Hi. While learning BVP, I came across a problem. It was mentioned that given the below left-focal BVP, a fixed–point problem can be formed for the system.

x'' = f (t, x), x (0) = 0, x(1) = 0.

where, f : [0, 1] × R → R is continuous and uniformly bounded. It was also mentioned that Schauder’s theorem can be applied to ensure the corresponding existence.

Can someone please help me in understanding this proof? Please let me know if the problem seems vague.

From this, can we infer that, in general, if

x'' = f (t, x), x(a) = x(b), x (a) = x (b) and f : [a, b] × R → R is continuous and uniformly bounded, then Schauder’s theorem can be applied to ensure that the system has at least one solution?

Regards, Salil


Fixed-point is the name of a method, not of a problem (so far we speak of nonlinear differential equations). Its starts from the observation that, if your equation was $x''=-g(t)$, then the solution would be $$x(t)=\int_0^t(1-t)sg(s)ds+\int_t^1t(1-s)g(s)ds,$$ which we write as $$x(t)=\int_0^1K(t,s)g(s)ds.$$ Therefore, your nonlinear problem is equivalent to the integral equation $$x(t)=-\int_0^1K(t,s)f(s,x(s))ds=:(Tx)(t).$$ This can be viewed as a fixed-point problem $Tx=x$. Hence the first idea of constructing a sequence of of 'approximate solutions' by $x^0\equiv0$ and then $x^{k+1}=Tx^k$, that is $$x^{k+1}(t)=-\int_0^1K(t,s)f(s,x^k(s))ds.$$ If $x^k$ converges pointwise boundedly to an $x$, then the convergence is uniform, thanks to the iteration, and $x$ is a fixed point. However, it is not always true that the sequence converges (it could oscillate, for instance), hence the second idea, which is of topological nature. Because $f$ is bounded, $T$ maps ${\mathcal C}(0,1)$ into a compact subset. In addition, $T$ is continuous. Then Schauder's fixed-point theorem ensures that $T$ admits a fixed point (not necessarily unique). Schauder's Theorem is a generalization of Brouwer's to infinite dimension (Banach spaces).

  • $\begingroup$ Ohh.. Thanks for the correction, Denis. Going by the notes which I was referring, I called it a 'problem' instead of 'method'. So, does that mean that the claim made in the second statement (x'' = f (t, x), x(a) = x(b), x (a) = x (b) and f : [a, b] × R → R has a solution) is true? $\endgroup$ – Salil Sep 23 '10 at 7:02
  • $\begingroup$ Yes, it does. And you can't say anything about uniqueness without an extra assumption. $\endgroup$ – Denis Serre Sep 23 '10 at 7:06

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