2
$\begingroup$

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]$^n$ ?

$\endgroup$
2
$\begingroup$

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 $(r_-,r_+)$.

$\endgroup$
  • $\begingroup$ ohh... that explains it. Thanks. Also, can you please shed some light on the apriori bounds part? That is quite confusing for me. $\endgroup$ – Salil Sep 21 '10 at 16:08
  • 1
    $\begingroup$ Isn't it a consequence of the answer above ? Take $f(y)=(y^2-1)^n$, $r_\pm=\pm1$. Then you obtain $-1<x'<1$. $\endgroup$ – Denis Serre Sep 21 '10 at 16:25
  • $\begingroup$ Damn... I am feeling stupid :-) Thank you very much for pointing that out!! $\endgroup$ – Salil Sep 21 '10 at 16:31
0
$\begingroup$

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 $r_{\pm}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.