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I want to solve the following:

Assume that $u(x,y)$ is a twice continuously differentiable function on a domain $D$. Let $(x_0,y_0) \in D$, and let $A_\epsilon(x_0,y_0)$ be the average of $u(x,y)$ on the circle centered at $(x_0,y_0)$ of radius $\epsilon$. Show that

$$ \lim_{\epsilon \to 0} \frac{A_\epsilon(x_0,y_0) - u(x_0,y_0)}{\epsilon^2} = \frac{1}{4} \Delta u(x_0,y_0), $$

where $\Delta$ is the Laplacian operator.

Now, I have written:

The second-order Taylor approximation of $u(x,y)$ centered at $(x_0,y_0)$ is:

$$ u(x,y) \approx u(x_0,y_0) + (y - y_0) u_x(x_0,y_0) + (x - x_0) u_y(x_0,y_0) + \frac{1}{2} (x - x_0)^2 u_{xx}(x_0,y_0) + \frac{1}{2} (y - y_0)^2 u_{yy}(x_0,y_0) + (x - x_0) (y - y_0) f_{xy}(x_0,y_0). $$

The second-order Taylor approximation must be a sum of polynomials in $x$ and $y$ of order 2 or less, \ie a linear combination of 1, $x$, $y$, $x^2$, $y^2$, and $xy$, and this approximation is true for sufficiently small $\epsilon$, so the problem is reduced to checking the limit

$$ \lim_{\epsilon \to 0} \frac{A_\epsilon(x_0,y_0) - u(x_0,y_0)}{\epsilon^2} $$

for each of those polynomials in the list taken as $u$ and then looking at the sum.

The functions $1$, $x$, $y$, and $xy$ are harmonic, and so have the mean value property. Therefore $A_\epsilon(x_0,y_0)$ is equal to $u(x_0,y_0)$ for those and the limit is zero.

I believe this is the correct direction, and I would like to finish it off by looking at the limit for $u = x^2$ and $u = y^2$. But I am not sure how to find $A_\epsilon(x_0,y_0)$ for those functions, so maybe this is not the correct direction of proof?

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  • $\begingroup$ I would suggest using polar coordinates centered in $(x_0,y_0)$, say $(r, \theta)$ and then compute the integral $A_{\epsilon}(x_0,y_0) = 1/(\pi \epsilon^2) \int_0^{2\pi} d\theta \int_0^{\epsilon} u(r, \theta) r dr $ as a series expansion in $\epsilon$ and use the result to try and compute the limit $\endgroup$ – Rogelio Molina Jun 4 '15 at 5:28
  • $\begingroup$ You can probably assume that $(x_0,y_0)=(0,0)$ to simplify the calculations... $\endgroup$ – DVD Jun 4 '15 at 5:41
  • $\begingroup$ @DVD: why is that assumption valid? $\endgroup$ – Marcus Emilsson Jun 4 '15 at 5:42
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    $\begingroup$ @Elizabeth Lin: Because the plane is homogeneous. $\endgroup$ – Rogelio Molina Jun 4 '15 at 5:48
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    $\begingroup$ The Laplacian commutes w/the plane translations/shifts... $\endgroup$ – DVD Jun 4 '15 at 6:00
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Consider $(x_0,y_0) = (0,0)$ (there is no loss of generality) and the average over the circle of radius $\epsilon$ centered in the origin:

$$ A_\epsilon =\frac{1}{\pi \epsilon^2}\int_{C(\epsilon)} u(x,y)dxdy $$ expand the function $u(x,y)$ in Taylor series, let $x_1=x, x_2=y$ and $a,b=1,2$:

$$ u(x,y) = u(0,0) + x_a \partial_au|_0 + \frac{1}{2}x_a x_b \partial_a \partial_b u|_0 + \cdots $$ now we integrate this series, to this end notice that the first order integral vanishes (on account of the symmetry of the circle)

$$ \int_{C(\epsilon)}x_a dx dy =0 $$ It will not be difficult to prove also that

$$ \int_{C(\epsilon)} x_a x_b dxdy = \frac{\pi \epsilon^2}{2} \delta_{ab} $$ Performing the integrals is now easy and you will get:

$$ \frac{A_\epsilon -u(0,0)}{\epsilon^2} = \frac{1}{4}(\partial_1^2 + \partial_2^2)u|_0 + O(\epsilon) $$ which leads the claimed result.

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