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Thread: Uniqueness Theorem

  1. #1
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    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!
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  2. #2
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    Wink Re: Uniqueness Theorem

    Quote Originally Posted by BrianMath View Post
    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,

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

    \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}

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

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

    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

    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}}

    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,

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

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