MHB Uniqueness of Solution for x' = f(x) = √(1-x^2), x(2) = 1

[onie mti]

given this equation
x' = f(x)= square root(1-x^2) x(2) = 1

I hae to show that teh solution is not unique

my work:
i tried to find the interval in which f(x) is defined,
i said:
1-x^2 ≥ 0 (because of the sqrt)
-x^2 ≥ -1
x^2≤ 1
x≤ ±1
my problem is if i take a number < 1 and substitute it on f(x) i get a negative answer> where am i going wrong

 

[alexmahone]

Re: initial value problem

1-x^2 ≥ 0 (because of the sqrt)
-x^2 ≥ -1
x^2≤ 1
x≤ ±1

That last line should be [math]-1<=x<=1[/math], which works.

The equation is [math]x'=\sqrt{1-x^2}[/math], [math]x(2)=1[/math].

Solving by separation of variables, I get

[math]x=\sin\left(t+\frac{\pi}{2}-2\right)[/math] (Assuming that the independent variable is [math]t[/math].)

Another solution would be the constant function [math]x=1[/math].

So, the solution is not unique.

 

[chisigma]

Re: initial value problem

That last line should be [math]-1<=x<=1[/math], which works.

The equation is [math]x'=\sqrt{1-x^2}[/math], [math]x(2)=1[/math].

Solving by separation of variables, I get

[math]x=\sin\left(t+\frac{\pi}{2}-2\right)[/math] (Assuming that the independent variable is [math]t[/math].)

The solution is unique.

May be that the last sentence is a little questionable... in fact the [constant] function $x=1$ also satisfies [among others] the IVP $\displaystyle x^{\ '} = \sqrt{1 - x^{2}},\ x(2)=1$...

Kind regards

$\chi$ $\sigma$

 

[alexmahone]

Re: initial value problem

May be that the last sentence is a little questionable... in fact the [constant] function $x=1$ also satisfies [among others] the IVP $\displaystyle x^{\ '} = \sqrt{1 - x^{2}},\ x(2)=1$...

Kind regards

$\chi$ $\sigma$

Yeah, I realized that later and edited my post. :p

Could you elaborate "[among others]". Are you saying that there are more than 2 solutions? Which ones?

 

[chisigma]

Re: initial value problem

Yeah, I realized that later and edited my post. :p

Could you elaborate "[among others]". Are you saying that there are more than 2 solutions? Which ones?

In general given an IVP...

$\displaystyle \frac{d x}{d t} = f(x,t),\ x(t_{0})=x_{0}\ (1)$

... il admits one and only one solution if f(*,*) satisfies the 'Lipschitz's conditions' on x in $(x_{0},t_{0})$, i.e. in a neighborhood of $(x_{0},t_{0})$ is...

$\displaystyle |f(x_{1},t) - f(x_{0},t)| \le K\ |x_{1} - x_{0}|\ (2)$

In Your case is $\displaystyle f(x,t) = \sqrt{1 - x^{2}}$, 'not Lipschitz' in $x =1$, so that more than one solution can exist. In particular f(*,*) is not 'anchored' at t [it is function of the x alone...] and that means that if x(t) is solution of the IVP, then x(t + a) with $a \in \mathbb{R}$ is also solution. Now You can verify that a solution is...

$\displaystyle x(t) =\begin{cases} \cos (t-2)\ \text{if}\ t \ge 2\\ 1\ \text{if}\ t< 2\end{cases}\ (3)$

... and that means that $x(t-a),\ a> 0$ is also solution of the IVP [the solution x=1 corresponds to $a = \infty$...]...

Kind regards

$\chi$ $\sigma$

 

[alexmahone]

Re: initial value problem

Now You can verify that a solution is...

$\displaystyle x(t) =\begin{cases} \cos (t-2)\ \text{if}\ t \ge 2\\ 1\ \text{if}\ t< 2\end{cases}\ (3)$

No, it isn't!

[math]x'=-\sin(t-2)[/math] whereas [math]\sqrt{1-x^2}=\sin(t-2)[/math] if [math]t\ge 2[/math].

Did you mean [math]x(t)=\sin(t-2)[/math] if [math]t\ge 2[/math]?