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Conserved quantities
If a pendulum swings back and forth without friction, its total energy is conserved. When the pendulum is at its heighest point, its kinetic energy is zero and
its potential energy is large. As it swings to its lowest point, the kinetic energy is
high and the potential energy is lower. As we will see below, the conservation of
energy can help us understand the motion of the pendulum.
A conserved quantity for an ODE, like the total energy, is a quantity that doesn’t
change over time. For example, consider the system of linear differential equations
x0 (t) = y(t)
y 0 (t) = −x(t).
For this system, the quantity x2 (t) + y 2 (t) is conserved. (You might want to take
a minute to think what this means geometrically.)
We can see that x2 (t) + y 2 (t) doesn’t change over time by computing its derivative:
d
2
2
0
0
dt [x + y ] = 2xx + 2yy .
Plugging in our equations for x0 and y 0 , we get
2xx0 + 2yy 0 = 2xy + 2y(−x) = 0.
Hence the quantity x2 + y 2 is constant in time. Geometrically, this means that
each solutions travels along a circle in the x − y plane. For example, if the initial
data lies on the circle of radius 10 around the origin, then the solution stays on this
circle for all time.
We also know how to solve this linear system. The general solution is given by
the formula
C1 (cos t, − sin t) + C2 (sin t, cos t).
One solution is 10(cos t, − sin t). As t increases, the solution moves around a
circle of radius 10. We can see this by solving the equation, but it also follows
because the quantity x2 + y 2 is conserved.
Sample problem. Consider the differential equation x0 = Ax, where A is the
matrix below.
A=
0 4
−1 0
.
Show that the quantity x2 + 4y 2 is conserved.
Conserved quantites don’t just appear in linear systems. Here’s another example.
x0 = − sin x + y + cos z
y 0 = sin x − sin y
z 0 = −y + sin y − cos z.
In this system, the quantity x+y+z is conserved. In other words, x(t)+y(t)+z(t)
doesn’t change over time. Geometrically that means that each solutions travels in
a plane of the form x + y + z = C.
To check that x(t) + y(t) + z(t) doesn’t change over time, we differentiate it.
d
dt [x + y + z] = (− sin x + y + cos z) + (sin x − sin y) + (−y + sin y − cos z) = 0.
Knowing about a conserved quantity is not as good as solving the differential
equation, but it gives us some information about the solution. Suppose we try to
solve the equation above with initial data x(0) = 0, y(0) = 1, z(0) = 1. It looks
1
2
to me very hard to solve the equation and hence very hard to find x(1), y(1), and
z(1). But, because of our conservation law, we know that x(1) + y(1) + z(1) =
x(0) + y(0) + z(0) = 2. So we get some information about x(1), y(1), and z(1).
Sample Problem. Suppose that x0 = cos y and y 0 = −2x cos y. Check that the
quantity x2 + y is conserved.
Here are some questions you might be wondering about: Does every system of
differential equations have a conserved quantity? Could it have several? If it does
have any, how can we find them?
Here are the answers. Some systems have no conserved quantities. Some have
a conserved quantity, and some have more than one. I don’t know any systematic
way to find them. If a problem comes from physics, it may be possible to find
conserved quantities using physical intuition. We’ll talk more about that below.
Other than that, the only way I know to find conserved quantities is trial and error.
We won’t ask you to find conserved quantities out of the blue.
The closest we might get is to give you the form of a conserved quantity with
some parameters in it and ask you to choose parameters to make the quantity
conserved. Even this kind of problem is at the extra credit level.
Sample extra credit problem: Suppose that x solves the ODE x0 = Ax, where
A is the matrix below.
A=
1
5
−2 −1
.
We compute the eigenvalues of A and find they are ±3i. Therefore, each solution
of the differential equation travels along a closed curve, returning infinitely often
to its initial value. In fact, each solution traces out an ellipse in the x1 − x2 plane.
Suppose we would like to find the equation for the ellipse. To do that, consider
the quantity ax21 + bx1 x2 + cx22 . Figure out how to choose a, b, and c so that
ax21 + bx1 x2 + cx22 does not change over time.
Now we turn to physical systems like springs and pendulums. In a system
modeled from physics, there is often a conserved energy when our model doesn’t
include friction. (We’ll talk about friction in a later section.)
Consider a spring described by the second-order ODE
x00 = −k 2 x.
This equation describes a spring with mass 1 and spring constant k.
The kinetic energy of the spring is (1/2)(x0 )2 . This formula may be familiar from
physics.
As we stretch the spring, we increase its potential energy, which is (1/2)k 2 x2 .
So the total energy is E = (1/2)(x0 )2 + (1/2)k 2 x2 .
The expression for E was essentially given to us by physicists, who found it using
physical intuition. Mathematically, we can check that E is conserved.
d
d
0 2
2 2
0 00
2
0
dt E = dt [(1/2)(x ) + (1/2)k x ] = x x + k xx .
00
Now we use the differential equation for x .
x0 x00 + k 2 xx0 = x0 (−k 2 x) + k 2 xx0 = 0.
This shows that the total energy of the spring is conserved.
For example, suppose we look at the initial value problem x00 = −4x, x(0) =
1, x0 (0) = 0. At time 0, the energy is E = (1/2)(x0 (0))2 + (1/2)4(x(0))2 = 2.
