# Existence and uniqueness of IVP solutions, vector-valued equations

The existence and uniqueness of a solution to the IVP $$\underline{y'}=f(x,\underline{y}),\enspace \underline y(0)=\underline{y_0},$$

where $$\underline{y} = (y_1, y_2,...,y_n)$$, is guaranteed if the function $$f$$ satisfies certain conditions. That is, $$f$$ must be a continuous bounded function, satisfying the Lipschitz condition on some rectangle $$R : = \vert x \vert \leq a, \enspace \vert \underline{y}\vert\leq b$$, where $$a,b >0$$. In this case, bounded means that $$\vert f(x,\underline{y})\vert\leq M$$, for some $$M>0$$.

My question is about the vector norm , $$\vert \underline{y} \vert$$. The proof of this theorem ( I'm reading Coddington's book ) when $$f$$ has only some real $$y$$ as input employs the absolute value. It is commented that the proof of the multivariable case is similar, except we replace real numbers with vector quantities when necessary.

Does the choice of the vector norm make a difference in the proof? And does it matter if we want to show local existence and uniqueness of IVP solutions? The book does not mention what vector norm is employed, but I suspect it is the "taxicab" norm $$\vert \underline{y} \vert =\vert y_1 \vert +\vert y_2 \vert+\cdots\vert y_n\vert$$

as the exercises given at the end of the chapter turn out correctly when using this norm.

Edit : The function $$f$$ here is vector valued, $$f(x\underline{y})=(f_1,f_2,...,f_n)$$

No, it makes no difference in the proof. All norms are equivalent in finite dimensions. You would find a difference when actually computing the numbers, maximum and Lipschitz constant, used. Also the box would be a different set, as $$|y|\le b$$ would change shape. But these quantitative differences do not change the qualitative claim of the existence of a unique local solution.