nyuuzyou/wikis huggingface dataset:
Necessary Condition for Integral Functional to have Extremum for given function

== Theorem ==
Let $S$ be a set of real mappings such that:
:$S = \set {\map y x: \paren {y: S_1 \subseteq \R \to S_2 \subseteq \R}, \paren {\map y x \in C^1 \closedint a b}, \paren {\map y a = A, \map y b = B} }$
Let $J \sqbrk y: S \to S_3 \subseteq \R$ be a functional of the form:
:$\ds \int_a^b \map F {x, y, y'} \rd x$ 
Then a necessary condition for $J \sqbrk y$ to have an extremum (strong or weak) for a given function $\map y x$ is that $\map y x$ satisfy Euler's equation:
:$F_y - \dfrac \d {\d x} F_{y'} = 0$
== Proof ==
From Condition for Differentiable Functional to have Extremum we have
:$\delta J \sqbrk {y; h} \bigg \rvert_{y = \hat y} = 0$
The variation exists if $J$ is a differentiable functional.
The endpoints of $\map y x$ are fixed.
Hence:
:$\map h a = 0$
:$\map h b = 0$.
From the definition of increment of a functional:
Using multivariate Taylor's Theorem, expand $\map F {x, y + h, y' + h'}$ $h$ and $h'$:
:$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y'+ h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$
Substitute this back into the integral: 
:$\ds \Delta J \sqbrk {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$
Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.
Suppose we expand $\ds \int_a^b \map \OO {h^2, h h', h'^2} \rd x$. 
Every term in this expansion will be of the form:
:$\ds \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial y'^n} h^m h'^n \rd x$
where $m, n \in \N: m + n \ge 2$
By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional:
:$\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$
Use lemma. 
Then for any $\map h x$ variation vanishes if:
:$F_y - \dfrac \d {\d x} F_{y'} = 0$
== Sources ==
* : $\S 1.4$: The Simplest Variational Problem. Euler's Equation
Category:Calculus of Variations
Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions

== Theorem ==
Let $\mathbf y$ be an $n$-dimensional real vector.
Let $J \sqbrk {\mathbf y}$ be a functional of the form:
$\ds J \sqbrk {\mathbf y} = \int_a^b \map F {x, \mathbf y, \mathbf y'} \rd x$
Let:
:$\mathbf y \in C^1 \closedint a b$
where $C^1 \closedint a b$ denotes that $\mathbf y$ is continuously differentiable in $\closedint a b$
Let $\mathbf y$ satisfy boundary conditions:
:$\map {\mathbf y} a = \mathbf A$ 
:$\map {\mathbf y} b = \mathbf B$
where $\mathbf A$, $\mathbf B$ are real vectors.
Then a necessary condition for $J \sqbrk {\mathbf y}$ to have an extremum (strong or weak) for a given $\mathbf y$ is that they satisfy Euler's equations:
$F_{\mathbf y} - \dfrac \d {\d x} F_{\mathbf y'} = 0$
== Proof ==
From Condition for Differentiable Functional of N Functions to have Extremum:
:$\ds \bigvalueat {\delta J \sqbrk {\mathbf y; \mathbf h} } {\mathbf y \mathop = \hat{\mathbf y} } = 0$
For the variation to exist it has to satisfy the requirement for a differentiable functional.
Note that the endpoints of $\map {\mathbf y} x$ are fixed. $\map {\mathbf h} x$ is not allowed to change values of $\map {\mathbf y} x$ at those points.
Hence $\map {\mathbf h} a = 0$ and $\map {\mathbf h} b = 0$.
We will start from the increment of a functional:
Using multivariate Taylor's theorem, one can expand $\map F {x, \mathbf y + \mathbf h, \mathbf y' + \mathbf h'}$ with respect to functions $\map {\mathbf h} x$ and $\map {\mathbf h'} x$:
We can substitute this back into the integral.
Note that the first term in the expansion and the negative one in the integral will cancel out.
Hence:
:$\ds \Delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \sum_{i \mathop = 1}^n \paren {F_{y_i} h_i + F_{y_i'} h_i' + \map \OO {h_i h_j,h_i h_j',h_i' h_j'} } \rd x$ for $i, j \in \openint 1 n$
By definition, the integral not counting in $\map \OO {h_i h_j,h_i h_j',h_i' h_j'}$ for $i, j \in \openint 1 n$ is a variation of functional:
:$\ds \delta J \sqbrk {\mathbf y; \mathbf h} = \int_a^b \paren {F_{\mathbf y} \mathbf h + F_{\mathbf y'} \mathbf h'} \rd x$
The variation vanishes if for all functions $h_i$ every term containing $h_i$ vanishes independently.
Therefore, we discover a set of Euler's Equations being satisfied simultaneously:
 
