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All around the Jacobi equation

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Edited by Valentin Fadeev, Sunday, 18 Sep 2011, 23:25

The Jacobi equation for a functional serving as a test for weak local extremum can be derived in quite different ways. The following geometrical approach was found in the book "Calculus of Variations" by L.E. Elsgoltz, 1958. It hangs upon the question of whether or not the stationary path can be included in a part of a one-parametric family of stationary paths (field of paths). There can be 2 options. Either no 2 paths of the family intersect, or all patghs of the family share 1 common point (but not more) in the given interval.

For example y equals cap c times sine of x will form a family of the first type in open delta comma a close , where delta greater than zero , a less than pi , the family of the second type in open zero comma a close , a less than pi . In the interval open zero comma a close , a greater than or equals pi no such family can be constructed.

Suppose we have a one-parametric family of stationary paths y equals y of x comma cap c . For example, we can fix one of the boundary points and use the gradient of the paths in this point as parameter C.

paths

The envelope of this family is found by eliminating C from the following system of equations:

equation sequence y equals y of x comma cap c times normal partial differential times y of x comma cap c divided by normal partial differential times cap c equals zero times open asterisk operator close

Along each path of the family normal partial differential times y of x comma cap c divided by normal partial differential times cap c is a function of only x. Denote this function as u of x for some given C. Then equation left hand side u sub x super prime equals right hand side normal partial differential squared times y of x comma cap c divided by normal partial differential times cap c times normal partial differential times x .

y equals y of x comma cap c are solutions of the E-L by assumption. Therefore:

cap f sub y of x comma y of x comma cap c comma y sub x super prime of x comma cap c minus d divided by d times x times cap f sub y super prime of x comma y of x comma cap c comma y sub x super prime of x comma cap c equals zero

Differentiating this equality by C and letting u equals normal partial differential times y of x comma cap c divided by normal partial differential times cap c we obtain:

cap f sub y times y times u plus cap f sub y times y super prime times u super prime minus d divided by d times x times open cap f sub y times y super prime times u plus cap f sub y super prime times y super prime times u super prime close equals zero

Rearranging we get:

open cap f sub y times y minus d divided by d times x times cap f sub y times y super prime close times u minus d divided by d times x times open cap f sub y super prime times y super prime times u super prime close equals zero

which is obviously a Jacobi equation.

Thus, if u has a zero somewhere in the interval, it follows from (*) above that this is a common point of the stationary path and the envelope. This is a point, conjugate to the left end of the interval.

It seemed to me at first that this proof serves only for theoretical purpose, as another way of deriving the Jacobi equation. However, the idea behind can be used to find the solution of the Jacobi equation, without actually solving the equation itself!

Consider the following example.

equation sequence cap s of y equals integral over zero under one y super prime two divided by y super four d x times y of zero equals one semicolon y of one equals one divided by two

equation left hand side y super prime times two times y super prime divided by y super four minus y super prime two divided by y super four equals right hand side cap c squared

y super prime divided by y squared equals cap c

equation left hand side negative one divided by y equals right hand side cap c times x plus cap c sub zero

(we can suppress plus minus , assuming it is absorbed by the constant)

y equals one divided by cap c sub one times x plus cap c sub two

Now apply boundary conditions:

equation sequence y of zero equals one times y of one equals one divided by two

y equals one divided by x plus one

If we only use the first condition, that is fix the left boundary, we get the one-parametric family:

y of x comma cap c equals one divided by cap c times x plus one

C being the gradient at x equals zero with a reverse sign. Now following the idea described above we can find u of x :

equation sequence u of x equals normal partial differential times y of x comma cap c divided by normal partial differential times cap c equals negative x divided by open cap c times x plus one close squared

