$\displaystyle \frac{dv}{dt}=k v^{2/3}$ where $\displaystyle k= 3^{2/3} (4 \pi)^{1/3}$

Why doesn't this equation satisfy the hypothesis of the uniqueness theorem?

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- Mar 26th 2009, 06:04 PMLenDifferential Equations
$\displaystyle \frac{dv}{dt}=k v^{2/3}$ where $\displaystyle k= 3^{2/3} (4 \pi)^{1/3}$

Why doesn't this equation satisfy the hypothesis of the uniqueness theorem? - Mar 26th 2009, 11:55 PMchisigma
For semplicity let's suppose that $\displaystyle k=1$, so that the equation is...

$\displaystyle v'= v^{\frac{2}{3}}$ (1)

Writing a generic first order equation in the form...

$\displaystyle v'= f(v,t)$ (2)

... with 'initial condition' $\displaystyle v(t_{0})=v_{0}$, it admits one and only one solution if and only if the so called 'Lipschitz conditions' are satisfied. One of these conditions requires that $\displaystyle f(*,*)$ must have bounded partial derivatives in a 'small region' around $\displaystyle (v_{0}, t_{0})$ and that is not for (1) if $\displaystyle v_{0}=0$.

The fact is self evident if we try to solve (1), for example, with the condition $\displaystyle v(0)=0$. Proceeding in the 'standard way' you find that 'the solution' is...

$\displaystyle v(t)=f(t)=(\frac{t}{3})^{3}$ (3)

All seems to be 'ok' but... but there is a little problem because it is not difficult to verify that this family of functions...

$\displaystyle f_{\tau}(t)= (\frac {t-\tau}{3})^{3} \cdot U(t-\tau)$ (4)

... where $\displaystyle \tau >0$ and $\displaystyle U(*)$ is the so called 'step function' also satisfies the (1) with 'initial condition' $\displaystyle v(0)=0$... just a little problem! (Headbang) ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$