Does the Initial Value Problem Always Have a Solution for All Time Intervals?

[sbashrawi]

Homework Statement

suppose that h $$\in$$C(R) and f$$\in$$C(R^2) satisfies
|f(t,x)|$$\leq$$h(t) |x| for all (t,x) $$\in$$R^2. Show that for any point , the IVP : x' = f( t, x) x( $$\tau$$)=$$\varsigma$$
has a solution which exists for all t in R

Homework Equations

The Attempt at a Solution

 

[kurt.physics]

Let g(t) = \int_\tau^t h(s)ds Note that g is continuous and strictly increasing. Define F: R^2 -> R by F(t,x) = \int_\tau^t f(s,x)ds It is easy to check that F is continuous and differentiable. Now consider the IVPy' = F(t,y) y(\tau) = \varsigmaBy existence and uniqueness theorem for IVPs, this has a unique solution , say y(t), which exists for all t \inR. We claim that x(t) = y(t)/ g(t) is a solution of the original IVP. Note that x(\tau) = \varsigma/g(\tau) = \varsigma So we just have to show that x'(t) = f(t,x(t)) for all t \in R We have x'(t) = (y'(t) g(t) - y(t) g'(t))/ g^2(t) But y'(t) = F(t,y(t)) and g'(t) = h(t) Also note that F(t, y(t)) = \int_\tau^t f(s,y(s))ds \leq \int_\tau^t h(s) |y(s)| ds = g(t) |y(t)| So x'(t) = (F(t,y(t)) g(t) - y(t) h(t))/ g^2(t) \leq h(t) |y(t)| g(t)/ g^2(t) = h(t) |y(t)|/ g(t) = h(t) |x(t)| Hence x'(t) \leq h(t