How Does the Lipschitz Condition Ensure Uniqueness in Differential Equations?

[evinda]

Hello! (Wave)

TheoremIf the vector field $\Phi(x,t)$ satisfies the Lipschitz condition as for $x$ and is continuous as for $t$ in a space $\Omega \times (a,b) \subset \mathbb{R}^{n+1}$, then there is at most one solution of the system $\frac{dx}{dt}=\Phi(x,t) (1)$, that satisfies the initial condition $x(t_0)=c (c=(c_1, \dots, c_n))$ where $(c, t_0) \in \Omega$.ProofWe suppose that there are two solutions $x(t)$ and $y(t)$, where $x(t_0)=y(t_0)=c$. We introduce the function $\sigma(t)$ which is equal to the square of the distance between $x(t)$ and $y(t)$

$$\sigma(t)= \sum_{k=1}^n [x_k(t)-y_k(t)]^2=|x(t)-y(t)|^2 \geq 0 (\star)$$

We find the derivative of $\sigma (t)$, taking into consideration that $x(t)$ and $y(t)$ are solutions of the system $(1)$

$$ (2)\sigma'(t)=2 \sum_{k=1}^n [x_k(t)-y_k(t)] [\Phi_k(x,t)-\Phi_k(y,t)]=2 [x-y][\Phi(x,t)-\Phi(y,t)]$$Using the Cauchy- Schwarz inequality we get $$\sigma'(t) \leq |\sigma'(t)|=2 |(x-y) \cdot (\Phi(x,t)-\Phi(y,t))| \leq 2 |x-y||\Phi(x,t)-\Phi(y,t)|\leq 2K |x-y|^2=2K \sigma(t) $$From $(2)$ it follows immediately that $\sigma' \leq 2K \sigma \Rightarrow (\sigma e^{-2Kt})' \leq 0$ and so for $t \geq t_0$, $\sigma (t) \leq \sigma(t_0) e^{2K(t-t_0)}$.

Since $x(t_0)=y(t_0)$ we have $\sigma (t_0)=0$ and so $\sigma (t) \leq 0$.

From $(\star)$ we get that $\sigma (t) \equiv 0$ and $|x(t)-y(t)|^2 \equiv 0$ for $t \geq t_0$.Consequently, $x(t) \equiv y(t)$ for $t \geq t_0$.

In a similar way we can look at the case $t<t_0$.

Obviously $-\sigma'(t) \leq |\sigma'(t)|$ and thus

$$- \sigma' \leq 2K \sigma \Rightarrow (\sigma e^{2Kt})' \geq 0$$

Therefore, for $t \leq t_0$ we have

$\sigma(t) e^{2Kt} \leq \sigma(t_0) e^{2Kt_0} \Rightarrow \sigma(t) \leq \sigma(t_0) e^{2K(t_0-t)}$

Consequently, $\sigma(t) \equiv 0$ for $t \leq t_0$, i.e. $x(t) \equiv y(t)$ for $t \leq t_0$.
Why do we use $- \sigma' \leq 2K \sigma \Rightarrow (\sigma e^{2Kt})' \geq 0$ in order to deduce that $x(t) \equiv y(t)$ for $t \leq t_0$ ?

Couldn't we also use this relation $(\sigma e^{-2Kt})' \leq 0$, which we also used for $t \geq t_0$ ?

 

[gtoguy97]

We use $- \sigma' \leq 2K \sigma \Rightarrow (\sigma e^{2Kt})' \geq 0$ in order to deduce that $x(t) \equiv y(t)$ for $t \leq t_0$ because it is a more direct way to reach the desired result. The relation $(\sigma e^{-2Kt})' \leq 0$, which we used for $t \geq t_0$, gives us the same result but requires more steps to get there.