Exercise · ZW-VECTOR-FIELD-011 · ★★★ Advanced

On \(\mathbb{R}\) take \(v(x)=x^2\), which is smooth but unbounded. With \(\varphi_0(x)=x\) for \(x>0\), on which interval of \(t\) is \(\varphi_t(x)\) defined?

Choose one option.

Show answer
  • B. Only for \(0\le t<1/x\)

Explanation

With \(v(x)=x^2\), the equation \(\dot{\varphi}_t=\varphi_t^2\) gives \(\varphi_t(x)=x/(1-xt)\), which blows up at \(t=1/x\). Without a bound the flow need not be global.

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Review the concept: Velocity field

A velocity field \(v_t\) is a vector field that changes with time: at each instant \(t\) it assigns to every point the velocity at which it moves. In LDDMM it guides the smooth motion of the deformation.

\[v_t(x)\]

More practice problems: velocity field

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