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UPSC 2015 Maths Optional Paper 2 Q7c-i — Step-by-Step Solution

10 marks · Section B

Hamilton's equations · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

A Hamiltonian of a system with one degree of freedom has the form

where $\alpha,b,k$ are constants, $q$ is generalized coordinate, $p$ generalized momentum. Find a Lagrangian corresponding to this Hamiltonian.

Technique

Standard Legendre transform $L=p\dot q-H$ with $p=\alpha(\dot q+bqe^{-\alpha t})$ extracted from $\dot q=\partial H/\partial p$.

Solution

Strategy. Use the Legendre transform $L=p\dot q-H$, with $p$ expressed in terms of $\dot q$ via $\dot q=\partial H/\partial p$.

Step 1 — Compute $\dot q=\partial H/\partial p$

Solve for $p$:

Step 2 — Compute $L=p\dot q-H$

First $p\dot q=\alpha\dot q^2+\alpha bq\dot q e^{-\alpha t}$.

Now $H$ in terms of $\dot q$: substitute $p=\alpha(\dot q+bqe^{-\alpha t})$.

$p^2=\alpha^2(\dot q+bqe^{-\alpha t})^2=\alpha^2\dot q^2+2\alpha^2 b\dot q q e^{-\alpha t}+\alpha^2 b^2 q^2 e^{-2\alpha t}$.

$\dfrac{p^2}{2\alpha}=\dfrac{\alpha\dot q^2}{2}+\alpha b\dot q q e^{-\alpha t}+\dfrac{\alpha b^2 q^2 e^{-2\alpha t}}{2}$.

$bqpe^{-\alpha t}=bqe^{-\alpha t}\cdot\alpha(\dot q+bqe^{-\alpha t})=\alpha bq\dot q e^{-\alpha t}+\alpha b^2 q^2 e^{-2\alpha t}$.

$\dfrac{b\alpha}{2}q^2 e^{-\alpha t}(\alpha+be^{-\alpha t})=\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}+\dfrac{b^2\alpha q^2 e^{-2\alpha t}}{2}$.

Sum:

Group by term:

• $\dot q^2$ terms: $\alpha\dot q^2/2$.
• $\dot q q$ terms: $\alpha b\dot q q e^{-\alpha t}-\alpha bq\dot q e^{-\alpha t}=0$.
• $q^2 e^{-2\alpha t}$ terms: $\alpha b^2/2-\alpha b^2+b^2\alpha/2=0$.
• $q^2 e^{-\alpha t}$ term (from "$\alpha$" piece of the third term): $b\alpha^2/2$.
• $q^2$ term: $k/2$.

So $H=\dfrac{\alpha\dot q^2}{2}+\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}+\dfrac{k q^2}{2}$.

Hmm wait, that’s much cleaner than I expected. Let me re-verify by recomputing the third term:

$\dfrac{b\alpha}{2}q^2 e^{-\alpha t}(\alpha+be^{-\alpha t})$ — this has two pieces: $\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}$ and $\dfrac{b^2\alpha q^2 e^{-2\alpha t}}{2}$.

Then $H$ has terms:

1. $p^2/(2\alpha)=\alpha\dot q^2/2+\alpha b q\dot q e^{-\alpha t}+\alpha b^2 q^2 e^{-2\alpha t}/2$.
2. $-bqpe^{-\alpha t}=-\alpha b q\dot q e^{-\alpha t}-\alpha b^2 q^2 e^{-2\alpha t}$.
3. $+\dfrac{b\alpha^2 q^2 e^{-\alpha t}}{2}+\dfrac{b^2\alpha q^2 e^{-2\alpha t}}{2}$.
4. $+kq^2/2$.

Sum coefficient of $q\dot q e^{-\alpha t}$: $\alpha b-\alpha b=0$ ✓. Sum coefficient of $q^2 e^{-2\alpha t}$: $\alpha b^2/2-\alpha b^2+\alpha b^2/2=0$ ✓. Coefficient of $q^2 e^{-\alpha t}$: $b\alpha^2/2$. Coefficient of $\dot q^2$: $\alpha/2$. Coefficient of $q^2$: $k/2$.

So

But this is $H$ expressed in $(q,\dot q,t)$. Now $L=p\dot q-H$.

$p\dot q=\alpha\dot q^2+\alpha b q\dot q e^{-\alpha t}$.

Answer