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Chapter 8 | 
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CALCULUS OF VARIATION
Problems 8.1 to 8.5 | Problems 8.6 to 8.8
problem8.1 | problem8.2 | problem8.3 | problem8.4 | problem8.5
8-1(16). A product is being produced at a steady rate of Po pounds per hour. It is necessary to change the production rate to P1, and minimize the cost resulting from raw material lost to off-specification product and overtime wages during the transition period. This is modeled by the following cost function.
C(t) = C1P'2 + C2tP'
where C1 and C2 are cost coefficients and P' = dP/dt. The total cost for the change in the production schedule is given by:
where Po = P(to) and P1 = P(t1) are known.
Determine the optimum way the production rate P(t) is to be changed to minimize the total cost.
Solution:
The Euler equation is solved for P(t) which minimizes CT.
C(t) = C1P'2 + C2tP'
The Euler equation is:
Substituting gives:
Integrating twice gives:
P = -C2t2/4C1 + k1t + k2
Solving for constants of integrating, k1 and k2, gives:
P1 = -C2t12/4C12 + k1t1 + k2
Po = -C2to2/4C1 + k1to + k2
P1 - Po = -(t12 - to2 )C2/4C1 + k1(t1 - to)
k1 = (P1 - Po/t1 - to) + (t1 + to)C2/4C1
k2 = Po + C2to2/4C1 - to [(P1 - Po)/(t1 - to)] + (t1 + to)C2/4C1]
k2 = Po - to [(P1 - Po)/(t1 - to) + C2t1/4C1]
P = -C2t2/4C1 + [(P1 - Po)/(t1 - to) + C2(t1 + to)/4C1 ]t + Po - to[(P1 - Po)/(t1 - to) + C2t1/4C1 ]
P = Po - C2t2/4C1 + [(P1 - Po)/(t1 - to) + C2(t1 + to)/4C1 ]t - to[(P1 - Po)/(t1 - to) + C2t1/4C1 ]
This is the equation of a parabola.
8-2(15). A classical problem in aerodynamics is to determine the optimum shape of a body of revolution which has the minimum drag. For a slender body of revolution at zero angle of attack in an inviscid hypersonic flow, the total drag is approximated by
where v and r are the free stream velocity and density respectively.
a. Obtain the differential equation and boundary conditions that are to be solved to obtain the optimum body shape.
b. Show the following is the solution to the differential equation obtained in part a
y = (d/2)(x/L)3/4
which according to Miele(15) means that the contours of a body of revolution having minimum drag for a given diameter, d, and a given length, L, is a parabola satisfying the 3/4 power law.
Solution:
y(L) = d/2
Flow direction
V_________________ O X L
a.
and F = y(y')3
The Euler equation is:
and
Fy = (y')3, Fy' = 3y(y')2
Substituting into the Euler equation gives:
(3y)(2y'y'') + 3y'(y')2 - (y')3 = 0
6yy'y'' + 3(y')3 - (y')3 = 0
3yy'' + (y')2 = 0
This is a nonlinear ordinary differential equation.
b. y = (d/2)(x/L)3/4
Let A = (d/2)/L3/4 contain the parameters of the equation, and:
y = Ax3/4
y' = 3/4 Ax -1/4
y" = -3/16 Ax -5/4
Substituting the above into the differential equation in part a gives:
3(Ax3/4)(-3/16 Ax -5/4) + (3/4 Ax -1/4)2 = 0
-9/16 A2x -1/2 + 9/16 A2x -1/2 = 0
Thus, this equation is a solution to the differential equation.
8-3(1). Determine the minimum surface of revolution by finding the curve y(x) with prescribed end points such that by revolving this curve around the x axis a surface of minimal area is obtained.
The integral to be minimized is:
a. Show that the Euler equation for F = F(y,y') only gives
b. Apply this result to the problem to obtain
y(1 + y'2) -½ = c1
c. Define the parametric variable y' = sinh t in order to have a more compact solution and obtain the following result.
x = c1t + c2
y = c1 cosh (t)
Which is the parametric form of a family of catenaries and c1 and c2 are boundary conditions for the end points of the curve.
Solution:
a. For F = F(y,y')
b.
Let F = y 1 + (y')2
Using Euler Equation
Simplifying gives:
y[1 + (y')2]-½ = C1
c. Given y' = sinh(t), then integrating gives:
y = C1 cosh t
Then using the above gives:
dx = dy/y' = C1 sinh(t) dt/sinh(t) = C1 dt
Integrating the above gives:
x = C1t + C2
and y1 = C1 cosh t
This is the parametric form of a family of catenaries, and the constants C1 and C2 are determined from end points of the curves.
8-4(9). Find the shape at equilibrium of a chain of length, L, which hangs from two points at the same level. The potential energy, E, of the chain is given by the following equation:
and is subject to the specified total length L by the following equation:
with boundary conditions of y(xo) = y(x1) = 0. To obtain the equilibrium shape of the chain, it is necessary to minimize the energy subject to the length restriction. Show that the following differential equation is obtained from the Euler equation.
y' = [k2(y + l)2 - 1]½
Make the substitution k(y + l) = cosh Q, and obtain the solution given below.
cosh k(x + a) = k(y + a)
This curve is the catenary, and the constants k, a, l can be obtained from the boundary conditions and the constraint on L.
Solution:
Using the Lagrange multiplier method, the following unconstrained extremum is obtained.
F = (y + l)[1 + (y')2]½
C = y' ¶F/¶y' - F
(y')2 = C2(y + l)2 - 1
y' = [C2(y + l)2 - 1]½
8-5(14). A simple optimal control problem related to an electro-mechanical system can be formulated as:
a. Obtain the differential equations to be solved for the optimal functions. Show that there are sufficient equations to determine the dependent variables.
b. What boundary conditions are required?
Solution:
a. Three optimal functions are to be determined: y1(t), y2(t), and y3(t). The Lagrangian function is:
F* = y22 - y32 + l1(y1' + y1 - y3) + l2(y2' - y1)
The Euler equations are:
Substituting the Euler equations become:
l1' - l1 + l2 = 0
l2' - 2y2 = 0
2y3 + l1 = 0
These are solved with the constraint equations.
y1' + y1 - y3 = 0
y2' - y1 = 0
There are five differential equations and five dependent variables: y1(t), y2(t), y3(t), 1(t) and 2(t) which area function of t.
b. Boundary conditions are required at t=0 and t=T as shown below.
y1(0) = y10 y1(T) = y1T
y2(0) = y20 y2(T) = y2T
y3(0) = y30 y3(T) = y3T