Brachistochrone Under Air Resistance
Christine Lind
2/26/05
SPCVC
Source of Information:
Collaborator:
Brachistochrone Setup:
• Initial PointP(x0,y0)
• Final PointQ(x1,y1)
• Resistance ForceFr
• Slope Angle
Geometric Constraints
• Parametric Approach:– Start by using arclength
(s) as the parameter
• Parametrized by arclength:
(curves parametrized by arclength have unit speed)
€
x'(s) − cos(ϕ (s)) = 0
y'(s) − sin(ϕ (s)) = 0
Energy Constraint?
• Normally we use conservation of energy to solve for velocity in terms of the other variables
• We have a Non-Conservative system, so what do we do?
Energy Constraint?
• Energy is lost to work done by the resistance force:
€
ΔE = ΔW
ΔT + ΔU = ΔW
Energy Constraint
• Non-conservative system:
• Constraint parametrized by time:
• Constraint parametrized by arclength:
€
dT
dt+
dU
dt=
dW
dt
€
T(t) = 12 mv 2
U(t) = −mgy
W (t) = Frvdτ0
t
∫
€
dv
dt− gsinϕ + R(v) = 0
€
vv'−gsinϕ + R(v) = 0€
d
dt= v
d
ds
⎛
⎝ ⎜
⎞
⎠ ⎟
Problem Formulation:
• Boundary & Initial Conditions:
• Minimize the time integral:
• Other constraints:How do we incorporate them?
€
x(0) = x0, y(0) = y0, v(0) = v0
x(l) = x1, y(l) = y1,
€
T =ds
v0
l
∫
€
x'−cosϕ = 0
y'−sinϕ = 0
vv'−gsinϕ + R(v) = 0
Lagrange Multipliers
• Introduce multipliers, vector:
• Create modified functional:
where
€
G(s,q,q') =1
v+ κ (x '−cosϕ ) + λ (y'−sinϕ ) + μ(vv'−gsinϕ + R(v))
€
κ(s), λ (s), μ(s)
€
q(s) = (v, x,y,ϕ ,κ ,λ ,μ)
€
F[q, l] = G(s,q,q')ds0
l
∫
Euler-Lagrange Equations
• System of E-L equations:
• Additional boundary conditions:
€
F[q, l] = G(s,q,q')ds0
l
∫
€
∂G
∂q−
d
ds
∂G
∂q'
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
€
−∂G
∂q's= 0
δq0 +∂G
∂q's= l
δq1 + G − q'∂G
∂q'
⎛
⎝ ⎜
⎞
⎠ ⎟s= l
δl = 0
7 Euler-Lagrange Equations:
€
1
v 2− μ v'+
dR
dv
⎛
⎝ ⎜
⎞
⎠ ⎟+ μv( )'= 0
κ sinϕ − λ cosϕ − gμ cosϕ = 0
κ '= 0
λ '= 0
x'−cosϕ = 0
y'−sinϕ = 0
vv'−gsinϕ + R(v) = 0
Note:
€
κ(s) = C1
λ (s) = C2
Note:Additional Constraints Appear as E-L equations!
Natural Boundary Conditions:
Note:v1 is not necessarily zero, so:
€
1
v1
−κ1 cosϕ1 − λ1 sinϕ1 + μ1 −gsinϕ1 + R(v1)( ) = 0
μ1v1 = 0
€
μ1 = μ(l) = 0
Lagrange Multipliers - Solved!
• Using:
• Determine the Lagrange Multipliers:
€
κ sinϕ − λ cosϕ − gμ cosϕ = 0s= l
1
v1
−κ1 cosϕ1 − λ1 sinϕ1 + μ1 −gsinϕ1 + R(v1)( ) = 0
€
κ(s) =cosϕ1
v1
, λ (s) =sinϕ1
v1
,
μ(s) =sin(ϕ −ϕ1)
gv1 cosϕ
Note:κ(s),(s) constants
μ(s)=μ((s))
First Integral
• Recall:
No explicit s-dependence!
• First Integral:€
G(s,q,q') =1
v+ κ (x '−cosϕ ) + λ (y'−sinϕ ) + μ(vv'−gsinϕ + R(v))
€
G(q,q') − q'∂G
∂q'= const.
€
1
v−κ cosϕ − λ sinϕ + μ(−gsinϕ + R(v)) = const.€
κ(s) =cosϕ1
v1
, λ (s) =sinϕ1
v1
,
μ(s) =sin(ϕ −ϕ1)
gv1 cosϕ
Parametrize by Slope Angle
€
κ(s) =cosϕ1
v1
, λ (s) =sinϕ1
v1
,
μ(s) =sin(ϕ −ϕ1)
gv1 cosϕ
Parametrize by Slope Angle
• Define f() to be the inverse function of (s):
• f() continuously differentiable, monotonic
• Now we minimize:€
s = f (ϕ ), f (ϕ 0) = 0
€
T =˙ f dϕ
vϕ 0
ϕ 1∫
€
ds = ˙ f dϕ( )
Still need constraints...
