Sunday, 21 August 2016

Help with functional equation F(x,x)+F(x,cx)F(cx,x)F(cx,cx)=0



How can we find F satisfying: exists a c such that F(x,x)+F(x,cx)F(cx,x)F(cx,cx)=0 for all x,y



Several quadratic polynomials in x,y satisfy the above property. I'm trying to generate more examples of functions that satisfy it. I will be very grateful for any other examples.



The property says there is one c that works for all x but given F we have freedom to find a c that works. For example:





  1. If F(x,y)=Ax2+By2+Cxy+Dx+Ey+F and A+C0 we set c=2DA+C and one can easily verify F satisfy the functional equation.

  2. Also if F(x,y)=(A+By)x2+Cy2+Dxy+Ex+Fy+H there are two values of c.

  3. As Jeb pointed in the comments if F(x,y) is even in x and also even in y then c=0 works.



My first idea was to try to transform the equation into a PDE by differentiating wrt x but we do not get a PDE because we are evaluating F at different points: x and cx. Honestly, I'm not even sure of what tags to use.


Answer



For any function, F(x,y), define the function g(x,c) satisfying




g(x,c)=F(x,x)+F(x,cx)F(cx,x)F(cx,cx)



We therefore require that c s.t. x, g(x,c)=0.



If we let x=z+c2 and ˆF(x,y)=F(x+c2,y+c2), then we have
ˆg(z,c)=g(z+c2,c)=ˆF(z,z)+ˆF(z,z)ˆF(z,z)ˆF(z,z)
Therefore, without loss of generality, we can consider the case of c=0, for which we require that ˆg(z,0)=0. In this case, we may split the even and odd components of our function in each of the two dimensions, thus giving us
ˆF(x,y)=A(x,y)+B(x,y)+C(x,y)+D(x,y)
satisfying
A(x,y)=A(x,y)=A(x,y)B(x,y)=B(x,y)=B(x,y)C(x,y)=C(x,y)=C(x,y)D(x,y)=D(x,y)=D(x,y)
From which (letting ˆg(x,c)=ˆg(x,0)=G(x)), we may determine that
G(x)=A(x,x)+A(x,x)A(x,x)A(x,x)+B(x,x)+B(x,x)B(x,x)B(x,x)+C(x,x)+C(x,x)C(x,x)C(x,x)+D(x,x)+D(x,x)D(x,x)D(x,x)=4B(x,x)
Therefore, the requirement is that B(x,x)=0. Recall that B(x,y) represents the component of ˆF(x,y) that is even in y and odd in x. This provides us with a "natural" method of creating a solution. Given a function ˆF(x,y), we can find B(x,y) as
B(x,y)=ˆF(x,y)ˆF(x,y)+ˆF(x,y)ˆF(x,y)4
and therefore, we may define a modified function
˜F(x,y)={3ˆF(x,x)+ˆF(x,x)ˆF(x,x)+ˆF(x,x)4x=yˆF(x,y)otherwise

Note that this leads to a result identical to that given by Ewan Delanoy, it is merely expressed in a different manner. For a more "natural" solution, we can require that B(x,y)=0, and thus define
˜F(x,y)=3ˆF(x,y)+ˆF(x,y)ˆF(x,y)+ˆF(x,y)4
From here, we may choose any c, and shift the function so that
F(x,y)=˜F(xc2,yc2)
and we have a solution for a given c. Notice that, for
F(x,y)=Ax2+By2+Cxy+Dx+Ey+F
with c=2D2A+C we have
ˆF(x,y)=Ax2+By2+Cxy2BDCD+E(2A+C)2A+Cy+C2F+4ACF+4A2FCDE2ADE+BD2AD2(2A+C)2
which, you can see, does not have the x term, which is the only term in a bivariate quadratic polynomial that is even in y and odd in x. Note the 2A+C on the denominator of c - the value given in the question is incorrect.


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