Coupled vector Euler–Lagrange equations
Consider
J [ x , y ] = ∫ 0 T [ 1 2 ( x ˙ 2 + y ˙ 2 ) + α x y ˙ − ω 2 2 ( x 2 + y 2 ) ] d t , J[x,y]
=
\int_0^T
\left[
\frac{1}{2}
\left(
\dot{x}^2+\dot{y}^2
\right)
+
\alpha x\dot{y}
-
\frac{\omega^2}{2}
\left(
x^2+y^2
\right)
\right]
\,\mathrm{d}t, J [ x , y ] = ∫ 0 T [ 2 1 ( x ˙ 2 + y ˙ 2 ) + αx y ˙ − 2 ω 2 ( x 2 + y 2 ) ] d t , subject to
x ( 0 ) = x 0 , y ( 0 ) = y 0 , x ( T ) = x f , y ( T ) = y f . x(0)=x_0,
\qquad
y(0)=y_0,
\qquad
x(T)=x_f,
\qquad
y(T)=y_f. x ( 0 ) = x 0 , y ( 0 ) = y 0 , x ( T ) = x f , y ( T ) = y f . (a) Derive the Euler–Lagrange equation for x ( t ) x(t) x ( t ) .
(b) Derive the Euler–Lagrange equation for y ( t ) y(t) y ( t ) .
(c) Write the resulting equations in the matrix form
q ¨ + A q ˙ + B q = 0 , q = [ x y ] . \ddot{\boldsymbol{q}}
+
A\dot{\boldsymbol{q}}
+
B\boldsymbol{q}
=
\boldsymbol{0},
\qquad
\boldsymbol{q}
=
\begin{bmatrix}
x\\y
\end{bmatrix}. q ¨ + A q ˙ + B q = 0 , q = [ x y ] . (d) For α = 0 \alpha=0 α = 0 , solve the complete boundary-value problem.
Weighted minimum-slope trajectory
Determine the extremal of
J [ x ] = 1 2 ∫ 0 T e β t x ˙ 2 ( t ) d t J[x]
=
\frac{1}{2}
\int_0^T
e^{\beta t}\dot{x}^2(t)\,\mathrm{d}t J [ x ] = 2 1 ∫ 0 T e βt x ˙ 2 ( t ) d t subject to
x ( 0 ) = x 0 , x ( T ) = x f . x(0)=x_0,
\qquad
x(T)=x_f. x ( 0 ) = x 0 , x ( T ) = x f . (a) Derive the Euler–Lagrange equation.
(b) Solve explicitly for x ∗ ( t ) x^*(t) x ∗ ( t ) .
(c) Evaluate J [ x ∗ ] J[x^*] J [ x ∗ ] .
(d) Show that the solution approaches the straight-line path as
β → 0. \beta\rightarrow 0. β → 0. Free terminal state
Consider
J [ x ] = ∫ 0 T ( x ˙ 2 + ω 2 x 2 ) d t , J[x]
=
\int_0^T
\left(
\dot{x}^2+\omega^2x^2
\right)
\,\mathrm{d}t, J [ x ] = ∫ 0 T ( x ˙ 2 + ω 2 x 2 ) d t , subject to
x ( 0 ) = x 0 , x(0)=x_0, x ( 0 ) = x 0 , where T T T is fixed and x ( T ) x(T) x ( T ) is free.
(a) Derive the Euler–Lagrange equation.
(b) Derive the natural boundary condition at t = T t=T t = T .
(c) Solve the resulting boundary-value problem.
(d) Determine x ( T ) x(T) x ( T ) explicitly in terms of x 0 x_0 x 0 , T T T , and ω \omega ω .
Free final time
Consider
J [ x , t f ] = ∫ 0 t f [ 1 + 1 2 ( x ˙ − a t ) 2 ] d t , J[x,t_f]
=
\int_0^{t_f}
\left[
1+
\frac{1}{2}
\left(
\dot{x}-at
\right)^2
\right]
\,\mathrm{d}t, J [ x , t f ] = ∫ 0 t f [ 1 + 2 1 ( x ˙ − a t ) 2 ] d t , subject to
x ( 0 ) = 0 , x ( t f ) = x f , x(0)=0,
\qquad
x(t_f)=x_f, x ( 0 ) = 0 , x ( t f ) = x f , where t f t_f t f is free.
(a) Derive the Euler–Lagrange equation.
(b) Determine the general extremal x ( t ) x(t) x ( t ) .
