1. ## Uniqueness Theorem

Hey can someone help me with this? the back of the book simply says, "This is true due to the uniqueness theorem."

Suppose that x is a solution to the initial value problem:

x'=(x^3-x)/(1+t^2*x^2) x(0)=1/2.

Show that 0<x(t)<1 for all t for which x is defined.

My work:
I set x'=f(t,x). I took the partial derivative of f(t,x) with respect to x. Unless I am mistaken, I believe both f(t,x) and its partial derivative are continuous over a small enough rectangle R containing the point (1/2,0). Since it is given that x is a solution, it means that x is unique.

Why does 0<x(t)<1 have to be true?

Thanks!

2. ## Re: Uniqueness Theorem

Originally Posted by BrianMath
Hey can someone help me with this? the back of the book simply says, "This is true due to the uniqueness theorem."

Suppose that x is a solution to the initial value problem:

x'=(x^3-x)/(1+t^2*x^2) x(0)=1/2.

Show that 0<x(t)<1 for all t for which x is defined.

My work:
I set x'=f(t,x). I took the partial derivative of f(t,x) with respect to x. Unless I am mistaken, I believe both f(t,x) and its partial derivative are continuous over a small enough rectangle R containing the point (1/2,0). Since it is given that x is a solution, it means that x is unique.

Why does 0<x(t)<1 have to be true?

Thanks!
Hi BrianMath,

$\displaystyle f(t,x)=\frac{x^3-x}{1+t^2 x^2}$

$\displaystyle \frac{\partial f(t,x)}{\partial x}=\frac{\partial}{\partial x}\left(\frac{x^3-x}{1+t^2 x^2}\right)=\frac{t^2 x^4+(t^2+3) x^2-1}{(t^2 x^2+1)^2}$

$\displaystyle \mbox{Let, }R=\left\{(t,x)~:~|t|\leq a\mbox{ and }\left|x-\frac{1}{2}\right|\leq \frac{1}{2}~;~a>0\right\}$

Clearly both $\displaystyle f(t,x)\mbox{ and }\frac{\partial f(t,x)}{\partial x}$ are continuous on $\displaystyle R$.

$\displaystyle M=max|f(t,x)|~;~(t,x)\in R$

$\displaystyle M=max\left|\frac{x^3-x}{1+t^2 x^2}\right|=max\frac{|x^3-x|}{1+t^2 x^2}~;~(t,x)\in R$

$\displaystyle M=max\left(\frac{x-x^3}{1+t^2 x^2}\right)=\frac{\frac{1}{\sqrt{3}}-\left(\frac{1}{\sqrt{3}}\right)^3}{1+0}=\frac{1}{ \sqrt{3}}\left(\frac{2}{3}\right)=\frac{2}{3\sqrt{ 3}}$

$\displaystyle h=min\left\{a,\frac{1}{2M}\right\}=min\left\{a, \frac{3 \sqrt{3}}{4}\right\}$

Therefore by the Picard existance and uniquenss theorem, the initial value problem has a unique solution on the rectangle,

$\displaystyle |t|\leq h\mbox{ and }\left|x-\frac{1}{2}\right|\leq\frac{1}{2}$

$\displaystyle \Rightarrow |t|\leq h\mbox{ and }0\leq x\leq 1$