Assume $f \in C(\mathbb{R})$ and $g\in C^1(\mathbb{R})$. Show that IVP problem $$y''+f(y)y'+g(y)=0$$$$y(a)=b , y'(a)=c $$ has a unique solution.

my strategy: if assume $y=x_1$ and $y'=x_2$ then we have $$ x_1' =x_2$$ $$x_2'=-f(x_1)x_2-g(x_1) $$

now if $$X=(x_1 , x_2)^T , H=(h_1 , h_2)^T$$ which $h_1(t,X)=x_2$ and $h_2(t,X)=-f(x_1)x_2-g(x_1)$. then $$X'=H(t,X)$$ $$ X(a)=(b,c) $$

Now I think if assumptions $f \in C(\mathbb{R})$ and $g\in C^1(\mathbb{R})$ conclude $H(t,x)$ is lipschitz continuous the problem is solved. Do these assumptions conclude lipschitz continuity?

Is there other approaches?


These assumptions guarantee local Lipschitz continuity. If $g(x)=x^2$, then $H$ cannot be globally Lipschitz continuous.

The same holds for the $f$-term: if $f$ is unbounded, then the derivative of $f(x_1)x_2$ is unbounded.


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