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I'm trying to solve the following problem:

Let $B=\{(x,y):x^2+y^2\le 1\}$, $\Delta f=x^2y^2$, here $\Delta$ is the Laplacian. Find $$\iint_B{\left( \frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}+\frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x} \right) \mathrm{d}x\mathrm{d}y}.$$

I tried to use Green's formula. Therefore, I define $\mathbf{F}=\left(\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}}\right)$, hence the integral becomes $\iint_B\mathbf{F}\cdot\nabla f\mathrm{d}x\mathrm{d}y$. If we can write $\mathbf{F}=\nabla\phi$, then by $\phi\Delta f=\nabla(\phi\nabla f)-\nabla\phi\nabla f$, we have $$ \iint_B\mathbf{F}\cdot\nabla f\mathrm{d}x\mathrm{d}y=\iint_B\phi\Delta f\mathrm{d}x\mathrm{d}y+\oint_{\partial D}\phi\nabla f\cdot\mathbf{n}\mathrm{d}s. $$ However, the problem is $\mathrm{curl}\mathbf{F}\ne 0$, hence such $\phi$ cannot exist.

The problem is also weird for not assuming any boundary conditions of $f$. How can I solve this problem?

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  • $\begingroup$ Are you sure one of those terms isn’t a negative? $\endgroup$ Commented yesterday
  • $\begingroup$ I'm sure the problem is presented as what I wrote, but I doubt the problem is acutually flawed. $\endgroup$ Commented yesterday
  • $\begingroup$ Your citing of Green's formula is not correct. It should mirror the integration by parts formula so it should be $(\text{original volume integral})=(\text{boundary integral})-(\text{volume integral})$ $\endgroup$ Commented yesterday

1 Answer 1

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Long Comment. Here, we investigate special cases.

Case 1. Let $B_r$ denote the disk of radius $r$ centered at the origin. Then for any integer $n \geq 0$ and with $f(x, y) = (x + iy)^n$,

$$\begin{align*} &\int_{B_1} \biggl( \frac{x}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial y} + \frac{y}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial x} \biggr) \, \mathrm{d}x\mathrm{d}y \\ &= \int_{0}^{1} \int_{\partial B_r} \biggl( \frac{x}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial y} + \frac{y}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial x} \biggr) \, \mathrm{d}s \mathrm{d}r \\ &= \int_{0}^{1} \int_{\partial B_r} \left( \frac{\partial f}{\partial y}, \frac{\partial f}{\partial x} \right) \cdot \mathbf{n} \, \mathrm{d}s \, \mathrm{d}r \\ &= \int_{0}^{1} \left( \int_{B_r} 2\frac{\partial^2 f}{\partial x \partial y} \, \mathrm{d}x\mathrm{d}y \right) \, \mathrm{d}r \\ &= \int_{0}^{1} \left( \int_{B_r} 2in(n-1)(x + iy)^{n-2} \, \mathrm{d}x\mathrm{d}y \right) \, \mathrm{d}r \\ &= \int_{0}^{1} \left( \int_{0}^{2\pi} \int_{0}^{r} 2in(n-1)\rho e^{i(n-2)\theta} \, \mathrm{d}\rho\mathrm{d}\theta \right) \, \mathrm{d}r \\ &= \frac{4\pi i}{3} \mathbf{1}[n = 2]. \end{align*}$$

Case 2. Now, define $f_0$ by

$$ f_0(x, y) = \frac{x^2y^4 + x^4y^2}{24} - \frac{x^6+y^6}{360}, $$

then $f_0$ satisfies $\Delta f_0 = x^2 y^2$ and

$$\begin{align*} \int_{B_1} \biggl( \frac{x}{\sqrt{x^2+y^2}} \frac{\partial f_0}{\partial y} + \frac{y}{\sqrt{x^2+y^2}} \frac{\partial f_0}{\partial x} \biggr) \, \mathrm{d}x\mathrm{d}y &= \int_{B_1} \frac{x y (x^4 + 5x^2y^2 + y^4)}{15\sqrt{x^2+y^2}} \, \mathrm{d}x\mathrm{d}y \\ &= 0. \end{align*}$$

Case 3. Finally, if $f$ is a continuous function on $\overline{B}_1$ such that

$$ f(x, y) = f_0(x, y) + \sum_{n=0}^{\infty} [a_n \operatorname{Re}[(x+iy)^n] + b_n \operatorname{Im}[(x+iy)^n] ] $$

holds in an appropriate sense (such as locally uniform convergence), then the above computations show that

$$ \int_{B_1} \biggl( \frac{x}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial y} + \frac{y}{\sqrt{x^2+y^2}} \frac{\partial f}{\partial x} \biggr) \, \mathrm{d}x\mathrm{d}y = \frac{4\pi}{3} b_2. $$

However, I don't have enough expertise to determine whether any solution of $\Delta f = x^2 y^2$ on $B_1$ admits the above decomposition or not.

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  • $\begingroup$ The integral in my problem is $\frac{x}{r}f_y+\frac{y}{r}f_x$. Though your integral is indeed a more natural one, I have double-checked that I copied the problem correct. $\endgroup$ Commented yesterday
  • $\begingroup$ @MathLearner, Oops, sorry that I misread your problem! I'll check if I can fix it. $\endgroup$ Commented yesterday
  • $\begingroup$ @MathLearner I was not able to answer the question in full generality, but it seems that the value of the integral does depend on $f$. $\endgroup$ Commented yesterday

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