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Show that {\begin{bmatrix}{\sqrt  {|x_{1}|}}\\x_{2}\\\end{bmatrix}} is continuous for all x, but does not satisfy a Lipschitz condition in any domain D which contains x=0.

f({\vec  x},t)={\begin{bmatrix}{\sqrt  {|x_{1}|}}\\x_{2}\\\end{bmatrix}}

If x_{{1,2}}\neq 0 then {\frac  {d}{dx_{1}}}{\sqrt  {|x_{1}|}}={\frac  {1}{2}}x_{1}^{{-1/2}}\, and {\frac  {d}{dx_{2}}}x_{2}=1\, are defined. If x_{{1,2}}=0\, then {\frac  {d}{dx_{1}}}0={\frac  {d}{dx_{2}}}0=0\, are defined, so f is continuous.

But {\frac  {\partial f}{\partial x}}={\begin{bmatrix}{\frac  {1}{2}}x_{1}^{{-1/2}}&0\\0&1\\\end{bmatrix}},\left|{\frac  {\partial f}{\partial x}}\right|={\frac  {1}{2}}x_{1}^{{-1/2}}+1\,

is only defined if x_{1}\neq 0, so f is Lipschitz only if x=0 is not in the domain.

Main Page : Ordinary Differential Equations : Lipshitz Conditions