:$F_{\mathbf y} - \dfrac \d {\d x} F_{\mathbf y'} = 0$
== Sources ==
* : $\S 2.9$: The Fixed End Point Problem for n Unknown Functions
Category:Calculus of Variations
Necessary Condition for Integral Functional to have Extremum for given function/Dependent on n Variables

== Theorem ==
Let $\mathbf x$ be an $n$-dimensional vector.
Let $\map u {\mathbf x}$ be a real function.
Let $R$ be a fixed region.
Let $J$ be a functional such that
:$\ds J \sqbrk u = \idotsint_R \map F {\mathbf x, u, u_{\mathbf x} } \rd x_1 \cdots \rd x_n$
Then a necessary condition for $J \sqbrk u$ to have an extremum (strong or weak) for a given mapping $\map u {\mathbf x}$ is that $\map u {\mathbf x}$ satisfies Euler's equation:
:$F_u - \dfrac {\partial} {\partial \mathbf x} F_{u_{\mathbf x} } = 0$
== Proof ==
By definition of increment of the functional:
:$\ds J \sqbrk {u + h} - J \sqbrk u = \idotsint_R \paren {F \sqbrk {x, u + h, u_{\mathbf x} + h_{\mathbf x} } - F \sqbrk {x, u, u_{\mathbf x} } } \rd x_1 \cdots \rd x_n$
Use multivariate Taylor's Theorem on $F$ around the point $\tuple {\mathbf x, u, u_{\mathbf x} }$:
:$F \sqbrk {\mathbf x, u + h, u_{\mathbf x} + h_{\mathbf x} } = F \sqbrk {\mathbf x, u, u_{\mathbf x} } + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} h + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } h_{\mathbf x} + \map \OO {h^2, h h_{\mathbf x}, h_{\mathbf x}^2}$
where $\OO$ stands for Big-O.
Then:
:$\ds \Delta J \sqbrk {u, h} = \idotsint_R \paren {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} h + \dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } h_{\mathbf x} + \map \OO {h^2, h h_{\mathbf x}, h_{\mathbf x}^2} } \rd x_1 \cdots \rd x_n$
By definition of variation of the functional:
By Green's theorem:
:$\ds \idotsint_R \dfrac {\partial} {\partial \mathbf x} \sqbrk {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u_{\mathbf x} } \map h {\mathbf x} } \rd x_1 \cdots \rd x_n = \idotsint_\Gamma \map h {\mathbf x} F_{u_{\mathbf x} } \sqbrk {\mathbf x, u, u_{\mathbf x} } \boldsymbol \nu \rd \sigma$
where:
:$\Gamma$ denotes the boundary of $R$
:$\boldsymbol \nu$ is an outward normal to $\Gamma$.
Since the region is fixed, so are its boundary points. 
Hence, the difference function $h$ has to vanish at the boundary.
In other words:
:$\forall \mathbf x \in \Gamma: \map h {\mathbf x} = 0$
This leaves only the first integral.
:$\ds \delta J = \idotsint_R \paren {\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} - \dfrac{\partial F_{u_{ \mathbf x } } \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial \mathbf x} } \map h {\mathbf x} \rd x_1 \cdots \rd x_n$
For arbitrary $h$ the first variation vanishes if the term in brackets vanishes:
:$\dfrac {\partial F \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial u} - \dfrac {\partial F_{u_{\mathbf x} } \sqbrk {\mathbf x, u, u_{\mathbf x} } } {\partial \mathbf x} = 0$
== Sources ==
* : $\S 7.35$: Variation of a Functional Defined on a Fixed Region
Category:Calculus of Variations
Necessary Condition for Integral Functional to have Extremum for given function/Lemma

