Consider the problem $$ (*)\begin{cases} x''(t)= F(x(t)) \\x(0)=P, x(1)=Q \end{cases}$$ where $P,Q\in\mathbb{R}^3$, $x=(x_1,x_2,x_3)$ and $F=-\nabla U$ for some potential $U:\mathbb{R}^3\rightarrow\mathbb{R}$. With the assumption of $U$ is upper bounded, I want to show that $(*)$ admits weak solution. This could be done by minimizing the functional $$\phi (x)= \int_0^1 \frac{1}{2} \Vert x' \Vert ^2 - U(x)\ dt $$on $A=\lbrace x\in H^1(0,1)^3: x(0)=P, x(1)=Q\rbrace$, where $\Vert x'\Vert^2=x_1'^2+ x_2'^2+ x_3'^2 $. One classical result in minimizing functional is essentially: $\phi$ coercive and weakly lower semicontinuous $\Rightarrow$ $\phi$ attains its minimum. But I don't know how to prove this two conditions for $\phi$. Any help/hint would be appreciated. Thanks in advance!

  • $\begingroup$ Can you define coercive in this context $\endgroup$ – George Dewhirst May 6 at 17:06
  • $\begingroup$ $\phi(x)\rightarrow +\infty $ as $\Vert x\Vert \rightarrow\infty$ $\endgroup$ – Senna May 6 at 17:08
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    $\begingroup$ I figure the integral is dt $\endgroup$ – George Dewhirst May 6 at 17:18
  • $\begingroup$ Yes, that was a typo. Thanks! $\endgroup$ – Senna May 6 at 17:21

I'm going to assume that $U$ is continuous (which is not that obvious because the condition $-\nabla U = f$ could be in the sense of distributions). Let's start by making the boundary conditions homogeneous: Take $x_0$ a fixed function that satisfies $$x_0(0) = P, x_0(1) = Q$$ for instance $x_0(t) = P + (P - Q)t$ will do. Now we consider the problem in the new variable $y = x - x_0$ which is in $H^1_0(0,1)^3$ $$\phi(y) = \int_0^1 \frac12 \|y' - (P - Q)\|^2 - U(y - x_0)dt$$

For the lower semi-continuity: the first part $$\int_0^1 \frac12 \|y' - (P - Q)\|^2dt$$ is clearly continuous in the strong topology and convex, so it is lower semi continuous in the weak topology. For the second part $$\int_0^1 -U(y - x_0)dx$$ we can simply use Fatou's lemma. Let say that $U$ is bounded by $U_0$, hence $U_0 - U$ is positive. Lets take $(y_n)_{n\in \mathbb N}\subseteq H^1_0(0,1)^3$ that converges weakly to $y\in H^1_0(0,1)^3$. Then by compact embedding, $y_n$ converges strongly in $C^0([0,1])^3$ so

$$\liminf \int_0^1 U_0 - U(y_n - x_0) dt \geq \int_0^1 \liminf U_0 - U(y_n - x_0) dt = \int_0^1 U_0 - U(y - x_0) dt $$ which proves sequential lower semi-continuity(which is enough for existence of minimisers). (Here you actually can prove that $U(y_n - x_0$ also converge uniformly so you can use the usual riemman convergece theorem instead of Fatou, but either way, it is good to now this way of proving it because sometimes the only thing that you can proof is point-wise convergence (a.e.))

The coercivity is trivial from the inequality $$ \phi(y) \geq C\|y\|_{H_0^1}^2 + \|P - Q\|^2 - 2\|y\|_{H_0^1} \|P - Q\|- U_0$$ where we used the Poincare inequality.

  • $\begingroup$ There is something that is confusing me. $$ \phi(y)\geq 1/2\int_0^1 ||y'||^2 +1/2||P-Q||^2-<y',P-Q>dt-U_0 ;$$ and $$ \phi(y)\geq 1/2\int_0^1 ||y'||^2dt +1/2||P-Q||^2 -\left(\int_0^1 ||y'||^2dt\right) ^{1/2} \left(\int_0^1 ||P-Q||^2dt\right) ^{1/2} -U_0 = $$ $$ =1/2||y||^2_{H^1_0} + C' - C||y||_{H^1_0} -U_0 $$ where I am using the equivalent norm $ ||y||^2=\int_0^1 ||y'||^2 dt$ on $H_0^1(0,1)^3$. So it seems that poincare ineq. Is not necessary? $\endgroup$ – Senna May 6 at 20:46
  • $\begingroup$ @R.N.Marley well, how do you prove that the norms are equivalent without Poincare? $\endgroup$ – themaker May 6 at 20:49
  • $\begingroup$ Yes it is necessary for that. But I mean that is not needed in the chain of inequalities? $\endgroup$ – Senna May 6 at 20:50
  • $\begingroup$ @R.N.Marley yeah, it is not necessary in that sense, I should mentioned it for clarity. $\endgroup$ – themaker May 6 at 21:28

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