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David Roberts
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I am working with surfaces in Euclidean 3-space. If we let X = X(u,v)$X = X(u,v)$ denote a parameterization of such a surface, then the mean curvature, H = H(u,v) $H = H(u,v)$, can be computed in terms of the coefficients for the first and second fundamental forms.

My question is this: Is it possible to express the mean curvature, H(u,v)$H(u,v)$, in terms of the support function for this surface? The support function is defined to be h = h(u,v) = <X, N>$h = h(u,v) = \langle X, N\rangle$ where N$N$ is a unit normal. (This function measures the oriented distance from a tangent plane to the origin.)

For curves in the plane there is a nifty result along these lines. If the curve has non-vanishing curvature its unit normal can be used for a parameterization, and in this situation the curvature satisfies 1/k = \pm (h''+ h)$1/k = \pm (h''+ h)$ where, again, h$h$ is the support function for the curve.

I'm hoping there is a similar result for convex surfaces in space, but, sadly, have been unable to find such a relationship. Any help would be greatly apprecaitedappreciated.

I am working with surfaces in Euclidean 3-space. If we let X = X(u,v) denote a parameterization of such a surface, then the mean curvature, H = H(u,v), can be computed in terms of the coefficients for the first and second fundamental forms.

My question is this: Is it possible to express the mean curvature, H(u,v), in terms of the support function for this surface? The support function is defined to be h = h(u,v) = <X, N> where N is a unit normal. (This function measures the oriented distance from a tangent plane to the origin.)

For curves in the plane there is a nifty result along these lines. If the curve has non-vanishing curvature its unit normal can be used for a parameterization, and in this situation the curvature satisfies 1/k = \pm (h''+ h) where, again, h is the support function for the curve.

I'm hoping there is a similar result for convex surfaces in space, but, sadly, have been unable to find such a relationship. Any help would be greatly apprecaited.

I am working with surfaces in Euclidean 3-space. If we let $X = X(u,v)$ denote a parameterization of such a surface, then the mean curvature, $H = H(u,v)$, can be computed in terms of the coefficients for the first and second fundamental forms.

My question is this: Is it possible to express the mean curvature, $H(u,v)$, in terms of the support function for this surface? The support function is defined to be $h = h(u,v) = \langle X, N\rangle$ where $N$ is a unit normal. (This function measures the oriented distance from a tangent plane to the origin.)

For curves in the plane there is a nifty result along these lines. If the curve has non-vanishing curvature its unit normal can be used for a parameterization, and in this situation the curvature satisfies $1/k = \pm (h''+ h)$ where, again, $h$ is the support function for the curve.

I'm hoping there is a similar result for convex surfaces in space, but, sadly, have been unable to find such a relationship. Any help would be greatly appreciated.

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Support Function and Mean Curvature

I am working with surfaces in Euclidean 3-space. If we let X = X(u,v) denote a parameterization of such a surface, then the mean curvature, H = H(u,v), can be computed in terms of the coefficients for the first and second fundamental forms.

My question is this: Is it possible to express the mean curvature, H(u,v), in terms of the support function for this surface? The support function is defined to be h = h(u,v) = <X, N> where N is a unit normal. (This function measures the oriented distance from a tangent plane to the origin.)

For curves in the plane there is a nifty result along these lines. If the curve has non-vanishing curvature its unit normal can be used for a parameterization, and in this situation the curvature satisfies 1/k = \pm (h''+ h) where, again, h is the support function for the curve.

I'm hoping there is a similar result for convex surfaces in space, but, sadly, have been unable to find such a relationship. Any help would be greatly apprecaited.