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isometry and orthogonality proof

EM waves - orthogonality - amplitude/phase angleGauss´s law proof “details”Application of Gauss Divergence TheoremIs the naive solution of this PDE/BVP unique?Proof using the orthogonality of the Hermite PolynomialsWhat Motivated the Definition of the Orthogonality of Functions?Riemannian connection under isometryProof that $det(AB) = det(A) det(B)$Proof of the fundamental theorem of line integralsSuppose $f:Uto mathbbR^m$ is differentiable and $U$ is convex. Show that $vert f(b)-f(a)vert leq Mvert b-a vert$

0

$begingroup$

If I have a relation (assuming $vecf$ is one-to-one with $det(nabla vecf)>0$) appicable to all points from the domain of $vecf$ which a regular region (a closed region with piecewise smooth boundary) in 3 dimensional Euclidean space:

$[vecf(vecx) - vecf(vecy)]cdot[vecf(vecx) - vecf(vecy)]=(vecx-vecy)cdot(vecx-vecy)$,

and apply $nabla_vecy$ holding $vecx$ constant, I get:

$[nabla_vecy vecf(vecy)]^rmT[vecf(vecx) - vecf(vecy)]=mathbbI (vecx-vecy)$.

If I then apply $nabla_vecx$ holding $vecy$ constant, I get:

$[nabla_vecy vecf(vecy)]^rmT nabla_vecxvecf(vecx) = mathbbI$,

where $mathbbI$ is identity. Does that mean that I can take $vecx=vecy$ in the last relation and get orthogonality of $nabla vecf(vecx)$ and from that that $nablavecf(vecx)$ is constant on the domain of $vecf$? My issue is with the fact that in the case $vecx=vecy$, I get $vec0=vec0$ in the first relation. Also, is the above proof correct?

proof-verification proof-explanation orthogonality isometry

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asked Mar 14 at 15:20

leosenkoleosenko

1878

$endgroup$

add a comment | 

0

$begingroup$

If I have a relation (assuming $vecf$ is one-to-one with $det(nabla vecf)>0$) appicable to all points from the domain of $vecf$ which a regular region (a closed region with piecewise smooth boundary) in 3 dimensional Euclidean space:

$[vecf(vecx) - vecf(vecy)]cdot[vecf(vecx) - vecf(vecy)]=(vecx-vecy)cdot(vecx-vecy)$,

and apply $nabla_vecy$ holding $vecx$ constant, I get:

$[nabla_vecy vecf(vecy)]^rmT[vecf(vecx) - vecf(vecy)]=mathbbI (vecx-vecy)$.

If I then apply $nabla_vecx$ holding $vecy$ constant, I get:

$[nabla_vecy vecf(vecy)]^rmT nabla_vecxvecf(vecx) = mathbbI$,

where $mathbbI$ is identity. Does that mean that I can take $vecx=vecy$ in the last relation and get orthogonality of $nabla vecf(vecx)$ and from that that $nablavecf(vecx)$ is constant on the domain of $vecf$? My issue is with the fact that in the case $vecx=vecy$, I get $vec0=vec0$ in the first relation. Also, is the above proof correct?

proof-verification proof-explanation orthogonality isometry

share|cite|improve this question

asked Mar 14 at 15:20

leosenkoleosenko

1878

$endgroup$

add a comment | 

$begingroup$

If I have a relation (assuming $vecf$ is one-to-one with $det(nabla vecf)>0$) appicable to all points from the domain of $vecf$ which a regular region (a closed region with piecewise smooth boundary) in 3 dimensional Euclidean space:

$[vecf(vecx) - vecf(vecy)]cdot[vecf(vecx) - vecf(vecy)]=(vecx-vecy)cdot(vecx-vecy)$,

and apply $nabla_vecy$ holding $vecx$ constant, I get:

$[nabla_vecy vecf(vecy)]^rmT[vecf(vecx) - vecf(vecy)]=mathbbI (vecx-vecy)$.

If I then apply $nabla_vecx$ holding $vecy$ constant, I get:

$[nabla_vecy vecf(vecy)]^rmT nabla_vecxvecf(vecx) = mathbbI$,

where $mathbbI$ is identity. Does that mean that I can take $vecx=vecy$ in the last relation and get orthogonality of $nabla vecf(vecx)$ and from that that $nablavecf(vecx)$ is constant on the domain of $vecf$? My issue is with the fact that in the case $vecx=vecy$, I get $vec0=vec0$ in the first relation. Also, is the above proof correct?

proof-verification proof-explanation orthogonality isometry

share|cite|improve this question

asked Mar 14 at 15:20

leosenkoleosenko

1878

$endgroup$

share|cite|improve this question

asked Mar 14 at 15:20

leosenkoleosenko

1878

share|cite|improve this question

asked Mar 14 at 15:20

leosenkoleosenko

1878

asked Mar 14 at 15:20

leosenkoleosenko

1878

asked Mar 14 at 15:20

leosenkoleosenko

1878