2
$\begingroup$

Let $X=\{u\in C^2([0,1],\mathbb R),u(0)=u'(0)=0 \}$. Let $F : X \rightarrow \mathbb R$ such as $F(u)=\frac{1}{2}\int_0^1(u')^2dt-\int_0^1\cos(u)dt$.

  1. For which norm $X$ is a Banach space ?
  2. Show that $F$ is differentiable in every point and calculate its differential.
  3. Find the critical point of $F$ and determine its nature.
  1. I had no problem to solve this question : I know that $C^2([0,1],\mathbb R)$ is a Banach space with the norm : $||f||=||f||_\infty+||f'||_\infty+||f''||_\infty$. Furthermore $X$ is closed in $C^2([0,1],\mathbb R)$ ans is a subspace, so $X$ is a Banach space.

  2. Let :
    $f_1:C^1([0,1],\mathbb R)\rightarrow\mathbb R,u\mapsto \int_0^1 u dt$
    $f_2:C^2([0,1],\mathbb R)\rightarrow C^1([0,1],\mathbb R),u\mapsto u'$
    $f_3:C^1([0,1],\mathbb R)\rightarrow C^1([0,1],\mathbb R),u\mapsto u^2$
    $f_4:C^2([0,1],\mathbb R)\rightarrow\mathbb R,u\mapsto \int_0^1 u dt$
    $f_5:C^2([0,1],\mathbb R)\rightarrow C^2([0,1],\mathbb R),u\mapsto \cos(u)$
    $f_1,f_2,f_4$ are linear and continuous : For example $|f_1(u)|\le ||u||_\infty\le||u||_\infty+||u'||_\infty$ so $f_1$ is continuous.
    Also $f_3$ is differentiable and we have : $f_3'(u)(h)=2uh$ and $f_5$ is differentiable with $f_5'(u)(h)=-h\sin(u)$.
    Thus we have : $$F'(u)(h)=\int_0^1u'h'dt+\int_0^1h\sin(u)dt$$

  3. I need to find the $u \in X$ such as $\forall h, F'(u)(h)=0$. I am not really sure how to solve this question. I know that $u=0$ is such a critical point but are there other such points ?
    To study the nature of the critical point in $u=0$, I want to show that it is a minima : indeed if $u$ is not constant on an interval the fist term of $F$is $>0$, so since $u \in C^2([0,1],\mathbb R)$, if $u$ is non constant the first term is $>0$. With a similar reasonning we need $u=0 \pmod{2\pi}$ to minimise the second term and the condition $x\in X$ impose that $u=0$ to minimise $F$.

All in all I want to know how to show that $0$ is the only critical point and I would like to know if there is a simplier way to determine the nature of this critical point.

1 Answers 1

0

With an integration by part $\forall u \in X$ such as $u(1)=0$ we have : $$F'(u)(h)=\int_0^1 (-u''+\sin(u))hdx$$ Thus all critical points $u\in X$ must verify the Cauchy problem : $$\begin{cases} -u''+\sin(u)=0 \\ u(0)=u'(0)=0\end{cases}$$

So with Cauchy–Lipschitz theorem $u=0$ is the unique solution to this differential equation.