Hi,

the problem:

Use each of the three theorems in this section to predict where this problem has a solution, and then solve the problem explicitly, comparing the theory to the fact:

$\displaystyle x'=x^2, x(0)=1$.

As stated in the problem, there are three theorems in my book dealing with existence and uniqueness of I.V.Ps.

The first one is a pure existence theorem and says;

If $\displaystyle f$ is continuous in a rectangle $\displaystyle R$ centered at $\displaystyle (t_0,x_0)$, say

$\displaystyle R = \{(t,x): |t-t_0|\leq \alpha, |x-x_0|\leq \beta\}$

then the initial-value problem has a solution $\displaystyle x(t)$ for $\displaystyle |t-t_0|\leq min(\alpha,\beta/M)$, where $\displaystyle M$ is the maximum of $\displaystyle |f(t,x)|$ in the rectangle $\displaystyle R$.

Attempt:

For this problem, the rectangle is;

$\displaystyle R = \{(t,x): |t|\leq \alpha, |x-1|\leq \beta\}$.

Hence $\displaystyle f(t,x)=x^2$ is bounded by, $\displaystyle (\beta+1)^2=M$.

Since $\displaystyle f$ is continuous, the I.V.P has a solution for $\displaystyle |t|\leq min(\alpha,\beta/(\beta+1)^2)$.

Now, $\displaystyle \beta/(\beta+1)^2$ has a maximum when $\displaystyle \beta=1$, so,

$\displaystyle |t|\leq min(\alpha,1/4)$.

Assuming $\displaystyle \alpha\geq 1/4$, we have that

$\displaystyle |t| \leq 1/4$.

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The explicit solutions is,

$\displaystyle x(t) = \frac{1}{1-t}$,

which is not defined if $\displaystyle t\neq 0$.

I can't say my prediction is very good...

Any suggestions are welcome, thanks.