== Theorem ==
Let $\map \alpha x, \map \beta x$ be real functions.
Let $\map \alpha x, \map \beta x$ be continuous on $\closedint a b$.
Let:
:$\forall \map h x \in C^1: \ds \int_a^b \paren {\map \alpha x \map h x + \map \beta x \map {h'} x} \rd x = 0$
subject to the boundary conditions:
:$\map h a = \map h b = 0$
Then $\map \beta x$ is differentiable.
Furthermore:
:$\forall x \in \closedint a b: \map {\beta'} x = \map \alpha x$
== Proof ==
Using Integration by Parts allows us to factor out $\map h x$:
Hence the problem has been reduced to:
:$\ds \int_a^b \paren {\map \alpha x - \map {\beta'} x} \map h x \rd x = 0$
Since If Definite Integral of $\map a x \map h x$ vanishes for any $C^0 \map h x$ then $C^0 \map a x$ vanishes, the conclusion is that in the considered interval $\closedint a b$ it holds that:
:$\map \alpha x = \map {\beta'} x$
== Sources ==
* : $\S 1.3$: The Variation of a Functional. A Necessary Condition for an Extremum
Category:Calculus of Variations
Necessary Condition for Standardized Random Variable to have Standard Normal Distribution

== Theorem ==
Let $Z$ be the standardized random variable of a random variable $X$:
:$Z = \dfrac {X - \mu} \sigma$
where:
:$\mu$ denotes the expectation of $X$
:$\sigma$ denotes the standard deviation of $X$.
In order for $Z$ to have a standard normal distribution it is necessary for:
:$X$ to have a normal distribution
:both $\mu$ and $\sigma$ to be known.
== Proof ==
== Sources ==
* 
* 
Category:Standardized Random Variables
Necessary Condition for Stationary Point to be Local Maximum

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Let $\xi$ be a local maximum.
Then there exists an open interval $\openint c d$ such that:
== Proof ==
== Also see ==
* Necessary Condition for Stationary Point to be Local Minimum
* Sufficient Condition for Stationary Point to be Local Maximum
* Sufficient Condition for Stationary Point to be Local Minimum
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Maximum, Minimum and Point of Inflection
* 
* 
Category:Stationary Points
Necessary Condition for Stationary Point to be Local Minimum

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Let $\xi$ be a local minimum.
Then there exists an open interval $\openint c d$ such that:
== Proof ==
== Also see ==
* Necessary Condition for Stationary Point to be Local Maximum
* Sufficient Condition for Stationary Point to be Local Minimum
* Sufficient Condition for Stationary Point to be Local Maximum
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Maximum, Minimum and Point of Inflection
* 
* 
Category:Stationary Points
Necessary Condition for Twice Differentiable Functional to have Minimum

== Theorem ==
Let $J \sqbrk y$ be a twice differentiable functional.
Let $\delta J \sqbrk {\hat y; h} = 0$.
Suppose, for $y = \hat y$ and all admissible $h$:
:$\delta^2 J \sqbrk {y; h} \ge 0$ 
Then $J$ has a minimum for $y=\hat y$ if 
== Proof ==
By definition, $ \Delta J \sqbrk y$ can be expressed as:
:$\Delta J \sqbrk {y; h} = \delta J \sqbrk {y; h} + \delta^2 J \sqbrk {y; h} + \epsilon \size h^2$
By assumption:
:$\delta J \sqbrk {\hat y; h} = 0$
Hence:
:$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$
Therefore, for sufficiently small $\size h$ both $\Delta J \sqbrk {\hat y; h}$ and $\delta^2 J \sqbrk {\hat y; h}$ will have the same sign.
 
