Continuous Dependency means that the solution of ODE with initial values would not differ very much from real solution when the errors of parameters are small enough.
and \(\pmb{f}(x,\pmb{y},\pmb{\lambda})\in C(G)\) and satisfies Lypschitz condition with respect to \(\pmb{y}\), i.e. \(\forall (x,\pmb{y}_1,\pmb{\lambda})\), \((x,\pmb{y}_2,\pmb{\lambda})\) we have
where \(\pmb{f}\) is bounded and continuous on region \(G\subset \mathbb{R}\times \mathbb{R}^n \times \mathbb{R}^m\). Assume for all initial points \((x_0,\pmb{y}_0)\), the solution \(\pmb{y}=\pmb{\phi}(x;x_0,\pmb{y}_0,\pmb{\lambda})\) to the ODE exists uniquely on interval \(I_0\), where \(I_0\subset \mathbb{R}\) is finite. Then \(\forall \varepsilon>0\), \(\exists \delta>0\), \(\forall (\xi, \pmb{\eta}, \pmb{\lambda}')\in G\) s.t.
We transfer the dependence on parameters into the dependence on initial values. In the following part, we only consider dependence on initial value.
By contradiction
Assume the above theorem does not hold, then \(\exists \varepsilon_0>0\), \(\forall \delta_n=1/n\), \(\exists(\xi_n,\pmb{\eta}_n, \pmb{\lambda}'_n)\in G\) with
it is easy to show that sequence \(\{\pmb{\phi}(x;\xi_n,\pmb{\eta}_n)\}\) is uniforly bounded and equicontinuous. Actually, if we assume \(M=\max_{(x,\pmb{y},\pmb{\lambda})\in G}|\pmb{f}|\) and denote \(\pmb{\psi}_n(x)=\pmb{\phi}(x;\xi_n,\pmb{\eta}_n)\), then
\[
|\pmb{\psi}_n(x)-\pmb{\eta}|\leq M|I_0|,\quad \forall n
\]
\[
|\pmb{\psi}_n(x_1)-\pmb{\psi}_n(x_2)|\leq M|x_1-x_2|,\quad \forall n
\]
Then by Ascoli-Arzelà Theorem, we have a uniforly convergent subsequence \(\{\pmb{\phi}(x;\xi_{n_j},\pmb{\eta}_{n_j})\}\) which converges to \(\pmb{\psi}(x)\) and they all satisfy
Here interval \(I_0\) is finite is necessary because in the proof we make use of the finite length of the interval. If a solution can satisfy the property on an infinite interval like \((\beta,\infty)\), then we call it Lyapunov stable.
Similar to the above part, if \(\pmb{f}(x,\pmb{y},\pmb{\lambda})\in C^1(G)\) in ODE \(\ref{eq1}\), then we can say the solution is continuous differential on a region. To be more specific, we have the following theorem.
Theorem for Continuous Differentiability
Assume \(\pmb{f}(x,\pmb{y},\pmb{\lambda})\in C(G)\) where
and has continuous partial derivatives with respect to \(\pmb{y}\) and \(\pmb{\lambda}\). Then the solution \(\pmb{y}=\pmb{\phi}(x,\pmb{\lambda})\) to ODE \(\ref{eq1}\) is continuous differential on region \(D\), where
\[
D=\{|x|\leq \alpha, |\pmb{\lambda}|\leq c\}
\]
Show that the partial derivatives can be the limit of divided difference.
Corollary 1
Assume \(\pmb{f}(x,\pmb{y})\in C(R)\) where
\[
R=\{(x,y): |x-x_0|\leq a, |\pmb{y}-\pmb{y}_0|\leq b\}
\]
and has partial derivative with respect to \(\pmb{y}\), i.e. \(\pmb{f}'_{\pmb{y}}(x,\pmb{y})\). Then for all \(\pmb{\eta}\), s.t. \(|\pmb{\eta}-\pmb{y}_0|<\frac{b}{2}\), initial problem
If \(\pmb{y}\) and \(\pmb{\lambda}\) is one dimension, then the above ODE becomes much simpler. And we can use method from Elementary Integration Method.
Q1. Assume \(\phi(x;x_0,y_0,\mu)\) is the solution to ODE
\[
y'=y+\mu(x^2+y^2),\quad y|_{x=x_0}=y_0
\]
Solve \(\frac{\partial y}{\partial \mu}\) when \(x_0=0\), \(y_0=1\), \(\mu=0\).