Download General solution method for 2x2 linear systems

General solution method
for 2x2 linear systems
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7/8/2012
To see how to do this, let's start with the general 2 × 2 system
of linear equations
dx
= a x+by
dt
where a, b, c, d ∈ \.
dy
= cx + d y
dt
Or, equivalently,
⎡ dx ⎤
⎢ dt ⎥ ⎡ a b ⎤ ⎡ x ⎤
⎢ ⎥=⎢
⎥⎢ ⎥
c
d
d
y
⎢ ⎥ ⎣
⎦ ⎣ y⎦
⎢⎣ dt ⎥⎦
or
G
G
X ′ = AX
G
G
The form X ′ = AX should remind you of differential equations
you have seen before. What were they?
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They were the first-order separable equations all having
exponential solutions.
This observation suggests we may expect the same behavior
from solutions to the matrix equation
dx
= a x+by
dt
dy
= cx + d y
dt
where a, b, c, d ∈ \.
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7/8/2012
Let's guess the solutions to the system to be
x(t ) = k1eλt and y(t ) = k2eλt , with unknowns
k1, k2 (not both zero) and λ , and see what happens.
G ⎡
λt ⎤
k
λ
e
G ⎡⎢ k1eλt ⎤⎥
dX ⎢ 1
⎥ and plug these into
=
Set X = ⎢
and
thus
dt ⎢ k λ eλt ⎥
k eλt ⎥
⎢⎣ 2
⎥⎦
⎢⎣ 2
⎥⎦
dx = a x + b y
dt
dy
= cx + d y
dt
to get
⎡ k λ eλt ⎤ ⎡
⎢ 1
⎥ = ⎢a
⎢
λt ⎥ ⎢ c
k
e
λ
⎢⎣ 2
⎥⎦ ⎣
λt
b ⎤⎥ ⎡⎢ k1e ⎤⎥ ⎡⎢a b ⎤⎥ λt ⎡⎢ k1 ⎤⎥
e ⎢ ⎥ or ...
=
⎢
⎥
λ
t
d ⎥⎦ ⎢k2e ⎥ ⎢⎣ c d ⎥⎦ ⎢⎣k2 ⎥⎦
⎣
⎦
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7/8/2012
... dividing out the exponential factor and switching left
and right-hand sides,
⎡a
⎢
⎢c
⎣
b ⎤⎥ ⎡⎢ k1 ⎤⎥ ⎡⎢ k1 ⎤⎥
= λ ⎢ ⎥ or
⎢
⎥
d ⎥⎦ ⎢⎣k2 ⎥⎦ ⎢⎣k2 ⎥⎦
AX = λ X , where X
⎡k ⎤
= ⎢ 1⎥
⎢⎣ k2 ⎥⎦
⎡a
and A = ⎢
⎢c
⎣
b ⎤⎥
.
d ⎥⎦
Our "guessed" solutions have allowed us to replace the
original differential equation problem with an algebraic one!
The purely algebraic equation, AX = λ X is THE key
equation in what follows!
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7/8/2012
We have ...
⎡a
⎢
⎢c
⎣
⎡k ⎤
⎡k ⎤ ⎡ ⎤
⎤
b ⎥ ⎢ 1 ⎥ ⎢ 1 ⎥ ⎢0⎥
−λ ⎢ ⎥ =
⎢
⎥
d ⎥⎦ ⎢k2 ⎥ ⎢k2 ⎥ ⎢⎣0⎥⎦
⎣ ⎦
⎣ ⎦
Since the right-hand-side is the zero vector, this is a
homogeneous algebraic system.
To proceed, we need some help from linear algebra.
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7/8/2012
Terminology ...
Given the 2x2 matrix
⎡a
A=⎢
⎢⎣ c
b⎤
⎥ we define the
d ⎥⎦
determinant of A as Det ( A) = ad − bc
and the trace of A as Tr ( A) = a + d
For simplicity we use D = Det ( A) and T = Tr ( A) .
In addition, we have the 2x2 identity matrix
I
⎡1
=⎢
⎢0
⎣
0⎤⎥
1⎥
⎦
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7/8/2012
We have ...
