Beltrami identity

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The Beltrami identity is an identity in the calculus of variations. It says that a function u which is an extremal of the integral

I(u)=\int_a^b f(x,u,u') \, dx

satisfies the differential equation

\frac{d}{dx}\left(f-u'\frac{\partial f}{\partial u'}\right)-\frac{\partial f}{\partial x}=0.

[edit] Proof

The Euler-Lagrange equation tells that

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

Now consider the total differential of functional f(x,u,u'). Substituting the Euler-Lagrange equation into it, we have

\begin{align} \frac{df}{dx} &= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial u} u' + \frac{\partial f}{\partial u'} u'' \\ & = \frac{\partial f}{\partial x} + \left(\frac{d}{dx}\frac{\partial f}{\partial u'}\right) u' + \frac{\partial f}{\partial u'} u''. \end{align}

Therefore,

\frac{d}{dx}\left(f-u'\frac{\partial f}{\partial u'}\right)-\frac{\partial f}{\partial x}=0.

[edit] Application

In case the functional f is independent of x, then the Beltrami identity can be simplified into

\begin{align} \frac{d}{dx}\left(f-u'\frac{\partial f}{\partial u'}\right) &=0 \\ f-u'\frac{\partial f}{\partial u'} &= \text{constant} \\ \end{align}

Using the above form is an easier approach to solve for the optimal function u than directly applying the Euler-Lagrange equation.