# well posed and Lipschitz condition

I encountered some problems about well posed and Lipschitz condition in the numerical analysis lecture.

Theorem: Suppose $D=\left\{\left(t,y\right)|0\leq t\leq 1,-\infty\le y\le \infty\right\}$ . If $f$ is continuous and satisfies a Lipschitz condition in the variable $y$ on the set $D$, then the initial value problem

Does $f$ satisfy a Lipschitz condition on $D=\left\{\left(t,y\right)|0\leq t\leq 1,-\infty\le y\le \infty\right\}$ . Can the above theorem be used to show that the initial problem $\frac{\text{d}y}{\text{d}t}=f(t,y)$, $0 \leq t \leq 1$, $y(0)=1$ is well posed.

• $f(t,y)=e^{t-y}$

• $f(t,y)=\frac{1+y}{1+t}$

• $f(t,y)=\cos(ty)$

• $f(t,y)=\frac{y^{2}}{1+t}$

I think four of them fulfill Lipschitz condition as I can partial derivative exist and is continuous but the well posed part I am not sure, what should I do? thanks.

• You also need that the first derivative is bounded to get this kind of global Lipschitz condition. Mar 5 '17 at 18:46
• how about the wellposed ?how to define if it is well posed? Mar 6 '17 at 2:21
• In this context "well-posed" should mean the existence of a solution at all, uniqueness you already covered. You want solution functions on $[0,1]$ without singularities or gaps. Mar 6 '17 at 6:49

According to that theorem, if $f(t,y)=\cos(ty)$ , then $\dfrac{dy}{dt}=f(t,y), \ y_0=\alpha$ is well-posed, because $f(t,y)=\cos(ty)$ is not only continuous in D but it is also satisfying Lipschitz Condition
• This argument is wrong, as the Lipschitz condition is only local, the first derivative is not bounded, as $e^{-y}\to\infty$ for $y\to-\infty$. A closer inspection or a transformation of the problem might save the well-posedness, but this effort goes above the half-cited theorem. Apr 16 '18 at 15:25