Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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