Transcript Document

Chapter day 4
Differential equations
Recall from AP Calculus
The number of rabbits in a population increases at a rate
that is proportional to the number of rabbits present (at
least for awhile.)
So does any population of living creatures. Other things
that increase or decrease at a rate proportional to the
amount present include radioactive material and money in
an interest-bearing account.
If the rate of change is proportional to the amount present,
the change can be modeled by:
dy
 ky
dt

dy
 ky
dt
1
dy  k dt
y
1
 y dy   k dt
Rate of change is proportional
to the amount present.
Divide both sides by y.
Integrate both sides.
ln y  kt  C

1
 y dy   k dt
Integrate both sides.
ln y  kt  C
ln y
e
e
kt C
y  e e
C
kt
Exponentiate both sides.
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.

ln y
e
e
kt C
y  e e
C
kt
Exponentiate both sides.
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.
y  e e
C kt
y  Ae
kt
Since
 eC
is a constant, let  e
C
A.

y  e e
C kt
y  Ae
kt
y0  Ae
Since
 eC
is a constant, let  e
C
A.
1
k 0
At
t  0 , y  y0
.
y0  A
y  y0 e
kt
This is the solution to our original initial
value problem.

So if we start with:
We end with:
dy
 ky
dt
y  y0 e
kt

What if we have a series of
differential equations?
dy1 = ky1
dt
dy2 = ky2
dt
dy3 = ky3
dt
We could solve of these individually
y1 =c1ekt
y2 =c2ekt
y3 =c3ekt
Provided that we have initial
conditions for each of these to solve
for the constants
If we define x
x’(t) =
[ ]
x1’(t)
x2’(t)
…
xn ’(t)
This yields the equation x’(t)= Ax
Which is easy to solve in the case of a
diagonal matrix.
[ ] [ ][ ]
x’1
x’2
x’3
=
3 0 0
0 -2 0
0 0 4
x1
x2
x3
We can solve each of these as a separate differential equation
x1’ = 3x1, x2’ = -2x2,
x3’ = 4x3
x1 (t) = b1e3t,
x2 (t) = b2e-2t,
x3 (t) = b3e4t,
This is the general solution. We can solve for the constants if given an
initial condition.
First order homogeneous linear
system of differential equations
x1’(t) = a11x1 (t) + a12 x2 (t) + … a1nxn (t)
x2’(t) = a21x1 (t) + a22 x2 (t) + … a2nxn (t)
…
xn ’(t) = an1x1 (t) + an2 x2 (t) + … annxn (t)
We could write this in matrix form as:
x1’ (t)
a11 a12 …. a12
x(t) = x2’ (t)
A = a21 a 22 … a2n
xn’ (t)
…
…
[] [ ]
an1 a n2 …anm
What if our system is not diagonal?
The system at the left can be written
as du/dt = Au with a as
du1 = -u1 +2 u2
dt
[ ]
A = -1 2
1 -2
du2 = u1 – 2u2
dt
[]
Initial condition u(0) =
1
0
How can we solve this system?
du/dt =Au
y= eAt
u(t) = c1eλ1 t x1 +c2e λ2 t x2+…+ cneλ n t xn
Check that each piece solves the given system
du/dt =Au
d (eλ t x1) = A eλ t x1
dt
λeλ t x1 = A eλ t x1
λx1 =Ax1
Key Formulas
Difference Equations
Differential Equations
du/dt =Au
y= eAt
Solve the differential equations
The system at the left can be written
as du/dt = Au with a as
[ ]
A = -1 2
1 -2
Start by computing the
eigenvalues and eigenvectors
What are the eigenvalues from inspection?
Hint: A is singular
The trace is -3
Solve the differential equations
Step 1 find the eigenvalues and
eigenvectors
We can a solve via finding the
determinant of A - λI
By inspection: the matrix is singular
det -1-λ 2
therefore 0 is an eigenvalue the trace is
1 -2-λ
-3 therefore the other eigenvalue is -3
[ ]
Calculate the eigenvector associated with λ = 0,-3
[ ]
[ ]
A = -1 2
1 -2
A+ 3I = 2
1
2
1
For λ = 0 find a basis for the kernel of A
[]
[]
2
1
For λ= -3 find a basis for the kernel of A+3I
1
-1
Solve the differential equations
The system at the left can be written
as du/dt = Au with a as
Note: the solutions of the equations
are going to be e raised to a
power.
[ ]
A = -1 2
1 -2
The form that we are expecting for the answer is
y = c1 e λ t x1 + c2 e λ t x2
