hfill \\ \end{gathered}$

Besides, $\displaystyle\begin{gathered} {I_2} = \int_{{B_R}({x^3})\backslash {B_\delta }({x^3})} {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} \hfill \\ \qquad\leqslant C\int_{{B_R}({x^3})\backslash {B_\delta }({x^3})} {\frac{{\left| {{x^1} - {x^2}} \right|dy}}{{{{\left| {{x^3} - y} \right|}^{n + 2 - \alpha }}}}} \hfill \\ \qquad= \frac{{C\delta }}{2}\int_\delta ^R {\int_{\partial {B_t}({x^3})} {\frac{{d{\sigma _y}}}{{{{\left| {{x^3} - y} \right|}^{n + 2 - \alpha }}}}} } dt \hfill \\ \qquad= \frac{{C\delta }}{2}\int_\delta ^R {\left[ {\int_{\partial {B_t}({x^3})} {d{\sigma _y}} } \right]\frac{{dt}}{{{t^{n + 2 - \alpha }}}}} \hfill \\ \qquad= \frac{{C\delta }}{2}\int_\delta ^R {n{t^n}{\omega _n}\frac{{dt}}{{{t^{n + 2 - \alpha }}}}} \hfill \\ \qquad= \frac{{nC\delta {\omega _n}}}{2}\int_\delta ^R {\frac{{dt}}{{{t^{2 - \alpha }}}}} \hfill \\ \qquad= \begin{cases}C\delta (\log R - \log \delta ),&\alpha = 1, \hfill \\ C\delta ({R^{\alpha - 1}} - {\delta ^{\alpha - 1}}),&\alpha \ne 1. \end{cases} \hfill \\\end{gathered}$

Hence if $\alpha \ne 1$, we get $\displaystyle {I_1} + {I_2} \leqslant C\left[ {\delta ({R^{\alpha - 1}} - {\delta ^{\alpha - 1}}) + {\delta ^{\alpha + 1}}} \right] \leqslant C\left( {\delta + {\delta ^{\alpha + 1}}} \right)$.

Obviously, for each $\beta \in (0,1)$ fixed, we can find a $C_\beta$ such that $\displaystyle C\left( {\delta + {\delta ^{\alpha + 1}}} \right) \leqslant {C_\beta }{\left( {\frac{\delta }{2}} \right)^\beta } = {C_\beta }{\left| {{x^1} - {x^2}} \right|^\beta }$.

In other words, $I \in C^{0,\beta}(\Omega)$.

The case $f \in C^{0,\beta}(\Omega)$. We will prove that $u \in C^{2,\beta}(\Omega)$ for any $\beta\in (0,1)$. Up to a constant factor, the second derivatives of $u$ are given by $\displaystyle {D_{ij}}u(x) = \int_\Omega {\left[ {{{\left| {x - y} \right|}^2}{\delta _{ij}} - n({x_i} - {y_i})({x_j} - {y_j})} \right]\frac{{f(y)}}{{{{\left| {x - y} \right|}^{n +4 - \alpha }}}}dy}$.

For simplicity, we denote $\displaystyle K(x - y) = \left[ {{{\left| {x - y} \right|}^2}{\delta _{ij}} - n({x_i} - {y_i})({x_j} - {y_j})} \right]\frac{1}{{{{\left| {x - y} \right|}^{n + 4 - \alpha }}}}$

which is nothing but $\displaystyle {D_j}\left( {\frac{{{x_i} - {y_i}}}{{{{\left| {x - y} \right|}^{n + 2 - \alpha }}}}} \right)$.

Obviously $\displaystyle \int_{\mathbb R^n} K(y)dy=0$.

We now write $\displaystyle {D_{ij}}u(x) = \int_\Omega {\left[ {f(y) - f(x)} \right]K(x - y)dy}$.

As before, we get the existence of $x^3$ with the following property $\displaystyle\left| {K({x^1} - y) - K({x^2} - y)} \right| \leqslant \frac{{c\left| {{x^1} - {x^2}} \right|}}{{{{\left| {{x^3} - y} \right|}^{n + 3 - \alpha }}}}$.

We again put $\displaystyle \delta = 2|x^1-x^2|$

and write $\displaystyle {D_{ij}}u({x^1}) - {D_{ij}}u({x^2}) = \int_\Omega {\left[ {f(y) - f({x^1})} \right]K({x^1} - y)dy} - \int_\Omega {\left[ {f(y) - f({x^2})} \right]K({x^2} - y)dy}$.

The last integral can be spitted into two parts $\displaystyle\int_\Omega {} = \int_{{B_\delta }({x^1})} {} + \int_{\Omega \backslash {B_\delta }({x^1})} {} = {I_1} + {I_2}$.

Obviously $\displaystyle {I_1} \leqslant C{\left\| f \right\|_{{C^\beta }}}\int_{{B_\delta }({x^1})} {\left[ {K({x^1} - y){{\left| {{x^1} - y} \right|}^\beta } - K({x^2} - y){{\left| {{x^2} - y} \right|}^\beta }} \right]dy} \leqslant C{\left\| f \right\|_{{C^\beta }}}{\delta ^\beta }$.

Moreover, $\displaystyle\begin{gathered} {I_2} = \int_{\Omega \backslash {B_\delta }({x^1})} {\left[ {f({x^2}) - f({x^1})} \right]K({x^1} - y)dy} + \hfill \\ \qquad\int_{\Omega \backslash {B_\delta }({x^1})} {\left[ {f(y) - f({x^2})} \right]\left[ {K({x^1} - y) - K({x^2} - y)} \right]dy} \hfill \\ \end{gathered}$

and the first integral vanishes. An estimate on $K$ will give us $\displaystyle\frac{1}{{{{\left| {{x^3} - y} \right|}^{n + 3 - \alpha }}}} \leqslant \frac{C}{{{{\left| {{x^1} - y} \right|}^{n + 3 - \alpha }}}}$

due to the domain of $y$. Thus $\displaystyle {I_2} \leqslant C\delta {\left\| f \right\|_{{C^\beta }}}\int_{\Omega \backslash {B_\delta }({x_1})} {\frac{1}{{{{\left| {{x^1} - y} \right|}^{n + 3 - \alpha - \beta }}}}dy} \leqslant C{\left\| f \right\|_{{C^\beta }}}{\delta ^{\alpha + \beta + 2}}$.

Hence there is a $C_\beta$ satisfying $\displaystyle C{\left\| f \right\|_{{C^\beta }}}\left( {{\delta ^\beta } + {\delta ^{\alpha + \beta + 2}}} \right) \leqslant {C_\beta }{\left( {\frac{\delta }{2}} \right)^\beta } = {C_\beta }{\left| {{x^1} - {x^2}} \right|^\beta }$.

The proof now follows.

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