# 4.3 Minimizing the energy of vector fields on surfaces of revolution  (Page 2/4)

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## Lemma 2

We may replace ${V}_{k}$ with another sequence so that all the ${\varphi }_{k}$ are continuous and differentiable and are $2\pi -$ periodic.

## Proof:

Let ${\varphi }_{1}$ be a linear, piecewise approximation of $\varphi$ ( $\varphi$ and ${\varphi }_{1}$ have the same values at the endpoints of the interval). Because $\varphi$ is piecewise continuous, the square integral of the approximation ${\varphi }_{1}$ will also approximate the square integral of $\varphi$ . The “edges” of ${\varphi }_{1}$ may be smoothed out by composing ${\varphi }_{1}$ with functions of the form ${e}^{\frac{1}{x}}$ while retaining its properties of approximation.

Let $\mathcal{H}$ be the following subset of the set of square integrable functions ${L}^{2}$ :

$\mathcal{H}=\left\{\varphi ,\in ,{\mathcal{L}}^{2},:,\text{there},\phantom{\rule{4.pt}{0ex}},\text{is},\phantom{\rule{4.pt}{0ex}},\text{a},\phantom{\rule{4.pt}{0ex}},\text{sequence},\phantom{\rule{4.pt}{0ex}},{\varphi }_{k},\in ,C,\phantom{\rule{4.pt}{0ex}},\text{such},\phantom{\rule{4.pt}{0ex}},\text{that},\int ,{|\varphi -{\varphi }_{k}|}^{2},\to ,0\right\}.$

We claim that every function $\varphi \in H$ has a weak derivative. To do this, suppose that $\varphi$ is a function which is an ${\mathcal{L}}^{2}$ limit of a sequence ${\varphi }_{k}\in C$ . Since $E\left(\varphi \right) , then we may show that the sequence ${\left({\varphi }_{k}\right)}_{t}$ converges weakly to a limit ${\varphi }_{t}$ . To do so, observe that if ${f}_{k}$ is a sequence of functions in ${\mathcal{L}}^{2}$ such that $||{f}_{k}{||}_{{\mathcal{L}}^{2}}$ is bounded, then there is a weakly convergent subsequence.

## Lemma 3

The derivatives ${\varphi }_{k,\theta }$ , ${\varphi }_{k,t}$ are a bounded sequence in ${\mathcal{L}}^{2}$ .

## Proof:

By Lemma 1, $E\left({\varphi }_{k}\right)\to E\left(\varphi \right)=E$ for a finite $E$ . Recall the definition of $E=\int \int co{s}^{2}\varphi +{\varphi }_{t}^{2}+{\varphi }_{\theta }^{2}dtd\theta$ . Because $E$ is the integral finite sum of these positive quantities, these positive quantities must in turn be finite and thus, bounded.

Then, because these derivatives ${\varphi }_{k,\theta }$ , ${\varphi }_{k,t}$ are a bounded sequence in ${L}^{2}$ , they have corresponding subsequences which converge weakly to functions ${\varphi }_{t},\phantom{\rule{0.166667em}{0ex}}{\varphi }_{\theta }$ .

Hence, we can define the weak energy $E\left(\varphi \right)$ for any function $\varphi \in \mathcal{H}$ . Let

${E}_{H}=\underset{\varphi \in \mathcal{H}}{inf}E\left(\varphi \right),$

then $\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{E}_{H}\le E$ . We will show that this inf is attained.

Take a sequence ${\varphi }_{k}\in \mathcal{H}$ so that $E\left({\varphi }_{k}\right)\to {E}_{\mathcal{H}}.$ Here we prove that a subsequence ${\varphi }_{k}$ have an ${\mathcal{L}}^{2}$ limit $\varphi \in \mathcal{H}.$

For this, note that by definition of $\mathcal{H}$ we can choose for each ${\varphi }_{k}$ a smooth function ${\varphi }_{k}^{\infty }\in \mathcal{C}$ so that

$\parallel {\varphi }_{k}-{\varphi }_{k}^{\infty }{\parallel }_{{\mathcal{L}}^{2}}<\frac{1}{k}.$

## Lemma 4

There is a continuous function $\varphi$ such that ${\varphi }_{k}$ converge uniformly to $\varphi$ .

## Proof:

$|{\varphi }_{k}\left(x\right)-{\varphi }_{k}\left(y\right)|=\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\left|{\int }_{y}^{x},{\varphi }_{k}^{\text{'}},\left(t\right),d,t\right|\le {\left({\int }_{y}^{x},d,t\right)}^{\frac{1}{2}}{\left({\int }_{y}^{x},\varphi ,{\text{'}}_{k}^{2},\left(t\right)\right)}^{\frac{1}{2}}$
$={|x-y|}^{\frac{1}{2}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{\left(\int ,\varphi ,{\text{'}}_{k}^{2},\left(t\right)\right)}^{\frac{1}{2}}\le \sqrt{\frac{|x-y|\left(E+1\right)}{2\pi }}$

Hence, the sequence ${\varphi }_{k}^{\infty }$ converges uniformly to some function $\varphi$ . By the triangle inequality, ${\varphi }_{k}\to \varphi$ in ${\mathcal{L}}^{2}$ , as well. Since $\mathcal{H}$ is a closed set in ${\mathcal{L}}^{2}$ , then $\varphi \in \mathcal{H}$ and accordingly, $\varphi$ has a weak derivative ${\varphi }_{t}$ .

