[Math] Variational Principles: Lagrange Multipliers

calculuscalculus-of-variations

I am trying to minimize the functional

$$I[\textbf{x}] = \int ||\dot{\textbf{x}}||^2 dt$$

subject to the constraint $\textbf{x}(t) \in \{\textbf{s} \in \mathbb{R}^3 : ||\textbf{s}|| = 1 \}$. The question states:

Use the Lagrange multiplier function formalism to obtain the following Euler-Lagrange equation:
$$\ddot{\textbf{x}} + ||\dot{\textbf{x}}||^2 \textbf{x} = \textbf{0}$$

I'm not totally sure what it means by the "Lagrange multiplier function formalism", but anyway, introducing a Lagrange multiplier, $\lambda(t)$, we get:

$$\int L(\textbf{x}, \dot{\textbf{x}}, \lambda) dt = \int ||\dot{ \textbf{x}}||^2 + \lambda(||\textbf{x}|| – 1) dt$$

and hence the Euler-Lagrange equations are:

$$\frac{d}{dt}\Big( \frac{\partial L}{\partial \dot{\textbf{x}}}\Big) – \frac{\partial L}{\partial \textbf{x}} = 2 \ddot{\textbf{x}} – \frac{\lambda}{||\textbf{x}||} \textbf{x} = \textbf{0}$$

and also $||\textbf{x}|| – 1 = 0$. I need to find what $\lambda(t)$ is but I'm not sure how to progress.

Thanks in advance for any help.

Best Answer

Note that $\mathbf{x}^2 = ||\mathbf{x}||^2 = 1$ can be differentiated twice to obtain $\dot{\mathbf{x}}^2+\mathbf{x}\cdot\ddot{\mathbf{x}}=0.$ But one can also obtain $\mathbf{x}\cdot \ddot{\mathbf{x}}$ from the Euler-Lagrange equation by an appropriate dot product. These together fix $\lambda$.

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