Energy Considerations

Vibration ODEs

1.6 Energy Considerations

solutions are just the real part of the expressions in polar form:

uⁿDI.1C!²t²/ⁿ⁼²cos.ntan¹.!t //; (1.44) vⁿD !I.1C!²t²/ⁿ⁼²sin.ntan¹.!t // : (1.45) The expression .1C!²t²/ⁿ⁼²causes growth of the amplitude, since a number greater than one is raised to a positive exponentn=2. We can develop a series expression to better understand the formula for the amplitude. Introducingp D

!tas the key variable and usingsympygives

>>> from sympy import *

>>> p = symbols(’p’, real=True)

>>> n = symbols(’n’, integer=True, positive=True)

>>> amplitude = (1 + p**2)**(n/2)

>>> amplitude.series(p, 0, 4) 1 + n*p**2/2 + O(p**4)

The amplitude goes like1C¹

2n!²t², clearly growing linearly in time (withn).

We can also investigate the error in the angular frequency by a series expansion:

>>> n*atan(p).series(p, 0, 4) n*(p - p**3/3 + O(p**4))

This means that the solution foruⁿcan be written as uⁿD

1C1

2n!²t²C_O.t⁴/

cos

!t1

3!tt²C_O.t⁴/

: The error in the angular frequency is of the same order as in the scheme (1.7) for the second-order ODE, but the error in the amplitude is severe.

byu⁰and integrating from0toT: ZT 0

u⁰⁰u⁰dtC ZT 0

!²uu⁰dtD0 :

Observing that

u⁰⁰u⁰D d dt

1

2.u⁰/²; uu⁰D d dt

1 2u²; we get

ZT 0

d dt

1

2.u⁰/²C d dt

1 2!²u²

dtDE.T /E.0/D0;

where we have introduced

E.t /D 1

2.u⁰/²C1

2!²u²: (1.46)

The important result from this derivation is that the total energy is constant:

E.t /DE.0/ : E.t/is closely related to the system’s energy

The quantityE.t /derived above is physically not the mechanical energy of a vibrating mechanical system, but the energy per unit mass. To see this, we start with Newton’s second lawF Dma(F is the sum of forces,mis the mass of the system, anda is the acceleration). The displacementuis related toathrough aDu⁰⁰. With a spring force as the only force we haveF D ku, wherekis a spring constant measuring the stiffness of the spring. Newton’s second law then implies the differential equation

kuDmu⁰⁰ ) mu⁰⁰CkuD0 :

This equation of motion can be turned into an energy balance equation by finding the work done by each term during a time intervalŒ0; T . To this end, we multiply the equation byduDu⁰dtand integrate:

ZT 0

muu⁰dtC ZT

0

kuu⁰dtD0 :

The result is

E.t /Q DE_k.t /CE_p.t /D0;

where

Ek.t /D 1

2mv²; vDu⁰; (1.47)

is thekinetic energyof the system, and E_p.t /D 1

2ku² (1.48)

is thepotential energy. The sumE.t /Q is the total mechanical energy. The deriva- tion demonstrates the famous energy principle that, under the right physical circumstances, any change in the kinetic energy is due to a change in potential energy and vice versa. (This principle breaks down when we introduce damping in the system, as we do in Sect.1.10.)

The equationmu⁰⁰CkuD0can be divided bymand written asu⁰⁰C!²uD0 for!Dp

k=m. The energy expressionE.t /D¹₂.u⁰/²C¹₂!²u²derived earlier is thenE.t /=m, i.e., mechanical energy per unit mass.Q

Energy of the exact solution Analytically, we haveu.t /DIcos!t, ifu.0/DI andu⁰.0/D0, so we can easily check the energy evolution and confirm thatE.t / is constant:

E.t /D 1

2I².!sin!t/²C1

2!²I²cos²!t D1

2!².sin²!tCcos²!t/D1 2!²: Growth of energy in the Forward Euler scheme It is easy to show that the energy in the Forward Euler scheme increases when stepping from time levelntonC1.

