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Is there a functional (i.e. infinite-dimensional) generalization of the second partial derivative test?
$begingroup$
For a smooth function $f: mathbb{R}^n to mathbb{R}$, we can (usually) test whether a critical point ${bf x}_0$ (at which ${bf nabla} f({bf x}_0) = {bf 0}$) is a local maximum, minimum, or saddle point via the second partial derivative test, which considers the signs of the eigenvalues of the Hessian matrix $H_{ij} := partial_i partial_j f$. This result can be generalized to successively more general domains of $f$:
If the domain of $f$ is an arbitrary Riemannian manifold, then the test still works, but we instead consider the eigenvalues of the tensor $H^i_{ j} = g^{ik} H_{kj}$, where $H_{kj}$ is the Hessian tensor $H_{ij} := nabla_i nabla_j f = partial_i partial_j f - Gamma_{ij}^k partial_k f$, where $Gamma_{ij}^k$ are the Christoffel symbols (of the second kind).
If the domain of $f$ is an arbitrary smooth manifold with a torsion-free connection, then the Hessian (as defined directly above) is a rank-$(2,0)$ tensor rather than a rank-$(1,1)$ tensor, so we can't talk about its eigenvalues. But we can still talk about the signature of the corresponding quadratic form by Sylvester's law of inertia, so I believe the test still works.
If the domain of $f$ is an arbitrary smooth manifold without a connection, then there's no natural way to define a Hessian tensor away from the critical points. But at a critical point the Hessian tensor becomes independent of the connection, so we can define it (in local coordinates) by the usual Euclidean-space formula $H_{ij} := partial_i partial_j f$ and use Sylvester's law of inertia as above.
Now if $f$ (which I will rename $S$) is a smooth functional $S[q]$, then its domain is an infinite-dimensional space of functions $q(t)$. In this case we can still talk about critical "points" of the functional, which are functions $q_0(t)$ at which $S[q]$ is stationary. For example, if $S$ takes the form
$$S[q] = int_a^b L left( q(t), frac{dq}{dt}, t right) dt$$ for some constants $a$ and $b$ and differentiable function $L: mathbb{R}^3 to mathbb{R}$, then the critical "points" $q_0(t)$ are given by the Euler-Lagrange equation
$$frac{partial L}{partial q} - frac{d}{dt} frac{partial L}{partial dot{q}} = 0.$$
Is there a functional generalization of the second partial derivative test that tests whether these critical functions $q_0(t)$ are local minima, local maxima, or saddle points of the functional $F[q]$?
(I don't know anything about quadratic forms on infinite-dimensional vector spaces, so I have no idea if there's any notion of a signature, etc.)
functional-analysis calculus-of-variations quadratic-forms stationary-point
asked Feb 3 at 5:50
tparkertparker
1,941834
$endgroup$
add a comment |
$begingroup$
For a smooth function $f: mathbb{R}^n to mathbb{R}$, we can (usually) test whether a critical point ${bf x}_0$ (at which ${bf nabla} f({bf x}_0) = {bf 0}$) is a local maximum, minimum, or saddle point via the second partial derivative test, which considers the signs of the eigenvalues of the Hessian matrix $H_{ij} := partial_i partial_j f$. This result can be generalized to successively more general domains of $f$:
If the domain of $f$ is an arbitrary Riemannian manifold, then the test still works, but we instead consider the eigenvalues of the tensor $H^i_{ j} = g^{ik} H_{kj}$, where $H_{kj}$ is the Hessian tensor $H_{ij} := nabla_i nabla_j f = partial_i partial_j f - Gamma_{ij}^k partial_k f$, where $Gamma_{ij}^k$ are the Christoffel symbols (of the second kind).
If the domain of $f$ is an arbitrary smooth manifold with a torsion-free connection, then the Hessian (as defined directly above) is a rank-$(2,0)$ tensor rather than a rank-$(1,1)$ tensor, so we can't talk about its eigenvalues. But we can still talk about the signature of the corresponding quadratic form by Sylvester's law of inertia, so I believe the test still works.
If the domain of $f$ is an arbitrary smooth manifold without a connection, then there's no natural way to define a Hessian tensor away from the critical points. But at a critical point the Hessian tensor becomes independent of the connection, so we can define it (in local coordinates) by the usual Euclidean-space formula $H_{ij} := partial_i partial_j f$ and use Sylvester's law of inertia as above.
