# Derivation of the Euler-Lagrange Equation for a Function of Several Dependent Variables

IMAGE CREDIT: NASA/JPL

SOURCE FOR CONTENT: Classical Dynamics of Particles and Systems. Thornton and Marion. 5th Edition.

Consider a functional

$\displaystyle \phi = \phi(y_{\mu},y_{\mu}^{\prime}; x), (1)$

where $\mu = 1,2,...,n$. By the method used in a previous section of the aforementioned text, we may write

$\displaystyle y_{\mu}(\alpha, x) = y_{\mu}(0,x) +\alpha \eta_{\mu}(x). (2)$

Additionally, we will find it useful to define

$\displaystyle y_{\mu}^{\prime}(\alpha,x) = y_{\mu}^{\prime}(0,x)+\alpha \eta_{\mu}^{\prime}(x). (3)$

Further we may also define an integral functional by way of integrating Eq.(1) over the interval $x_{1}\leq x \leq x_{2}$, and introducing a variational parameter $\alpha$ we have

$\displaystyle J(\alpha) = \int_{x_{1}}^{x_{2}} \phi(y_{\mu},y_{\mu};x)dx. (4)$

Two necessary conditions that are used to derive the Euler-Lagrange equation include

$\displaystyle \frac{\partial J(\alpha)}{\partial \alpha}\bigg\|_{\alpha=0}=0, (5)$

and

$\displaystyle \eta_{\mu}(x_{1})=\eta_{\mu}(x_{2})=0. (6)$

Let us take the derivative of $J(\alpha)$ with respect to $\alpha$ yielding

$\displaystyle \frac{\partial J}{\partial \alpha}\bigg\|_{\alpha=0}=\frac{\partial}{\partial \alpha}\int_{x_{1}}^{x_{2}}\phi(y_{\mu},y_{\mu}^{\prime};x)dx. (7)$

Carrying out the derivative operator on the right-hand-side of Eq.(7) we get

$\displaystyle \frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi \partial_{\alpha}y_{\mu}+\partial_{y_{\mu}^{\prime}}\phi \partial_{\alpha}y_{\mu}^{\prime}\bigg\}dx. (8)$

From Eqs.(2) and (3) we see that

$\displaystyle \partial_{\alpha}y_{\mu}= \eta_{\mu}(x), (9)$

and

$\displaystyle \partial_{\alpha}y_{\mu}^{\prime} = \eta_{\mu}^{\prime}(x). (10)$

Thus Eq.(8) becomes…

$\displaystyle \frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi \eta_{\mu}(x)+\partial_{y_{\mu}^{\prime}}\phi \eta_{\mu}^{\prime}(x)\bigg\}dx. (11)$

Consider the second term under the summation. We may make use of integration by parts to obtain the following

$\displaystyle \frac{\partial J}{\partial \alpha}=\int_{x_{1}}^{x_{2}}\sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi +\frac{d}{dx}(\partial_{y_{\mu}^{\prime}}\phi) \bigg\}\eta_{\mu}(x)dx. (12)$

By the necessary condition (Eq.(5)), it follows that

$\displaystyle 0 = \sum_{\mu}\bigg\{\partial_{y_{\mu}}\phi +\frac{d}{dx}(\partial_{y_{\mu}^{\prime}}\phi) \bigg\}. (13)$

Additionally, in Eq.(13) above, we have also made use of the condition that $\eta_{\mu}(x_{1})=\eta_{\mu}(x_{2})=0$. Since $\eta_{\mu}(x) \neq 0$ for any $x_{1}\leq x \leq x_{2}$, then the terms in the brackets must vanish, yielding the Euler-Lagrange Equation for several dependent variables.

Update: the next posts will be those discussed on my Facebook page. Namely, I intend to continue with my Research Series and my series in Tensor Calculus and General Relativity with various ancillary posts in my Astrophysics Series.

Clear Skies!