E^nC1D1

2.v^nC1/²C1

2!².u^nC1/² D1

2.vⁿ!²t uⁿ/²C1

2!².uⁿCt vⁿ/² D.1Ct²!²/Eⁿ:

1.6.2 An Error Measure Based on Energy

The constant energy is well expressed by its initial valueE.0/, so that the error in mechanical energy can be computed as a mesh function by

eⁿ_ED 1 2

u^nC1uⁿ¹ 2t

2

C1

2!².uⁿ/²E.0/; nD1; : : : ; Nt1; (1.49) where

E.0/D 1 2V²C1

2!²I²;

ifu.0/DI andu⁰.0/DV. Note that we have used a centered approximation to u⁰:u⁰.t_n/ŒD_2tuⁿ.

A useful norm of the mesh functione_Eⁿ for the discrete mechanical energy can be the maximum absolute value ofe_Eⁿ:

jje_Eⁿjj_`¹ D max

1n<N_tjeⁿ_Ej:

Alternatively, we can compute other norms involving integration over all mesh points, but we are often interested in worst case deviation of the energy, and then the maximum value is of particular relevance.

A vectorized Python implementation ofe_Eⁿ takes the form

# import numpy as np and compute u, t dt = t[1]-t[0]

E = 0.5*((u[2:] - u[:-2])/(2*dt))**2 + 0.5*w**2*u[1:-1]**2 E0 = 0.5*V**2 + 0.5**w**2*I**2

e_E = E - E0

e_E_norm = np.abs(e_E).max()

The convergence rates of the quantitye_E_normcan be used for verification.

The value ofe_E_normis also useful for comparing schemes through their ability to preserve energy. Below is a table demonstrating the relative error in total energy for various schemes (computed by thevib_undamped_odespy.pyprogram). The test problem isu⁰⁰C4²u D 0withu.0/ D 1andu⁰.0/ D 0, so the period is 1 andE.t /4:93. We clearly see that the Crank-Nicolson and the Runge-Kutta schemes are superior to the Forward and Backward Euler schemes already after one period.

Method T t maxˇˇe_Eⁿˇˇ=e⁰_E

Forward Euler 1 0.025 1:67810⁰

Backward Euler 1 0.025 6:23510¹

Crank-Nicolson 1 0.025 1:22110²

Runge-Kutta 2nd-order 1 0.025 6:07610³

Runge-Kutta 4th-order 1 0.025 8:21410³

However, after 10 periods, the picture is much more dramatic:

Method T t maxˇˇe_Eⁿˇˇ=e⁰_E

Forward Euler 10 0.025 1:78810⁴

Backward Euler 10 0.025 1:00010⁰

Crank-Nicolson 10 0.025 1:22110²

Runge-Kutta 2nd-order 10 0.025 6:25010²

Runge-Kutta 4th-order 10 0.025 8:28810³

The Runge-Kutta and Crank-Nicolson methods hardly change their energy error withT, while the error in the Forward Euler method grows to huge levels and a relative error of 1 in the Backward Euler method points toE.t / ! 0ast grows large.

Running multiple values oft, we can get some insight into the convergence of the energy error:

Method T t maxˇˇeⁿ_Eˇˇ=e⁰_E

Forward Euler 10 0.05 1:12010⁸

Forward Euler 10 0.025 1:78810⁴

Forward Euler 10 0.0125 1:37410²

Backward Euler 10 0.05 1:00010⁰

Backward Euler 10 0.025 1:00010⁰

Backward Euler 10 0.0125 9:92810¹

Crank-Nicolson 10 0.05 4:75610²

Crank-Nicolson 10 0.025 1:22110²

Crank-Nicolson 10 0.0125 3:12510³

Runge-Kutta 2nd-order 10 0.05 6:15210¹

Runge-Kutta 2nd-order 10 0.025 6:25010²

Runge-Kutta 2nd-order 10 0.0125 7:63110³

Runge-Kutta 4th-order 10 0.05 3:51010²

Runge-Kutta 4th-order 10 0.025 8:28810³

Runge-Kutta 4th-order 10 0.0125 2:05810³

A striking fact from this table is that the error of the Forward Euler method is re- duced by the same factor astis reduced by, while the error in the Crank-Nicolson method has a reduction proportional tot²(we cannot say anything for the Back- ward Euler method). However, for the RK2 method, halvingtreduces the error by almost a factor of 10 (!), and for the RK4 method the reduction seems propor- tional tot²only (and the trend is confirmed by running smaller time steps, so for tD3:910⁴the relative error of RK2 is a factor 10 smaller than that of RK4!).