⎧⎡
⎪⎢a
⎨
⎪⎩ ⎢⎣ c
b ⎤⎥ ⎡⎢1 0⎤⎥ ⎫⎪ ⎡⎢ k1 ⎤⎥ ⎡⎢0⎤⎥
−λ
⎬⎢ ⎥ =
d ⎥⎦ ⎢⎣0 1⎥⎦ ⎪⎭ ⎢k2 ⎥ ⎢⎣0⎥⎦
⎣
⎦
or
⎡a − λ
⎢
⎢ c
⎣
⎡k ⎤ ⎡ ⎤
⎤
b ⎥ ⎢ 1 ⎥ ⎢0⎥
=
⎢
⎥
d − λ ⎥⎦ ⎢k2 ⎥ ⎢⎣0⎥⎦
⎣ ⎦
This homogeneous algebraic system always has the
trivial solution
⎡k ⎤ ⎡ ⎤
⎢ 1 ⎥ = ⎢0⎥
⎢ k ⎥ ⎢0⎥
⎢⎣ 2 ⎥⎦ ⎣ ⎦
But we're interested in non-trivial solutions ...
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7/8/2012
According to a theorem in linear algebra, this homogeneous
algebraic system has non-trivial solutions if and only if
⎡a − λ
Det ⎢
⎢ c
⎣
b ⎤⎥
= 0 ⇒ ( a − λ ) ( d − λ ) − bc = 0
d − λ ⎥⎦
This is a quadratic in λ :
or
λ 2 − ( a + d ) λ + ad − bc = 0
λ 2 −T λ + D = 0
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7/8/2012
The quadratic λ 2 − T λ + D = 0 is called the
characteristic equation of the matrix A.
The characteristic equation has solutions
T ± T 2 − 4D
λ=
2
These roots are called the eigenvalues of the matrix A.
So, by using the theorem regarding homogeneous algebraic
systems, we have been able to isolate λ and determine its
values independent of the unknowns k1 and k2
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7/8/2012
T + T 2 − 4D
T − T 2 − 4D
.
Let λ1 =
and λ2 =
2
2
Now, for each λ , we must find k1 and k2 which are the
components of the vector
⎡k ⎤
⎢ 1 ⎥.
⎢k ⎥
⎢⎣ 2 ⎥⎦
The vector
⎡k ⎤
⎢ 1⎥
⎢k ⎥
⎢⎣ 2 ⎥⎦
is called
an eigenvector associated with the eigenvalue λ.
So we must consider the two cases, λ1 and λ2 .
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7/8/2012
G ⎡⎢ k1 ⎤⎥
Start with λ1. Let X1 = ⎢ ⎥ . Substituting gives
⎢⎣ k2 ⎥⎦
⎡a − λ
1
⎢
⎢
⎢⎣ c
b ⎤⎥ ⎡⎢ k1 ⎤⎥ ⎡0⎤
⎢ ⎥.
=
d − λ1 ⎥⎥ ⎢⎢k2 ⎥⎥ ⎢⎣0⎥⎦
⎦⎣
⎦
This matrix equation can be written in algebraic form
( a − λ1 ) k1 + bk2 = 0 which we can solve for k and k .
1
2
ck1 + ( d − λ1 ) k2 = 0
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7/8/2012
⎡j ⎤
G
Then for λ2 . Let X 2 = ⎢ 1 ⎥ . Substituting gives
⎢j ⎥
⎣ 2⎦
⎡a − λ
2
⎢
⎢
⎢⎣ c
b ⎥⎤ ⎡ j1 ⎤ ⎡0⎤
⎢ ⎥ = ⎢ ⎥.
d − λ2 ⎥⎥ ⎢⎣ j2 ⎥⎦ ⎢⎣0⎥⎦
⎦
Again, this matrix equation can be written in algebraic form
( a − λ2 ) j1 + bj2 = 0 which we can solve for j and j .
1
2
cj1 + ( d − λ2 ) j2 = 0
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7/8/2012
We now have λ1
and λ2
G ⎡⎢ k1 ⎤⎥
and its associated eigenvectors X1 = ⎢ ⎥
⎢⎣ k2 ⎥⎦
⎡j ⎤
G
and its associated eigenvectors X 2 = ⎢ 1 ⎥ .
Note that in
⎢j ⎥
⎣ 2⎦
⎡k ⎤
⎢ 1⎥ k
⎢k ⎥ 1
⎢⎣ 2 ⎥⎦
and k2 are not both zero and in
⎡j ⎤
⎢ 1⎥
⎢j ⎥
⎣ 2⎦
j1 and j2 are not both zero.
G
G
Since each of X1 and X 2 represent an infinite number of vectors,
we select just one representative vector from each when forming the
general solution.
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7/8/2012
The general solution is a linear combination of the two
eigenvalue/eigenvector pairs ...
G
G
G
Y ( t ) = α1 exp ( λ1t ) ⋅ X1 + α 2 exp ( λ2t ) ⋅ X 2
where α1 and α 2 are constants of "integration".
