3
$\begingroup$

Consider the energy functional $e(\cdot)$ \begin{align*} e(f,Q)&=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, \end{align*} over the space of \begin{equation*} {\mathcal E}:=\left\{ (Q, f) : \begin{array}{l} Q \in W^{1,4}([a,b],{\bf SO}(n)),\\ Q(a)=Q(b)=I,\\ f \in W^{1,4}[a,b],\\ \dot f>0 \mbox{ ${\cal L}^1$-a.e. on $(a,b)$},\\ f(a)=a, f(b)=b. \end{array} \right\} \end{equation*} It is easy to show that, energy functional $e(\cdot)$ is coercive when $Q\in W^{1,2}$ and $f\in W^{1,2}$, in another words, there exists $d=d(n, a, b)>0$ such that \begin{equation*} e(f, Q) \ge d ( \|Q\|^2_{W^{1,2}} + \|f\|^2_{W^{1,2}}). \end{equation*} Now my question is: The energy functional $e(\cdot)$ is coercive on the space $\mathcal E$ or not? In another words could we find $\gamma=\gamma(n, a, b)>0$ such that \begin{equation*} e(f, Q) \ge \gamma ( \|Q\|^4_{W^{1,4}} + \|f\|^4_{W^{1,4}}). \end{equation*} for all $(f, Q)\in \mathcal E$.

$\endgroup$

4 Answers 4

2
$\begingroup$

No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$ --- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that $$ e(f,Q)\geq e(f,I) $$ for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve $$ \min_{f\in \mathcal{X}} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt, $$ with $\mathcal{X}=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand-> textbook question. The infimum will be in $\bar{\mathcal{X}}$ in general, and $f$ satisfies either the Euler-Lagrange equation, or is constant. To find a simple answer, let us assume that $0<a<b$. Then $f>0$ in $\mathcal{X}$, and if you set $g=f^4$, you find that your problem is also $$ \min_{g\in \mathcal{Y}} \int_a^b \left( g + \frac{1}{16} \left(\frac{dg}{dt}\right)^2\right) \,dt, $$ with $\mathcal{Y}=\{f \in W^{1,2}(a,b)~:~g(a)=a^4,g(b)=b^4, g^\prime>0~a.e.\}$, a convex, open set. Now, the Euler-Lagrange equation is simply $$1+\frac{1}{8}g^{\prime\prime}=0.$$ You can solve it by hand, and playing with it, you find for example that when $$a> \frac{\sqrt{6}}{8}\frac{\left(3554+2(33)^{3/2}\right)^{1/3}}{\sqrt{7}\left(3554+2(33)^{3/2}\right)^{2/3}+(33)^{3/2}+116\left(3554+2(33)^{3/2}\right)^{1/3}+1777)},$$ the solution with $g(a)=a^4$ and $g(b)=b^4$ is strictly increasing and therefore is in $\mathcal{Y}$, whereas otherwise the positivity of the gradient constraint comes into play for suitable $b$s.

$\endgroup$
3
  • $\begingroup$ Indeed what I'm trying to do is to show that the energy functional $e(f, Q)$ admit local minimizers. I would be grateful if you let me know yours ideas. $\endgroup$
    – MSSHD
    Mar 10, 2014 at 14:29
  • $\begingroup$ @MSSHD see updated answer $\endgroup$
    – username
    Mar 10, 2014 at 16:35
  • $\begingroup$ What about $W^{1,4}$, Actually I would like to prove we have the minimizer in $W^{1,4}$ not in $W^{1,2}$. $\endgroup$
    – MSSHD
    Mar 11, 2014 at 5:36
1
$\begingroup$

that is a good idea !

i guess that your conjecture may has a counter-example !

see : http://www.stanford.edu/class/math220b/handouts/calcvar.pdf

that is because :

$e(f, Q) \ge d ( \|Q\|^2_{W^{1,2}} + \|f\|^2_{W^{1,2}})$$\Longrightarrow$$f^4(1+\vert{\nabla{Q}}\vert^2+f^2\vert{\nabla{f}}^2\vert)$$\ge$$d(Q^2+\vert{\nabla{Q}}\vert^2+f^2+\vert{\nabla{f}}^2\vert)$

we select $d=f^4$ to get :