Now if $f$ (which I will rename $S$) is a smooth functional $S[q]$, then its domain is an infinite-dimensional space of functions $q(t)$. In this case we can still talk about critical "points" of the functional, which are functions $q_0(t)$ at which $S[q]$ is stationary. For example, if $S$ takes the form
$$S[q] = int_a^b L left( q(t), frac{dq}{dt}, t right) dt$$ for some constants $a$ and $b$ and differentiable function $L: mathbb{R}^3 to mathbb{R}$, then the critical "points" $q_0(t)$ are given by the Euler-Lagrange equation
$$frac{partial L}{partial q} - frac{d}{dt} frac{partial L}{partial dot{q}} = 0.$$
Is there a functional generalization of the second partial derivative test that tests whether these critical functions $q_0(t)$ are local minima, local maxima, or saddle points of the functional $F[q]$?
(I don't know anything about quadratic forms on infinite-dimensional vector spaces, so I have no idea if there's any notion of a signature, etc.)
functional-analysis calculus-of-variations quadratic-forms stationary-point
asked Feb 3 at 5:50
tparkertparker
1,941834
$endgroup$
add a comment |
$begingroup$
For a smooth function $f: mathbb{R}^n to mathbb{R}$, we can (usually) test whether a critical point ${bf x}_0$ (at which ${bf nabla} f({bf x}_0) = {bf 0}$) is a local maximum, minimum, or saddle point via the second partial derivative test, which considers the signs of the eigenvalues of the Hessian matrix $H_{ij} := partial_i partial_j f$. This result can be generalized to successively more general domains of $f$:
If the domain of $f$ is an arbitrary Riemannian manifold, then the test still works, but we instead consider the eigenvalues of the tensor $H^i_{ j} = g^{ik} H_{kj}$, where $H_{kj}$ is the Hessian tensor $H_{ij} := nabla_i nabla_j f = partial_i partial_j f - Gamma_{ij}^k partial_k f$, where $Gamma_{ij}^k$ are the Christoffel symbols (of the second kind).
If the domain of $f$ is an arbitrary smooth manifold with a torsion-free connection, then the Hessian (as defined directly above) is a rank-$(2,0)$ tensor rather than a rank-$(1,1)$ tensor, so we can't talk about its eigenvalues. But we can still talk about the signature of the corresponding quadratic form by Sylvester's law of inertia, so I believe the test still works.
If the domain of $f$ is an arbitrary smooth manifold without a connection, then there's no natural way to define a Hessian tensor away from the critical points. But at a critical point the Hessian tensor becomes independent of the connection, so we can define it (in local coordinates) by the usual Euclidean-space formula $H_{ij} := partial_i partial_j f$ and use Sylvester's law of inertia as above.
Now if $f$ (which I will rename $S$) is a smooth functional $S[q]$, then its domain is an infinite-dimensional space of functions $q(t)$. In this case we can still talk about critical "points" of the functional, which are functions $q_0(t)$ at which $S[q]$ is stationary. For example, if $S$ takes the form
$$S[q] = int_a^b L left( q(t), frac{dq}{dt}, t right) dt$$ for some constants $a$ and $b$ and differentiable function $L: mathbb{R}^3 to mathbb{R}$, then the critical "points" $q_0(t)$ are given by the Euler-Lagrange equation
$$frac{partial L}{partial q} - frac{d}{dt} frac{partial L}{partial dot{q}} = 0.$$
Is there a functional generalization of the second partial derivative test that tests whether these critical functions $q_0(t)$ are local minima, local maxima, or saddle points of the functional $F[q]$?
(I don't know anything about quadratic forms on infinite-dimensional vector spaces, so I have no idea if there's any notion of a signature, etc.)
functional-analysis calculus-of-variations quadratic-forms stationary-point
asked Feb 3 at 5:50
tparkertparker
1,941834
$endgroup$
For a smooth function $f: mathbb{R}^n to mathbb{R}$, we can (usually) test whether a critical point ${bf x}_0$ (at which ${bf nabla} f({bf x}_0) = {bf 0}$) is a local maximum, minimum, or saddle point via the second partial derivative test, which considers the signs of the eigenvalues of the Hessian matrix $H_{ij} := partial_i partial_j f$. This result can be generalized to successively more general domains of $f$:
If the domain of $f$ is an arbitrary Riemannian manifold, then the test still works, but we instead consider the eigenvalues of the tensor $H^i_{ j} = g^{ik} H_{kj}$, where $H_{kj}$ is the Hessian tensor $H_{ij} := nabla_i nabla_j f = partial_i partial_j f - Gamma_{ij}^k partial_k f$, where $Gamma_{ij}^k$ are the Christoffel symbols (of the second kind).
