Let \(\varphi(x,y)\) have continuous second partial derivatives and suppose on \(\mathbb R^2\setminus\{0\}\) that \[ \varphi_{xx}+\varphi_{yy}+\varphi_x=0. \] Writing \(r=\sqrt{x^2+y^2}\), suppose also that as \(r\to0\), \[ r\varphi_x\longrightarrow \frac{x}{2\pi r}, \qquad r\varphi_y\longrightarrow \frac{y}{2\pi r}. \] Let \(C_R\) be the circle \(x^2+y^2=R^2\). Show that \[ \int_{C_R}e^x(-\varphi_y\,dx+\varphi_x\,dy) \] is independent of \(R\), and evaluate it.
Solution
Define the one-form \[ \omega \coloneqq e^x\left( -\varphi_y\,dx+\varphi_x\,dy \right). \]
1. The form \(\omega\) is closed on\[ \RR^2\setminus\{0\}. \]
Write \[ \omega=P\,dx+Q\,dy \] with \[ P=-e^x\varphi_y, \qquad Q=e^x\varphi_x. \] Then \[ \begin{aligned} Q_x-P_y &= \partial_x(e^x\varphi_x) -\partial_y(-e^x\varphi_y)\\ &= e^x\varphi_x +e^x\varphi_{xx} +e^x\varphi_{yy}\\ &= e^x \left( \varphi_{xx}+\varphi_{yy}+\varphi_x \right)\\ &=0 \end{aligned} \] by the differential equation.
2. If\[ 0<R_1<R_2, \]then\[ \int_{C_{R_2}}\omega = \int_{C_{R_1}}\omega, \]where both circles have counterclockwise orientation.
Apply Green’s theorem to the annulus \[ \{(x,y):R_1\leq r\leq R_2\}. \] Its positively oriented boundary consists of the outer circle \(C_{R_2}\) counterclockwise and the inner circle \(C_{R_1}\) clockwise. Therefore \[ \int_{C_{R_2}}\omega -\int_{C_{R_1}}\omega = \iint (Q_x-P_y)\,dx\,dy. \] Step 1 makes the right-hand side zero.
3. The integral is independent of \(R>0\).
Step 2 shows that any two positive radii give the same value.
4. Parametrize \(C_R\) counterclockwise by\[ x=R\cos\theta, \qquad y=R\sin\theta, \qquad 0\leq\theta\leq2\pi. \]Then\[ \int_{C_R}\omega = \int_0^{2\pi} e^{R\cos\theta} \left( R\varphi_x\cos\theta +R\varphi_y\sin\theta \right) \,d\theta. \]
Along the parametrization, \[ dx=-R\sin\theta\,d\theta, \qquad dy=R\cos\theta\,d\theta. \] Substitute these expressions into \[ e^x(-\varphi_y\,dx+\varphi_x\,dy). \]
5. As \(R\to0\),\[ R\varphi_x(R\cos\theta,R\sin\theta) \longrightarrow \frac{\cos\theta}{2\pi} \]and\[ R\varphi_y(R\cos\theta,R\sin\theta) \longrightarrow \frac{\sin\theta}{2\pi}, \]uniformly in \(\theta\).
The hypotheses state \[ r\varphi_x-\frac{x}{2\pi r}\longrightarrow0, \qquad r\varphi_y-\frac{y}{2\pi r}\longrightarrow0 \] as \((x,y)\to(0,0)\) through the punctured plane. By the definition of a limit in \(\RR^2\), the error estimates hold simultaneously for all points with sufficiently small \(r\), hence uniformly on each circle \(r=R\). Substituting \[ \frac{x}{r}=\cos\theta, \qquad \frac{y}{r}=\sin\theta \] gives the two limits.
6. One has\[ \lim_{R\to0} \int_{C_R}\omega =1. \]
By steps 4 and 5, \[ e^{R\cos\theta} \left( R\varphi_x\cos\theta +R\varphi_y\sin\theta \right) \] converges uniformly to \[ \frac{ \cos^2\theta+\sin^2\theta }{2\pi} = \frac1{2\pi}. \] Therefore the integrals converge to \[ \int_0^{2\pi}\frac{d\theta}{2\pi} =1. \]
7. For every \(R>0\),\[ \boxed{ \int_{C_R} e^x(-\varphi_y\,dx+\varphi_x\,dy) =1 }. \]
Step 3 says the integral is constant as a function of \(R\), and step 6 computes its limit as \(R\to0\). The constant must therefore equal \(1\).
8. Q.E.D.
Steps 3 and 7 prove radius-independence and evaluate the common value.