# How to prove that a parametrized linear differential equation solution induces a lipschitz function with respect to the parameter?

Let's consider the following linear differential equation:

$X'_{\lambda}(t) = A_{\lambda}(t) X_{\lambda}(t)$ where: $X_{\lambda}(t) = \begin{pmatrix} x_{\lambda}(t) \\ x'_{\lambda}(t) \end{pmatrix}$ and $A_{\lambda}(t) = \begin{bmatrix} 0 & 1 \\ p(t) - \lambda & 0 \end{bmatrix}$ where $p$ is a continuous real function and $\lambda \in \mathbb{R}$ and $X_{\lambda}(0) = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

$\newcommand{\norm}[1]{\lVert #1 \rVert}$ $\newcommand{\abs}[1]{\lvert #1 \rvert}$

Let us assume $\norm{\cdot}$ is the usual norm over vectors and for matrices:

For all matrix $A$ of size $n \times p$ : $\norm{A} = \max\limits_{i \in [[1, n]]} \sum\limits_{j=1}^p \abs{a_{ij}}$.

Let be $R > 0$.

Also, let us define $c = \sup \{ \norm{A_{\lambda}(t)} \mid (t, \lambda) \in [-R, R]^2 \}$.

I am trying to show that: $\norm{X_{\lambda}(t) - X_{\mu}(t)} \leq Re^{2cR} \abs{\lambda - \mu}$ for all $(\lambda, \mu, t) \in [-R, R]^3$.

• For all $t \in [-R, R], \norm{X_\lambda}(t) \leq e^{c\abs{t}}$.
• For all $(s, t, \lambda) \in [-R, R]^3, \norm{X_{\lambda}(t) - X_{\lambda}(s)} \leq ce^{cR} \abs{t - s} (*)$.

What I have tried:

• Creating a new differential equation which is verified by $X_{\lambda}$ and $X_{\mu}$, but I could not find one.
• Using $(*)$ and $X_{\lambda}(0) = X_{\mu}(0)$, but I cannot get $\abs{\lambda - \mu}$ this way in my inequality…
• A hint: try finding an integral inequality satisfied by $\lVert X_\lambda(t)-X_\mu(t)\rVert$ and applying Grönwall's inequality. – user539887 Apr 21 '18 at 17:06
• @user539887 Thank you, it did the job. Do you want me to post the details as the answer, or would you do it? – Raito Apr 21 '18 at 17:55
• Yes, please do it – user539887 Apr 21 '18 at 19:57

Let us denote $$M = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$.

Let's fix $$(\lambda, t, \mu) \in [-R, R]^3$$.

We have, by the fundamental theorem of analysis, because $$X_{\lambda}$$ and $$X_{\mu}$$ are twice continuously differentiable.$$\newcommand{\norm}[1]{\lVert #1 \rVert}\newcommand{\abs}[1]{\lvert #1 \rvert}$$

\begin{align*} X_{\lambda}(t) - X_{\mu}(t) & = X_{\lambda}(t) - X_{\lambda}(0) - (X_{\mu}(t) - X_{\mu}(0)) \\ & = \int_0^t X'_{\lambda}(s) - X'_{\mu}(s) \textrm{d}s \\ & = \int_0^t A_{\lambda}(s) X_{\lambda}(s) - A_{\mu}(s) X_{\mu}(s) \textrm{d}s \\ & = \int_0^t A_{\lambda}(s)(X_{\lambda}(s) - X_{\mu}(s)) \textrm{d}s + (\lambda - \mu)M\int_0^t X_{\mu}(s) \textrm{d}s \end{align*}

By taking norms: \begin{align*} \norm{X_{\lambda}(t) - X_{\mu}(t)} \leq c \int_0^t \norm{X_{\lambda}(s) - X_{\mu}(s)} \textrm{d}s + \abs{\lambda - \mu}R e^{cR} \end{align*}

By Gronwall's lemma:

$$\begin{equation*} \norm{X_{\lambda}(t) - X_{\mu}(t)} \leq Re^{cR}e^{cR} \abs{\lambda - \mu} = Re^{2cR} \abs{\lambda - \mu} \end{equation*}$$