2

I'm working this function:

$f(t,x)=\frac{4t^3x}{t^4+x^2}$ for $(t,x)\neq(0,0);$

$f(t,x)=0$ for $(t,x)=(0,0)$

As I proved continuity using the line $x=t^2$, I need to study if it's globally Lipschitz. I tried to derivate on $x$, and I got:

$\frac{df(t,x)}{dx}=\frac{4t^3(t^4+x^2)-8t^3x^2}{(t^4+x^2)^2}=\frac{4t^3(t^4-x^2)}{(t^4+x^2)^2}$

So if $(t,x)\to (0^+,0^+), \frac{df(t,x)}{dx}\to +\infty$. Would it be a proof to negate the global Lipschitz condition? And this would be enough to study the local Lipschitz condition?

Thank you very much

Update: Fixed $\frac{df(t,x)}{dx}$ to avoid mistakes

Pablo Luis
  • 77

1 Answers1

1

Using the quotient rule You should get $$\frac{\partial f(t,x)}{\partial x}=\frac{4t^3(t^4+x^2)-4t^3 2x}{(t^4+x^2)^2}=\frac{4t^7-4t^3x^2}{(t^4+x^2)^2}$$ and thus $$\lim_{|x|\rightarrow \infty}\frac{\partial f(t,x)}{\partial x}=0.$$ Furthermore for each $t\neq 0$ the denominator of the derivative $\frac{\partial f(t,x)}{\partial x}$ is bounded from below, thus the derivative is bounded and so the function $f_t:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R},x\mapsto f(t,x)$ is Lipschitz-continuous ( by a mean-value theorem-argument) with a Lipschitzconstant $L_t$. (For $t=0$ this map $f_0$ is the zero map.) But one can see that $$\lim_{t\rightarrow 0}L_t=\infty$$ and thus one can't establish a global Lipschitz-condition of the form $$|f(t,x_1)-f(t,x_2)|\leq L|x_1-x_2|$$ $\textbf{ for all }$ $t,x_1,x_2\in\mathbb{R}$. If one considers restrictions $f_{|I\times\mathbb{R}}$ where $I$ is an interval that does not contain $0$ this is possible and the corresponding boundary value problem has a unique solution by Picard- Lindelöff's theorem. (The counterexample $x_1(t)=t^2,x_2(t)=-t^2$ doesn't work then because if $I=[a,b]$ then $x_1(a)\neq x_2(a)$ in this case (since $a\neq 0$).

Peter Melech
  • 4,353
  • 1
    My question is: if $f(t,x)$ is globally Lipschitz-continuous, why can we find two different solutions on $x(t)=t^2$ and $x(t)=-t^2$ when, by Lipschitz Theorem, if $f(t,x)$ is globally Lipschitz-continuous, then the solution should be unique? – Pablo Luis Dec 21 '18 at 14:49
  • Good point! Didn't notice that at a first glance. I edited my answer! – Peter Melech Dec 21 '18 at 15:24
  • In that case, we are saying that our function is locally Lipschitz-continuous but not globally, since we can't bound that constant if $t\to 0$ – Pablo Luis Dec 21 '18 at 15:52
  • Exactly! The condition of global Lipschitz-continuity has to be read quite carefully! – Peter Melech Dec 21 '18 at 16:01