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I could really do with a few hints with this. (As soon as possible!)

**1. Homework Statement**

For a repeated line of cells, two 'equations of motion' can be written:

[tex]C_{n}\dot{U_{n+1}} = I_{n}-I_{n+1}[/tex]

[tex]L_{n}\dot{I_{n}} = U_{n} - U_{n+1}[/tex]

where [tex]C_{n}[/tex] is the capacitance, [tex]U_{n+1}[/tex] the voltage after the nth cell, [tex]I_{n}-I_{n+1}[/tex] the charging current, [tex]L_{n}[/tex] the inductance.

Work out the Lagrangian that generated these equations.

You should find that

[tex]E = \sum_{n}\left( (1/2)\L_{n}\dot{Q_{n}}^{2} + (1/2)C_{n}U_{n+1}^{2} \right)[/tex]

and

[tex]H = (1/2)\sum_{n}\left( \frac{P_{n}^{2}}{L_n}} + \frac{(Q_{n+1}-Q_{n})^{2}}{C_{n}} \right)[/tex]

**2. Homework Equations**

Note that

[tex]Q_{n} = -\int I_{n} dt => C_{n}U_{n+1} = Q_{n+1} - Q_{n}[/tex]

**3. The Attempt at a Solution**

I have suggested a Lagrangian of

[tex]L = (1/2)L_{n}\dot{Q_{n}}^{2} + Q_{n}(U_{n+1}-U_{n}) + (1/2)C_{n}\dot{U_{n+1}^{2}} + U_{n+1}(I_{n}-I_{n+1})[/tex]

(which can be turned into a sum for all the repeated circuits)

but it doesn't seem convincing, despite the fact that you can recover the original 'equations of motion' using Euler-Lagrange equations (using Q and U).

I'm not sure how the energy is being derived from the Lag. in this case (in mechanics, it was always the case that L = T - U, hence E = T + U), and it seems clear that you aren't going to get either that expression for the energy or that Hamiltonian from my guess-work Lagrangian.

Any suggestions?

Many thanks!