 there exists $h = h_0$ such that:
:$\delta^2 J \sqbrk {\hat y; h_0} < 0$ 
Then, for any $\alpha \ne 0$:
Therefore, $\Delta J \sqbrk {\hat y; h}$ can be made negative for arbitrary small $\size h$.
However, by assumption $\Delta J \sqbrk {\hat y; h}$ is a minimum of $\Delta J \sqbrk {y; h}$ for all sufficiently small $\size h$.
This is a contradiction.
Thus, a function $h_0: \delta^2 J \sqbrk {\hat y; h_0} < 0$ does not exist.
In other words:
:$\delta^2 J \sqbrk {\hat y; h} \ge 0$ 
for all $h$.
== Sources ==
* : $\S 5.24$: Quadratic Functionals. The Second Variation of a Functional 
Category:Calculus of Variations
Necessary Conditions for Existence of Skolem Sequence

== Theorem ==
A Skolem sequence of order $n$ can only exist if $n \equiv 0, 1 \pmod 4$.
== Proof ==
Let $S$ be a Skolem sequence of order $n$.
Let $a_i$ and $b_i$ be the positions of the first and second occurrences respectively of the integer $i$ in $S$, where $1 \le i \le n$.
We can thus conclude that:
:$b_i - a_i = i$
for each $i$ from 1 to $n$.
Summing both sides of this equation we obtain:
:$\ds \sum_i^n b_i - \sum_i^n a_i = \sum_i^n i = \frac {n \paren {n + 1} } 2$
Now the $a_i$ and the $b_i$ represent the positions in the sequence from $1$ to $2 n$.
Hence:
:$\ds \sum_i^n b_i + \sum_i^n a_i = \frac {2 n \paren {2 n + 1} } 2 = n \paren {2 n + 1}$
Summing the previous two equations we obtain the identity:
:$\ds \sum_i^n b_i = \frac {n \paren {5 n + 3} } 4$
The of this last equality is a sum of positions and thus must be an integer.
We conclude that the must also be an integer.
This occurs exactly when:
$n \equiv 0, 1 \pmod 4$
== Sources ==
* 
Category:Combinatorics
Necessary Precision for x equal to log base 10 of 2 to determine Decimal expansion of 10 to the x

== Theorem ==
Let $b = 10$.
Let $x \approx \log_{10} 2$.
Let it be necessary to calculate the decimal expansion of $x$ to determine the first $3$ decimal places of $b^x$.
An infinite number of decimal places of $x$ would in fact be necessary.
== Proof ==
This is a trick question:
:How many decimal places of accuracy of $x$ are needed to determine the first $3$ decimal places of $b^x$?
We have that $b^x = 10^{\log_{10} 2} = 2$.
Let $x_a < x < x_b$, where $x_a$ and $x_b$ are ever closer approximations to $x$.
Then:
:$x_a$ begins $1 \cdotp 999 \ldots$
:$x_b$ begins $2 \cdotp 000 \ldots$
and it will not be possible to achieve the full expansion of $b^x$.
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $11$
Category:Common Logarithms
Category:2
Category:10
Necessary Proposition is Strictly Implied by Every Proposition/Examples/Grass is Blue

== Example of the Paradoxes of Strict Implication ==
The proposition:
:Grass is blue strictly implies that $2 + 2 = 4$
is an example of the Paradoxes of Strict Implication:
:Necessary Proposition is Strictly Implied by Every Proposition.
== Sources ==
* 
* 
Category:Paradoxes of Strict Implication
Necessary and Sufficient/Examples/Arbitrary Example 1

== Examples of Necessary and Sufficient ==
Let $x$ be an integer.
For $x$ to be divisible by $6$, it is:
:necessary but not sufficient for $x$ to be even
:sufficient but not necessary for $x$ to be divisible by $12$
:necessary and sufficient for $x$ to be both even and divisible by $3$.
== Sources ==
* 
* 
Category:Examples of Necessary and Sufficient
Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint

