Functionals leading to special cases

When the integrand F of the functional in our typical calculus of variations problem does not depend explicitly on x, for example if

$$\int_{0}^{1}$$

extremals satisfy an equation called the Beltrami identity which can be easier to solve than the Euler-Lagrange equation.

Theorem 2   If I(Y) is an extremum of the functional

I = $$\int_{a}^{b}$$ F(y, y') dx

defined on all functions y∈C²[a, b] such that y(a) = A, y(b) = B then Y(x) satisfies

$${\frac{{\partial F}}{{\partial y'}}}$$ (3)

for some constant C.

Definition 6 ()   is called the Beltrami identity or Beltrami equation.

Proof. Consider

$${\frac{{{\rm d}}}{{{\rm d} x}}} \left(\vphantom{ F- y'\frac{\partial F}{\partial y'} }\right. {\frac{{\partial F}}{{\partial y'}}} \left.\vphantom{ F- y'\frac{\partial F}{\partial y'} }\right) {\frac{{{\rm d} F}}{{{\rm d} x}}} {\frac{{\partial F}}{{\partial y'}}} {\frac{{{\rm d}}}{{{\rm d} x}}} \left(\vphantom{ \frac{\partial F}{\partial y'} }\right. {\frac{{\partial F}}{{\partial y'}}} \left.\vphantom{ \frac{\partial F}{\partial y'} }\right)$$ (4)

Using the chain rule to find the x-derivative of F(y(x), y'(x)) gives

$${\frac{{{\rm d} F}}{{{\rm d} x}}}$$ = y'$${\frac{{\partial F}}{{\partial y}}}$$ + y'^′$${\frac{{\partial F}}{{\partial y'}}}$$

so that () is equal to

y'$${\frac{{\partial F}}{{\partial y}}}$$ + y′′$${\frac{{\partial F}}{{\partial y'}}}$$ - y^′′$${\frac{{\partial F}}{{\partial y'}}}$$ - y'$${\frac{{{\rm d}}}{{{\rm d} x}}}$$$${\frac{{\partial F}}{{\partial y'}}}$$ = y'$$\left(\vphantom{ \frac{\partial F}{\partial y} - \frac{{\rm d}}{{\rm d} x}\frac{\partial F}{\partial y'} }\right.$$$${\frac{{\partial F}}{{\partial y}}}$$ - $${\frac{{{\rm d}}}{{{\rm d} x}}}$$$${\frac{{\partial F}}{{\partial y'}}}$$$$\left.\vphantom{ \frac{\partial F}{\partial y} - \frac{{\rm d}}{{\rm d} x}\frac{\partial F}{\partial y'} }\right)$$

Since Y is an extremal, it is a solution of the Euler-Lagrange equation and so this is zero for y = Y. If something has zero derivative it is a constant, so Y is a solution of

$${\frac{{\partial F}}{{\partial y'}}}$$

for some constant C. $\qedsymbol$

Exercise 5 (Exercise 1 revisited)   : Use the Beltrami identity to find an extremal of

I(y) = $$\int_{0}^{1}$$(y' - y)² dx,      y(0) = 0, y(1) = 2,

Answer:

y = f (x) = 2$${\frac{{\sinh{x}}}{{\sinh{1}}}}$$

(again).