Is it a conservative field?

(i) Let \mathbb{D} \subset \mathbb{R}^2 be the unit disk and \partial \mathbb{D}  be its positive oriented boundary. Evaluate the line integral

\displaystyle \mathcal{J} = \ointctrclockwise \limits_{\partial \mathbb{D}} (x-y^3, x^3-y^2)\, {\rm d}(x, y)

(ii) Can you deduce if the function

f(x, y) =(x-y^3, x^3-y^2)

is a conservative field using the above question?

Solution

(i) We are invoking Green’s theorem. Thus:

\begin{aligned} \ointctrclockwise \limits_{\partial \mathbb{D}} \left ( x-y^3, x^3-y^2 \right ) {\rm d}\left ( x, y \right ) &=\iint \limits_{\mathbb{D}} \left ( \frac{\partial }{\partial x}Q(x, y) - \frac{\partial }{\partial y} P(x, y)\right )\, {\rm d}(x, y) \\ &= \iint \limits_{\mathbb{D}} \left ( 3x^2 + 3y^2 \right )\, {\rm d}(x, y)\\ &= 3\int_{0}^{1}\int_{0}^{2\pi} \, {\rm d} \theta \, {\rm d} \rho\\ &= 6\pi \neq 0 \end{aligned}

(ii) If f was a conservative field then then the line integral over all closed curves would have to be 0. But in the previous question we found one closed curve whose line integral is not 0. Thus f  is not a conservative field.

The exercise can also be found in the Jom Forum here.

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Existence of constant (2)

Let f:[0, 1] \rightarrow \mathbb{R} be a continous function such that f(0)=0 and

(1)   \begin{equation*} \int_{0}^{1} f(x) \, {\rm d}x = \int_{0}^{1} x f(x) \, {\rm d}x\end{equation*}

Prove that there exists a c\in (0, 1) such that

\displaystyle \int_{0}^{c} x f(x) \, {\rm d}x = \frac{c}{2} \int_{0}^{c} f(x) \, {\rm d}x

Solution

Let I(x) = \bigintsss_{0}^{x} f(t) \, {\rm d}t and G(x) = \bigintsss_{0}^{x} I(t) \, {\rm d}t. Integrating by parts (1) reveals that

\displaystyle \int_{0}^{1} I(t) \, {\rm d}t =0 \implies G(1) =0

Now let us consider the function \displaystyle K(x) = \frac{G(x)}{x^2} . It holds that K(1)=0. As we can also see using two consecutive DeL’ Hospital’s Rules , it also holds that \lim \limits_{x \rightarrow 0} K(x)=0. So, by Rolle’s theorem there exists a c \in (0, 1) such that

\displaystyle G\left ( c \right ) = \frac{c}{2} I(c)

However integration by parts reveals that

\displaystyle G(c) = c I(c) - \int_{0}^{c} x f(x) \, {\rm d}x

and thus \displaystyle \int_{0}^{c} x f(x) \, {\rm d}x = \frac{c}{2} \int_{0}^{c} f(x) \, {\rm d}x which is the desired output.

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Existence of constant (1)

Let f:[0, 1] \rightarrow \mathbb{R} be a continous function such that

(1)   \begin{equation*} \int_{0}^{1} f(x) \, {\rm d}x = \int_{0}^{1} x f(x) \, {\rm d}x\end{equation*}

Prove that there exists a c \in (0, 1) such that

\displaystyle c f(c) = 2\int_{c}^{0} f(x) \, {\rm d}x

Solution

Let F be a primitive of f. Consider the function G(x)=x^2 F(x). Trivially G(0)=0. Now, we note that:

    \begin{align*} \int_0^1 F(x) \, {\rm d}x &=\int_{0}^{1} (x)' F(x) \, {\rm d}x \\ &= \left [ x F(x) \right ]_0^1 - \int_{0}^{1} x f(x) \, {\rm d}x\\ &= F(1) - \int_{0}^{1} x f(x) \, {\rm d}x \\ &= 0 \end{align*}

because F is of the form F(x)= \bigintsss_{0}^{x} f(t) \, {\rm d}t. Thus

\displaystyle F(1) = \int_{0}^{1} f(t) \, {\rm d}t = \int_{0}^{1} t f(t) \, {\rm d}t

due to the initial assumptions. Applying the Integral Mean Value Theorem  we have that there exists an m \in (0, 1) such that

\displaystyle \int_{0}^{1} F(x) \, {\rm d}x =0 = F(m)

Thus G(0)=0=G(m). The conclusion now follows from Rolle’s theorem.

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An inequality with integrals

Let f:\mathbb{R} \rightarrow (0, +\infty) be a continuous function. Prove that

\displaystyle \int_{0}^{1} f(x) \, {\rm d} x \geq 2^{\bigintsss_0^1 \log_2 f(x) \, {\rm d}x}

Solution

The function \log_2 is concave. Jensen’s inequality in its integral form states that

\displaystyle \int_{0}^{1} \varphi\left ( f(x) \right ) \, {\rm d}x \leq \varphi\left (\int_{0}^{1} f(x) \, {\rm d}x \right )

whenever \varphi is concave. Taking \varphi(x)=\log_2(x) the result follows immediately.

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Constant function

Let f:[0, 1] \rightarrow \mathbb{R} be a continous function such that \bigintsss_0^1 f(x) \, {\rm d}x=1 and

(1)   \begin{equation*}\int_0^1 \left(1-f(x) \right)e^{-f(x)}\, {\rm d}x\leq 0 \end{equation*}

Prove that f(x)=1 forall x\in \mathbb{R}.

Solution

Consider the function g(x)=xe^x-x , \; x \in \mathbb{R}  which is differentiable in \mathbb{R}. We can easily see that g has a global minimum at x_0=0 that is equal to g(0)=0. Visually we have that:

Rendered by QuickLaTeX.com

Clearly as we can see it holds that g(x) \geq 0 forall x \in \mathbb{R}. Also:

(2)   \begin{equation*} \int_{0}^{1}f(x)\, {\rm d}x =1 \Leftrightarrow \int_{0}^{1}\left ( 1-f(x) \right )\, {\rm d}x =0\end{equation*}

Thus (1) gives us:

\begin{aligned} \int_{0}^{1}\left ( 1-f(x) \right )e^{-f(x)}\, {\rm d}x\leq 0 &\Rightarrow e \int_{0}^{1}\left ( 1-f(x) \right )e^{-f(x)}\, {\rm d}x\leq 0 \\ &\Rightarrow \int_{0}^{1}\left ( 1-f(x) \right )e^{1-f(x)}\, {\rm d}x\leq 0\\ &\overset{(2)}{\Rightarrow }\int_{0}^{1}\left ( 1-f(x) \right )e^{1-f(x)}\, {\rm d}x - \\ &\quad \quad \quad - \int_{0}^{1}\left ( 1-f(x) \right )\, {\rm d}x \leq 0\\ &\Rightarrow \int_{0}^{1}\bigg [ \left ( 1-f(x) \right )e^{1-f(x)} - \\&\quad \quad \quad -\left ( 1-f(x) \right )  \bigg]\, {\rm d}x\leq 0 \\ &\Rightarrow \int_{0}^{1}g\left ( 1-f(x) \right )\, {\rm d}x\leq 0 \end{aligned}

Therefore g \left( 1- f(x) \right)=0 . Thus f(x)=1 forall  x \in \mathbb{R}.

The exercise can also be found in mathematica.gr

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