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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.

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  • $\begingroup$ You also need that the first derivative is bounded to get this kind of global Lipschitz condition. $\endgroup$ Mar 5 '17 at 18:46
  • $\begingroup$ how about the wellposed ?how to define if it is well posed? $\endgroup$
    – Ccnoc
    Mar 6 '17 at 2:21
  • $\begingroup$ 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. $\endgroup$ Mar 6 '17 at 6:49
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Brother what remains here is for you to show whether each of these functions is continuous on D or not. If a function is continuous in D, then it is well-posed given that it satisfies Lipschitz condition. For more details read a theorem in Burden and Faires, Numerical Analysis, Ninth Edition, Chapter 5 page 263. Good Luck.

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

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  • $\begingroup$ 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. $\endgroup$ Apr 16 '18 at 15:25
  • $\begingroup$ No it is not wrong, brother because the definition of D is not correct as requested by the one asking the question brother. $\endgroup$ Apr 16 '18 at 17:12
  • $\begingroup$ I have done corrections @LutzL, please vote upwards, my answer is now 100% correct. $\endgroup$ Apr 16 '18 at 17:18

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