- #1
LumenPlacidum
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I'm working on a proof of the Peano Existence Theorem. It says that:
For a continuous ordinary differential equation dx/dt = f(x, t), where x is in R^n and f is continuous on |t-t_0|<=a, ||x-x_0||<=b
when you have an initial value, x(t_0) = x_0, then there is a solution on |t-t_0|<= c, 0<c<=a.
Wikipedia has a pretty relevant statement of the theorem, though I'm trying it for dimensions higher than one. My confusion is more easily described by their statement at http://en.wikipedia.org/wiki/Peano_existence_theorem
They say that the function f is defined so that it satisfies the ODE on its interval of definition, then why can't you just use any continuous function y(x) as the z(x) in that explanation? It just seems like the second line of the statement of the theorem shows that such a solution exists.
If anyone has some insights to offer, I'd be much obliged.
For a continuous ordinary differential equation dx/dt = f(x, t), where x is in R^n and f is continuous on |t-t_0|<=a, ||x-x_0||<=b
when you have an initial value, x(t_0) = x_0, then there is a solution on |t-t_0|<= c, 0<c<=a.
Wikipedia has a pretty relevant statement of the theorem, though I'm trying it for dimensions higher than one. My confusion is more easily described by their statement at http://en.wikipedia.org/wiki/Peano_existence_theorem
They say that the function f is defined so that it satisfies the ODE on its interval of definition, then why can't you just use any continuous function y(x) as the z(x) in that explanation? It just seems like the second line of the statement of the theorem shows that such a solution exists.
If anyone has some insights to offer, I'd be much obliged.