1
2
The eigenvalues are already telling us about the form of the solutions
A negative eigenvalue will mean that that portion goes to zero as x
goes to infinity. An eigenvalue of zero will mean that we will have an
e0 which will be a constant. We will call this type of system a steady
state.
Solve the differential equations
Solve by plugging in eigenvalues into expected equation and for λ1 and
λ2. and the corresponding eigenvectors in x1 and x2
[ ]
A = -1 2
1 -2
y = c1 e0t 2 + c2 e -3t 1
1
-1
[]
We find c1 and c2 by
using the initial condition
Recall:
Initial condition u(0) = 1
0
c1 = 1/3
c2 = 1/3
[]
Plugging in zero for t
and the initial conditions yields:
[] [] [ ] [ ]
1
0
= c1
2
1
+ c2 1
-1
Solve the differential equations
The general solution is
y = 1/3
2 + 1/3 e -3t
1
[]
[]
1
-1
We are interested in hat happens as time goes to infinity
Recall our initial condition was 1 all of our quantity was in u1
0
[]
Then as time progressed there was flow from u1 to u2. As time
approaches infinity we end with the steady state 2/3
1/3
[]
The solution to y’ = ky is y = y0ekt
The solution to x’ = Au
is u = c0eAt
Applications of Linear Algebra
• Flow of water, electricity or money through
a net work that continues over time.
Applications of Differential Equations
• Differential Equations are the language in
which the laws of nature are expressed.
Understanding properties of solutions of
differential equations is fundamental to
much of contemporary science and
engineering. Ordinary differential equations
(ODE's) deal with functions of one variable,
which can often be thought of as time
- MIT
http://en.wikipedia.org/wiki/Differential
_equation
• Many fundamental laws of physics and chemistry can be formulated
as differential equations. In biology and economics, differential
equations are used to model the behavior of complex systems. The
mathematical theory of differential equations first developed together
with the sciences where the equations had originated and where the
results found application. However, diverse problems, sometimes
originating in quite distinct scientific fields, may give rise to identical
differential equations. Whenever this happens, mathematical theory
behind the equations can be viewed as a unifying principle behind
diverse phenomena. As an example, consider propagation of light and
sound in the atmosphere, and of waves on the surface of a pond. All of
them may be described by the same second-order partial differential
equation, the wave equation, which allows us to think of light and
sound as forms of waves, much like familiar waves in the water.
Conduction of heat, the theory of which was developed by Joseph
Fourier, is governed by another second-order partial differential
equation, the heat equation. It turned out that many diffusion
processes, while seemingly different, are described by the same
equation; the Black–Scholes equation in finance is, for instance,
related to the heat equation.
All of the following are stated in terms
of differential equations
•
•
•
•
•
•
•
•
•
Newton’s second law of dynamics
Euler –Lagrange theorem Classical mechanics
Radioactive decay – nuclear physics
Newton’s Law of cooling – Thermodynamics
Maxwell’s equation – electro magnetism
Einstein’s field equation – General relativity
The Shroedinger equation – Quantum mechanics
Mathusian growth model – Economics
Verhulst equation – biological population growth
•
http://en.wikipedia.org/wiki/Differential_equation
• http://ocw.mit.edu/courses/mathematics/18-03differential-equations-spring-2010/
Homework: wkst 8.4 1-9 odd, 2 and 8
What if the matrix is not diagonal?
• White book p. 520 ex 3, 4, 5