# Equation of motion through the Lagrangian with Lagrange multipliers

I ask for advice, I'm a little confused. I have such a Lagrangian.

$$L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-\lambda(2x^2+3y^2-1^2)$$

Here $$\lambda(2x^2+3y^2-1^2)$$ is the constraint on the phase variables.

I need to derive the equation of motion given the constraints and solve them numerically with the help of NDSolve.

We do this in accordance with the classic formula:

$$\frac{d}{dt}(\frac{dL}{d\dot{q}})-\frac{dL}{dq}=0$$

Where $$q=[x,y]$$ are generalized coordinates. I'm not sure about the Lagrange multiplier as a generalized coordinate.

Clear["Derivative"]

ClearAll["Global*"]

T = 1/2 m (x'[t]^2 + y'[t]^2);(*Kinetic Energy*)

f = \[Lambda] (2 x[t]^2 + 3 y[t]^2 - 1^2);(*Constraint*)

L = T - f;(*Lagrangian*)

D[D[L, x'[t]], t] - D[L, x[t]];

D[D[L, y'[t]], t] - D[L, y[t]];

D[D[L, \[Lambda]'[t]], t] - D[L, \[Lambda][t]];


Question: how are the Lagrange multipliers included in this system when compiling the ODE system and numerically solving it?

EDIT (with correct initial conditions and another constraint):

Clear["Derivative"]

ClearAll["Global*"]

q = Function[t, {x[t], y[t]}];(*coordinates*)

zwang = x[t] + x[t] y[t] + y[t] - 1;(*constraints*)

Solve[zwang == 0, y[t]];

Q = \[Lambda][t] D[zwang, {q[t]}];(*gen.forces*)

L = 1/2 (x'[t]^2 + y'[t]^2);

eqn = D[D[L, {q'[t]}], t] -
D[L, {q[t]}] - \[Lambda][t] D[zwang, {q[t]}];

sol = Solve[
Join[eqn, {D[zwang, {t, 2}]}] == 0, {x''[t],
y''[t], \[Lambda][t]}][[1]];

x0 = 1/2;

y0 = 1/3;

s = NDSolve[{sol[[1, 1]] == sol[[1, 2]], sol[[2, 1]] == sol[[2, 2]],
sol[[3, 1]] == sol[[3, 2]], x[0] == x0, y[0] == y0, x'[0] == 0,
y'[0] == 0}, {x, y, \[Lambda]}, {t, 500}];

Plot[Evaluate[zwang /. s], {t, 0, 500}, PlotRange -> All]

• [Lambda] is a constant, the derivative would be zero. Instead you need to use the equation of the constrain: x^2+y^2-1==0 Oct 14, 2021 at 8:14
• @DanielHuber Yes, I agree. As for the second remark, constraint equation does not contain derivatives. This means that the system of ODE will include algebraic constraints, with such systems, as far as I know, acceptable in Mathematica. Is that correct?
– dtn
Oct 14, 2021 at 8:17
• Which of the constraints is the correct one: 2 x[t]^2 + 3 y[t]^2 ==1^2 or x[t]^2 + y[t]^2 ==1^2 ? Oct 14, 2021 at 8:24
• Have a look at: farside.ph.utexas.edu/teaching/336k/lectures/node90.html Oct 14, 2021 at 9:03
• @DanielHuber Thanks for this very interesting link! Oct 14, 2021 at 12:35

As far as I remember Lagrangeformalism (classic form) is only valid for generalized coordinates, which fullfill the constraints . In your example x[t],y[t] aren't generalized coordinates!

For the constraint 2 x[t]^2 + 3 y[t]^2 ==1^2 possible generalized coordinate q[t] would be x[t]->Cos[q[t]]/Sqrt[2],y[t]->Sin[q[t]]/Sqrt[3]

2 x[t]^2 + 3 y[t]^2 ==1^2 /.{x[t]->Cos[q[t]]/Sqrt[2],y[t]->Sin[q[t]]/Sqrt[3]}// Simplify
(*True*)

L= 1/2 m (3 x'[t]^2 + y'[t]^2) /. {x -> ( Cos[q[#]]/Sqrt[2] &), y  -> (Sin[q[#]]/Sqrt[3] &)} //Simplify
(*-(1/24) m (-11 + 7 Cos[2 q[t]]) Derivative[1][q][t]^2*)


The equations of motion follow to

D[D[L, q'[t]], t] - D[L, q[t]]==0
(*7/12 m Sin[2 q[t]] Derivative[1][q][t]^2 -1/12 m (-11 + 7 Cos[2 q[t]]) (q^\[Prime]\[Prime])[t]*)


Hope it helps!

addendum Lagrange with constraints (thanks @DanielHuber's comment )

q = Function[t, {x[t], \[CurlyPhi][t]}]; (*coordinates*)
zwang = x[t] - r \[CurlyPhi][t] (* constraints *)
Q = \[Lambda][t] D[zwang, {q[t]}] (* gen. forces*)

L = 1/2 m x'[t]^2 + 1/2 \[Theta] \[CurlyPhi]'[t]^2 +m g Sin[\[Alpha]]  x[t]
eqn = D[D[L, {q'[t]}], t] -D[L, {q[t]}] - \[Lambda][t] D[zwang, {q[t]}]


From equations eqn and zwang the Lagrange multiplier  \[Lambda][t]is evaluated, substitution gives the equations of motion

sol = Solve[Join[eqn, {D[zwang, {t, 2}]}] ==0, {x''[t], \[CurlyPhi]''[t], \[Lambda][t]}][[1]]
(*{(x^\[Prime]\[Prime])[t] -> (g m r^2 Sin[\[Alpha]])/(m r^2 + \[Theta]), (\[CurlyPhi]^\[Prime]\[Prime])[t] -> (g m r Sin[\[Alpha]])/(m r^2 + \[Theta]), \[Lambda][t] -> -((g m \[Theta] Sin[\[Alpha]])/(m r^2 + \[Theta]))}*)

Q /. sol (* Reibkraft und -moment*)


• I answered the question you asked for. In this new example substitute x[t]-> (1 - y[t]^2)/(1 + 3 y[t])) , Lagrangian follows to L=(m (5 + 12 y[t] + 38 y[t]^2 + 60 y[t]^3 + 45 y[t]^4) Derivative[1][y][ t]^2)/(1 + 3 y[t])^4 with generalized coordinate y[t] Oct 14, 2021 at 8:53
• I tried to solve the constraints Solve[x + 3 x y + y^2 \[Minus] 1 == 0, x]. Oct 14, 2021 at 8:55
• My modified answer shows how to proceed! You could check your results on your own I think. Oct 16, 2021 at 11:04
• In your edit the initial conditions doesn't fullfill the constraint: zwang /. t -> 0 /. {x[0] -> 1/2, y[0] -> 1/4} (*-1/8 !=0 *) Oct 18, 2021 at 8:44
• Fine! But still in your edit you are using wrong initial conditions... Oct 18, 2021 at 9:20