Legendre's Condition/Lemma 1
 == Lemma == Let $y = \map y x$ be a real function, such that: :$\map y a = A$ :$\map y b = B$ Let $J \sqbrk y$ be a functional, such that: :$\ds J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$ where: :$F \in C^2 \closedint a b$   all its variables. Then: :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\map P {x, \map y x} h'^2 + \map Q {x, \map y x} h^2} \rd x$ where: {{begin-eqn}} {{eqn | l = \map P {x, \map y x}       | r = \frac 1 2 F_{y'y'} }} {{eqn | l = \map Q {x, \map y x}       | r = \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy} } }} {{end-eqn}} == Proof == The minimising function $y$ has fixed end-points. Therefore, consider an increment of a functional with $h$ such that: :$h \in C^1 \closedint a b: \paren {\map h a = 0} \land \paren {\map h b = 0}$ Then: {{begin-eqn}} {{eqn | l = \Delta J \sqbrk {y; h}       | r = J \sqbrk {y + h} - J \sqbrk y       | c = {{Defof|Increment of Functional}} }} {{eqn | r = \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x       | c = form of $J$ }} {{eqn | r = \int_a^b \paren {F + \paren {F_y h + F_{y'} h'} + \frac 1 2 \paren {\overline F_{yy} h^2 + \overline F_{yy'} h h' + \overline F_{y'y'} h'^2} - F} \rd x       | c = Taylor's Theorem }} {{eqn | r = \int_a^b \paren {F_y h + F_{y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\overline F_{yy} h^2 + \overline F_{yy'} h h' + \overline F_{y'y'} h'^2} \rd x       | c = cancel $F$ }} {{end-eqn}} where omitted variables are $\paren {x, y, y'}$, and the overbar indicates derivatives being taken along some intermediate curves: {{begin-eqn}} {{eqn | l = \overline {\map {F_{yy} } {x,y,y'} }       | r = \map {F_{yy} } {x, y + \theta h, y' + \theta h'} }} {{eqn | l = \overline {\map {F_{yy'} } {x,y,y'} }       | r = \map {F_{yy'} } {x, y + \theta h, y' + \theta h'} }} {{eqn | l = \overline {\map {F_{y'y'} } {x,y,y'} }       | r = \map {F_{y'y'} } {x, y + \theta h, y' + \theta h'} }} {{end-eqn}} with $0 < \theta < 1$. If $\overline F_{yy}$, $\overline F_{yy'}$, $\overline F_{y'y'} $ are to be replaced by $F_{yy}$, $F_{yy}$, $F_{y'y'}$ evaluated at the point $\tuple {x, \map y x, \map {y'} x}$, then: :$\ds \Delta J \sqbrk {y; h} = \int_a^b \paren {\map {F_y} {x, y, y'} h + \map {F_{y'} } {x, y, y'} h'} \rd x + \frac 1 2 \int_a^b \paren {\map {F_{yy} } {x, y, y'} h^2 + 2 \map {F_{yy'} } {x, y, y'} h h'+ \map {F_{y'y'} } {x, y, y'} h'^2} \rd x + \epsilon$ where: :$\ds \epsilon = \int_a^b \paren {\epsilon_1 h^2 + \epsilon_2 h h' + \epsilon_3 h'^2}$ By continuity of $F_{yy}$, $F_{yy}$, $F_{y'y'}$: :$\size h_1 \to 0 \implies \epsilon_1, \epsilon_2, \epsilon_3 \to 0$ {{explain|What does $\size h_1$ mean?}} Thus, $\epsilon$ is an infinitesimal of the order higher than 2  $\size h$. {{Stub|Expand on steps including $ \epsilon $}} The first and second term on the  of $\Delta J \sqbrk {y; h}$ are $\delta J \sqbrk {y; h}$ and $\delta^2 J \sqbrk {y; h}$ respectively. Integrate the second term of $\delta^2 J \sqbrk {y; h}$ by parts: {{begin-eqn}} {{eqn | l = \int_a^b 2 F_{yy'} h h' \rd x       | r = \int_a^b 2 F_{yy'} h \rd h }} {{eqn | r = \int_a^b F_{yy'} \rd h^2 }} {{eqn | r = \bigintlimits {F_{yy'} h^2} {x \mathop = a} {x \mathop = b} - \int_a^b \map {\frac \d {\d x} } {F_{yy'} } h^2 \rd x }} {{eqn | r = -\int_a^b \map {\frac \d {\d x} } {F_{yy'} } h^2 \rd x }} {{end-eqn}} {{explain|Why does $\bigintlimits {F_{yy'} h^2} {x \mathop {{=}} a} {x \mathop {{=}} b}$ vanish?}} Therefore: :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {\frac 1 2 F_{y'y'} h'^2 + \frac 1 2 \paren {F_{yy} - \frac \d {\d x} F_{yy'} } h^2} \rd x$ {{qed|lemma}} {{explain|Review use of square brackets. If they are being used purely for parenthesis, better to replace with round ones, as square ones have conventional meanings, so use of them for parenthesis may confuse.}} == Mistake == {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin}}: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition p. 102 states that :$P = \dfrac 1 2 F_{y'y'} \quad Q = \dfrac 1 2 \paren {F_{yy'} - \dfrac \d {\d x} F_{yy'} }$ This is a mistake, since the second variation should contain both pure and mixed partial derivatives of the order 2. However, $F_{yy} $ is missing and could not have been lost during derivation of the proof. It should be: :$Q = \dfrac 1 2 \paren {F_{yy} - \dfrac \d {\d x} F_{yy'} }$ == Sources == {{BookReference|Calculus of Variations|1963|I.M. Gelfand|author2 = S.V. Fomin|prev = Legendre's Condition|next = Legendre's Condition/Lemma 2}}: $\S 5.25$: The Formula for the Second Variation. Legendre's Condition