Derivation of the Euler Equation

Unlike static optimization problems, the basic problem of the calculus of variations is to find a function, $y(x)$, such that some integral

$$J = \int_{x_{1}}^{x_{2}} f\{y(x),y^{'}(x);x\}dx$$

is either a maximum or a minumum. In the above equation, $f()$ is the functional, $y^{'}(x) \equiv dy/dx$, and $x$ is the independent variable, which is commonly separated by a semicolon to denote its independence. That is, knowing $f()$ in terms of general dependent variables on $x$, we have a problem in the calculus of variations that we may feel justified in solving using its techniques.

For $y(x)$ to be an extremum it is required that any neighboring function is not so maximized or minimized. We define a neighboring function by adding on arbitrary amounts to $y(x)$, represented parametrically as $y=y(\alpha,x)$. For $\alpha=0$,$y=y(0,x)=y(x)$ is the function we pursue. Then

$$y(\alpha,x)=y(0,x)+\alpha\eta(x)$$

where $\eta(x)$ is some arbitary function of $x$ that has a continuous first derivative and vanishes at $x_{1}$ and $x_{2}$.

Now that we have defined variations on our solution, we can find the conditions such that our solution is an extremum by substituting in our parametrically defined $y(\alpha,x)$ in our functional:

$$J(\alpha) = \int_{x_{1}}^{x_{2}} f\{y(\alpha,x),y^{'}(\alpha,x);x\}dx$$

For $J(\alpha)$ to be an extremum, it must be independent of $\alpha$ in first order, or

$$\frac{\partial J}{\partial\alpha}\mid_{\alpha=0}=0$$

for all $\eta(x)$'s.

Taking the derivative of $J(\alpha)$,

$$\frac{\partial J}{\partial\alpha}=\int_{x_{1}}^{x_{2}} f\{y(\alpha,x),y^{'}(\alpha,x);x\}dx$$

and since the limits of integration are fixed, we may internalize the differential operation into the integrand. Hence,

$$\frac{\partial J}{\partial\alpha}=\int_{x_{1}}^{x_{2}}\left(\... ...partial y^{'}} \frac{\partial y^{'}}{\partial \alpha}\right)dx$$

by the chain rule.

From our definition of $y(\alpha,x)$, we can solve several of the above differential operations:

$$\frac{\partial y}{\partial \alpha}=\eta(x)$$

and

$$\frac{\partial y^{'}}{\partial \alpha}=\frac{\partial \eta}{\partial x}$$

.

We substitute these back into the integrand to get

$$\frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\left(... ...l f}{\partial y^{'}} \frac{\partial \eta}{\partial x}\right)dx$$

If we take a close look at this equation, we may notice that if we can somehow reduce $\frac{\partial \eta}{\partial x}$ to $g() \eta(x)$ then we can factor out $\eta(x)$ and we will be left with something that must then be equal to zero. We can do this using integration by parts:

$$\int u dv = u v - \int v du$$

$$\int_{x_{1}}^{x_{2}}\frac{\partial f}{\partial y^{'}} \frac{\... ...d}{dx}\left(\frac{\partial f}{\partial y^{'}}\right)\eta(x) dx$$

We have already defined $\eta(x)$ to be zero at the limits of integration, and therefore the first terms disappears. The second suites our hopes of being able to factor out $\eta{x}$, as when we substitute it back into the initial equation

$$\frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\left(... ...ft(\frac{\partial f}{\partial y^{'}} \right)\eta(x) \right) dx$$

so that we then factor out the arbitrary function, leaving us with

$$\frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\left(... ...rac{d}{dx} \frac{\partial f}{\partial y^{'}} \right)\eta(x) dx$$

$J$'s independence from $\alpha$ has now been established, and all that remains is to understand that since $\eta(x)$ is arbitrary, the remaining portion of the integrand must be equal to zero as a necessary condition for $J$ to be an extremum. Thus,

$$\frac{\partial f}{\partial y}-\frac{d}{dx} \frac{\partial f}{\partial y^{'}} = 0$$

.

This is the Euler equation.