So now we know that we can solve the equations of motion which have these ones here numerically.
However, as you have seen, the solution is pretty confusing.
So there are a lot of oscillations here and these are not harmonic anymore.
So we could just say, OK, this is not the solution, but we don't really understand what's going on
here.
And to get more of an understanding what these oscillations mean, we must solve the so-called eigenvalue
problem.
And first of all, I will teach you a bit about the mathematics, and I will explain to you why this
is an eigenvalue problem.
So if you find this too mathematical, no problem.
You can just skip ahead.
But let me say you, the the main topic or the main message of this lecture, and this is that we have
this equation of motion, which are three coupled equations.
And you see our one depends on our two as well, and our two depends on our one and our three as well
and so on.
So it's not so easy to solve the problem.
However, it would be much more easy if this matrix here would be a diagonal matrix, or even if it
would just be a scalar, because in this case, we would just have that.
The first coordinate on one would only depend on our one or, as I have written down here, we may accomplish
this by doing some transformation and then we have some new coordinates.
Q and Q one depends only linearly on Q1.
So here the derivative and here not so.
This would, of course, be really nice because then we would have three uncoupled oscillators as we
also had before and one of our examples, and then we could just solve each of them individually.
So this would be really, really simple.
And then we would have a frequency which would be given by the square root of K over m times this value
lambda, which resembles this matrix.
So it turns out that this is the case if the lambdas here are the eigenvalues of this matrix.
So this is why we will continue and solve the eigenvalues of this matrix, and this is why the eigenvalues
are so important and why they can be identified with the frequency.
So now I will talk a bit more about mathematics.
And as I said, if you are not interested in the and this and if you find it too difficult, it's no
problem.
You can just go ahead to the next lecture is not really not important.
It's just that you understand better what's going on here.
So the thing is to transform this equation to that one, we must apply some transformation and this
is typically done by multiplying with the unitary matrix.
So you mean this matrix and unitary means that the inverse times to normal matrix is the same as the
normal matrix times the inverse and is equal to one.
And this is really the key thing here, because of course, you can always multiply one.
For example, you can see matrix times one times this vector.
This is still the same thing.
So what we do is we write this equation vector our second derivative sequence of minus K over m times
two Matrix A, which is this one times two vector.
And now we can multiply here times one because this doesn't change anything and we say one is equal
to you, times you inverse.
So these two equations are still the same.
And now we can just change the brackets here.
So this means we multiply u minus one times r we multiply u minus one times, eight times u and we have
ten here missing u.
And so we can see multiply from the left with you.
Two The power of minus one.
So then we have here u minus one times u, which is one.
So we can leave it out and we must, of course, multiply it also to the left hand side.
So we have u minus one.
And then this vector R. And the second derivative.
And so now I I introduce that we must find an I introduce new variables Q, which are u minus one times
R.
And then the same applies, of course, for the second derivative because this is not time dependent
here.
So we can just write that u minus one second or the derivative of R is second order the relative of
Q.
And then if we want to say that this is equal to lambda, then this one here you two the power of minus
one times eight times you must be equal to lambda.
And this is how we indirectly define you.
So the thing is, you can show mathematically that it is possible to find such matrices you.
Transformation here is correct on these, this equation is correct.
And if you use this, then you can see that you can transform this matrix equation to this equation.
So it means that lambda is defined like this or not, lambda as the out is more, you could say you
is defined by eight and lambda.
And so this is the equation that we have used for the definition, and now you can multiply with you
from the left.
So this one disappears and goes here.
And you can also say, if you multiply here, another vector, that you have to solve this equation
here.
And so it turns out that we must solve the eigenvalues of the matrix a and that's the eigenvalues of
the Matrix.
Eight are lambda because this year's and IGen eigenvalue equation.
So we have transformed, in short this equation to that one, and we figured out that lambda already
eigenvalues of this matrix.
And furthermore, we have figured out from our physical understanding that the eigenvalues lambda can
be identified with the frequency not directly, but by the square root K over m times lambda relation.
So in the following, we will analyze this matrix and determine the eigenvalues and then calculate the
EIGEN frequencies for this matrix.
And then we will compare these IGen frequencies with the frequencies that we see in our numerical solution.