This question below, which was given in an exam, goes as follows:
A planar curve $y(x)$ is such that its curvature, $$k(x)=y''(x)/(1+y'(x)^2)^{3/2}$$ is equal to its height $y(x)$. Write the equation as a pair of first order equations, and show that the system is Lipschitz in $y$.
Now my working is as follows: Let $$u=y, v=y'$$ $$\implies u'=v, v'=(1+v^2)^{3/2}u$$ which answers the first part of the question. Now to show that it is Lipschitz in $y$: $$|u_1'-u_2'|=|v_1-v_2|\leq |u_1-u_2|+|v_1-v_2|$$
My problem is for the second part: \begin{align} |v_1'-v_2'|=&|(1+v_1^2)^{3/2}u_1-(1+v_2^2)^{3/2}u_2|\\ =&|(1+v_1^2)^{3/2}u_1-(1+v_1^2)^{3/2}u_2+(1+v_1^2)^{3/2}u_2-(1+v_2^2)^{3/2}u_2|\\ \leq &|1+v_1^2|^{3/2}|u_1-u_2|+|u_2|||(1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}| \end{align}
Now I can always bound the first term, but for the second term, I have no idea how to simplify in such a way to get $K|v_1-v_2|$ for some constant $K$. Any hints? Am I going about this the wrong way?
Apply either a mean value theorem or the binomial theorem as in $$ (1+v_1^2)^{3/2}-(1+v_2^2)^{3/2}=\frac{(1+v_1^2)^3-(1+v_2^2)^3}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}} \\ =\frac{(v_1-v_2)(v_1+v_2)\Bigl((1+v_1^2)^2+(1+v_1^2)(1+v_2^2)+(1+v_2^2)^2\Bigr)}{(1+v_1^2)^{3/2}+(1+v_2^2)^{3/2}} $$ where you get the simple difference as factors and can bound the other factors.