The quantity $$ (\mathbf{B} \cdot \nabla) \mathbf{A} $$ represents the rate of change of $\mathbf{A}$ in the direction of $\mathbf{B}$. I am interested in the case where $\mathbf{B} = (\mathbf{A} \cdot \nabla) \mathbf{A}$, which is essentially the rate of change of $\mathbf{A}$ in the direction of its own rate of change. Can anything be said about the quantity $$ \left\{ \left[ (\mathbf{A} \cdot \nabla) \mathbf{A} \right] \cdot \nabla \right\} \mathbf{A} $$ Can it be simplified? Does there exist an identity for it? If we expand using the Cartesian components of $\mathbf{A}$: \begin{align} \left\{ \left[ (\mathbf{A} \cdot \nabla) \mathbf{A} \right] \cdot \nabla \right\} \mathbf{A} &= \left[ (A^i \partial_i \mathbf{A}) \cdot \nabla \right] \mathbf{A} \\ &= A^i (\partial_i A^j) \partial_j \mathbf{A} \\ &= A^i (\partial_i A^j) (\partial_j A^k) e_k \end{align} where $e_k$ are the Cartesian basis vectors, I don't see any obvious way to proceed here. Does anyone see a way to continue and derive a nice identity from this? Ideally I am looking for a way to express the result in terms of $\mathbf{A}$ and $(\mathbf{A} \cdot \nabla) \mathbf{A}$ if possible.
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- 1$\begingroup$ I don't think there's any particularly nice final form. You could use $(\mathbf A \cdot \nabla) \mathbf A = \frac 12 \nabla(A^2) - \mathbf A \times (\nabla \times \mathbf A)$ but I think that's about it. $\endgroup$kipf– kipf2025-08-27 12:55:39 +00:00Commented Aug 27 at 12:55
- $\begingroup$ By definition, it's the square of the Jacobian of $A$ applied to $A$, $$J_A^2A,$$ is that what you're looking for? $\endgroup$Nicholas Todoroff– Nicholas Todoroff2025-08-27 13:52:53 +00:00Commented Aug 27 at 13:52
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This answer is just elaborating on Nicholas' comment under the OP. The Jacobian matrix of a vector $\mathbf{A}$ is defined as, $$ (J_{\mathbf{A}})_{ij} = \partial_j A^i $$ So, \begin{align} ( \mathbf{B} \cdot \nabla ) \mathbf{A} &= B^j \partial_j \mathbf{A} \\ &= B^j (J_{\mathbf{A}})^T_{\cdot,j} \\ &= J_{\mathbf{A}} \mathbf{B} \end{align} Therefore, \begin{align} \left\{ \left[ (\mathbf{A} \cdot \nabla) \mathbf{A} \right] \cdot \nabla \right\} \mathbf{A} &= \left( J_{\mathbf{A}} \mathbf{A} \cdot \nabla \right) \mathbf{A} \\ &= J_{\mathbf{A}}^2 \mathbf{A} \end{align} Following this further, nesting the directional derivative $n$ times should produce the result $J_{\mathbf{A}}^n \mathbf{A}$.