If the domain of $f$ is an arbitrary smooth manifold with a torsion-free connection, then the Hessian (as defined directly above) is a rank-$(2,0)$ tensor rather than a rank-$(1,1)$ tensor, so we can't talk about its eigenvalues. But we can still talk about the signature of the corresponding quadratic form by Sylvester's law of inertia, so I believe the test still works.
If the domain of $f$ is an arbitrary smooth manifold without a connection, then there's no natural way to define a Hessian tensor away from the critical points. But at a critical point the Hessian tensor becomes independent of the connection, so we can define it (in local coordinates) by the usual Euclidean-space formula $H_{ij} := partial_i partial_j f$ and use Sylvester's law of inertia as above.
Now if $f$ (which I will rename $S$) is a smooth functional $S[q]$, then its domain is an infinite-dimensional space of functions $q(t)$. In this case we can still talk about critical "points" of the functional, which are functions $q_0(t)$ at which $S[q]$ is stationary. For example, if $S$ takes the form
$$S[q] = int_a^b L left( q(t), frac{dq}{dt}, t right) dt$$ for some constants $a$ and $b$ and differentiable function $L: mathbb{R}^3 to mathbb{R}$, then the critical "points" $q_0(t)$ are given by the Euler-Lagrange equation
$$frac{partial L}{partial q} - frac{d}{dt} frac{partial L}{partial dot{q}} = 0.$$
Is there a functional generalization of the second partial derivative test that tests whether these critical functions $q_0(t)$ are local minima, local maxima, or saddle points of the functional $F[q]$?
(I don't know anything about quadratic forms on infinite-dimensional vector spaces, so I have no idea if there's any notion of a signature, etc.)
functional-analysis calculus-of-variations quadratic-forms stationary-point
functional-analysis calculus-of-variations quadratic-forms stationary-point
asked Feb 3 at 5:50
tparkertparker
1,941834
asked Feb 3 at 5:50
tparkertparker
1,941834
asked Feb 3 at 5:50
tparkertparker
1,941834
asked Feb 3 at 5:50
tparkertparker
1,941834
asked Feb 3 at 5:50
tparkertparker
1,941834
1,941834
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Suppose that your functional $S$ is defined on an infinite-dimensional (real) Banach space $X$. Further, we need some differentiability of $S$, and the right tool here is 'twice Fréchet differentiable'.
This ensures that $S$ satisfies a Taylor expansion
$$S(x + h) = S(x) + S'(x),h + frac12 , S''(x) [h,h] + o(|h|^2).$$
Here, the second derivative $S''(x)$ is a bounded, bilinear form on $X$.
Now, if $bar x$ is a local minimizer, you have
- $S'(bar x) = 0$
$S''(bar x) [h,h] ge 0$ for all $h in X$ (note that this generalizes 'positive semi-definite').
Similarly, if
$S'(bar x) = 0$ and- there exists $alpha > 0$ such that for all $h in X$ we have $S''(bar x) [h,h] ge alpha , |h|^2$ (note that this generalizes 'positive definite'),
then $bar x$ is a local minimizer.
answered Feb 3 at 18:25
gerwgerw
20.1k11334
$endgroup$
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
add a comment |
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1 Answer
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$begingroup$
Suppose that your functional $S$ is defined on an infinite-dimensional (real) Banach space $X$. Further, we need some differentiability of $S$, and the right tool here is 'twice Fréchet differentiable'.
This ensures that $S$ satisfies a Taylor expansion
$$S(x + h) = S(x) + S'(x),h + frac12 , S''(x) [h,h] + o(|h|^2).$$
Here, the second derivative $S''(x)$ is a bounded, bilinear form on $X$.
Now, if $bar x$ is a local minimizer, you have
- $S'(bar x) = 0$
$S''(bar x) [h,h] ge 0$ for all $h in X$ (note that this generalizes 'positive semi-definite').
