Conditions for uniqueness and Global/Local existence of solution

Suppose the function f satisfies the following conditions $\exists$ finite constants $T,r,h$ and $K,S,T$ Lipschitz Condition for local existence.

$$\Vert f(t,x)-f(t,y)\Vert \leq k \Vert X-Y\Vert \quad \forall [X,Y] \in B_{r} \quad \forall \,t \in [0,T]$$ (4.1)

$$\Vert f(t,x_{0}\Vert \leq h \quad \forall t \in [0,T]$$ (4.2)

where $B_{r}= [ X \in \mathbb{R}^{n}: \Vert X-X_{0}\Vert \leq r ]$ then it has exactly one solution over $[0,\infty]$ whenever $\delta$ is sufficiently small to satisfy

$$h \delta e^{k\delta} < r$$ (4.3)

$$\delta < \min [T, \frac{\rho}{k}, \frac{r}{(k r +h)}]$$ (4.4)

Lipschitz Condition for Global existence.

$$\Vert f(t,x)-f(t,y)\Vert \leq k\Vert X-Y\Vert \quad \forall [X,Y] \in \mathbb{R}^{n},\;\forall \; t \in [0,\infty]$$ (4.5)

Corollary 4.1.1   Consider,if $f(t,x)$ is conditionally differentiable then $f$ is Lipschitz.

$$f_{1} = \dot x_{1}$$ (4.6)

$$\dot x_{1} = x_{2}$$ (4.7)

$$f_{2} = \dot x_{2}$$ (4.8)

$$\dot x_{2} = u\mp F_{c}$$ (4.9)

$$\Vert f_{i}(t,x) - f_{i}(t,y)\Vert \leq k_{i}\Vert X-Y\Vert$$ (4.10)

then prove that

$$\Vert x_{2}-y_{2}\Vert \leq k\Vert X-Y\Vert$$ (4.11)

where $X = \begin{bmatrix}x_{1} \\ x_{2} \\ \end{bmatrix}$ and $Y = \begin{bmatrix}y_{1} \\ y_{2} \end{bmatrix}$
if $f(t,x), g(t,x)$ are Lipschitz then

• $f + g$ is Lipschitz
• $f \times g$ is Lipschitz where $f,g \in \mathbb{R}^{n}$

$$\Vert f_{1}(t,x) - f_{1}(t,y)\Vert \leq k_{1}\Vert X-Y\Vert$$

$$\Vert g_{1}(t,x) - g_{1}(t,y)\Vert \leq k_{2}\Vert X-Y\Vert$$

$$\Vert f_{1}(t,x) \times g_{1}(t,y) - f_{1}(t,y) \times g_{1}(t,y)\Vert \leq k_{3}\Vert X-Y\Vert$$

$$F_{1} = \Vert f_{1}(t,x)\times g_{1}(t,y) - f_{1}(t,y)\times g_{1}(t,y)\Vert$$

$$F_{1} = \Vert f_{1}(t,x)\times g_{1}(t,y)-f_{1}(t,y)\times g_{1}(t,x)$$

$$\quad + f_{1}(t,y)\times g_{1}(t,x) - f_{1}(t,y)\times g_{1}(t,y)\Vert$$

$$F_{1} \leq \Vert f_{1}(t,x)- f_{1}(t,y)\Vert\times \Vert g_{1}(t,x)\Vert$$

$$\quad + \Vert f_{1}(t,y)\Vert\times \Vert g_{1}(t,x) - g_{1}(t,y)\Vert$$

$$F_{1} \leq k_{1}\Vert X-Y\Vert g_{1}(t,x) + k_{2} \Vert X-Y\Vert\Vert f_{1}(t,y)\Vert$$

$$F_{1} = {k_{1}\Vert g_{1}(t,x)\Vert + k_{2} \Vert f_{1}(t,y)}\Vert\Vert X-Y\Vert$$