Okay. To finish off our series on
systems of linear differential equations,
we're gonna consider the case where the
matrix is not diagonalizable, so here's
our example. The matrix is 3, 1, -1, 1,
we'll take initial conditions 5 and 2,
and this is the same matrix that we
used for non-diagonalizable difference
equations, so the eigenvalue is 2,
it has algebraic multiplicity 2,
and geometric multiplicity 1,
the only eigenvector is 1, -1 up
to scale, but there's also a power vector
0, 1, so if you apply A minus twice the
identity to this you get 0, you apply
A minus twice the identity to this,
you get the first eigenvector. Now the
key observation is that our solution is
still of the form e to the A t times
your initial condition, so let's put a box
around that, and the point is that if you
take the derivative of that, the
derivative of e to the A t by definition
is A e to the A t, and so you wind up with
A times e to the A t times x of 0, so
that's A times x of t, so the derivative
of this expression is A times itself,
and if you plug in time 0, well, e to the
A t, by definition at time zero is the
identity, times x of zero, gives you
x of 0, and it works. And as always,
we're gonna let y be the coordinates
of x in the B-basis, we're now- B isn't
the basis of eigenvectors, it's the
basis of one eigenvector and one power
vector. Okay, the only difference from the
diagonalizable case is how we go from
y of 0 to y of t, and we do this by just
multiplying through by e to the A t,
you see x of 0 is y1 of 0 b1, plus y2 of
0 b2. If we hit both sides of the equation
by e to the A t, well we need to
understand what e to the A t does
to the first eigenvector, and what it does
to this second power vector.
b1 is an eigenvector, so we know
what e to the A t does to an eigenvector,
it just multiplies by e to the eigenvalue
times t, so that gives us e to the 2t b1.
b2 is a power vector, so e to the A t
acting on b2 is a bit more complicated.
You rewrite e to the A t as e to the 2t
times e to the A minus 2i t, and you
expand this out in a power series,
so you get e to the 2t times b2 plus
t A minus 2i b2, plus t squared over
2, A minus 2i squared, b2, plus
t cubed over 3 factorial, A minus 2i
cubed b2, plus so on forever.
But, it's a power vector. Power vector
means- and it's power vector of
degree 2, so A minus 2i squared
acting on b2, is 0. This term is 0, the
next term is 0, all the remaining terms
are 0, and you're just left with,
well, A minus 2i b2, we said that
that was b1, so you wind up with
e to the 2 t b2, plus t e to the 2t,
times b1.
So if we plug that back in to our
expression for e to the A t, well if
we want to know y1 and y2 at time
t, together they make x of t, and
that's e to the A t x of 0, which we've
just been working out, and we saw
that it's y1 times e to the 2t b1,
and then y2 of 0, times e to the 2t
b2, plus t e to the 2t, b1, and now the
only way that this linear combination
can be equal to this linear combination
is if this is equal to the first
expression, and if this is equal to the
second expression, so there we go.
y1 of t is e to the 2t y1 of 0, plus t
e to the 2t y2 of 0, and y2 of t is
e to the 2t, y2 of zero. Now, we
can write this in matrix form,
and this matrix here is not the
exponential- it's not e to the dt, because
it's not what you get when exponentiating
a diagonal matrix, but it is what you get
when exponentiating 2, 1, 0, 2, so what
you get when you exponentiate the
Jordan form, of the matrix.
Okay, so now we've got our equations
for y of t in terms of y of 0, and we can
go around our square, as always, we
want to start off with x of 0, turn it
into y of 0, y of t, and then x of t,
so our change of basis matrix in one
direction is 1, 0, -1, 1, that's P, our
change of basis matrix in the other
direction is the inverse of this
matrix, that's P inverse, and so we
start off with 5 2, and we multiply
through by the change of basis
matrix, and we get 5 7. Then we use
our formula for converting y of 0
into y of t, and we get some terms with
e to the 2t in them, and we get some
terms that go as t e to the 2t. Then,
we multiply through by the other change
of basis matrix, and there we go.
We've got our solution. It's got e to the
2t in them, and it's got t e to the 2t in
them, and you see that if you plug in t
equals 0, these terms die, and you get
5 e to the 0, 2 e to the 0, and that's
5 2, and that was our initial condition.
Okay, so that's how you do it but with
power vectors. You could also have done
it with diagonalization and perturbations,
we could have looked at a matrix that
was just just a perturbation of the
original ones, for example if you added
epsilon to the lower-right-hand number,
you would get a matrix that's
diagonalizable. Since it's diagonalizable,
you can compute it's exponential by
diagonalization, so you would convert
x of 0 by taking P epsilon inverse,
to get y of 0, you multiply by the
exponential of the eigenvalues to
get y of t, you multiply by the basis
vectors to get x of t, and then you
would have to take a limit as epsilon
goes to 0, and some of the terms would
involve 0 over 0, and you resolve this
indeterminate form by using L'Hopital's
Rule, and that's where the t e to the 2t
would come from, so you can do it by
perturbation, but in terms of getting
the general form of the expressions,
it's really more insightful to use power
vectors, and that shows how when you
have a non-diagonalizable matrix,
you wind up with terms that look not
just like e to the lambda t, but you
wind up with lambda e to the lambda
t expressions. If you had a power vector
of degree three, you would also have a
lambda squared e to the lambda t,
degree four, lambda cubed e to the
lambda t. So solutions to systems
of differential equations always
involve exponentials unless it's non-
diagonalizable, which in case it can
involve powers of t times exponentials.
