# Proof that Lipschitz condition guarantees well posedness of initial value problems

In the proof of the theorem which states that the Lipschitz condition guarantees well posed-ness of an initial value problem $y'=f(x,y)$, $y(x_0)=y_0$, I came across this

Let the perturbed problem be

$$z'=f(x,z) + \delta(t) \\ z(x_0)=y_0 + \epsilon_0$$

Let $\epsilon(x)=z(x)-y(x)$ and let $f$ satisfy the Lipschitz condition. Then we have

$$\epsilon '=z'-y'=f(x,z)-f(x,y)+\delta(x)\\ \implies |\epsilon'(x)| \leq |f(x,z)-f(x,y)| + |\delta(x)|\\$$

Let $\delta(x)$ have the maximum value $\epsilon$. Therefore, we get,

$$|\epsilon'(x)| \leq |f(x,z)-f(x,y)| + \epsilon\\ \implies |\epsilon'(x)| \leq L|\epsilon(x)| + \epsilon$$

I don't understand the next step i.e. how is the above inequality integrated to give

$$|\epsilon(x)| \leq \frac{\epsilon}{L} \left[ (L+1)e^{Lt}-1\right]$$

How is the inequality integrated?

• Is this in a homework? Or did you come up with it yourself? Apr 2 '14 at 17:45
• @ellya this is not homework...came across this in a textbook... Gear(1971)...i'll update the exact title later Apr 2 '14 at 18:43

It is more convenient to use the integral form: $$y(x)=y_0+\int_0^x f\big(s,y(s)\big)\,ds$$ and $$z(x)=y_0+\varepsilon_0+\int_0^x \big(\,f\big(s,z(s)\big)+\delta(s)\big)\,ds$$ and hence $$z(x)-y(x)=\varepsilon_0+\int_0^x\delta(s)\,ds+\int_0^x\big(f\big(s,z(s)\big)-f\big(s,y(s)\big)\big)\,ds,$$ which implies that (for $x>0$) $$\lvert z(x)-y(x)\rvert \le \lvert \varepsilon_0\rvert +\varepsilon x+ L\int_0^x\big|\,z(s)-y(s)\big|\,ds. \tag{1}$$ Now, let $w(x)=\int_0^x \big|\,z(s)-y(s)\big|\,ds$, so the above becomes $$w'(x) \le \lvert \varepsilon_0\rvert +\varepsilon x+ L w(x)$$ or $$\left(\mathrm{e}^{-Lx}w(x)\right)'=\mathrm{e}^{-Lx}\big(w'(x)-Lw(x)\big)\le \mathrm{e}^{-Lx}\big(\lvert\varepsilon_0\rvert +\varepsilon x\big).$$ Integrating the above in $[0,x]$ we obtain $$\mathrm{e}^{-Lx}w(x)-w(0)=\frac{\lvert\varepsilon_0\rvert}{L}(1-\mathrm{e}^{-Lx}) +\varepsilon\mathrm{e}^{-Lx}\left(-\frac{x}{L}+\frac{1}{L^2}\right)-\frac{\varepsilon}{L^2}$$ or $$\int_0^x\lvert z(s)-y(s)\rvert\,ds\le \frac{\lvert\varepsilon_0\rvert}{L}(\mathrm{e}^{Lx}-1) +\varepsilon\left(-\frac{x}{L}+\frac{1}{L^2}\right)-\frac{\varepsilon \mathrm{e}^{Lx}}{L^2}.$$ Using $(1)$ we obtain $$\lvert z(x)-y(x)\rvert \le \lvert\varepsilon_0\rvert(\mathrm{e}^{Lx}-1) +\varepsilon\left(-x+\frac{1}{L}\right)-\frac{\varepsilon \mathrm{e}^{Lx}}{L}+\lvert\varepsilon_0\rvert +\varepsilon x =\cdots.$$

Multiply through by $|\epsilon (x )|$ to get:

$|\epsilon'(x )\epsilon (x )|\leq L|\epsilon (x )|^2+\epsilon|\epsilon (x )|$

$\frac{1}{2}|(\epsilon^2(x))'|\leq L |\epsilon^2(x)|+\epsilon|\epsilon (x)|\leq L|\epsilon(x)^2|+\epsilon^2$

$\Rightarrow \frac {1}{2}\frac {d}{dx}(|\epsilon ^ 2( x ) | e ^ { -2Lx })\leq \epsilon^2e^{-2Lx}\Rightarrow\frac {1}{2}|\epsilon ^ 2( x ) | e ^ { -2Lx}\leq \frac{1}{2}|\epsilon ^ 2 ( 0) |+\frac{1}{2L}(\epsilon ^ 2)(1-e^{-2Lx})$

$\Rightarrow |\epsilon ^ 2( x ) |\leq|\epsilon ^ 2( 0) |e^{2Lx}+\frac{1}{L}(\epsilon^2)(e^{2Lx}-1)\leq\epsilon^2e^{2Lx}+\frac{1}{L}(\epsilon^2)(e^{2Lx}-1)=\frac{\epsilon^2}{L^2}(L^2+Le^{2Lx}-L)$

still working on this, but it's quite close to the desired estimate.