Gronwall's Lemma - Continuously Dependence of Initial Data

Solution 1:

suppose we write the equation in matrix form $$x' = Ax + f(x), x(0) = x_0$$ and the same for $y.$ suppose $w = e^{-At}(x- y).$ then $w$ satisfies $$w' = -Aw + e^{-At}(x'-y')=-Aw + e^{-At}(Aw-f(x) - f(y)) =f(y+w)-f(w),$$ and $$w' = f(y+w)-f(w), \quad w(0) = x_0-y_0 \tag 1$$ the integral representation of the solution of $(1)$ is $$w = w(0) + \int_0^te^{-As}\left(f(x+w) - f(x)\right) \, ds\tag 2$$ taking the norm of $(2),$ we have $$|w| \le |w(0|+\int_0^t\|e^{-sA}\||f(x+w)-f(x)|\, ds$$

now we need to use the fact $$|f(x+w) - f(x) | = \left|\pmatrix{\sin (x_2+w_2)-\sin(x_2)\\\cos(x_1+w_1)-\cos(x_1)}\right|\le L|w|$$ and gronwalls lemma to conclude $$|w| \le |w(0|e^{Lt} \to \|e^{-At}(x- y) \| \le |w(0|e^{Lt}\\ \implies |x-y| \le \|e^{-At}(x-y) \| \, \| e ^{At} \| \le e^{Lt}\, \| e ^{At} \| \, |x(0) - y(0)|$$

that is $$ |x-y|\le e^{Lt}\, \| e ^{At} \| \, |x(0) - y(0)|$$ showing continuous dependence on the initial condition $x(0).$

Solution 2:

To begin, some broad concerns:

Suppose we have a general, non-linear, autonomous ordinary differential equation

$\dot x = f(x), \tag{1}$

where $f(x)$ is defined on some convex, open domain $\Omega \subset \Bbb R^n$. (These hypothesis on $\Omega$ are perhaps not the most general possible, but such $\Omega$ are more than adequate for the present purposes.) We assume $f(x)$ is possessed of a Lipschitz constant $L$ in $\Omega$; that is, for any $x, y \in \Omega$, we have

$\Vert f(y) - f(x) \Vert \le L \Vert y - x \Vert; \tag{2}$

under such circumstances we also say that $f(x)$ is Lipschitz continuous on $\Omega$; I think it's pretty clear that such a function is continuous in the ordinary sense as well; indeed. if

$\Vert y - x \Vert < \delta, \tag{3}$

then

$\Vert f(y) - f(x) \Vert \le L \Vert y - x \Vert < L\delta; \tag{4}$,

if we choose $\epsilon$ freely and set

$\delta = \dfrac{\epsilon}{L}, \tag{5}$

then (4) becomes

$\Vert f(y) - f(x) \Vert \le \epsilon; \tag{6}$

in this manner ordinary continuity is seen to be a consequence of the stronger continuity a la Lipschitz. We shall return to these notions momentarily . . .

We return from the preceding brief excursus to once again take up our main line, the continuity of solutions to (1) with respect to variations in the initial data. We can in fact use the Gronwall inequality to establish such continuity as follows: in the usual manner we cast (1) into integral form:

$x(t) - x(t_0) = \displaystyle \int_{t_0}^t \dot x(s)ds = \int_{t_0}^t f(x(s))ds, \tag{7}$

or

$x(t) = x(t_0) + \displaystyle \int_{t_0}^t f(x(s)) ds. \tag{8}$

If now $y(t)$ is another solution to (1) initialzed at $t = t _0$, we also have

$y(t) = y(t_0) + \displaystyle \int_{t_0}^t f(y(s)) ds; \tag{9}$

subtracting (9) from (8):

$x(t) - y(t) = (x(t_0) - y(t_0)) + \displaystyle \int_{t_0}^t (f(x(s)) - f(y(s)))ds; \tag{10}$

taking norms:

$\Vert x(t) - y(t) \Vert = \left \Vert x(t_0) - y(t_0) + \displaystyle \int_{t_0}^t (f(x(s)) - f(y(s)))ds \right \Vert$ $\le \Vert x(t_0) - y(t_0) \Vert + \left \Vert \displaystyle \int_{t_0}^t (f(x(s)) - f(y(s)))ds \right \Vert$ $\le \Vert x(t_0) - y(t_0) \Vert + \displaystyle \int_{t_0}^t \Vert f(x(s)) - f(y(s)) \Vert ds; \tag{11}$

we use the Lipschitz relation (2) to eliminate the explicit dependence on $f$:

$\Vert x(t) - y(t) \Vert \le \Vert x(t_0) - y(t_0) \Vert + \displaystyle \int_{t_0}^t L \Vert x(s) - y(s) \Vert ds; \tag{12}$

(12) is in perfect form suitable for application of the Gronwall lemma; we immediately conclude that

$\Vert x(t) - y(t) \Vert \le \Vert x(t_0) - y(t_0) \Vert \exp \left( \displaystyle \int_{t_0}^t Ls ds \right) = \Vert x(t_0) - y(t_0) \Vert e^{L(t - t_0)}. \tag{13}$

A few observations on (13): first, we see, taking $t = t_0$, that (13) is consistent with the initial values $x(t_0)$, $y(t_0)$ of $x(t)$, $y(t)$, respectively, for it becomes

