Hamiltonian formulism of mechanics (part 2)

-Following four forms of generator are possible : F₁(q,Q,t) , F₂(q,P,t) , F₃(p,Q,t) , F₄(p,P,t)

  Legendre transforms are
    g(u,y)=f(x,y)-ux
    where u=(∂ F/∂ x)
 (i) F₁=F₁(q,Q,t) putting it in equation 1 and finding derivative of F w.r.t. we get
          p_j = ∂F₁(q,Q,t)/∂q_j

          P_j = - ∂F₁(q,Q,t)/∂Q_j

          K=H+∂F₁(q,Q,t)/∂t
(ii) putting u = -P_j , x = Q_j , y = q_j , g = F₂(q,P,t) , f=F₁ and using Legender transforms we get
          F₂(q_j,P_j,t) = F₁(q_j,Q_j,t)+ΣP_jQ_j
         Putting this in 1 and solving we get
          p_j = ∂F₂(q_j,P_j,t)/∂q_j

          Q_j =  ∂F₂(q_j,P_j,t)/∂P_j

          K=H+∂F₂(q_j,P_j,t)/∂t

 (iii) F₃ = F₃(Q,p,t) is the third form
            Connect it to the first form using legender transforms
            Since p_j = ∂F₁(q_j,P_j,t)/∂q_j and u=∂f/∂x
             this implies that u=p_j , x=q_j , y=Q_j , g=F₃ , f=F₁
              Thus,  F₃ = F₁(Q,q,t)-Σp_jq_j
              or, F₁(Q,q,t)=F₃+Σp_jq_j
              putting these in equation 1
            q_j = -∂F₃(p_j,Q_j,t)/∂p_j
            P_j = -∂F₃(p_j,Q_j,t)/∂Q_j

             K=H+∂F₃(p_j,Q_j,t)/∂t
(iv)    Fourth form is F₄(p,P,t) , connecting F₄ by F₁ through legender transforms and solving we get
         q_j = -∂F₄(p_j,P_j,t)/∂p_j

         Q_j = ∂F₄(p_j,P_j,t)/∂P_j
         K=H+∂F₄(p_j,P_j,t)/∂t