Predict where I.V.P has a solution by using theorem(s)

**Mollier** · April 1st 2011, 10:10 PM

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:

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

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

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

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

Attempt:
For this problem, the rectangle is;

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

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

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

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

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

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

$|t| \leq 1/4$ .

--------------------------------------------------------------------------------------

The explicit solutions is,

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

which is not defined if $t\neq 0$ .
I can't say my prediction is very good...

Any suggestions are welcome, thanks.

**FernandoRevilla** · April 2nd 2011, 01:30 AM

I've looked through your process and all is correct. However, why do you say the following?

Originally Posted by Mollier

The explicit solutions is, $x(t) = \frac{1}{1-t}$ , which is not defined if $t\neq 0$ .

The function $x(t)=1/(1-t)$ is not defined at $t=1$ but it is defined in $|t-0|\leq 1/4$ and satifies the I.V.P.

**Mollier** · April 3rd 2011, 07:35 PM

Originally Posted by FernandoRevilla

I've looked through your process and all is correct. However, why do you say the following?

The function $x(t)=1/(1-t)$ is not defined at $t=1$ but it is defined in $|t-0|\leq 1/4$ and satifies the I.V.P.

Hi,

thank you for taking the time to look through my attempt.

I was hoping that the theorem would give me a stronger indication that something is wrong when $t=1$ . I do not know why I would expect such a thing though.

Again, thanks.

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