How can the Hamilton-Jacobi equation be derived and solved?

[GleefulNihilism]

Well, I had a couple problems on my final I was hoping to go over- hope nobody minds. Here's the third.

Homework Statement

The Hamiltonian of the System is H = (1/2)*(p1*q2 + 2*p1*p2 + q1^2)
A.) Derive the Hamilton-Jacobi Equation
B.) Find the Solution.

Homework Equations

H = (q1, q1, . . ., qk , dS/dq1, dS/dq2, . . ., dS/dqk , t ) + dS/dt = 0 where S is the generator.

The Attempt at a Solution

Putting the Hamiltonian in proper formulation, it becomes.

(1/2)*(dS/dq1*q2 + 2*dS/dq1*dS/dq2 + q1^2)

So (1/2)*(dS/dq1*q2 + 2*dS/dq1*dS/dq2 + q1^2) + dS/dt = 0

B.

Let's try S = -q1^2 * t - q2^2 / 4

then (q2 / 2)*(-2*q1*t) + (-q2 / 2)*(-2*q1*t) + q1^2 + (-q1^2) = 0

Therefore S = -q1^2 * t - q2^2 / 4 + A is the generator, where A is a constant.

Let S = S2(q2) + S1(q1,t) + A as well.

Then (q2 / 2)*(dS1/dq1) + (dS1/dq1)*(dS2/dq2) + q1^2 + (dS1/dt) = 0

Dividing all terms by (dS1/dq1) creates:

(q2 / 2) + (dS2/dq2) + q1^2*(dS1/dq1)^-1 + (dS1/dt)*(dS1/dq1)^-1 = 0

(-q2 / 2) + (-dS2/dq2) = q1^2*(dS1/dq1)^-1 + (dS1/dt)*(dS1/dq1)^-1

Which can only be true in general if both sides are equal to a constant we'll call C.

(-q2 / 2) + (-dS2/dq2) = C implies dS2/dq2 = -(q2/2 + C). But from earlier we know that S2 = -q2^2 / 4.

So -q2 / 2 = -(q2 / 2 + C) and C = 0, q2 = q2

A similar argument with q1^2*(dS1/dq1)^-1 + (dS1/dt)*(dS1/dq1)^-1 = C creates:

q1^2*(dS1/dq1)^-1 + q1' = C implies (q1^2)/(C - q1') = dS1/dq1.

But from earlier we know S1 = -q1^2 * t so. . .

(q1^2)/(C - q1') = -2*q1*t, which is nasty- and I doubt it's separable.

 

[GleefulNihilism]

Still no bites?

Well, I just want to know how the professor got these answers- don't think I need to be lead by the hand.

A.) (1/2)*( (dS0 / dq1)*q2 + 2*(dS0 / dq1)*(dS0 / dq2) + q1^2) = E

Where S(q,t) = -E*t + S0(q)

B.) S(q,t) = -E*t + (1/2a)*(2*E*q1 - q1^3 / 3) + (1/2)*(2*a*q2 - q2^2 / 2)

Where E is energy and a is an arbitrary constant.