Ngô Quốc Anh

October 4, 2012

Integral representation of radially symmetric smooth solutions of bilaplacian equations

Filed under: PDEs — Ngô Quốc Anh @ 8:16

This entry came from one of my calculation concerning to my recent work. Let us consider the following PDE $\displaystyle \Delta^2 u = -u^{-q} \quad \text{ in }\mathbb R^3$

with $q>3$.

We assume throughout this entry that the function $u$ is radially symmetric and also smooth. In this manner, we can write $u=u(r)$ where $r \geqslant 0$ is the radius. Using the calculation the previous entry, we obtain the following ODE $\displaystyle\frac{1}{{{r^4}}}({r^4}{u^{(3)}})' = - {u^{ - q}}.$

The aim of this entry is to transform the above ODE to its corresponding integral equation. To understand the procedure, we do it step by step.

First, we multiply both sides by $r^4$ and integrate over $[0,r]$ to gain the following $\displaystyle {r^4}{u^{(3)}}(r) = - \int_0^r {{s^4}u{{(s)}^{ - q}}ds} .$

Next, we divide both sides by $r^4$ and integrate over $[o,r]$ again to get $\displaystyle u''(r) - u''(0)= - \int_0^r {\left( {\frac{1}{{{t^4}}}\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)dt} = \frac{1}{3}\int_0^r {\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)d(\frac{1}{{{t^3}}})} .$

By integration by parts, we get $\displaystyle u''(r) = u''(0)+\frac{1}{{3{r^3}}}\int_0^r {{s^4}u{{(s)}^{ - q}}ds} - \frac{1}{3}\int_0^r {tu{{(t)}^{ - q}}dt}.$

We now repeat the above procedure to get $\displaystyle\begin{gathered} u'(r) = u'(0) + u''(0){\kern 1pt} r + \frac{1}{3}\int_0^r {\left( {\frac{1}{{{t^3}}}\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)dt} - \frac{1}{3}\int_0^r {\left( {\int_0^t {su{{(s)}^{ - q}}ds} } \right)dt} \hfill \\ \qquad= u'(0) + u''(0){\kern 1pt} r - \frac{1}{6}\int_0^r {\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)d(\frac{1}{{{t^2}}})} - \frac{1}{3}\int_0^r {\left( {\int_0^t {su{{(s)}^{ - q}}ds} } \right)dt} . \hfill \\ \end{gathered}$

Since $\displaystyle\int_0^r {\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)d(\frac{1}{{{t^2}}})} = \frac{1}{{{r^2}}}\int_0^r {{s^4}u{{(s)}^{ - q}}ds} - \int_0^r {{t^2}u{{(t)}^{ - q}}dt}$

and $\displaystyle\int_0^r {\left( {\int_0^t {su{{(s)}^{ - q}}ds} } \right)dt} = t\int_0^t {su{{(s)}^{ - q}}ds} - \int_0^r {{t^2}u{{(t)}^{ - q}}dt}$

we obtain $\displaystyle u'(r) = u'(0) + u''(0)r - \frac{1}{{6{r^2}}}\int_0^r {{s^4}u{{(s)}^{ - q}}ds} + \frac{1}{2}\int_0^r {{t^2}u{{(t)}^{ - q}}dt} - \frac{t}{3}\int_0^t {su{{(s)}^{ - q}}ds}.$

Again by integrating, $\displaystyle\begin{gathered} u(r) = u(0) + u'(0)r + \frac{u''(0)}{2} {r^2} - \frac{1}{6}\int_0^r {\frac{1}{{{t^2}}}\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)dt} \hfill \\ \qquad+ \frac{1}{2}\int_0^r {\left( {\int_0^t {{s^2}u{{(s)}^{ - q}}ds} } \right)dt} - \frac{1}{3}\int_0^r {t\left( {\int_0^t {su{{(s)}^{ - q}}ds} } \right)dt} \hfill \\ \qquad =u(0) + u'(0)r +\frac{u''(0)}{2}{r^2} + \frac{1}{6}\int_0^r {\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)d(\frac{1}{t})} \hfill \\ \qquad+ \frac{r}{2}\int_0^r {{s^2}u{{(s)}^{ - q}}ds} - \frac{1}{2}\int_0^r {{t^3}u{{(t)}^{ - q}}dt} - \frac{1}{6}\int_0^r {\left( {\int_0^t {su{{(s)}^{ - q}}ds} } \right)d({t^2})}. \hfill \\ \end{gathered}$

Since $\displaystyle\int_0^r {\left( {\int_0^t {{s^4}u{{(s)}^{ - q}}ds} } \right)d(\frac{1}{t})} = \frac{1}{r}\int_0^r {{s^4}u{{(s)}^{ - q}}ds} - \int_0^r {{t^3}} u{(t)^{ - q}}dt$

and $\displaystyle\int_0^r {\left( {\int_0^t {su{{(s)}^{ - q}}ds} } \right)d({t^2})} = {r^2}\int_0^r {su{{(s)}^{ - q}}ds} - \int_0^r {{t^3}} u{(t)^{ - q}}dt$

we eventually arrive at $\displaystyle\begin{gathered} u(r) = u(0) + u'(0)r +\frac{u''(0)}{2}{r^2} + \frac{r}{2}\int_0^r {{t^2}u{{(t)}^{ - q}}dt} \hfill \\ \qquad- \frac{1}{2}\int_0^r {{t^3}u{{(t)}^{ - q}}dt} + \frac{1}{{6r}}\int_0^r {{t^4}u{{(t)}^{ - q}}dt} - \frac{{{r^2}}}{6}\int_0^r {tu{{(t)}^{ - q}}dt} . \hfill \\ \end{gathered}$

By definition, $u'(0)=\lim_{r \to 0}\frac{u(r)-u(0)}{r}=0$, therefore, $\displaystyle\begin{gathered} u(r) = u(0) + \frac{u''(0)}{2}{r^2} + \frac{r}{2}\int_0^r {{t^2}u{{(t)}^{ - q}}dt} - \frac{1}{2}\int_0^r {{t^3}u{{(t)}^{ - q}}dt} \hfill \\ \qquad+ \frac{1}{{6r}}\int_0^r {{t^4}u{{(t)}^{ - q}}dt} - \frac{{{r^2}}}{6}\int_0^r {tu{{(t)}^{ - q}}dt} . \hfill \\ \end{gathered}$

We can rewrite $u(r)$ as follows $\displaystyle\begin{gathered} u(r) = u(0) + \overbrace {\left( {\frac{u''(0)}{2} - \frac{1}{6}\int_0^\infty {tu{{(t)}^{ - q}}dt} {\kern 1pt} } \right)}^\beta {r^2} + \frac{r}{2}\int_0^r {{t^2}u{{(t)}^{ - q}}dt} \hfill \\ \qquad - \frac{1}{2}\int_0^r {{t^3}u{{(t)}^{ - q}}dt} + \frac{1}{{6r}}\int_0^r {{t^4}u{{(t)}^{ - q}}dt} + \frac{{{r^2}}}{6}\int_r^\infty {tu{{(t)}^{ - q}}dt} \hfill \\ \end{gathered}$

provided the integral in the round bracket converges. By using $\displaystyle u''(r) = u''(0)+\frac{1}{{3{r^3}}}\int_0^r {{s^4}u{{(s)}^{ - q}}ds} - \frac{1}{3}\int_0^r {tu{{(t)}^{ - q}}dt}$

we know that $\displaystyle 2\beta =\lim_{r \to \infty}\left( u''(r) - \frac{1}{{3{r^3}}}\int_0^r {{s^4}u{{(s)}^{ - q}}ds} \right).$