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 ?