# Find a simple and smooth curve $C$ such that $\displaystyle \int_C\vec{F}\cdot d\vec{r}$ gets its maximum value

I've been trying to solve this problem for a while, but for too long couldn't I continue my partial solution. I would be glad if you could shed some light on my solution.

The task: Given the vector field $$\vec{F}(x,y)\equiv(P(x,y),Q(x,y))$$ such that $$P(x,y)=y^3-3y+xy^2$$ and $$Q(x,y)=3x-x^3+x^2y$$, which is defined on $$D=\{(x,y)\ |\ x^2+y^2\leq2\}$$, find a simple and smooth curve $$C$$ from $$A(1,1)$$ to $$B(-1,-1)$$ which is inside $$D$$ such that:

$$\int_C\vec{F}\cdot d\vec{r}$$

gets its maximum value.

My solution

Let $$C$$ and $$C_0$$ be two curves that satisfy the requirements of the question, such that they are the boundaries of a closed area $$S\subseteq D$$. We will choose $$C$$ to be positively oriented, whereas $$C_0$$ will be negatively oriented.

Now, as $$\vec{F}\in C^1$$, we are able to use Green's Theorem. We can see that:

$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=6-3(x^2+y^2)$$ So according to the theorem, defining $$\Gamma\equiv C\ \cup\ C_0$$:

$$\oint_{\Gamma}\vec{F}\cdot d\vec{r}=\int_C\vec{F}\cdot d\vec{r}-\int_{C_0}\vec{F}\cdot d\vec{r}=\iint_Sr(6-3r^2)drd\theta$$

Accordingly:

$$\int_C\vec{F}\cdot d\vec{r}=\int_{C_0}\vec{F}\cdot d\vec{r}+\iint_Sr(6-3r^2)drd\theta$$

Since $$\displaystyle \iint_Sr(6-3r^2)drd\theta$$ is always positive inside $$D$$ $$(0\leq r\leq\sqrt2)$$, the obvious would be to choose $$C_0$$ such that $$S$$ is the biggest area we can fit inside $$D$$. The problem is, that this might cause $$\displaystyle \int_{C_0}\vec{F}\cdot d\vec{r}$$ to be relatively small.

That is where I got stuck.

P.S.: I also looked at the field, geometrically. It seems that inside $$r=\sqrt2$$ it looks different than the outside. I bet that's a thing I should take into account, but I don't know how.

Thanks!

Without loss of generality, consider a fixed curve $$C_0$$. Specifically, take $$C_0$$ to be the straight line from $$B$$ to $$A$$.

As you noted, $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$$ is non-negative over the domain. Thus, maximizing $$\iint_S (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dA$$ is a matter of taking the maximal area "above" the line parameterized. In particular, if there is a $$C$$ such that the interior region $$S$$ that results from appending $$C$$ to $$C_0$$ is $$S = \{(x,y) \in D: y \geq x\}$$, then it must be that $$\int_C \vec F \cdot d \vec r$$ is maximal.

Thus, the best we could do is $$C = \{(x,y): x^2 + y^2 = 2, y \geq x\}$$.

• Why did you choose $C_0$ to be the straight line from $B$ to $A$? How do you know I can't choose another curve and get a bigger value? – Amit Zach Jun 6 at 9:12
• @AmitZach the point here is that our choice of $C_0$ does not affect $\int_C$ – Omnomnomnom Jun 6 at 9:13
• I still can't understand why. Do you mind elaborate on this? – Amit Zach Jun 6 at 9:15
• In principle, we could evaluate $\int_C F \cdot dr$ by parameterizing $C$. This would give us a formula that has nothing to do with $C_0$. – Omnomnomnom Jun 6 at 9:16
• So why not, for example, choose $C_0$ to be the arc of the circle from $B$ to $A$? That way S would be the biggest (it would be the whole circle), and we will get a bigger value, am I right? – Amit Zach Jun 6 at 9:21

I know this question already has a nice answer from somebody who must be really hungry @Omnomnomnomnom, but let me point out something very interesting:

With some knowledge of calculus of variations it is possible to reframe this problem in terms of an integral to be maximized in terms of two smooth functions. To wit, we would like to maximize/minimize

$$I[x(t), y(t)]=\int_A^B \mathbf{F\cdot dr}=\int_0^{1}(P(x(t), y(t))\frac{dx}{dt}+Q(x(t), y(t))\frac{dy}{dt})dt$$

where $$A=(x(0), y(0)), B=(x(1), y(1))$$. Applying standard calculus of variation procedures we calculate that after an integration by parts:

$$I[x(t)+\epsilon\delta x(t), y(t)+\zeta \delta y(t)]-I[x(t), y(t)]=\int_0^1 dt\Big[\Big(\frac{\partial P}{\partial x}\dot{x}+\frac{\partial Q}{\partial x}\dot{y}-\dot{P}\Big)\epsilon\delta x +\Big(\frac{\partial P}{\partial y}\dot{x}+\frac{\partial Q}{\partial y}\dot{y}-\dot{Q}\Big)\zeta\delta y\Big]$$

Setting all to zero after some final simplification we obtain that curiously, the integral has a critical point for curves that satisfy the following condition:

$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big|_{(x,y)=(x(t),y(t))}=0$$

Apparently the circle $$x^2+y^2=2$$ is the one that extremizes the line integral above, surprise! $$\textbf{#your_professor_knows_things}$$.

P.S 1: The above is just an extremum and NOT a local minimum of the line integral, and this problem has been cleverly restricted to reflect that.

P.S 2: One can prove that for generic vector fields in this special case the extrema are neither minima or maxima, but instead always have directions where you can increase or decrease the value of the integral. Hint: Consider the Hessian

$$H=\pmatrix{a_1&-(a_1+a_2)/2\\-(a_1+a_2)/2& a_2}$$

with $$a_1=\Big(\frac{\partial^2 Q}{\partial x^2}-\frac{\partial^2 P}{\partial y\partial x}\Big)\dot{y}~~,~~ a_2=\Big(\frac{\partial^2 P}{\partial y^2}-\frac{\partial^2 Q}{\partial y\partial x}\Big)\dot{x}$$

It's determinant is negative or zero and therefore it's eigenvalues have opposite signs (in the special case where it's zero the curve is a saddle since it has a flat direction).

• Interesting (and funny!), thank you! – Amit Zach Jun 6 at 11:58