Putnam 2019, Problem B-4

Let \(\mathcal{F}\) be the set of functions \(f(x,y)\) that are twice continuously differentiable for \(x \geq 1\), \(y \geq 1\) and that satisfy the following two equations (where subscripts denote partial derivatives):
\[
\begin{cases}
x f_x + y f_y = xy \ln(xy), \\
x^2 f_{xx} + y^2 f_{yy} = xy.
\end{cases}
\]
For each \(f \in \mathcal{F}\), let
\[m(f) = \min_{s \geq 1} \left( f(s+1, s+1) - f(s+1, s) - f(s, s+1) + f(s, s) \right).\]
Determine \(m(f)\), and show that it is independent of the choice of \(f\).

Solution:
**Solution 1.** We compute that $m(f) = 2\ln 2 - \frac{1}{2}$. Label the given differential equations by (1) and (2). If we write, e.g., $x\frac{\partial}{\partial x}(1)$ for the result of differentiating (1) by $x$ and multiplying the resulting equation by $x$, then the combination $x\frac{\partial}{\partial x}(1) + y\frac{\partial}{\partial y}(1) - (1) - (2)$ gives the equation $2xyf_{xy} = xy\ln(xy) + xy$, whence $f_{xy} = \frac{1}{2}(\ln(x) + \ln(y) + 1)$.

Now we observe that
$$f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s)$$
$$= \int_s^{s+1} \int_s^{s+1} f_{xy} dy dx$$
$$= \frac{1}{2} \int_s^{s+1} \int_s^{s+1} (\ln(x) + \ln(y) + 1) dy dx$$
$$= \frac{1}{2} + \int_s^{s+1} \ln(x) dx.$$

Since $\ln(x)$ is increasing, $\int_s^{s+1} \ln(x) dx$ is an increasing function of $s$, and so it is minimized over $s \in [1,\infty)$ when $s = 1$. We conclude that
$$m(f) = \frac{1}{2} + \int_1^2 \ln(x) dx = 2\ln 2 - \frac{1}{2}$$
independent of $f$.

**Remark.** The phrasing of the question suggests that solvers were not expected to prove that $\mathscr{F}$ is nonempty, even though this is necessary to make the definition of $m(f)$ logically meaningful. Existence will be explicitly established in the next solution.

**Solution 2.** We first verify that
$$f(x,y) = \frac{1}{2}(xy\ln(xy) - xy)$$
is an element of $\mathscr{F}$, by computing that
$$xf_x = yf_y = \frac{1}{2}xy\ln(xy)$$
$$x^2 f_{xx} = y^2 f_{yy} = xy.$$

(See the following remark for motivation for this guess.) We next show that the only elements of $\mathscr{F}$ are $f + a\ln(x/y) + b$ where $a,b$ are constants. Suppose that $f + g$ is a second element of $\mathscr{F}$. As in the first solution, we deduce that $g_{xy} = 0$; this implies that $g(x,y) = u(x) + v(y)$ for some twice continuously differentiable functions $u$ and $v$. We also have $xg_x + yg_y = 0$, which now asserts that $xg_x = -yg_y$ is equal to some constant $a$. This yields that $g = a\ln(x/y) + b$ as desired.

We next observe that
$$g(s+1,s+1) - g(s+1,s) - g(s,s+1) + g(s,s) = 0,$$
so $m(f) = m(f+g)$. It thus remains to compute $m(f)$. To do this, we verify that
$$f(s+1,s+1) - f(s+1,s) - f(s,s+1) + f(s,s)$$
is nondecreasing in $s$ by computing its derivative to be $\ln(s+1) - \ln(s)$ (either directly or using the integral representation from the first solution). We thus minimize by taking $s = 1$ as in the first solution.

**Remark.** One way to make a correct guess for $f$ is to notice that the given equations are both symmetric in $x$ and $y$ and posit that $f$ should also be symmetric. Any symmetric function of $x$ and $y$ can be written in terms of the variables $u = x+y$ and $v = xy$, so in principle we could translate the equations into those variables and solve. However, before trying this, we observe that $xy$ appears explicitly in the equations, so it is reasonable to make a first guess of the form $f(x,y) = h(xy)$. For such a choice, we have
$$xf_x + yf_y = 2xyh' = xy\ln(xy)$$
which forces us to set $h(t) = \frac{1}{2}(t\ln(t) - t)$.