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I'm pretty confused about this topic in Differential Equations. It's a simple topic, but I just can't get the gist of it.

Here are two examples that I would like for you to explain to me

  1. $\dfrac{dy}{dx} = y^\frac{1}{3},\quad y(0) = 1$
  2. $\dfrac{dy}{dx} = y^\frac{1}{3},\quad y(0) = 0$

for number 1 a unique solution exist near $x = 0$, but for 2 uniqueness is not guaranteed. Please explain, boggles my mind.

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2 Answers

up vote 2 down vote accepted

We know that Picard gives sufficient conditions for the existence of a unique solution:

Theorem: Let $R$ be a rectangular region in the $xy$-plane defined by $$a\leq x\leq b, ~~c\leq y\leq d$$ that contains the point $(x_0,y_0)$ in its interior. If $f(x,y)$ and $\partial f/\partial y$ are continuous on $R$ , then there exists an interval $I$ centered at $x_0$ and a unique function $y(x)$ defined on $I$ satisfying the initial-value problem $$y'=f(x,y),~~y(x_0)=y_0$$

Here you have the OE, $y'=\sqrt[3]{y}$. So $f(x,y)=\sqrt[3]{y}$ and then $\partial f/\partial y=\frac{1}{3y^{2/3}}$. Now, I think you will find the answer by your own and by using the Theorem. Just look at the origin and the point $(0,1)$ in the following graph:

enter image description here

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Nice graph, nice answer! +1 –  amWhy Jan 31 '13 at 0:09
    
@amWhy: Thanks amWhy. ;-) –  B. S. Jan 31 '13 at 6:33
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Observe, for number 2 for example, that applying the Mean Value Theorem we have $$\frac{| \sqrt[3]{y} - \sqrt[3]{x} |}{|x-y|}= \frac{1}{3\sqrt[3]{c^2}}=f^{\prime}(c), $$ for a certain $c \in (x,y) \subset B(0,\varepsilon)$, $\varepsilon >0$, so as $x \to y$, this tends to $f^{\prime}(0) = \infty$. Hence, there exists some $M>0$, as big as you want, such that $$| \sqrt[3]{y} - \sqrt[3]{x} | > M |y - x |,$$ for $x$, $y$ in a neighbourhood of zero. Hence, $\sqrt[3]{y}$ is not a Lipschitz function near zero and thus uniqueness cannot be granted. Note that this does not happen in case 1 because if you take values around one instead of zero the derivative is not going to be infinity. This is due to Picard's Theorem. The existance is granted in both cases because of the weaker version, the Peano's Theorem, which states that if the function is continuous around the initial condition, then the solution does exist. However, you need Lipschitz-ness for the uniqueness.

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+1 an Illuminating answer. –  B. S. Feb 1 '13 at 9:46
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