## Lemma 5

The sequence ${\left({\varphi }_{k}\right)}_{t}$ converges weakly to ${\varphi }_{t}$ .

## Proof:

We take advantage of the fact that for every $f\in {\mathcal{L}}^{2}$ , there is a sequence of test functions ${f}_{m}\in {C}_{c}^{\infty }\left(\left(0,1\right)\right)$ such that ${f}_{m}\to f$ in ${\mathcal{L}}^{2}$ .

So then, take any $f\in {\mathcal{L}}^{2}.$ We need to show that

$\left|\int ,{\left({\varphi }_{k}\right)}_{t},f,-,\int ,{\varphi }_{t},f\right|\to 0.$

For this, take using the triangle inequality:

$\left|\int ,{\left({\varphi }_{k}\right)}_{t},f,-,\int ,{\varphi }_{t},f\right|\le \int |{\left({\varphi }_{k}\right)}_{t}||f-{f}_{m}|+\left|\int ,\left({\left({\varphi }_{k}\right)}_{t}-{\varphi }_{t}\right),{f}_{m}\right|+\int |{\varphi }_{t}||f-{f}_{m}|$

and using Cauchy-Schwartz:

$\le \parallel {\left({\varphi }_{k}\right)}_{t}{\parallel }_{{\mathcal{L}}^{2}}\parallel f-{f}_{m}{\parallel }_{{\mathcal{L}}^{2}}+\left|\int ,\left({\left({\varphi }_{k}\right)}_{t}-{\varphi }_{t}\right),{f}_{m}\right|+\parallel {\varphi }_{t}{\parallel }_{{\mathcal{L}}^{2}}{\parallel f-{f}_{m}\parallel }_{{\mathcal{L}}^{2}}$

using the definition of the sequence ${\varphi }_{k}:$

$\le \left({E}_{H}+1\right)\parallel f-{f}_{m}{\parallel }_{{\mathcal{L}}^{2}}+\left|\int ,\left({\left({\varphi }_{k}\right)}_{t}-{\varphi }_{t}\right),{f}_{m}\right|+\parallel {\varphi }_{t}{\parallel }_{{\mathcal{L}}^{2}}{\parallel f-{f}_{m}\parallel }_{{\mathcal{L}}^{2}}$

and using the definition of the weak derivative, since ${f}_{m}$ is a smooth test function:

$=\left({E}_{H}+1\right)\parallel f-{f}_{m}{\parallel }_{{\mathcal{L}}^{2}}+\left|\int ,\left({\varphi }_{k}-\varphi \right),{\left({f}_{m}\right)}_{t}\right|+\parallel {\varphi }_{t}{\parallel }_{{\mathcal{L}}^{2}}{\parallel f-{f}_{m}\parallel }_{{\mathcal{L}}^{2}}$

and using Cauchy-Schwartz again we have:

$\left|\int ,{\left({\varphi }_{k}\right)}_{t},f,-,\int ,{\varphi }_{t},f\right|\le$
$\left({E}_{H}+1\right)\parallel f-{f}_{m}{\parallel }_{{\mathcal{L}}^{2}}+\parallel {\varphi }_{k}{-\varphi \parallel }_{{\mathcal{L}}^{2}}\parallel {\left({f}_{m}\right)}_{t}{\parallel }_{{\mathcal{L}}^{2}}+\parallel {\varphi }_{t}{\parallel }_{{\mathcal{L}}^{2}}{\parallel f-{f}_{m}\parallel }_{{\mathcal{L}}^{2}}.$

Letting $k\to \infty$ FIRST, and then $m\to \infty ,$ we get the result.

By Lemma 3, since ${\left({\varphi }_{k}\right)}_{t}\to {\varphi }_{t}$ weakly, we get

$\int {\varphi }_{t}^{2}\le \underset{k\to \infty }{lim}\int {\left({\varphi }_{k}\right)}_{t}^{2}.$

We may show that $\int co{s}^{2}{\varphi }_{k}\to \int co{s}^{2}\varphi$ . Because $f\left(x\right)=co{s}^{2}\left(x\right)$ is continuous, $co{s}^{2}\left(x\right)\to cos\left(y\right)$ as $|x-y|\to 0$ ; thus, because $\parallel {\varphi }_{k}-\varphi \parallel \to 0$ by definition of weak convergence, $co{s}^{2}{\varphi }_{k}\to co{s}^{2}\varphi$ .

We have shown that the components $\int {\varphi }_{t}^{2}$ and $\int co{s}^{2}\varphi$ allow $\varphi$ to attain the minimum ${E}_{H}$ . Now, note that if $\eta$ is a test function, then $\varphi +\eta \in \mathcal{H},$ and so we automatically get

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