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Suppose $\Omega$ is a bound set in $\mathbb{R}^2$ and $\bar\Omega$ its closure. Assume $f\in C^2(\Omega)\cap C^0(\bar\Omega)$. Moreover, assume $f$ satisfies the partial differential equation$$\frac{\partial^2f(x,y)}{\partial x^2}+\frac{\partial^2f(x,y)}{\partial y^2}=[1+f(x,y)^2]^{-1}$$ Using necessary conditions for maxima in the interior, prove that $f$ takes on its maximum on the boundary of $\Omega$.

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  • $\begingroup$ Which necessary conditions for an interior maximum are known to you? $\endgroup$
    – Thomas
    May 3, 2016 at 17:38
  • $\begingroup$ I am studying out of Postmodern Analysis - Jurgen Jost 3rd edition so the definition within the book of minima and maxima. @Thomas $\endgroup$
    – user336424
    May 3, 2016 at 17:42
  • $\begingroup$ I don't know that book. I was trying to push you a bit into a more explicit statement about first and second derivatives of a two times differentiable function at a point where an interior maximum is attained. $\endgroup$
    – Thomas
    May 3, 2016 at 17:53

1 Answer 1

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Hint: if $f$ takes on an interior maximum in some point, then in that point ($\nabla f=0$) and the second derivative is $\le 0$. Assuming you know a little bit of linear algebra, this has an implication for the trace of the second derivative, which is what you have on the left hand side of your PDE.

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  • $\begingroup$ All I know is what it states in the book, Let $f:(a,b)\rightarrow \mathbb{R}$ be twice differentiable and let $x_0 \in (a,b)$ with $f'(x_0)=0, \space f''(x_0)>0$. The $f$ has a strict local minimum at $x_0$. Conversely, if $f$ has a local minimum at $x_0 \in (a,b)$ and if it is twice differentiable there, then $f''(x_0)\geq 0$... That is what is says, but I'm unsure how to apply it to this problem $\endgroup$
    – user336424
    May 3, 2016 at 18:35
  • $\begingroup$ @user336424 well, this is about a local maximum, so if one would be attained, you'd have $f^{\prime\prime}(x_0) \le 0$. Now note that the right hand side of your equation is strictly positive, and, again, that the left hand side is the trace of the second derivative, which is the sum of the eigenvalues. $\endgroup$
    – Thomas
    May 3, 2016 at 18:41
  • $\begingroup$ I appreciate the help. I'm not super sure about a solution, but I'll take your advice on the hint and in these comments and try to work something out $\endgroup$
    – user336424
    May 3, 2016 at 18:46
  • $\begingroup$ @user336424 It's not too difficult, and if you do it yourself you'll gain more than if I tell you how it works. If the two dimensions confuse you start by considering a real function on $[0,1]$, say, with zero boundary values and strictly positive second derivative. Now assume that function takes on an interior maximum and look at the derivatives in that point. $\endgroup$
    – Thomas
    May 3, 2016 at 19:28
  • $\begingroup$ I understand. As a CS major, I don't dabble in Pure Mathematics that much. I just sort of wanted to understand the reasoning behind coding certain functions and not just straight regurgitate code $\endgroup$
    – user336424
    May 3, 2016 at 19:46

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