We have $$y' = f(x,y), \; y(x_0) = y_0 \;(*)$$ Let $f(x,y)$ be continuous and satisfy the general Lipschitz condition $$|f(x,y_1) - f(x,y_2)| \leq L(x)|y_1-y_2|$$ for all $(x,y_1), \; (x,y_2)$ in $\bar{S}$, where the function $L(x)$ is such that the integral $\int_{x_0-a}^{x_0+a} L(t)dt$ exist. Show that $(*)$ has at most one solution in $|x-x_0| \leq a$

The way I did it is

Asuume that $y_1, \; y_2$ are two solution of DE. Let $u = |(y_1 - y_2)|^2$ with $u(x_0) = 0$, then $$u' = |2(y_1-y_2)(y'_1 - y'_2)|$$ $$ = |2(y_1 - y_2)(f(x,y_1) - f(x,y_2))|$$ $$\leq 2|y_1 - y_2||f(x,y_1) - f(x,y_2)|$$ $$\leq 2L(x)|y_1 - y_2|$$

Edit: someone suggest me to do this one, it works out, but I still don't understand why

continued from above, we have $$u \leq 2L(x)|y_1 - y_2|^2$$ $$ \leq 2L(x)u$$ $$u' - 2L(x)u \leq 0$$ $$(ue^{-2\int^x L(t)dt })' \leq 0$$ Hence, $u = 0$, therefore $y_1 = y_2$ in $x_0 \leq x \leq x_0+a$

And the proof $x_0 -a$ should be the same as above.

Can someone explain to me why does t his way make sense? Is there another way to prove it . Thank you.

  • $\begingroup$ When you define $u=|(y_1-y_2)|^2$, you can drop the absolute value function. Then when you computed $u'$, either you should've added the absolute value in both sides or in none. When you continue to $u\leq 2L(x) |y_1-y_2|^2$, it likely should have been $|u'|\leq 2L(x) |y_1-y_2|$. $\endgroup$ – Mefitico May 24 '18 at 17:18

Hint: Assume there are two solutions $y_1,y_2$ and then try to prove that

$$ |y_1-y_2| < \epsilon. $$

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  • $\begingroup$ Oh, I forgot to include that this function $y_1,\; y_2$ are two solutions of DE, and this function is always continuous $\endgroup$ – Alexander Mar 16 '15 at 6:38
  • $\begingroup$ I edited in my post. But, then after assume that, I have to prove that $u = 0$ in order to make $y$ has at most one solution. But, I don't know how to prove it from above. $\endgroup$ – Alexander Mar 16 '15 at 6:41

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