# Ngô Quốc Anh

## February 11, 2011

### The implicit function theorem: An ODE example

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 23:35

We want to continue the series of notes involving some applications of  the implicit function theorem. As in the previous note, here we consider the solvability of the following ODE

$x''+\mu x+f(x)=0, \quad J= [0,1]$

with the following boundary conditions

$x(0)=0=x(1)$.

We assume that $f \in C^1(\mathbb R)$ and $f(0)=0$. For the sake of convenience, let us recall the standard implicit function theorem

Theorem (implicit function theorem). Let $X, Y, Z$ be Banachspaces, $U\subset X$ and $V\subset Y$ neighbourhoods of $x_0$ and $y_0$ respectively, $F: U\times V\to Z$ continuous and continuously differentiable with respect to $y$. Suppose also that

$F(x_o, y_o) = 0,\quad F_y^{-1}(x_o,y_o) \in L(Z, Y).$

Then  there exist balls $\overline B_r(x_o) \subset U$, $\overline B_r (y_o) \subset V$ and exactly one map $T: B_r(x_o) \to B_r (y_o)$ such that

$Tx_o = y_o$ and $F(x, Tx) = 0$ on $B_r(x_o)$.

This map $T$ is continuous.

Now we let $X = \mathbb R$, $Z = C(J)$ and

$Y = C_0^2(J) = \{y \in C^2(J): y(0) = y(1) = 0\}$

with norm

$|y| = |y''|_0$

and

$F(\mu, x) = D^2 x + \mu x + f(x)$,

where $D^2 x = x''$. We now calculate Fréchet derivative of $F$ with respect to $x$ in the direction $y$. Since $D^2 \in L(Y, Z)$, we have

${D^2}(x + y) - {D^2}(x) - {D^2}y = 0$.

Besides,

$\mu(x+y)-\mu x -\mu y=0$

and

$f(x+y)-f(x)-f'(x)y \approx 0$.

Therefore,

${F_x}(\mu ,x)y = {D^2}y + \mu y + f'(x)y$

and

${F_x}(\mu ,0)y = {D^2}y + (\mu + f'(0))y$.

Hence ${F_x}(\mu_0,0)$ is a homeomorphism if and only if

$\mu_0+f'(0)\ne m^2\pi^2$

for all $m\in \mathbb N$. Then the implicit function theorem tells us that there is an interval $(\mu_0 - r, \mu_0 + r)$ such that $x =0$ is the only ‘small’ solution of our ODE for $\mu$ in this interval.