Proof. Let $u,v:\mathbin{\mathbb{R}}^{2}\rightarrow\mathbin{\mathbb{R}}$ be solutions of (33), and define $f$ by

$$f(x,y)=\int_{0}^{x}u(s,0){\rm d}s+\int_{0}^{y}v(x,t){\rm d}t.$$

Then $\frac{\partial f}{\partial y}(x,y)=v(x,y)$ and $f(0,0)=0$ are immediate, and

	$\displaystyle\frac{\partial f}{\partial x}(x,y)$	$\displaystyle=u(x,0)+\int_{0}^{y}\frac{\partial v}{\partial x}(x,t){\rm d}t$
		$\displaystyle=u(x,0)+\int_{0}^{y}\frac{\partial u}{\partial y}(x,t){\rm d}t$
		$\displaystyle=u(x,0)+\bigl[u(x,t)\bigr]^{y}_{0}=u(x,y),$

as $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$. This proves the first line of (39), and the second follows by substituting $\frac{\partial f}{\partial x}=u$ and $\frac{\partial f}{\partial y}=v$ into the first equation of (33). The converse is easy. $\square$