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


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


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.


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


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


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


{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.

Fréchet derivative

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