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16 hours ago comment added Paruru Can you elaborate a bit more? I understood that $(\Delta f_1, \Delta f_2)^\prime = \tilde A ((\nabla f_1)^\prime, (\nabla f_2)^\prime)^\prime$ holds for some $\tilde A.$ Does this fit in the standard framework of elliptic PDE?
17 hours ago history edited LSpice CC BY-SA 4.0
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17 hours ago history edited Paruru CC BY-SA 4.0
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18 hours ago comment added mlk It is a bit confusing, because in a usual elliptic operator the matrix would apply to the derivatives, not the components of $f$. But this means by product rule your PDE is equivalent to $A \Delta f = \text{ lower order terms}.$ So if $A$ is invertible everywhere we can multiply with $A^{-1}$ and get $\Delta f = \text{slightly different lower order terms}$ at which point uniqueness should follow from standard arguments for elliptic PDE.
22 hours ago history asked Paruru CC BY-SA 4.0