Finally setting cap c equals one by virtue of the boundary conditions:

u of x equals negative x divided by open x plus one close squared

Now we move on to dervie the Jacobi equation through the coefficients of the second variation:

equation sequence cap p of x equals normal partial differential squared times cap f divided by normal partial differential times y super prime two equals two divided by y super four equals two left parenthesis x plus one times right parenthesis super four

equation sequence cap q of x equals normal partial differential squared times cap f divided by normal partial differential times y squared minus d divided by d times x times open normal partial differential squared times cap f divided by normal partial differential times y times normal partial differential times y super prime close equals 20 times y super prime two divided by y super six plus eight times d divided by d times x times open y super prime divided by y super five close equals 20 times y super prime two divided by y super six plus eight times y super double prime divided by y super five minus 40 times y super prime two divided by y super six equals eight times y super double prime divided by y super five minus 20 times y super prime two divided by y super six

equation sequence cap q of x equals 16 times open x plus one close super five divided by open x plus one close cubed minus 20 times open x plus one close super six divided by open x plus one close super four equals negative four times open x plus one close squared

Inserting the above results into the equation:

left parenthesis cap p times u super prime times right parenthesis super prime minus cap q times u equals zero

two times open open x plus one close super four times u super prime close super prime plus four times open x plus one close squared times u equals zero

open open x plus one close super four times u super prime close super prime plus two times open x plus one close squared times u equals zero

sum with, 3 , summands open x plus one close super four times u super double prime plus four times open x plus one close cubed times u super prime plus two left parenthesis x plus one times right parenthesis squared times u equals zero

sum with, 3 , summands u super double prime plus four divided by x plus one times u super prime plus two divided by open x plus one close squared times u equals zero

Instead of solving the equation which can be technically demanding, we shall verify if the expression for u of x found above is a solution. We do not need a general solution in this case. All non-trivial solutions of a homogeneous equation of the second order satisfying the condition u of x sub zero equals zero differ from each other only by a constant multiplier and thus have the same zeros.

u equals negative x divided by open x plus one close squared

equation sequence u super prime equals negative one divided by open x plus one close squared plus two times x divided by open x plus one close cubed equals negative x minus one plus two times x divided by open x plus one close cubed equals x minus one divided by open x plus one close cubed

equation sequence u super double prime equals one divided by open x plus one close cubed minus three times x minus one divided by open x plus one close super four equals x plus one minus three times x plus three divided by open x plus one close super four equals negative two times x minus two divided by open x plus one close super four

negative two times x plus two divided by open x plus one close super four times open x plus one close super four plus four times x minus one divided by open x plus one close cubed times open x plus one close cubed minus two times x divided by open x plus one close squared times open x plus one close squared equals zero

sum with, 3 , summands negative two times x plus four plus four times x minus four minus two times x equals zero

So we indeed have a solution.

For the integrand does not depend explicitly on x we can simplify the Jacobi equation by exchanging the roles of the variables:

equation sequence cap s of y equals integral over zero under one y super prime two divided by y super four d x equals integral over zero under one open d times y close squared divided by open d times x close squared times one divided by y super four d x equals integral over one under one divided by two one divided by x super prime times y super four d y equals negative integral over one divided by two under one one divided by x super prime times y super four d y equals cap s of x

cap g of y comma x comma x super prime equals negative one divided by x super prime times y super four

Euler-Lagrange equation has the first integral:L

cap g sub x super prime equals cap c

equation left hand side negative one divided by x super prime two times y super four equals right hand side negative c squared

x equals cap c sub one divided by y plus cap c sub two

cap c sub one plus cap c sub two equals zero

Again we leave one arbitrary constant to form a family:

x of y comma cap c equals cap c times open one divided by y minus one close

equation sequence u of y equals normal partial differential times x of y comma cap c divided by normal partial differential times cap c equals one divided by y minus one

equation sequence cap p of y equals normal partial differential squared times cap g divided by normal partial differential times x super prime two equals negative two divided by x super prime three times y super four equals negative two divided by negative one divided by y super six times y super four equals two times y squared

Since the transformed integrand does not depend explicitly on the dependent variable, Q will vanish and the Jacobi equation has the first integral:

cap p times u sub y super prime equals cap c

equation left hand side two times y squared times u sub y super prime equals right hand side negative two times cap c sub one

u of y equals cap c sub one divided by y plus cap c sub two

cap c sub one plus cap c sub two equals zero

equation left hand side cap c sub one equals right hand side negative one

u of y equals one minus one divided by y
Thus, up to the sign we get the same expression. (I am not sure where I may be loosing the sign, but it obviously has litle effect on the argument)

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