== Theorem ==
Let $\mathbf p$ be continuously differentiable.
The boundary conditions
:$\bigvalueat {\map {\mathbf y} a} {x \mathop = a} = \bigvalueat {\map {\boldsymbol \psi} {\mathbf y} } {x \mathop = a}$
are self-adjoint :
:$\forall i, k \in \N: 1 \le i, k \le N: \valueat {\dfrac {\partial p_i \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} } {x \mathop = a} = \valueat {\dfrac {\partial p_k \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_i} } {x \mathop = a}$
== Proof ==
=== Necessary Condition ===
By assumption the boundary conditions are self-adjoint.
Then there exists $\map g {x, \mathbf y}$ such that:
:$\map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } = \dfrac {\partial \map g {x \mathbf y} } {\partial y_i}$
Because $\mathbf p \in C^1$:
:$g \in C^2$
Differentiate both sides $y_k$:
:$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} = \dfrac {\partial^2 \map g {x, \mathbf y} } {\partial y_k \partial y_i}$
By the Schwarz-Clairaut Theorem, partial derivatives commute, hence indices can be mutually replaced:
:$\dfrac {\partial \map {p_i} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_k} = \dfrac {\partial \map {p_k} {x, \mathbf y, \map {\boldsymbol \psi} {\mathbf y} } } {\partial y_i}$
Fixing $x = a$ provides the result.
=== Sufficient condition ===
By assumption:
:$\valueat {\dfrac {\partial p_i} {\partial y_j} } {x \mathop = a} = \valueat {\dfrac {\partial p_j} {\partial y_i} } {x \mathop = a}$
Then:
:$\exists \map g {x, \mathbf y} \in C^2: \valueat {\dfrac {\partial p_i} {\partial y_j} } {x \mathop = a} = \valueat {\dfrac {\partial p_j} {\partial y_i} } {x \mathop = a} = \valueat {\dfrac {\partial^2 g} {\partial y_i \partial y_j} } {x \mathop = a}$
In other words:
:$\bigvalueat {p_i} {x \mathop = a} = \valueat {\dfrac {\partial g} {\partial y_i} } {x \mathop = a}$
Hence, the boundary conditions are self-adjoint.
== Sources ==
* : $\S 6.31$: Consistent Boundary Conditions. General Definition of a Field
Category:Calculus of Variations
Necessary and Sufficient Condition for Diagonal Operator to be Invertible

== Theorem ==
Let $\mathbb F \in \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\mathbb F$.
Let $\ell^2$ be the $2$-sequence space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space on $\ell^2$.
Let $\Lambda \in \map {CL} {\ell^2}$ be the diagonal operator such that:
:$\forall \mathbf x := \tuple {x_1, x_2, \ldots} \in \ell^2 : \map \Lambda {\mathbf x} = \tuple {\lambda_1 \cdot x_1, \lambda_2 \cdot x_2, \ldots}$
Then $\Lambda$ is invertible in $\map {CL} {\ell^2}$ $\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} } > 0$ where $\inf$ denotes the infimum.
== Proof ==
=== Necessary Condition ===
Suppose $\ds \inf_{n \mathop \in \N_{> 0} } \sequence {\size {\lambda_n} } > 0$.
Then:
Moreover:
Let $V : \ell^2 \to \ell^2$ be a mapping such that:
:$\ds \forall \mathbf a := \sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, \ldots} : \map V {\mathbf a} = \tuple {\frac {a_1} {\lambda_1}, \frac {a_2} {\lambda_2}, \ldots}$
Then $V \in \map {CL} {\ell^2}$.
Furthremore:
Therefore:
:$\Lambda V = V \Lambda = I$
Hence, $\Lambda$ is invertible.
Finally, the unique inverse of $\Lambda$ is $\Lambda^{-1} = V$
=== Sufficient Condition ===
Suppose $\Lambda$ is invertible in $\map {CL} {\ell^2}$.
Then there is a unique inverse $\Lambda^{-1}$ of $\Lambda$:
:$\ds \exists \Lambda^{-1} \in \map {CL} {\ell^2} : \Lambda \circ \Lambda^{-1} = \Lambda^{-1} \circ \Lambda = I$
Furthermore, Invertible Operator is not Zero Operator.
Thus:
:$\exists \mathbf x \in \ell^2 : \map {\Lambda^{-1} } {\mathbf x} \ne \mathbf 0$
We have that P-Sequence Space with P-Norm forms Normed Vector Space.
Hence:
Let $\mathbf x := \mathbf e_k = \tuple {\underbrace{0, \ldots, 0}_{k - 1}, 1, 0, \ldots}$
Then:
Therefore:
:$\ds \inf_{k \mathop \in \N_{> 0} } \size {\lambda_k} \ge \frac 1 {\norm {\Lambda^{-1} } } > 0$
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Diagonal Operator
Category:Inverse Mappings
Necessary and Sufficient Condition for First Order System to be Field for Second Order System

