# Link between deformation and velocity gradient: $L_v = \dot{M}_t \cdot M_t^{-1}$

I'm studying fluid dynamics and I'm really struggling with notations and passages in some proofs that simply vanish.

In this question I need some guide-lines to show how to link the defomation map of a fluid and the velocity gradient: Let $$B_0$$ be a body that after a certain amount of time $$t$$ takes a different shape $$B_t$$. Supposing the change of shape is smooth enought into time, I used to call $$M_t$$ the deformation map which takes $$B_0 \to B_t$$.

Now, a part the clarifications about eulerian and lagrangian descripition of motion that is almost clear to me, I struggling to prove a theorem that states: $$L_v = \dot{M}_t \cdot M_t^{-1}$$ where $$L_v$$ is used ad a notation of the gradient of velocity (which one: the eulero's or lagrange's one?) and the dot operator is the convective/material derivative.

Can anyone help me to dissolve my doubts and give me a sketch of proof (in the case the theorem I stated is true)?

$$M_t=\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}$$, so $$\begin{split} \dot{M}_t M_t^{-1} = \left[\frac{D}{\mathrm{d}t}\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}\right]\left[\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}\right]^{-1}\\ =\left[\frac{\partial\mathbf{v}(t)}{\partial\mathbf{x}(0)}\right]\cdot\left[\frac{\partial\mathbf{x}(0)}{\partial\mathbf{x}(t)}\right]=\frac{\partial\mathbf{v}(t)}{\partial\mathbf{x}(t)}=L_\mathbf{v} \end{split}$$ This is an Eulerian quantity because we are taking derivative with respect to the current $$\mathbf{x}(t)$$ not a fixed reference $$\mathbf{x}(0)$$.
• I miss the passage $\left[\frac{D}{\mathrm{d}t}\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}\right]\left[\frac{\partial\mathbf{x}(t)}{\partial\mathbf{x}(0)}\right]^{-1} =\left[\frac{\partial\mathbf{v}(t)}{\partial\mathbf{x}(0)}\right]\cdot\left[\frac{\partial\mathbf{x}(0)}{\partial\mathbf{x}(t)}\right]$. Can you be more specific please? Commented Mar 3, 2021 at 10:50
• That is just $\frac{D}{\mathrm{d}t}\mathbf{x}(t)=\mathbf{v}(t)$ (since $\mathbf{x}(0)$ is fixed you can swap $\frac{\partial}{\partial\mathbf{x}(0)}$ and material derivative), and the chain rule to get $[\frac{\partial\mathbf{a}}{\partial\mathbf{b}}]^{-1}=\frac{\partial\mathbf{b}}{\partial\mathbf{a}}$. Commented Mar 3, 2021 at 11:47