While learning differential equations, I was reading some notes, and it was mentioned that for Dirichlet BVP

x'' = f (t, x), x(0) = 0 = x(1). Suppose f : [0, 1] × R → R is continuous and there is a constant R > 0 such that f (t, R) ≥ 0, f (t, −R) ≤ 0, for all t ∈ [0, 1]. It can be shown that there is at lease one solution, but the proof is missing.

Can someone please help me out with this?

How to find such an constant R, for some problem say, x'' = $x^3$ + t, x(0) = 0 = x(1) to show that this has at least one solution?

Regards, Salil


The assumption tells you that $x_+\equiv R$ is a super-solution, and $x_-\equiv -R$ a sub-solution, of your problem. This means that $x_-''\ge f(t,x_-)$, $x_+''\le f(t,x_+)$, and $x_-\le0\le x_+$ at $t=0,1$.

It is a general fact that when a BVP for a second-order differential equation admits a sub- and a super-solution $x_\pm$, and if $x_-\le x_+$, then the problem admits a solution $x$ such that $x_-\le x\le x_+$.

Idea : Take $\lambda>0$ large enough that $x\mapsto g(t,x):=f(t,x)-\lambda x$ be non-increasing over $[-R,R]$, for every $t\in[0,1]$. Rewrite the problem as $$x''-\lambda x=g(t,x).$$With $x(0)=x(1)=0$, this is equivalent to an integral equation $$x(t)=-\int_0^1K_\lambda(t,s)g(s,x(s))ds=:(Tx)(t).$$ The kernel $K_\lambda$ is bounded and continuous. Let us define the convex subset $C$ of ${\mathcal C}(0,1)$ by the conditions $$x(0)=x(1)=0,\qquad x_-\le x \le x_+.$$ Because of the monotonicity of $g$, and because $Tx_-\ge x_-$, $Tx_+\le x_+$, $T$ maps $C$ into itself. In addition, $T(C)$ is relatively compact (a consequence of Ascoli-Arzela). By Schauder fixed point theorem, it admits a fixed point $\bar x$. This $\bar x$ is a solution of your problem.

  • $\begingroup$ So, does this imply that x'' = x^3 + t, x(0) = 0 = x(1) has a solution? If yes, how? $\endgroup$ – Salil Sep 23 '10 at 8:01
  • 1
    $\begingroup$ You just take $R=1$. You should do the application by yourself. MO is not the place to discuss these details. $\endgroup$ – Denis Serre Sep 23 '10 at 12:58
  • $\begingroup$ ok... thanks Denis! I will do that. $\endgroup$ – Salil Sep 23 '10 at 14:54

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.