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Consider $x'=f(x)$ such that $(x_1,x_2)\mapsto(-x_1+2x_2,-2x_1-x_2)$.

Show that for two solutions $x(t)$ and $y(t)$ of the above differential equation, we have:

$$\lVert x(t)-y(t)\rVert \leq e^{-t}\lVert x(0)-y(0)\rVert .$$

Please can you show me how to apply the Gronwall lemma to this?

This part that is most confusing to is how to use the lemma and how to use the lemma with $$(x_1,x_2)\mapsto(-x_1+2x_2,-2x_1-x_2).$$

P.s, I dont understand why this has been voted down!! The guidelines for the site state that if a questions hasnt been answered or answered properly you can ask it again!

Thanks very much.

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that wasnt an accepted answer though, plus the answer looks very different to what mine is ment to look like!! In which case you are allowed to ask the question again, that person did not get their answered properly otherwise i would be using theirs! I read the guidelines before i posted this as i had already been through all other gronwall questions and seen that one. – sarah Mar 19 '13 at 1:31

This looks suspiciously like one of the questions in my differential equations workshops this year. Anyway, let $z=||x-y||^2.$ Then $\frac{d}{dt}z=\frac{d}{dt}||x-y||^2$ which is equal to $$ =2 \begin{pmatrix} x_1-y_1 \\ x_2-y_2 \end{pmatrix} \begin{pmatrix} -x_1+2x_2+y_1-2y_2\\ -2x_1-x_2+2y_1+y_2 \end{pmatrix} \\ =2(x_1-y_1)(-x_1+2x_2+y_1-2y_2)+2(x_2-y_2)(-2x_1-x_2+2y_1+y_2) \\ =-2x^2_1+4x_1y_1-2x^2_2-2y^2_1-2y^2_2+4x_2y_2 \\ =-2(x_1-y_1)^2-2(x_2-y_2)^2 =-2z. $$ Applying Gronwall's inequality gives $$ z\leq e^{-2t}z(0)\Rightarrow ||x(t)-y(y)||\leq e^{-t}||x(0)-y(0)||. $$

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I think this is how your would answer the question but not sure and dont quite get all of it either

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it would have been better if you had posted the link as a comment – Giovanni Aug 27 '15 at 16:44

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