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7/8/2012
IMPORTANT NOTE:
In the systems
( a − λ1 ) k1 + bk2 = 0
ck1 + ( d − λ1 ) k2 = 0
and
( a − λ2 ) j1 + bj2 = 0
cj1 + ( d − λ2 ) j2 = 0
we have selected the eigenvalues λ1 and λ2 to force the
determinant of each of the two systems to be zero.
As a result, if the eigenvalues have been determined correctly,
each system reduces to a single equation. If this is not true,
there is something wrong with the eigenvalues!
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7/8/2012
dx
= −2 x + y
dt
As an example, consider the system
dy
= −4 x + 3 y
dt
(*)
which can be written as a matrix equation
⎡ dx ⎤
⎢ dt ⎥ ⎡ −2
⎢ ⎥=⎢
⎢ dy ⎥ ⎣ −4
⎢⎣ dt ⎥⎦
1⎤ ⎡ x ⎤
⎥ ⎢ ⎥.
3⎦ ⎣ y ⎦
We want to find the general solution of this system.
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7/8/2012
From
⎡ dx ⎤
⎢ ⎥ ⎡ −2
⎢ dt ⎥ = ⎢
⎢ dy ⎥ ⎢⎣ −4
⎢ ⎥
⎣ dt ⎦
⎡ −2
We have A = ⎢
⎢⎣ −4
1⎤ ⎡ x ⎤
⎥ ⎢ ⎥.
3⎥⎦ ⎢⎣ y ⎥⎦
1⎤
⎥ and D = ( −2 )( 3) − ( −4 )(1) = −2 and
3⎥⎦
T = −2 + 3 = 1 in the characteristic equation λ 2 − T λ + D = 0.
To find the eigenvalues we have λ 2 − λ − 2 = ( λ − 2 )( λ + 1) = 0.
The eigenvalues are λ = 2 and λ = −1.
Now we must find the eigenvectors associated with the eigenvalues.
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7/8/2012
We must consider each eigenvalue in turn to find an
associated eigenvector. Start with the eigenvalue λ1 = 2.
Plug into the system
( −2 − 2) k1 + k2 = 0
−4k1 + (3 − 2) k2 = 0
−4k1 + k2 = 0
⇒
so the two
−4k1 + k2 = 0
equations collapse into one (as they should).
− 4k1 + k2 = 0 ⇒ k2 = 4k1
G ⎡ k1 ⎤
⎡1 ⎤
⎢
⎥
Our infinite set of eigenvectors is then X1 = ⎢ ⎥ = k1 ⎢ ⎥ .
⎢ 4⎥
4k
⎣ ⎦
⎢⎣ 1 ⎥⎦
Now since k1 is arbitrary but not zero we pick k1 = 1 to get
G ⎡1 ⎤
a representative eigenvector for this eigenvalue, X1 = ⎢ ⎥ .
⎢ 4⎥
⎣ ⎦
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7/8/2012
Now for the eigenvalue λ2 = −1.
Plug into the system ...
( −1) ⎞⎟⎠ j1 + j2 = 0 − j1 + j2 = 0
so again the two
⇒
−4 j1 + 4 j2 = 0
−4 j1 + ⎛⎜ 3 − ( −1) ⎞⎟ j2 = 0
⎝
⎠
⎛ −2 −
⎜
⎝
equations collapse into one (as they should).
− j1 + j2 = 0 ⇒ j2 = j1
G ⎡ j1 ⎤
⎡1⎤
Our infinite set of eigenvectors is then X 2 = ⎢ ⎥ = j1 ⎢ ⎥ .
⎢j ⎥
⎢1⎥
⎣ ⎦
⎣ 1⎦
Now since j1 is arbitrary and not zero we pick j1 = 1 to get
G ⎡1⎤
a representative eigenvector for this eigenvalue, X 2 = ⎢ ⎥ .
⎢1⎥
⎣ ⎦
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7/8/2012
By the linearity Principle we can now write down the general
solution for our system of differential equations.
We have:
⎛ ⎡ ⎤
⎞
⎛ ⎡ ⎤
⎞
G
G
G
G
1
1
Y (t ) = α1 X1 + α 2 X 2 or Y (t ) = α1 ⎜⎜ k1 ⎢ ⎥ e2t ⎟⎟ + α 2 ⎜⎜ k1 ⎢ ⎥ e−t ⎟⎟
⎟
⎜ ⎢⎣ 4⎥⎦
⎟
⎜ ⎢⎣1⎥⎦
⎝
⎠
⎝
⎠
Combining constants gives our general solution:
G
⎡1 ⎤
⎡1⎤
2
t
Y (t ) = K1 ⎢ ⎥ e + K2 ⎢ ⎥ e−t
⎢⎣ 4⎥⎦
⎢⎣1⎥⎦
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End Presentation
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