$\Vert x(t_0) - y(t_0) \Vert \le \Vert x(t_0) - y(t_0) \Vert e^{L(t_0 - t_0)} = \Vert x(t_0) - y(t_0) \Vert; \tag{14}$

second, if $x(t_0) = y(t_0)$, we find

$\Vert x(t) - y(t) \Vert \le 0, \tag{15}$

whence

$\Vert x(t) - y(t) \Vert = 0 \tag{16}$

for all $t$ for which $x(t)$, $y(t)$ are defined, demonstrating, from yet another point of view, that Lipschitz continuity implies unicity of solutions; third, we see from (13) that, choosing any $\epsilon > 0$ and $T \ge t_0$, and setting $\delta = \epsilon e^{-L(T - t_0)}$, for $\Vert x(t_0) - y(t_0) \Vert < \delta$ we have, since $e^{L(t - t_0)}$ is monotonically increasing in $t$,

$\Vert x(t) - y(t) \Vert \le \Vert x(t_0) - y(t_0) \Vert \exp \left ( \displaystyle \int_{t_0}^t Ls ds \right ) = \Vert x(t_0) - y(t_0) \Vert e^{L(t - t_0)}$ $< \delta e^{L(T - t_0)} = \epsilon e^{-L(T - t_0)}e^{L(T - t_0)} = \epsilon, \tag{17}$

holding uniformly throughout the interval $[t_0, T]$; thus the map $x(0) \mapsto x(t)$ sending $x(0) \in \Omega$ to $x((t) \in C([t_0, T], \Omega)$ is continuous in the topology of uniform convergence on $[t_0, T]$; this effectively establishes continuity with respect to initial conditions on any bounded interval.

In the event that $f(x)$ has a bounded derivative $Df(x)$ in $\Omega$, we may always take

$K = \sup_{x \in \Omega} \Vert Df(x) \Vert \tag{18}$

as a Lipschitz constant for $f(x)$, viz.: for $x, y \in \Omega$, let the path $\gamma: [0, 1] \to \Omega$ be defined by

$\gamma(t) = (1 - s)x + sy; \tag{19}$

note that $\gamma(0) = x$ and $\gamma(1) = y$; also,

$\gamma'(t) = y - x, \tag{20}$

and we have, from the vector version of the fundamental theorem of calculus,

$\Vert f(y) - f(x) \Vert = \left \Vert \displaystyle \int_0^1 f'(\gamma(t)) dt \right \Vert = \left \Vert \displaystyle \int_0^1 Df(\gamma(t)) \gamma'(t) dt \right \Vert$ $\le \displaystyle \int_0^1 \Vert Df(\gamma(t)) \Vert \Vert \gamma'(t) \Vert dt = \int_0^1 \Vert Df(\gamma(t)) \Vert \Vert y - x \Vert dt$ $= \Vert y - x \Vert \displaystyle \int_0^1 \Vert Df(\gamma(t)) \Vert dt \le \Vert y - x \Vert \int_0^1 K dt = K \Vert y - x \Vert, \tag{21}$

which shows that $f(x)$ is Lipschitz continuous in $\Omega$ with Lipschitz constant $K = \sup_{x \in \Omega} \Vert Df \Vert$, as claimed. Furthermore, it is clear that any $L \ge K$ will also serve as a Lipschitz constant for $f$: $\Vert f(y) - f(x) \Vert \le K \Vert y - x \Vert \le L \Vert y - x \Vert$; we thus see that if $Df(x)$ is bounded in $\Omega$, $f(x)$ is Lipschitz conitnuous there.

We apply these considerations to the given system

$x'(t) + 2x(t) -y(t) = \sin{(y(t))}, \tag{22}$

$y'(t)+y(t)-x(t) = \cos{(x(t))}, \tag{23}$

with

$x(0) = x_0, \quad y(0) = y_0; \tag{24}$

setting

$r(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}, \tag{25}$

$B = \begin{bmatrix} -2 & 1 \\ 1 & -1 \end{bmatrix}, \tag{26}$

and

$b(r) = b(x, y) = \begin{pmatrix} \sin y \\ \cos x \end{pmatrix}, \tag{27}$

we write (22)-(23) as

$r' = Br + b(r) \tag{28}$

with

$r(0) = \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}. \tag{29}$

The derivative of $Br + b(r)$ is

$D(Br + b(r)) = DB(r) + Db(r) = B + \begin{bmatrix} 0 & \cos y \\ -\sin x & 0 \end{bmatrix}$ $= \begin{bmatrix} -2 & 1 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 0 & \cos y \\ -\sin x & 0 \end{bmatrix} = \begin{bmatrix} -2 & 1 + \cos y \\ 1 - \sin x & -1 \end{bmatrix}. \tag{30}$

It is easy to estimate $\Vert D(Br + b(r)) \Vert$: we see from (30) that

$\vert (D(Br + b(r)))_{ij} \vert \le 2 \tag{31}$

for $1 \le i, j \le 2$; thus by Lemma 2 of my answer to this question, we have

$\Vert D(Br + b(r)) \Vert \le 2^{3/2} \cdot 2 = 4\sqrt{2}; \tag{32}$

since (31) holds globally on all of $\Bbb R^2$, it follows that $Br + b(r)$ is Lipschitz continuous on $\Bbb R^2$; we can in fact take $C = 4\sqrt{2}$ as the Lischitz constant for $f(x)$, although as we have seen there may be a better (i.e., smaller) one. But the mere existence of $C$ is sufficient for the present purposes.

Since $Br + b(r)$ is Lipschitz continuous, we may directly and immeiadately apply the results developed above and conclude that, for any $T \ge 0$, any two solutions $r_1(t)$, $r_2(t)$ of (21)-(22) satisfy

$\Vert r_1(t) - r_2(t) \Vert \le \Vert r_1(0) - r_2(0) \Vert e^{Ct} \le \Vert r_1(0) - r_2(0) \Vert e^{CT}; \tag{33}$

for $t \in [0, T]$; thus as we have seen the continuity of $r(t)$ with respect to $r(0)$ is established.