Ngô Quốc Anh

December 4, 2014

Equations satisfied by standard bubbles and their derivatives in the Euclidean space

Filed under: Uncategorized — Ngô Quốc Anh @ 21:37

This note is purely involved calculation. In \mathbb R^n, let denote by V_{(x,\varepsilon)} (y) the standard bubbles given by

\displaystyle V_{(x,\varepsilon)} (y)= \left( \frac{\varepsilon}{\varepsilon^2+|y-x|^2}\right)^\frac{n-2}{2}.

I am trying to derive some PDE for which the bubbles V_{(x,\varepsilon)} solves.

1. First, we try to calculate \Delta V_{(x,\varepsilon)}. Clearly,

\begin{array}{lcl} {\partial _{{y_i}}}{V_{(x,\varepsilon )}}(y) &=& \displaystyle \frac{{n - 2}}{2}{\left( {\frac{\varepsilon }{{{\varepsilon ^2} + |y - x{|^2}}}} \right)^{ - 1}}{V_{(x,\varepsilon )}}(y){\partial _{{y_i}}}\left( {\frac{\varepsilon }{{{\varepsilon ^2} + |y - x{|^2}}}} \right) \hfill \\ &=& \displaystyle -\frac{{n - 2}}{2}{\left( {\frac{\varepsilon }{{{\varepsilon ^2} + |y - x{|^2}}}} \right)^{ - 1}}{V_{(x,\varepsilon )}}(y)\frac{{2\varepsilon ({y_i} - {x_i})}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}} \hfill \\ &=& \displaystyle -(n - 2)\frac{{{y_i} - {x_i}}}{{{\varepsilon ^2} + |y - x{|^2}}}{V_{(x,\varepsilon )}}(y).\end{array}

Taking derivative again gives

\begin{array}{lcl} \partial _{{y_i}{y_i}}^2{V_{(x,\varepsilon )}}(y) &=& \displaystyle -(n - 2) {\partial _{{y_i}}}\left( {\frac{{{y_i} - {x_i}}}{{{\varepsilon ^2} + |y - x{|^2}}}{V_{(x,\varepsilon )}}(y)} \right) \hfill \\ &=& \displaystyle -(n - 2) {V_{(x,\varepsilon )}}(y){\partial _{{y_i}}}\left( {\frac{{{y_i} - {x_i}}}{{{\varepsilon ^2} + |y - x{|^2}}}} \right) - (n - 2)\frac{{{y_i} - {x_i}}}{{{\varepsilon ^2} + |y - x{|^2}}}{\partial _{{y_i}}}{V_{(x,\varepsilon )}}(y) \hfill \\ &=& \displaystyle -(n - 2) {I_1} - (n - 2) {I_2}.\end{array}

Notice that

\displaystyle {\partial _{{y_i}}}\left( {\frac{{{y_i} - {x_i}}}{{{\varepsilon ^2} + |y - x{|^2}}}} \right) = \frac{1}{{{\varepsilon ^2} + |y - x{|^2}}} - \frac{{2|{y_i} - {x_i}{|^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}}.

Then we have

\displaystyle (n - 2){I_1} = (n - 2)\left( {\frac{1}{{{\varepsilon ^2} + |y - x{|^2}}} - \frac{{2|{y_i} - {x_i}{|^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}}} \right){V_{(x,\varepsilon )}}(y)

and

\displaystyle (n - 2){I_2} = - {(n - 2)^2}\frac{{{{({y_i} - {x_i})}^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}}{V_{(x,\varepsilon )}}(y).

Hence,

\begin{array}{lcl} \partial _{{y_i}{y_i}}^2{V_{(x,\varepsilon )}}(y) &= \displaystyle - (n - 2){V_{(x,\varepsilon )}}(y)\left( {\frac{1}{{{\varepsilon ^2} + |y - x{|^2}}} - 2\frac{{{{({y_i} - {x_i})}^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}} - (n - 2)\frac{{{{({y_i} - {x_i})}^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}}} \right) \hfill \\ &= \displaystyle - (n - 2)\left( {\frac{1}{{{\varepsilon ^2} + |y - x{|^2}}} - n\frac{{{{({y_i} - {x_i})}^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}}} \right){V_{(x,\varepsilon )}}. \hfill \\ \end{array}

Therefore,

\begin{array}{lcl} \Delta {V_{(x,\varepsilon )}} &=& \displaystyle\sum\limits_i {\partial _{{x_i}{x_i}}^2{V_{(x,\varepsilon )}}} \hfill \\ &=& \displaystyle - (n - 2)\left( {\frac{n}{{{\varepsilon ^2} + |y - x{|^2}}} - n\frac{{|y - x{|^2}}}{{{{({\varepsilon ^2} + |y - x{|^2})}^2}}}} \right){V_{(x,\varepsilon )}} \hfill \\ &=& \displaystyle - n(n - 2){\left( {\frac{\varepsilon }{{{\varepsilon ^2} + |y - x{|^2}}}} \right)^2}{V_{(x,\varepsilon )}} \hfill \\ &=& \displaystyle - n(n - 2){({V_{(x,\varepsilon )}})^{\frac{{n + 2}}{{n - 2}}}}.\end{array}

Thus, V_{(x,\varepsilon)} solves

\displaystyle\Delta {V_{(x,\varepsilon )}} + n(n - 2){({V_{(x,\varepsilon )}})^{\frac{{n + 2}}{{n - 2}}}} = 0.

2. Second, we try to find equations satisfied by \partial_\varepsilon V_{(x,\varepsilon)}. Clearly, by differentiating both sides with respect to \varepsilon, we obtain

\displaystyle \Delta {\partial _\varepsilon }{V_{(x,\varepsilon )}} + n(n - 2){({V_{(x,\varepsilon )}})^{\frac{4}{{n - 2}}}}{\partial _\varepsilon }{V_{(x,\varepsilon )}} = 0.

Hence \partial_\varepsilon V_{(x,\varepsilon)} solves the linearized equation

\displaystyle \Delta V + n(n - 2){({V_{(x,\varepsilon )}})^{\frac{4}{{n - 2}}}}V = 0.

3. Third, we try to find equations satisfied by \partial_{x_i} V_{(x,\varepsilon)}. Clearly by differentiating again, we also obtain

\displaystyle \Delta {\partial _{{x_i}}}{V_{(x,\varepsilon )}} + n(n - 2){({V_{(x,\varepsilon )}})^{\frac{4}{{n - 2}}}}{\partial _{{x_i}}}{V_{(x,\varepsilon )}} = 0.

Hence \partial_{x_i} V_{(x,\varepsilon)} also solves the linearized equation

\displaystyle \Delta V + n(n - 2){({V_{(x,\varepsilon )}})^{\frac{4}{{n - 2}}}}V = 0.

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