Necessary Condition for Integral Functional to have Extremum for given Function/Dependent on N-th Derivative of Function

From ProofWiki
Jump to navigation Jump to search


Let $\map F {x, y, z_1, \ldots, z_n}$ be a function in differentiability class $C^2$ with respect to all its variables.

Let $y=\map y x \in C^n\openint a b$ such that:

$\map y a=A_0,\map {y'} a=A_1,\ldots, \map {y^{\paren{n-1} } } a=A_{n-1}$


$\map y b=B_0,\map {y'} b=B_1,\ldots, \map {y^{\paren {n-1} } } b=B_{n-1}$

Let $J \sqbrk y$ be a functional of the form

$\displaystyle J\sqbrk y=\int_a^b \map F {x,y,y',\ldots,y^{\paren n} }\rd x$

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:

$F_y-\dfrac \d {\d x} F_{y'}+\dfrac {\d^2} {\d x^2} F_{y''}-\cdots+\paren {-1}^n \dfrac {\d^n} {\d x^n} F_{y^{\paren n} }=0$


From Condition for Differentiable Functional to have Extremum we have

$\delta J\sqbrk {y h}\bigg\rvert_{y \mathop=\hat y}=0$

For the variation to exist it has to satisfy the requirement for a differentiable functional.

Note that the endpoints of $\map y x$ are fixed.

$\map h x$ is not allowed to change values of $\map y x$ at those points.

Hence $\map {h^{\paren i} } a=0$ and $\map {h^{\paren i} } b=0$ for $i=\openint 1 n$.

We will start from the increment of a functional:

\(\displaystyle \Delta J\sqbrk {y;h}\) \(=\) \(\displaystyle J\sqbrk {y+h}-J\sqbrk y\) definition
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b\map F {x,\ldots,y^{\paren i}+h^{\paren i},\ldots}\rd x-\int_a^b\map F {x,\ldots,y^{\paren i},\ldots}\rd x\) where $i=\paren {0,1,\ldots,n}$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {\map F {x,\ldots,y^{\paren i}+h^{\paren i},\ldots}-\map F {x,\ldots,y^{\paren i},\ldots} }\rd x\) bringing under the same integral

Using multivariate Taylor's theorem, one can expand $\map F {x,\ldots,y^{\paren i}+h^{\paren i},\ldots}$ with respect to $h^{\paren i}$:

$\displaystyle \map F {x,\ldots,y^{\paren i}+h^{\paren i},\ldots}=\map F {x,\ldots,y^{\paren i}+h^{\paren i},\ldots}\bigg\rvert_{h^{\paren i}=0,i=\openint 0 n}+\sum_{i \mathop=0}^n\frac {\partial \map F {x,\ldots,y^{\paren i}+h^{\paren i}, \ldots} } {\partial y^{\paren i} } \bigg\rvert_{h^{\paren i}=0,i=\openint 0 n} h^{\paren i}+\map {\mathcal O} {h^{\paren i} h^{\paren j},i,j=\openint 0 n}$

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.

$\displaystyle\Delta J\sqbrk {y;h}=\int_a^b\sum_{i \mathop=0}^n \paren {\map F {x,\ldots,y^{\paren i},\ldots}_{y^{\paren i} }h^{\paren i}+\map {\mathcal O} {h^{\paren i} h^{\paren j},i,j=\openint 0 n} }\rd x$

Terms in $\map {\mathcal O} {h^{\paren i}h^{\paren j},i,j=\openint 0 n}$ represent terms of order higher than 1 with respect to $h^{\paren i}$.

Now, suppose we expand $\displaystyle\int_a^b\map {\mathcal O} {h^{\paren i}h^{\paren j},i,j=\openint 0 n}\rd x$.

By definition, the integral not counting in $\map {\mathcal O} {h^{\paren i}h^{\paren j},i,j=\openint 0 n}$ is a variation of functional:

$\displaystyle\delta J \sqbrk {y;h}=\int_a^b\sum_{i\mathop=0}^n F_{y^{\paren i} }h^{\paren i}\rd x$

Application of Generalized Integration by Parts, together with boundary values for $h^{\paren i}$ yields:

$\displaystyle\int_a^b h\sum_{i\mathop=0}^n\paren {-1}^i\frac {\d^i} {\d x^i} F_{y^{\paren i} }\rd x$

Since If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes, then for any $\map h x$ the variation vanishes if:

$F_y-\dfrac \d {\d x} F_{y'}+\dfrac {\d^2} {\d x^2} F_{y''}-\cdots+\paren {-1}^n \dfrac {\d^n} {\d x^n} F_{ y^{\paren n} }=0$