(c) Derive the free-final-time condition.
(d) Reduce the unknown final time to a scalar algebraic equation.
(e) State how all positive admissible roots should be checked.
Mass–spring boundary-value problem and singular terminal times
For
L = 1 2 m x ˙ 2 − 1 2 k x 2 , L
=
\frac{1}{2}m\dot{x}^2
-
\frac{1}{2}kx^2, L = 2 1 m x ˙ 2 − 2 1 k x 2 , consider the endpoint conditions
x ( 0 ) = x 0 , x ( T ) = x f . x(0)=x_0,
\qquad
x(T)=x_f. x ( 0 ) = x 0 , x ( T ) = x f . (a) Derive the Euler–Lagrange equation.
(b) Solve the differential equation for m > 0 m>0 m > 0 and k > 0 k>0 k > 0 .
(c) Determine the constants in the general solution.
(d) Show that a unique solution exists when
sin ( k m T ) ≠ 0. \sin\left(\sqrt{\frac{k}{m}}\,T\right)\neq 0. sin ( m k T ) = 0. (e) Analyze separately the case
sin ( k m T ) = 0 \sin\left(\sqrt{\frac{k}{m}}\,T\right)=0 sin ( m k T ) = 0 and determine when the boundary-value problem has no solution or infinitely many solutions.
First variation and the natural boundary condition
Let
J [ x ] = ∫ t 0 t f L ( x , x ˙ , t ) d t , J[x]
=
\int_{t_0}^{t_f}
L(x,\dot{x},t)\,\mathrm{d}t, J [ x ] = ∫ t 0 t f L ( x , x ˙ , t ) d t , where t 0 t_0 t 0 , t f t_f t f , and x ( t 0 ) x(t_0) x ( t 0 ) are fixed, but x ( t f ) x(t_f) x ( t f ) is free.
Starting from
x ϵ ( t ) = x ( t ) + ϵ η ( t ) , x_\epsilon(t)
=
x(t)+\epsilon\eta(t), x ϵ ( t ) = x ( t ) + ϵη ( t ) , derive the complete first variation. Use integration by parts to show that stationarity requires
∂ L ∂ x − d d t ( ∂ L ∂ x ˙ ) = 0 \frac{\partial L}{\partial x}
-
\frac{\mathrm{d}}{\mathrm{d}t}
\left(
\frac{\partial L}{\partial\dot{x}}
\right)
=
0 ∂ x ∂ L − d t d ( ∂ x ˙ ∂ L ) = 0 and
∂ L ∂ x ˙ ∣ t f = 0. \left.
\frac{\partial L}{\partial\dot{x}}
\right|_{t_f}
=
0. ∂ x ˙ ∂ L ∣ ∣ t f = 0. Endpoint constrained to a curve
Let the initial endpoint be fixed and let the terminal endpoint satisfy
ψ ( x f , t f ) = x f 2 + c 2 t f 2 − R 2 = 0. \psi(x_f,t_f)
=
x_f^2+c^2t_f^2-R^2
=
0. ψ ( x f , t f ) = x f 2 + c 2 t f 2 − R 2 = 0. (a) Starting from
δ x f = δ x ( t f ) + x ˙ ( t f ) δ t f , \delta x_f
=
\delta x(t_f)
+
\dot{x}(t_f)\delta t_f, δ x f = δ x ( t f ) + x ˙ ( t f ) δ t f , derive the linearized endpoint constraint
ψ x δ x f + ψ t δ t f = 0. \psi_x\,\delta x_f
+
\psi_t\,\delta t_f
=
0. ψ x δ x f + ψ t δ t f = 0. (b) Compute ψ x \psi_x ψ x and ψ t \psi_t ψ t .
(c) Express δ x ( t f ) \delta x(t_f) δ x ( t f ) in terms of δ t f \delta t_f δ t f , assuming ψ x ≠ 0 \psi_x\neq 0 ψ x = 0 .
(d) Substitute this relation into the terminal contribution
∂ L ∂ x ˙ ∣ t f δ x ( t f ) + L ( t f ) δ t f \left.
\frac{\partial L}{\partial\dot{x}}
\right|_{t_f}
\delta x(t_f)
+
L(t_f)\delta t_f ∂ x ˙ ∂ L ∣ ∣ t f δ x ( t f ) + L ( t f ) δ t f and derive the corresponding scalar transversality relation.