== Theorem ==
Let $\mathbf y$, $\mathbf f$, $\boldsymbol \psi$ be N-dimensional vectors.
Let $\boldsymbol\psi$ be continuously differentiable.
Then $\forall x \in \closedint a b$ the first-order system of differential equations:
:$\mathbf y' = \map {\boldsymbol \psi} {x, \mathbf y}$
is a field for the second-order system
:$\mathbf y = \map {\mathbf f} {x, \mathbf y, \mathbf y'}$
 $\boldsymbol \psi$ satisfies:
:$\ds \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{i \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_i} \psi_i = \map {\mathbf f} {x, \mathbf y, \boldsymbol \psi}$
That is, every solution to Hamilton-Jacobi system is a field for the original system.
== Proof ==
=== Necessary condition ===
Differentiate the first-order system $x$:
This can be rewritten as the following system of equations:
:$\mathbf y = \map {\mathbf f} {x, \mathbf y, \mathbf y'}$
:$\ds \frac {\partial \boldsymbol \psi} {\partial x} + \sum_{i \mathop = 1}^N \frac {\partial \boldsymbol \psi} {\partial y_i} \psi_i = \map {\mathbf f} {x, \mathbf y, \mathbf y'}$
By assumption, the first-order system is valid in $\closedint a b$.
For the second-order system to be valid in the same interval, the corresponding Hamilton-Jacobi equation has to hold as well.
=== Sufficient condition ===
== Sources ==
* : $\S 6.31$: Consistent Boundary Conditions. General Definition of a Field
Category:Calculus of Variations
Necessary and Sufficient Condition for First Order System to be Mutually Consistent

== Theorem ==
Let $\mathbf y$, $\boldsymbol \psi$ be $n$-dimensional vectors.
Let $g$ be a twice differentiable mapping.
Let
:$(1): \quad \map {\boldsymbol \psi} {x, \mathbf y} = \map {\mathbf y'} {x, \mathbf y}$
:$(2): \quad \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } = \map {g_{\mathbf y} } {x, \mathbf y}$
where $\mathbf p$ is a momentum.
Then the boundary conditions defined by $(1)$ are mutually consistent the mapping $\map g {x, \mathbf y}$ satisfies the Hamilton-Jacobi equation:
:$(3): \quad \dfrac {\partial g} {\partial x} + \map H {x, \mathbf y, \dfrac {\partial g} {\partial \mathbf y} } = 0$
== Proof ==
=== Necessary condition ===
Take the partial derivative of $(3)$ $x$:
:$(4): \quad \frac {\partial^2 \map g {x, \mathbf y} } {\partial \mathbf y \partial x} = -\dfrac \partial {\partial \mathbf y} \map H {x, \mathbf y, \dfrac {\partial g} {\partial \mathbf y} }$
By the Schwarz-Clairaut Theorem, the order of partial derivatives of $g$ can be exchanged:
:$(5): \quad \dfrac {\partial^2 \map g {x, \mathbf y} } {\partial \mathbf y \partial x} = \dfrac {\partial \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x}$
Since $\map g {x, \mathbf y}$ depends on $\mathbf y$ only in a direct way, it follows that:
:$\dfrac {\partial g} {\partial \mathbf y} = g_{\mathbf y}$
Hence, from $(2)$, $(4)$ and $(5)$ it follows that:
:$\dfrac {\partial \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } } {\partial x} = -\dfrac \partial {\partial \mathbf y} \map H {x, \mathbf y, \mathbf p \sqbrk {x, \mathbf y, \map {\boldsymbol \psi} {x, \mathbf y} } }$
These are the consistency equations.
=== Sufficient condition ===
== Sources ==
* : $\S 6.32$: The Field of a Functional
Category:Calculus of Variations
Necessary and Sufficient Condition for Hypersurface in Oriented Riemannian Manifold to be Orientable

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be an oriented Riemannian manifold.
Let $M$ be a hypersurface in $\tilde M$.
Let $g$ be the induced metric on $M$.
Then $M$ is orientable there exists a unit global normal vector $N$ for $M$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Riemannian Geometry