Putnam 2010, Problem A-3

Suppose that the function \( h : \mathbb{R}^2 \to \mathbb{R} \) has continuous partial derivatives and satisfies the equation
\[
h(x,y) = a \frac{\partial h}{\partial x}(x,y) + b \frac{\partial h}{\partial y}(x,y)
\]
for some constants \( a, b \). Prove that if there is a constant \( M \) such that \( |h(x,y)| \leq M \) for all \( (x,y) \in \mathbb{R}^2 \), then \( h \) is identically zero.

Solution:
If $a = b = 0$, then the desired result holds trivially, so we assume that at least one of $a, b$ is nonzero. Pick any point $(a_0, b_0) \in \mathbb{R}^2$, and let $L$ be the line given by the parametric equation $L(t) = (a_0, b_0) + (a, b)t$ for $t \in \mathbb{R}$. By the chain rule and the given equation, we have $\frac{d}{dt}(h \circ L) = h \circ L$. If we write $f = h \circ L : \mathbb{R} \to \mathbb{R}$, then $f'(t) = f(t)$ for all $t$. It follows that $f(t) = Ce^t$ for some constant $C$. Since $|f(t)| \leq M$ for all $t$, we must have $C = 0$. It follows that $h(a_0, b_0) = 0$; since $(a_0, b_0)$ was an arbitrary point, $h$ is identically $0$ over all of $\mathbb{R}^2$.