$1+\vert{\nabla{Q}}\vert^2+f^2\vert{\nabla{f}}^2\vert$$\ge$$Q^2+\vert{\nabla{Q}}\vert^2+f^2+\vert{\nabla{f}}^2\vert$$\ge$$4Q\nabla{Q}+4f\nabla{f}$$\Longrightarrow$$f\nabla{f}(4-f\nabla{f})+Q\nabla{Q}(4-\frac{\nabla{Q}}{Q})\le1$

on the other hand ,

$f^4(1+\vert{\nabla{Q}}\vert^2+f^2\vert{\nabla{f}}^2\vert)$$-\gamma(Q^4+\nabla{Q}^4+f^4+\nabla{f}^4)\ge0$$\Longrightarrow$$1+\vert{\nabla{Q}}\vert^2+f^2\vert{\nabla{f}}^2\vert$$\ge$$4Q^2\nabla{Q}^2+4f^2\nabla{f}^2$$\Longrightarrow$$3f^2\nabla{f}^2+Q^2\nabla{Q}^2(4-\frac{1}{Q^2})\le1$

by comparing the two inequality above, we have :

$3f^2\nabla{f}^2+Q^2\nabla{Q}^2(4-\frac{1}{Q^2})\le$$f\nabla{f}(4-f\nabla{f})+Q\nabla{Q}(4-\frac{\nabla{Q}}{Q})$

$\Longrightarrow$$4f\nabla{f}(f\nabla{f}-1)+4Q\nabla{Q}(Q\nabla{Q}-1)\le0$

now, it is clear that your problem is related with the convexity and the concavity of the function you defined on $[a,b]$ !

$\endgroup$
0
$\begingroup$

The Euler-Lagrange Equations of the $e(f, Q)$ over the admissible space $\mathcal E$ arise as the following system \begin{align*} \left \{ \begin {array}{ll} (i)\ \ \frac{d}{dr} \bigg[ f^4 Q^t \frac{d}{dr} Q \bigg] ={\bf 0},\\ \\ (ii)\ \frac{d}{dr} \bigg[ f^2 \dot f \bigg] = 2f^3 +3f^5 \dot f^2 + 2f^3 |dQ|^2, \end{array} \right. \end{align*} Now as the functional $e(\cdot, \cdot)$ is coercive on $\mathcal E$ when we choose the $f, Q$ in $W^{1,2}$ instead of $W^{1,4}$ and an application of direct methods shows the system has a solution. Well, could we claim this result in $\mathcal E$ by $f, Q$ in $W^{1,4}$?

$\endgroup$
1
  • $\begingroup$ you just need to show that the minimiser you obtain is in $W^{1,4}$. Note that $\mathcal{E}$ is not closed, so it is an infimum and not a minimum in general, cf. detailed answer above. $\endgroup$
    – username
    Mar 11, 2014 at 10:01
0
$\begingroup$

It is evident that your functional $e(f, Q)$ has minimizer in the space $W^{1,2}$ since you have coercivity in this space and an application of direct method of calculus of variation givee you the required result in $W^{1,2}$. So the system of Euler-Lagrange equation you wrote in below, has solution in $W^{1,2}$. Now from the first equation (equation (i) you can compute $\dot{Q}$ and $Q$, in fact we have $$ \dot{Q} = \frac{1}{f^4}Q C$$ in which $C$ is a (skew symmetric) constant matrix. therefore $|\dot{Q}|^4 < \infty$ and hence $Q \in W^{1,4}$ (in fact by this argument you can show that the solution $Q$ is smooth). Similarly form the equation (ii) you can compute $f''$ in terms of $f$ and $f'$ and $|\dot{Q}|$ which shows that $f''$ exists (note that $f$ is bounded on [a,b]) so $f'$ is continuous and consequently for the solution $f\in W^{1,2}$ of the above system we have $f\in W^{1,4}$. In fact by the above argument one can show that the solution $f$ of the above system is also smooth (similar to the solution $Q$). Therefore the solution (Q,f) which first was proved to be in the space $W^{1,2}$ is in the space $W^{1,4}$ and moreover by applying the system of equation (i) and (ii) similarly we could prove that this solution $(Q, f)$is a $C^{\infty}$ solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.