Problem 2, Lab 13

(a) Answer: A=H, p=0

(b) p=0, A=1

(c) Use pplane, web site at Missouri, search string "free pplane phase portrait"

    dx/dt = 0.5*(1-y-x),  dy/dt = 0.1*(1-y)

    Don't print it. Draw a replica by hand.

(d) A(t) = equil sol + constant/integrating factor = H + c / exp(epsilon * t)
    Then find c = A(0)-H to satisfy A(t) = H + c / exp(epsilon * t) at t=0.

(e) Let q(t) = p(t)*exp(k*t). Product rule implies q' = p' * exp(k*t) + k * p * exp(k*t). Use
    p' = k*(H-A-p) to obtain 

                   q' = k * (H-A-p) * exp(k*t) + k * p * exp(k*t)
                   q' = k * (H-A) * exp(k*t) - p * k * exp(k*t) + k * p * exp(k*t)              
                   q' = k * (H-A) * exp(k*t)
                   q' = k * ((-c) / exp(epsilon * t)) * exp(kt)  where c = A(0)-H, by part (d)
                   q' = k * (H-A(0)) * exp(-epsilon * t)) * exp(k*t)
                   q' = k * (H-A(0)) * exp(k*t-epsilon*t)

(f) Quadrature applied to the equation in (e) gives 

                   q(t) = constant + integral of q' over [0,t]   Fundamental Theorem of Calculus
                   q(t) = c[1] + k * (H-A(0)) * exp(k*t-epsilon*t)/(k-epsilon) 
                   q(t) = c[1] + k * (H-A(0)) * exp(a*t)/ a,  where a=k-epsilon

(g) The the results are consistent, because the pplane portrait already chose many values for p[0] and A[0].

(h) Because a = k - epsilon = 0.5 - 0.1 = 0.4 > 0, then 

          p(t) = q(t) * exp(-k*t) 
          p(t) = c[1] * exp(-k*t) + k * (H-A(0)) *exp(-k*t+a*t)/ a
          p(t) = c[1] * exp(-k*t) + k * (H-A(0)) *exp(-epsilon*t)/ a
          A(t) = H + (A(0)-H) * exp(-epsilon*t)    by part (d)

The limits of each variable are 0, H, respectively, because of the negative exponentials.
REPORT: Limit p(t) = 0,  Limit A(t) = H. This is verified from the pplane phase portrait.