Modified Functional
• Transform modified problem in terms of :
€
H(ϕ ,p, ˙ p ) =˙ f
v+ κ ( ˙ x − ˙ f cosϕ ) + λ ( ˙ y − ˙ f sinϕ ) + μ(v ˙ v − ˙ f gsinϕ + ˙ f R(v))
€
p(ϕ ) = (v,s = f (ϕ ),x, y,κ ,λ ,μ)
€
L[p,ϕ 0,ϕ1] = H(ϕ ,p, ˙ p )dϕϕ 0
ϕ 1∫
€
x(ϕ 0) = x0, y(ϕ 0) = y0,
x(ϕ1) = x1, y(ϕ1) = y1,
v(ϕ 0) = v0, f (ϕ 0) = 0
€
G(s,q,q') =1
v+ κ (x'−cosϕ ) + λ (y '−sinϕ ) + μ(vv'−gsinϕ + R(v))
⎛
⎝ ⎜
⎞
⎠ ⎟
7 Euler-Lagrange Equations
€
˙ f
v 2− μ ˙ v + ˙ f
dR
dv
⎛
⎝ ⎜
⎞
⎠ ⎟+
d μv( )dϕ
= 0
d
dϕ
1
v−κ cosϕ − λ sinϕ + μ(−gsinϕ + R(v))
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
˙ κ = 0˙ λ = 0
˙ x − ˙ f cosϕ = 0
˙ y − ˙ f sinϕ = 0
v ˙ v − ˙ f gsinϕ + ˙ f R(v) = 0
€
1
v 2− μ v'+
dR
dv
⎛
⎝ ⎜
⎞
⎠ ⎟+ μv( )'= 0
κ sinϕ − λ cosϕ − gμ cosϕ = 0
κ '= 0
λ '= 0
x'−cosϕ = 0
y'−sinϕ = 0
vv'−gsinϕ + R(v) = 0
(Old Equations)
First Integral!
Same Natural B.C.’s & Lagrange Multipliers
€
1
v1
−κ1 cosϕ1 − λ1 sinϕ1 + μ1 −gsinϕ1 + R(v1)( ) = 0
μ1v1 = 0
€
κ(ϕ ) =cosϕ1
v1
, λ (ϕ ) =sinϕ1
v1
,
μ(ϕ ) =sin(ϕ −ϕ1)
gv1 cosϕ
Solve for v()
• Using Lagrange Multipliers and First Integral:
• Obtain:
€
d
dϕ
1
v−κ cosϕ − λ sinϕ + μ(−gsinϕ + R(v))
⎛
⎝ ⎜
⎞
⎠ ⎟= 0
€
1
v−
cosϕ1
v1 cosϕ+
sin(ϕ −ϕ1)
gv1 cosϕR(v) = 0
Solve for Initial Angle 0
• Evaluate at 0:
• Obtain Implicit Equation for
initial slope angle:€
1
v−
cosϕ1
v1 cosϕ+
sin(ϕ −ϕ1)
gv1 cosϕR(v) = 0
€
cosϕ 0 −v0
v1
+ v0R(v0)sin(ϕ 0 −ϕ1)
gv1
= 0
Solving for f()
• Rearrange E-L equation:
• Obtain ODE:
( Recall that we already have v(), 0, & initial condition f(0) = 0 )
€
˙ f
v 2− μ ˙ v + ˙ f
dR
dv
⎛
⎝ ⎜
⎞
⎠ ⎟+
d μv( )dϕ
= 0
€
˙ f (ϕ ) =μv 3
μv 2 dR
dv−1
Solving for x() and y()
• Integrate the E-L equations
• Obtain
€
˙ x − ˙ f cosϕ = 0
˙ y − ˙ f sinϕ = 0
€
x(ϕ ) = x0 + ˙ f (ϑ )cosϑ dϑϕ 0
ϕ
∫
y(ϕ ) = y0 + ˙ f (ϑ )sinϑ dϑϕ 0
ϕ
∫
Seems like we are done...
• What about parameters 1 & v1?
Appear everywhereeverywhere, due to:
• How can we solve for them?€
κ(ϕ ) =cosϕ1
v1
, λ (ϕ ) =sinϕ1
v1
, μ(ϕ ) =sin(ϕ −ϕ1)
gv1 cosϕ
Newton’s Method...
• Use the equations for x() and y() and the corresponding boundary conditions:
• Now we really are done!
€
M(ϕ1,v1) = x0 + ˙ f (ϑ ,ϕ1,v1)cosϑ dϑϕ 0
ϕ 1∫ − x1 = 0
N(ϕ1,v1) = y0 + ˙ f (ϑ ,ϕ1,v1)sinϑ dϑϕ 0
ϕ
∫ − y1 = 0
Example: Air Resistance
Example: Air Resistance
• Take R(v) = k v(k - coefficient
of viscous friction)– Newtonian fluid– first order approx. for air resistance
• Let x0 = 0, y0 = 0, v0 = 0,
€
cosϕ 0 −v0
v1
+ v0R(v0)sin(ϕ 0 −ϕ1)
gv1
= 0
cosϕ 0 −v0
v1
+ kv02 sin(ϕ 0 −ϕ1)
gv1
= 0 0 = /2
Solve for v()
€
1
v−
cosϕ1
v1 cosϕ+
sin(ϕ −ϕ1)
gv1 cosϕR(v) = 0
1−cosϕ1
v1 cosϕv +
sin(ϕ −ϕ1)
gv1 cosϕkv 2 = 0
€
v =g
2k
cosϕ1
sin(ϕ −ϕ1)1− 1− 4
kv1
g
sin(ϕ −ϕ1)cosϕ
cos2 ϕ1
⎛
⎝ ⎜
⎞
⎠ ⎟
Quadratic Formula:
( take the negative root to satisfy v(0) = 0 )
Many Calculations...
Results:
(Straight Line)
(Cycloid)
Conclusions
• Different approach to the Brachistochrone – parametrization by the slope angle – use of Lagrange Multipliers
• Gained:– analytical solution for non-conservative
velocity-dependent frictional force
• Lost ( due to definition s = f() ):– ability to descibe free-fall and cyclic motion
Questions ?