Maximizing u(x,y): Using the Harmonic Function

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my function is u(x,y) = xy(x2 - y2 + 2).

Using the fact that this fxn is harmonic, how do I find the max of u on [0,1]x[0,1]?
 
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Well, do you know the theorem that says if a function is harmonic on a set, then any extreme values must occur on the boundary of the set?

The boundary of [0,1]x[0,1] consists of 4 line segments but the function, reduced to those segments, is still harmonic so any extreme values must occur at the endpoints of those segments.
 
so I guess the max would have to be at (1,1) then. So would i just let u(x,y) = u(1,1) and use the value that comes out?
 
Yes! The max and min must occur at the four "corners" (1,1), (1, 0), (0, 1) and (0, 0). You can see that u(1,1)= 2 while the others are all 0.
 
Thank you so much. That was very helpful.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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