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 Post subject: Do such functions exist? Posted: Thu Nov 26, 2015 8:58 am
 Administrator  Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Find all smooth functions $g$ with domain $\mathbb{R}^2 \setminus \{(0, 0) \}$ such that:

$$\nabla g = \left( { - \frac{y}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$$

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Imagination is much more important than knowledge. Top   Post subject: Re: Do such functions exist? Posted: Thu Nov 26, 2015 5:42 pm

Joined: Thu Nov 26, 2015 5:16 pm
Posts: 1
Good evening.

Let's suppose vector field $\displaystyle \vec{F} = \bigtriangledown f = \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right ) = \left(P(x,y),Q(x,y)\right )$

$\displaystyle P(x,y) = \frac{-y}{x^2+y^2} \Rightarrow f(x,y) = \int P(x,y) dx= \int -\frac{y}{x^2+y^2}dx + g(y)$
$g(y)$ is a function of $y$.

$\displaystyle f(x,y) = \int \frac{-1}{y} \frac{1}{\left(\frac{x}{y} \right)^2+1} dx + g(y) = g(y) - \arctan\left(\frac{x}{y}\right)$
I differentiate:
$\displaystyle \frac{\partial f}{\partial y} = g'(y) + \frac{x}{y^2} \frac{1}{1+\left(\frac{x}{y} \right )^2} =\frac{x}{x^2+y^2} \Leftrightarrow g'(y) = 0 \Rightarrow g(y) = c \in \mathbb{R}$

So, the function is: $\displaystyle f(x,y) = c - \arctan\left(\frac{x}{y} \right ), (x,y) \in \mathbb{R}^2 - (0,0)$

Top   Post subject: Re: Do such functions exist? Posted: Tue Aug 30, 2016 12:09 pm

Joined: Fri Aug 12, 2016 4:33 pm
Posts: 15
Such function NO exist.

Top   Post subject: Re: Do such functions exist? Posted: Tue Aug 30, 2016 1:55 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
Indeed such function does not exist. Reason being that if we consider $\mathcal{C}(R)$ to be a circle of center $0$ and radius $R$ then:

$$0= \oint \limits_{\mathcal{C}(R)} \nabla g \, {\rm d} r >0$$

which obviously is an obscurity.

@S.F.Papadopoulos: What kind of approach do you have?

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