Planar locus-to-locus formulation
Let
x ( t ) = [ x ( t ) y ( t ) ] . \boldsymbol{x}(t)
=
\begin{bmatrix}
x(t)\\y(t)
\end{bmatrix}. x ( t ) = [ x ( t ) y ( t ) ] . The initial state must lie on a circle of radius r 0 r_0 r 0 centered at
( a 0 , b 0 ) (a_0,b_0) ( a 0 , b 0 ) , and the final state must lie on a circle of radius r f r_f r f
centered at ( a f , b f ) (a_f,b_f) ( a f , b f ) .
(a) Write the two scalar endpoint constraints.
(b) Collect them in the vector form
ϕ ( x ( t 0 ) , t 0 , x ( t f ) , t f ) = 0 . \boldsymbol{\phi}
\bigl(
\boldsymbol{x}(t_0),
t_0,
\boldsymbol{x}(t_f),
t_f
\bigr)
=
\boldsymbol{0}. ϕ ( x ( t 0 ) , t 0 , x ( t f ) , t f ) = 0 . (c) Derive the first-order variations of both endpoint constraints.
(d) Identify the normal vector to each circle at the corresponding endpoint.
General fixed point-to-point boundary mapping
Express
x ( t 0 ) = x 0 , x ( t f ) = x f \boldsymbol{x}(t_0)=\boldsymbol{x}_0,
\qquad
\boldsymbol{x}(t_f)=\boldsymbol{x}_f x ( t 0 ) = x 0 , x ( t f ) = x f in the form
ϕ ( x ( t 0 ) , t 0 , x ( t f ) , t f ) = 0 . \boldsymbol{\phi}
\bigl(
\boldsymbol{x}(t_0),
t_0,
\boldsymbol{x}(t_f),
t_f
\bigr)
=
\boldsymbol{0}. ϕ ( x ( t 0 ) , t 0 , x ( t f ) , t f ) = 0 . Then:
(a) determine the dimension of ϕ \boldsymbol{\phi} ϕ when
x ∈ R n \boldsymbol{x}\in\mathbb{R}^n x ∈ R n ;
(b) compute the Jacobians of ϕ \boldsymbol{\phi} ϕ with respect to
x ( t 0 ) \boldsymbol{x}(t_0) x ( t 0 ) and x ( t f ) \boldsymbol{x}(t_f) x ( t f ) ;
(c) write the first-order variation δ ϕ \delta\boldsymbol{\phi} δ ϕ ;
(d) show that fixed endpoint states imply
δ x 0 = 0 \delta\boldsymbol{x}_0=\boldsymbol{0} δ x 0 = 0 and
δ x f = 0 \delta\boldsymbol{x}_f=\boldsymbol{0} δ x f = 0 .
Moving upper limit and free-final-time condition
Consider
J [ x , t f ] = ∫ t 0 t f L ( x , x ˙ , t ) d t , J[x,t_f]
=
\int_{t_0}^{t_f}
L(x,\dot{x},t)\,\mathrm{d}t, J [ x , t f ] = ∫ t 0 t f L ( x , x ˙ , t ) d t , with fixed t 0 t_0 t 0 , fixed x ( t 0 ) x(t_0) x ( t 0 ) , fixed terminal value
x ( t f ) = x f x(t_f)=x_f x ( t f ) = x f , and free t f t_f t f .
(a) Use the Leibniz rule to show that variation of the upper limit contributes
L ( t f ) δ t f . L(t_f)\delta t_f. L ( t f ) δ t f . (b) Derive the complete first variation.
(c) Use
δ x ( t f ) = − x ˙ ( t f ) δ t f \delta x(t_f)
=
-\dot{x}(t_f)\delta t_f δ x ( t f ) = − x ˙ ( t f ) δ t f to eliminate the terminal trajectory variation.
(d) Derive
[ L − x ˙ ∂ L ∂ x ˙ ] t f = 0. \left[
L
-
\dot{x}
\frac{\partial L}{\partial\dot{x}}
\right]_{t_f}
=
0. [ L − x ˙ ∂ x ˙ ∂ L ] t f = 0. (e) Show that this condition is equivalent to
H ( t f ) = 0 , H = x ˙ ∂ L ∂ x ˙ − L . H(t_f)=0,
\qquad
H
=
\dot{x}
\frac{\partial L}{\partial\dot{x}}
-
L. H ( t f ) = 0 , H = x ˙ ∂ x ˙ ∂ L − L .