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!

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