3
Hence the energy will be equal to 2 at all times. For example, if at any time t, the
spring reaches the origin, so that x(t) = 0, then we find that (x0 (t))2 = 4, and so
x0 (t) = ±2.
Instead of using conservation of energy, we can also solve this initial value problem as we did in class. The solution is x(t) = cos(2t). We find that x(t) = 0
when t = π/4, and again at t = π/4 + π/2, and again at intervals of π/2. Also,
x0 (t) = −2 sin(2t). Therefore, at time t = π/4, when the spring comes back to the
origin for the first time, it has velocity x0 (π/4) = −2. This result agrees with what
we saw using conservation of energy.
Conservation of energy applies to many non-linear problems involving springs or
pendulums. For example, consider the pendulum governed by
θ00 = − sin(θ).
In this equation θ is the angle the pendulum makes; θ = 0 means the pendulum
is pointing straight down.
The kinetic energy of the pendulum is (1/2)(θ 0 )2 .
The potential energy can be found using physical reasoning. The force that pulls
the pendulum down is gravity, and so the potential energy is proportional to how
high the pendulum is. The height of the pendulum is 1 − cos θ. We included this
paragraph to make the formula for the total energy look plausible, but you’re not
responsible for this kind of physical reasoning.
Total energy E = (1/2)(θ 0 )2 + (1 − cos θ).
We check that E is conserved.
d
d
0 2
0 00
0
dt E = dt [(1/2)(θ ) + (1 − cos θ)] = θ θ + sin(θ)θ .
00
Again, we use the differential equation for θ .
θ0 θ00 + sin(θ)θ0 = θ0 [− sin(θ)] + sin(θ)θ 0 = 0.
For example, if we drop the pendulum with initial conditions θ(0) = π/4, θ 0 (0) =
0. We can use conservation of energy to determine what its velocity will be when
it reaches a vertical position (θ = 0). To √
do this, note that its energy at time
0 is (1/2)02 + (1 − cos(π/4)) = 1 − (1/2) 2. If θ(t) = 0, then at time t the
energy will be√(1/2)(θ 0 (t))2 + (1 − cos 0) = (1/2)(θ 0 (t))2 . Therefore, we see that
θ0 (t) = ±(2 − 2)1/2 .
Remark: For this non-linear problem, I don’t think there is any explicit formula
for the solution θ(t). But conservation of energy still allows us to say some useful
things about the solution.
Sample problem. Suppose that x obeys the following nonlinear equation.
x00 = −x − x3 .
This equation models a “hard” spring. As x moves away from its equilibrium
position x = 0, the force pulling it back increases faster than linearly.
Check that the energy E = (1/2)(x0 )2 + (1/2)x2 + (1/4)x4 is conserved.
Suppose we solve x00 = −x − x3 with initial values x(0) = 2, x0 (0) = 0. When
the spring reaches x = 0 for the first time, what will its velocity be?
Friction
If we consider a spring with friction, then the energy is no longer conserved.
Instead, the friction gradually saps the energy. We describe the situation mathematically, beginning with a linear spring.
x00 = −(1/10)x0 − x.
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The term −(1/10)x0 on the right-hand side is the force of friction.
The energy of the system is defined to be E(t) = (1/2)(x0 )2 + (1/2)x2 . If x
obeyed the frictionless equation x00 = −x, then this energy would be conserved.
Because of the friction term, the energy is not conserved but decreases. To verify
d
E.
that the energy decreases, we compute dt
d
d
0 2
2
0 00
0
dt E = dt [(1/2)(x ) + (1/2)x ] = x x + xx .
00
As usual, we use the equation for x .
x0 x00 + xx0 = x0 (−(1/10)x0 − x) + xx0 = −(1/10)(x0 )2 .
So we see that E 0 (t) is always less than or equal to zero. Also, the faster the
velocity of the object, the more negative E 0 is - the faster the object is going, the
more friction takes energy out of the system.
Sample problem. Hard spring with friction.
Suppose that x00 = −x−x3 −(1/5)x0 . Show that the “energy” E(t) = (1/2)(x0 )2 +
(1/2)x2 + (1/4)x4 is decreasing over time.
Philosophical summary. Conserved quantities are an important mathematical
tool for understanding differential equations. It often happens that a differential
equation is hard or impossible to solve and yet still possesses a simple conserved
quantity. The conserved quantity gives us useful information about the solutions,
even though it usually doesn’t give us enough information to find the solutions
exactly.
From the physical point of view, the conservation of energy and other conservation laws are even more important than in mathematics. Physicists regard these
laws as fundamental features of the universe, more fundamental than any particular differential equation. For example, Einstein’s theory of general relativity refines
Newton’s theory of gravity. General relativity gives (slightly) different predictions
from the Newtonian theory, and it has a very different mathematical formulation.
Nevertheless, in the theory of general relativity, there is still a law of conservation of
energy. Another example is given by quantum mechanics - a physical theory that
describes subatomic particles such as elections, protons, etc. The theory is very
strange at the first encounter (and second and third...). In quantum mechanics,
an electron does not even have any definite position or velocity. The mathematical
formulation of quantum mechanics is very different from anything in 19th century
physics and also very different from general relativity. But in quantum mechanics
too, there is a law of conservation of energy, which plays a central role in the theory.