Hi All,


I am learning Differential Equations, and came across a specific problem of Dirichlet BVP, which says that:


Given x'' = f(x'), x(0) = 0 = x(1), If f(0) $\neq $ 0 and f has two zeros of opposite sign (say, $r^+$ $\gt$ 0 and $r^−$ $\lt$ 0) then all solutions to Dirichlet BVP have derivatives satisfying


$r^−$ $\lt$ x'(t) $\lt$ $r^+$ , $\forall$t $\epsilon$ [0, 1].


Is this true? How can be this shown?


Also, can someone please tell me how to establish apriori bounds on the derivative of solutions on [0, 1] for the BVP x'' = [(x')$^2$ − 1]$^n$ ?



Specific Dirichlet BVP solutions
Say that $f$ is of class ${\mathcal C}^2$. Set $y:=x'$ and differentiate. You get $y''=f'(y)y'$. This is a linear ODE in $y'$, if we think of $f'(y)$ as a given function $g(t)$. Since $y$ is not $\equiv0$ (because $f(0)\ne0$), Cauchy-Lipschitz tells you that $y'=f(y)$ does not vanish over $[0,1]$. In particular, $y$ is strictly monotonous, say increasing. Because $\int\_0^1y(t)dt=x(1)-x(0)=0$, $y$ vanishes at some $t\_0$. Then $x''(t\_0)=f(0)$ shows that $f(0)>0$ and therefore ($r\_\pm$ are consecutive zeroes of $f$), $f$ remains $>0$ over $(r\_-,r\_+)$. Since $f(y)$ does not vanish, $y$ does not take the values $r\_\pm$. But $y$ takes the value $0$ (at $t\_0$). Therefore $y=x'$ remains in $(r\_-,r\_+)$.


