Question

show that the addition of the type dg/dx to the integrand function leaves the euler equation in the same form

Answer 1

Answer on Question #82242 – Math – Functional Analysis

Question

show that the addition of the type dg/dx to the integrand function leaves the euler equation in the same form

Solution

We consider the functional

$$J[y] = \int_{x_0}^{x_1} f(x, y, y') dx.$$

Suppose $J[y]$ has an extremum for the curve $y$. Then we have $\delta J[y] = 0$, and therefore, the Euler-Lagrange equation

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

Now let's take a new function

$$F(x, y, y') = f(x, y, y') + \frac{dg(x, y)}{dx}.$$

Note, that $g(x, y)$ must be a function of only $(x, y)$, and not of $y'$. The new functional will be

$$\begin{aligned} J_1[y] &= \int_{x_0}^{x_1} F(x, y, y') dx = \int_{x_0}^{x_1} f(x, y, y') dx + \int_{x_0}^{x_1} \frac{dg(x, y)}{dx} dx = \int_{x_0}^{x_1} f(x, y, y') dx + \int_{x_0}^{x_1} dg(x, y) \\ &= \int_{x_0}^{x_1} f(x, y, y') dx + g(x_0, y(x_0)) - g(x_1, y(x_1)). \end{aligned}$$

But $g(x_0, y(x_0))$ and $g(x_1, y(x_1))$ are constants, and their variations are null. Therefore, we have

$$\delta J_1[y] = \delta \int_{x_0}^{x_1} F(x, y, y') dx = \delta \int_{x_0}^{x_1} f(x, y, y') dx = \delta J[y].$$

That means

$$\delta J[y] = 0 \Rightarrow \delta J_1[y] = 0$$

and so the new Euler-Lagrange equations are

$$\frac{\partial F}{\partial y} - \frac{d}{dx} \frac{\partial F}{\partial y'} = 0.$$

As we can see this addition to the integrand function indeed leaves the Euler-Lagrange equation in the same form.

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