Suppose \(u,v:\mathbb R^2\to\mathbb R\) are continuously differentiable and \[ \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}. \] Show that there is a function \(f:\mathbb R^2\to\mathbb R\) such that \[ u=\frac{\partial f}{\partial x}, \qquad v=\frac{\partial f}{\partial y}. \]
Prove that there is no function \(f:\mathbb R^2\setminus\{0\}\to\mathbb R\) such that \[ \frac{\partial f}{\partial x}=-\frac{y}{x^2+y^2}, \qquad \frac{\partial f}{\partial y}=\frac{x}{x^2+y^2}. \]
Solution
1. For part 1, define\[ f(x,y) = \int_0^x u(t,0)\,dt + \int_0^y v(x,s)\,ds. \]
The functions \(u\) and \(v\) are continuously differentiable, so both integrals are well defined for every \((x,y)\in\RR^2\) and may be differentiated by the fundamental theorem of calculus and differentiation under the integral sign.
2. The function in step 1 satisfies\[ \frac{\partial f}{\partial y}(x,y)=v(x,y). \]
The first integral in step 1 is independent of \(y\), while the fundamental theorem of calculus gives \[ \frac{\partial}{\partial y} \int_0^y v(x,s)\,ds = v(x,y). \]
3. The same function satisfies\[ \frac{\partial f}{\partial x}(x,y)=u(x,y). \]
Differentiating step 1 with respect to \(x\) gives \[ \frac{\partial f}{\partial x}(x,y) = u(x,0) + \int_0^y \frac{\partial v}{\partial x}(x,s)\,ds. \] By the hypothesis \[ \frac{\partial v}{\partial x} = \frac{\partial u}{\partial y}, \] the right-hand side is \[ u(x,0) + \int_0^y \frac{\partial u}{\partial y}(x,s)\,ds = u(x,y), \] again by the fundamental theorem of calculus.
4. The function \(f\) of step 1 satisfies \(u=f_x\) and \(v=f_y\).
Steps 2 and 3 give exactly \[ u=f_x, \qquad v=f_y. \]
5. For part 2, suppose for contradiction that such an \(f\) exists,and define\[ h(t)=f(\cos t,\sin t), \qquad 0\le t\le2\pi. \]Then\[ h'(t)=1 \]for every \(t\).
The prescribed partial derivatives are continuous on \(\RR^2\setminus\{0\}\), so the chain rule applies along the unit circle. There, \[ x^2+y^2=1, \] and therefore \[ \begin{aligned} h'(t) &= f_x(\cos t,\sin t)(-\sin t) + f_y(\cos t,\sin t)\cos t\\ &= (-\sin t)(-\sin t) + (\cos t)(\cos t)\\ &= 1. \end{aligned} \]
6. The identity in step 5 is impossible.
By the fundamental theorem of calculus, \[ h(2\pi)-h(0) = \int_0^{2\pi}h'(t)\,dt = 2\pi. \] But \[ (\cos0,\sin0) = (\cos2\pi,\sin2\pi) = (1,0), \] so \(h(2\pi)=h(0)\), a contradiction.
7. Consequently, no function satisfying the equations in part 2exists.
The assumption of existence led to the contradiction in step 6.
8. Q.E.D.
Step 4 proves part 1, and step 7 proves part 2.