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UPSC 2018 Maths Optional Paper 2 Q7c — Step-by-Step Solution

20 marks · Section B

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

Question

The Hamiltonian of a mechanical system is given by H=p1q1−aq12+bq22−p2q2H=p_1q_1-aq_1^2+bq_2^2-p_2q_2, where a,ba,b are the constants. Solve the Hamiltonian equations and show that p2−bq2q1=\dfrac{p_2-bq_2}{q_1}= constant.

Technique

Hamilton’s canonical equations; the qq‘s give pure exponentials; first-order linear ODEs (integrating factor) for the pp‘s; the bBe−tbB e^{-t} term in p2p_2 cancels bq2bq_2, leaving p2−bq2=Cet∝q1p_2-bq_2=Ce^{t}\propto q_1.

Solution

Setup. Hamilton’s canonical equations:

q˙i=∂H∂pi,p˙i=−∂H∂qi,i=1,2.\dot q_i=\frac{\partial H}{\partial p_i},\qquad \dot p_i=-\frac{\partial H}{\partial q_i},\qquad i=1,2.

With H=p1q1−aq12+bq22−p2q2H=p_1q_1-aq_1^2+bq_2^2-p_2q_2.

Step 1 — Write the four equations

q˙1=∂H∂p1=q1,q˙2=∂H∂p2=−q2,\dot q_1=\frac{\partial H}{\partial p_1}=q_1,\qquad \dot q_2=\frac{\partial H}{\partial p_2}=-q_2, p˙1=−∂H∂q1=−(p1−2aq1)=2aq1−p1,\dot p_1=-\frac{\partial H}{\partial q_1}=-(p_1-2aq_1)=2aq_1-p_1, p˙2=−∂H∂q2=−(2bq2−p2)=p2−2bq2.\dot p_2=-\frac{\partial H}{\partial q_2}=-(2bq_2-p_2)=p_2-2bq_2.

Step 2 — Solve the qq-equations

q˙1=q1 ⇒ q1=A et,q˙2=−q2 ⇒ q2=B e−t,\dot q_1=q_1\ \Rightarrow\ \boxed{q_1=A\,e^{t}},\qquad \dot q_2=-q_2\ \Rightarrow\ \boxed{q_2=B\,e^{-t}},

where A,BA,B are constants.

Step 3 — Solve the pp-equations

p1p_1: p˙1+p1=2aq1=2aAet\dot p_1+p_1=2aq_1=2aAe^{t}. Integrating factor ete^{t}: (p1et)′=2aAe2t(p_1e^{t})'=2aAe^{2t}, so p1et=aAe2t+Dp_1e^{t}=aAe^{2t}+D, giving

p1=aAet+D e−t.p_1=aAe^{t}+D\,e^{-t}.

p2p_2: p˙2−p2=−2bq2=−2bBe−t\dot p_2-p_2=-2bq_2=-2bBe^{-t}. Integrating factor e−te^{-t}: (p2e−t)′=−2bBe−2t(p_2e^{-t})'=-2bBe^{-2t}, so p2e−t=bBe−2t+Cp_2e^{-t}=bBe^{-2t}+C, giving

p2=C et+bB e−t.\boxed{p_2=C\,e^{t}+bB\,e^{-t}.}

Step 4 — The conserved combination

From Step 2, q1=Aetq_1=Ae^{t} and q2=Be−tq_2=Be^{-t}, so bq2=bBe−tbq_2=bBe^{-t}. Then

p2−bq2=(Cet+bBe−t)−bBe−t=Cet.p_2-bq_2=\big(Ce^{t}+bBe^{-t}\big)-bBe^{-t}=Ce^{t}.

Therefore

p2−bq2q1=CetAet=CA=constant.\frac{p_2-bq_2}{q_1}=\frac{Ce^{t}}{Ae^{t}}=\frac{C}{A}=\text{constant}.

Answer

  p2−bq2q1=CA=const.  \boxed{\;\frac{p_2-bq_2}{q_1}=\frac{C}{A}=\text{const}.\;}
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