A functional equation with a quadratic solution

Question

I have the following problem. I have a function $v(x, \theta)$ that can be expressed in two ways, for all $x, \theta \in \Re$:

1. $v(x, \theta) = u(x - \theta)$, where $u$ is strictly concave and symmetric about $0$, and

2. $v(x, \theta) = g_1 (x) f_1 (\theta) + g_2(x) f_2 (\theta) + f_3 (\theta)$

It is clear that $v(x, \theta) = - (x - \theta)^2$ is a solution, as I can set $g_1 = - x^2$, $g_2 = 2x$, $f_1 = 1$, $f_2 = \theta$, and $f_3 = - \theta^2$. Similarly, it is easy to check that $v(x, \theta) = - k_1 (x - \theta)^2 + k_2$ is a solution for all $k_1 > 0$ and $k_2 \in \Re$.

I have a heuristic argument that these are the only solutions, but I have been unable to prove it, or find a reference that would provide an answer.

So the question is to find all functions $v(x, \theta)$ that satisfy (1) and (2).

(I am happy to impose sufficient smoothness on $v(x, \theta)$ if that helps find answers, and I am also willing to assume $f_1$ and $f_2$ are monotonic).

Any guidance would be appreciated!

Answer 1

Let us slightly generalize your functional equation as $u(x,y)=v(x-y)$, $$u(x,y)=\sum_{j=1}^3f_j(x)g_j(y).$$ In your case, $f_3=1$. According to a theorem of Rubel and Gauchman, for a sufficiently smooth function $u$, the necessary and sufficient condition for such representation of sums of products of smooth functions is that the determinant $$\det\left(\frac{\partial^{i+j}u}{\partial x^i\partial y^j}\right)_{i,j=0,\ldots,3}\equiv0.$$ Using $u(x,y)=v(x-y)$ we obtain $u_x=v'$ and $u_y=-v'$, and in general, $$\frac{\partial^{i+j}}{\partial x^i\partial y^j}=(-1)^jv^{(i+j)}.$$ Substituting this to our determinant we obtain the Wrosnkian $$W(v,-v',v'',-v''')\equiv 0.$$ This means that $v,v',v'',v'''$ are linearly dependent, therefore $v$ satisfies a linear differential equation of at most $3$-d order with constant coefficients. So is a generalized exponential sum: $$v(x)=c_1e^{\lambda_1 x}+c_2e^{\lambda_2 x}+c_3e^{\lambda_3 x},$$ if the roots of the characteristic equation are distinct, or $$v(x)=c_1e^{\lambda_1 x}+(c_2+c_3)e^{\lambda_2 x},$$ if there are two roots, or $$v(x)=(c_1+c_2x+c_3x^2)e^{\lambda x},$$ if there is only one root. These are all possible forms of $v$. I leave it to you to determine exactly which of these functions satisfy your specific equation (with $f_3=1$) and which of them are concave or whatever other conditions you want to impose. Solutions that you found correspond to case 3, with $\lambda=0$. Solution that I gave in the comment belongs to case 1 with $c_3=0$ or to case 2 with $c_3=0$.

Ref. MR1024482 Gauchman, Hillel; Rubel, Lee A. Sums of products of functions of x times functions of y. Linear Algebra Appl. 125 (1989), 19–63.

Answer 2

The condition that $u(x-y)=g_1(x)f_1(y) + g_2(x)f_2(y) + f_3(y)$ immediately implies that $u(x)$ satisfies a kind of addition theorem and also linear recurrence relations with constant coefficients. Since there are at most $3$ linearly independent terms on the right side, one recursion is as follows: $u(x)=c_1u(x-1)+c_2u(x-2)+c_3u(x-3)\;$ for some constants $c_1,c_2,c_3.$ The theory of solutions to such recursions is well-known. Of course, you want $u(x)$ to satisfy condition (1) so that restricts the solutions. Aside from the quadratic solutions that you found, and which correspond to double root of $1$, there are solutions $u(x)=-k_1(r^x+r^{-x})+k_2$ where $r\ne 0$ is real, and the $r^x+r^{-x}$ can also be written in the form $\cosh(cx).$