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Let $D_1$ denote the disc of radius $1$ with center at the origin of the cartesian $xy$ coordinates. Similarly, let $D_2$ be the disc of radius $2$ with the origin as its center. Do there exist two non-constant functions $f(x,y)$ and $g(x,y)$ that satisfy the following three conditions?:

1) $f(x,y)$ is zero on the boundary of $D_1$ and positive inside $D_1$. Similarly, $g(x,y)$ is zero on the boundary of $D_2$ and positive inside $D_2$.

2) $\int_{D_1}f(x,y)dxdy=\int_{D_2}g(x,y)dxdy$.

3) $\int_{D_1}(f(x,y))^2dxdy=\int_{D_2}(g(x,y))^2dxdy$.

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2 Answers 2

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An example of a continuous pair: Define $f(x)=2(1-x^2)^{1/3},\ g(x)=3(1-x^2/4)^7.$ These define functions on the discs via the radius $r$, so that for area one is e.g. integrating $rf(r).$ Then $$\int_0^1 rf(r)\ dr = \int_0^2 r g(r)\ dr=\frac34,\\ \int_0^1 rf(r)^2\ dr = \int_0^2 r g(r)^2\ dr=\frac65.$$ (Each is multiplied by $2\pi$ for the resulting area.) It actually took some fooling around to get the constants to work out, since without the multipliers the function $g(x)$ was taken to be of the form $f(x/2)$ to keep the algebra simpler. Also inserting the $x^2$ inside before raising to the power was so that the $r$ of $r f(r)$ would give an integral solvable by a simple substitution, so that the forms for the integrals in terms of the powers was manageable.

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Let $f$ be constant $a$ on the interior of $D_1$, $g$ be constant $b$ on $D_1$ and constant $c$ on $D_2 \setminus D_1$. You are asking for $a=b+3c, a^2=b^2+3c^2$ Two equations in three variables will have solutions.

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Would it be two equations in three variables? –  Mark Fantini Jan 7 at 5:20
@Fantini: thanks. Fixed. –  Ross Millikan Jan 7 at 5:21
Thanks, but can you find some non-constant (and probably continuous) solutions? –  Goodarz Mehr Jan 7 at 5:50
Have you tried using Euler's formula? –  CAGT Jan 7 at 6:20
Sure, you can find non-constant solutions. Given any two functions $f,g$ that look different and let $F=\int f, FF=\int f^2, G=\int g, GG=\int g^2$. Now solve $aF=bG, a^2FF=b^2GG$ for $a,b$and you have an answer in $af, bg$. An integral over an area is a very weak constraint on a function. –  Ross Millikan Jan 7 at 13:51

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