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

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

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

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

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

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.

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

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

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

The function $\displaystyle x(t)=1/(1-t)$ is not defined at $\displaystyle t=1$ but it is defined in $\displaystyle |t-0|\leq 1/4$ 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 $\displaystyle t=1$. I do not know why I would expect such a thing though.

Again, thanks.