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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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F(v) = -k|v|v = -k|/I/j
Summarizing, we have the models
ODE Architect only accepts first-order ODEs, so that’s why we use the first-order system form.
/ = -g -
0 no drag
k / •
— ó VISCOUS
m
k
—1/1/ Newtonian m
or, in system form,
y' = v
0 no drag k
U VISCOUS
v =e m k
— |u|u Newtonian m
To observe different rise times and fall times, you can set y(0) = 0, v(0) = v0 and see what happens for various positive values of v0. See Figure 5.2 for graphs of y( t) with viscous damping, four different initial velocities, k/ m =
2 sec-1, and g = 32 ft/sec2. In this setting v is the rate of change of y, so v is positive as the ball rises and negative as it falls.
¦ Indiana Newton
You notice that Indiana Newton is about to jump from a ledge onto a boxcar of a speeding train. His timing has to be perfect. He also gets to choose his drag: none, viscous, or Newtonian. If you knew the train’s position at all times, and how long it takes Indy to drop from the ledge to the top of the boxcar, then you could give him good advice about which drag to choose.
The initial value problem that models Indy’s situation is
y = v y(0) = h
v' = —g- F(v)/m v(0) = 0
where m is his mass, F(v) is a drag function, g is the acceleration due to gravity, and h is the height of the ledge above the boxcar. His life is in your hands! Figure 5.3 shows Indy’s free-fall solution curves y(t) from a height of 100 ft with three different drag functions.
Indiana Newton
83
Time (seconds)
Figure 5.2: Height vs. time of a whiffle ball thrown straight up four times with viscous damping and different initial velocities. Does the ball take longer to rise or to fall?
Time (seconds)
Figure 5.3: Indianajumps with no drag (left curve), viscous drag —0.2/ (middle), Newtonian drag —0.02|/|y (right).
84
Chapter 5
¦ Ski Jumping
Check that this force is perpendicular to velocity.
The origin of the xy-plane is at the edge ofthe ski jump (x-horizontal, y-vertical). The edge is horizontal so X(0) = V0 > 0, but y (0) = 0.
When a ski jumper is aloft she is subject to gravitational, drag, and lift forces. She can diminish the drag and increase the lift by her posture, ski angle, and choice of clothing. Drag acts opposite to velocity and its magnitude is usually taken to be proportional to the skier’s velocity R;:
Drag force = — 8R; = —SX'i — 8yj
The lift force is what makes ski jumping fun. The lift force is that force which acts perpendicular to the velocity and enables the jumper to soar. Its magnitude is usually taken to be proportional to the speed, so
Lift force = —ky'i + X/j
Newton’s second law in the i- and j-directions gives us
mX'= —8X — Xyf x/(0) =v0, x(0) = 0
mf = -mg + Xx — 8y /(0) =0, y(0) = 0
where m is the skier’s mass and 8, X, and v0 are positive constants. Integration of each of these ODEs yields
mx — mv0 = —8x — Xy
my = —mgt + Xx — 8 y
Divide by the mass to get the system IVP
Ó = —ax — by + v0 x(0) = 0
y = —gt + bx — ay y(0) = 0
where a = 8/m and b = X/m are the drag and lift coefficients, respectively.
When Newtonian drag and lift occur, 8 and X are not constants, so we can no longer integrate once to get x and /, and we must treat the original
second-order ODE differently:
x = v x(0) = 0
v' = —8v/m — Xw/m v(0) = v0
y = w y(0) = 0
w' = —g + Xv/m — 8w/m w(0) = 0
where v and w are the velocities in the i- and j-directions, respectively.
We have assumed that the bottom edge of the ski jump is horizontal, but everything can be modified to accommodate a tilt in the launch angle (see the chapter cover figure and Exploration 5.4, Problem 1).
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