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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function and $g:\mathbb{R}\rightarrow\mathbb{R}$ be a Lipschitz function. Would you help me to prove that the system of differential equation

$$ x'=g(x)$$ $$y'=f(x)y $$

with initial value $x(t_0)=x_0$ and $y(t_0)=y_0$ has a unique solution.

Could I prove the uniqueness solution of $x'=g(x)$, $x(t_0)=x_0$ by Gronwall Inequality first then use the result to prove the second?

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  • $\begingroup$ My answer is suppose $(x_1(t),y_1(t))$ and $(x_2(t),y_2(t))$ is solution. First, write $x_1(t)=x_0+\int_{t_0}^t x'(s) ds$ the do the same for $x_2(t)$ and get $|x_1-x_2|=|\int_{t_0}^t g(x_1)-g(x_2) ds| \leq K \int_{t_0}^t |x_1-x_2| ds$. I doubt about this separation process. $\endgroup$ – beginner Oct 18 '12 at 9:28
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Your reasonning is correct. Since $g$ is Lipschitz and the first equation of the system involves only $x$, there is a unique solution $x(t)$ such that $x(t_0)=x_0$.

The second equation becomes $$ y'=f(x(t))\,y,\quad y(t_0)=y_0. $$ It is a linear equation and has a unique solution, given by $$ y(t)=y_0e\,^{\int_{t_0}^t f(x(s))ds}. $$

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  • $\begingroup$ No simultaneous method to show directly that (x_1,y_1)=(x_2,y_2)? $\endgroup$ – beginner Oct 18 '12 at 9:51
  • $\begingroup$ Since $f$ is assumedcontinuous, I do not see a way of proving uniqueness simultaneously (this does not mean that it can't be done). You first show uniqueness of $x$, and then of $y$. $\endgroup$ – Julián Aguirre Oct 18 '12 at 10:00

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