# Specific Dirichlet BVP solutions

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]$^n$ ?

-

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 $(r_-,r_+)$.
Isn't it a consequence of the answer above ? Take $f(y)=(y^2-1)^n$, $r_\pm=\pm1$. Then you obtain $-1<x'<1$. – Denis Serre Sep 21 '10 at 16:25
This is just to record that the condition $f\in C^2$ in the accepted answer can be weakened to Lipschitz continuity of $f$, and that $r_{\pm}$ do not have to be consecutive zeros. With $y=x'$, let us write the equation as $$y'=f(y).$$ First of all, $y\not\equiv0$ because $f(0)\neq0$. On the other hand, if $f(\eta)=0$ for some $\eta\neq0$ then $y$ cannot take the value $\eta$ because if $y(b)=\eta$ for some $b\in[0,1]$ then by uniqueness we would have $y\equiv\eta$. Now, if $y(t)>0$ for all $t\in(0,1)$ then by definition $x$ would be strictly increasing on $(0,1)$, and likewise $y<0$ on $(0,1)$ implies that $x$ would be strictly decreasing on $(0,1)$, meaning that it would not be possible to satisfy the boundary conditions $x(0)=x(1)=0$. Hence $y$ must change sign on $(0,1)$, i.e., there is $a\in(0,1)$ such that $y(a)=0$. The claim follows from the fact that $y$ is a continuous function that takes the value $0$, but cannot take the values $r_{\pm}$.