0
$\begingroup$

I am reading the text Elliptic Partial Differential Equations of Second Order by D.Gilbard and N.Trudinger, I am struggling with a particular proof they have given. (p.21)

In the text they are seeking to prove that the function defined on the ball $B=B_R(0)$, by

\begin{equation} u(x)= \begin{cases} \int_\limits{{\partial B}}K(x,y)\phi(y)ds_y \qquad x \in B\\ \phi(y) \qquad \qquad \qquad \qquad \qquad \quad x \in \partial B \end{cases} \end{equation} is continuous on $\partial B$. Here the integral is done over the $y$ variable, $\phi$ is a continuous function on $\partial B$ and $K(x,y)$ is the Poisson kernel;

$$K(x,y)=\frac{R^2-|x|^2}{s_nR|x-y|^n}\qquad x \in B \quad y \in \partial B$$ $s_n$ is the surface area of the unit $n$ sphere.

The proof goes like this;

Let $x_0 \in \partial B$. By continuity of $\phi$ for $\epsilon>0$ we can take $\delta>0$ such that $$|x-x_0|<\delta \implies |\phi(x)-\phi(x_0)|< \epsilon$$as usual. Now if we let $|x-x_0| <\frac{\delta}{2}$ and assume that $|\phi(y)|<M$ on $\partial B$, the author derives the estimate $$|u(x)-u(x_0)|\leq \epsilon + \frac{2M(R^2-|x^2|)R^{n-2}}{\left(\frac{\delta}{2}\right)^n} < 2\epsilon$$

Where the second inequality above follows by taking $|x-x_0|$ sufficiently small.

What I am struggling to understand is how this second inequality follows. By taking $|x-x_0|$ to be sufficiently small I assume what they mean is to introduce $\delta'$ such that $|x-x_0|<\delta'<\frac{\delta}{2}$ so the above estimate holds, and make $\delta'$ smaller as this will squeeze the size of $|x-x_0$| down. This would make sense as then $(2\epsilon,\delta')$ would serve as the required epsilon and delta to give the continuity of $u$.

However I just can't see how they've gotten the upper bound of $2\epsilon$ above by squeezing $|x-x_0|$?

Any help would be appreciated.

$\endgroup$
1
$\begingroup$

$x_0 \in \partial B$ so $|x_0| =R$. Hence $R^{2}-|x|^{2}$ can be made as small as wish by making $|x-x_0|$ small.

$\endgroup$
  • $\begingroup$ Ah how obvious! I am kicking myself. Thank you for your answer Kavi. $\endgroup$ – valcofadden Jul 20 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.