Ngô Quốc Anh

March 16, 2011

Cofactor matrix has divergence-free rows

Filed under: PDEs — Ngô Quốc Anh @ 15:45

In this entry, we prove the following interesting result

Let \mathbf{u} : \mathbb R^n \to \mathbb R^n be a smooth function. Then

\displaystyle \sum_{i=1}^n (\mbox{cof}D\mathbf{u})_{i,x_i}^k=0

for each k=\overline{1,n} fixed.

For simplicity, let us write \mathbf{u}=(u^1,...,u^n) \in \mathbb R^n. Then

\displaystyle D\mathbf{u} = \left( {\begin{array}{*{20}{c}} {u_{{x_1}}^1}&{u_{{x_2}}^1}& \cdots &{u_{{x_2}}^1} \\ {u_{{x_1}}^2}&{u_{{x_2}}^2}& \cdots &{u_{{x_2}}^2} \\ \vdots & \vdots & \ddots & \vdots \\ {u_{{x_1}}^n}&{u_{{x_2}}^n}& \cdots &{u_{{x_n}}^n} \end{array}} \right).

Proof. We recall from this entry that for each matrix A=(a_{ij}),

\displaystyle \frac{\partial}{\partial a_{ij}}\mbox{det}(A)=(\mbox{cof}A)_i^j.

This and the fact that

\displaystyle A^{-1}=\frac{1}{\mbox{det}(A)}\mbox{cof}A

or equivalently

\displaystyle \mbox{det}(A)I=A\mbox{cof}A

i.e.,

\displaystyle \mbox{det}(A)\delta_{ij}=\sum_{k=1}^n a_{ik}\mbox{cof}A_j^k.

give us after differentiating with respect to x_j

\displaystyle\frac{d}{{d{x_j}}}{\text{det}}({\text{cof}}D{\mathbf{u}}){\delta _{ij}} = \frac{d}{{d{x_j}}}\sum\limits_{l = 1}^n {u_{{x_l}}^i} ({\text{cof}}D{\mathbf{u}})_j^l = \sum\limits_{l = 1}^n {\left[ {u_{{x_l}{x_j}}^i({\text{cof}}D{\mathbf{u}})_j^l + u_{{x_l}}^i{{{\text{(}}({\text{cof}}D{\mathbf{u}})_j^l)}_{{x_j}}}} \right]} .

Summing with respect to j we get

\displaystyle\sum\limits_{j = 1}^n {\frac{d}{{d{x_j}}}{\text{det}}({\text{cof}}D{\mathbf{u}}){\delta _{ij}}} = \sum\limits_{j = 1}^n {\sum\limits_{l = 1}^n {\left[ {u_{{x_l}{x_j}}^i({\text{cof}}D{\mathbf{u}})_j^l + u_{{x_l}}^i{{{\text{(}}({\text{cof}}D{\mathbf{u}})_j^l)}_{{x_j}}}} \right]} } .

Again, it is not hard to see that

\displaystyle\sum\limits_{j = 1}^n {\frac{d}{{d{x_j}}}{\text{det}}({\text{cof}}D{\mathbf{u}}){\delta _{ij}}} = \sum\limits_{j = 1}^n {\sum\limits_{l = 1}^n {u_{{x_l}{x_j}}^i({\text{cof}}D{\mathbf{u}})_j^l} }.

Thus

\displaystyle\sum\limits_{j = 1}^n {\sum\limits_{l = 1}^n {u_{{x_l}}^i{{{\text{(}}({\text{cof}}D{\mathbf{u}})_j^l)}_{{x_j}}}} } = \sum\limits_{l = 1}^n {u_{{x_l}}^i\sum\limits_{j = 1}^n {{\text{(cof}}D{\mathbf{u}})_{j,{x_j}}^l} } = 0

which completes the proof.

Source: L. Evans, Partial differential equations, AMS, 1998.

1 Comment »

  1. If the Jacobian matrix is not invertible in some point, then the proof is not complete.

    Comment by Alex — March 22, 2015 @ 19:54


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