Similarly, if
$S'(bar x) = 0$ and- there exists $alpha > 0$ such that for all $h in X$ we have $S''(bar x) [h,h] ge alpha , |h|^2$ (note that this generalizes 'positive definite'),
then $bar x$ is a local minimizer.
answered Feb 3 at 18:25
gerwgerw
20.1k11334
$endgroup$
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
add a comment |
$begingroup$
Suppose that your functional $S$ is defined on an infinite-dimensional (real) Banach space $X$. Further, we need some differentiability of $S$, and the right tool here is 'twice Fréchet differentiable'.
This ensures that $S$ satisfies a Taylor expansion
$$S(x + h) = S(x) + S'(x),h + frac12 , S''(x) [h,h] + o(|h|^2).$$
Here, the second derivative $S''(x)$ is a bounded, bilinear form on $X$.
Now, if $bar x$ is a local minimizer, you have
- $S'(bar x) = 0$
$S''(bar x) [h,h] ge 0$ for all $h in X$ (note that this generalizes 'positive semi-definite').
Similarly, if
$S'(bar x) = 0$ and- there exists $alpha > 0$ such that for all $h in X$ we have $S''(bar x) [h,h] ge alpha , |h|^2$ (note that this generalizes 'positive definite'),
then $bar x$ is a local minimizer.
answered Feb 3 at 18:25
gerwgerw
20.1k11334
$endgroup$
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
add a comment |
$begingroup$
Suppose that your functional $S$ is defined on an infinite-dimensional (real) Banach space $X$. Further, we need some differentiability of $S$, and the right tool here is 'twice Fréchet differentiable'.
This ensures that $S$ satisfies a Taylor expansion
$$S(x + h) = S(x) + S'(x),h + frac12 , S''(x) [h,h] + o(|h|^2).$$
Here, the second derivative $S''(x)$ is a bounded, bilinear form on $X$.
Now, if $bar x$ is a local minimizer, you have
- $S'(bar x) = 0$
$S''(bar x) [h,h] ge 0$ for all $h in X$ (note that this generalizes 'positive semi-definite').
Similarly, if
$S'(bar x) = 0$ and- there exists $alpha > 0$ such that for all $h in X$ we have $S''(bar x) [h,h] ge alpha , |h|^2$ (note that this generalizes 'positive definite'),
then $bar x$ is a local minimizer.
answered Feb 3 at 18:25
gerwgerw
20.1k11334
$endgroup$
Suppose that your functional $S$ is defined on an infinite-dimensional (real) Banach space $X$. Further, we need some differentiability of $S$, and the right tool here is 'twice Fréchet differentiable'.
This ensures that $S$ satisfies a Taylor expansion
$$S(x + h) = S(x) + S'(x),h + frac12 , S''(x) [h,h] + o(|h|^2).$$
Here, the second derivative $S''(x)$ is a bounded, bilinear form on $X$.
Now, if $bar x$ is a local minimizer, you have
- $S'(bar x) = 0$
$S''(bar x) [h,h] ge 0$ for all $h in X$ (note that this generalizes 'positive semi-definite').
Similarly, if
$S'(bar x) = 0$ and- there exists $alpha > 0$ such that for all $h in X$ we have $S''(bar x) [h,h] ge alpha , |h|^2$ (note that this generalizes 'positive definite'),
then $bar x$ is a local minimizer.
answered Feb 3 at 18:25
gerwgerw
20.1k11334
answered Feb 3 at 18:25
gerwgerw
20.1k11334
answered Feb 3 at 18:25
gerwgerw
20.1k11334
answered Feb 3 at 18:25
gerwgerw
20.1k11334
20.1k11334
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
add a comment |
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
Operationally, how do we calculate the second Fréchet derivative for a particular functional $S$ (e.g. of the form given in the question, where the EL equation gives the first Fréchet derivative)?
$endgroup$
– tparker
Feb 3 at 20:27
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
$begingroup$
You just take another derivative w.r.t. $q$. But for your functional, you have to choose $X = H^1(a,b)$ and then, $S$ has to be quadratic w.r.t. $dq/dt$, otherwise the differentiability fails.
$endgroup$
– gerw
Feb 4 at 6:29
add a comment |
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