To let us consider our first multidimensional example of a differential equation, and let's see what
is different to the one dimensional case.
So first of all, we must load all of these modules as previously and especially we have to load also
from PSI Pi and the integrated module because we can use the same command that we used previously for
the one dimensional examples.
So the first example will be a ball in a bowl.
So physically or better to the same?
Mathematically, this is nothing different to two uncoupled harmonic oscillators because this ball in
a ball which is two dimensional, can be described by the following differential equation.
So we have here to force.
So this is the mass times, the acceleration.
So the double of R.
So the position vector and into second time derivative, which is the acceleration, which is also a
vector, of course.
And then we have here several terms.
First of all, we have a damping term, which we also had in the previous case of the one dimension.
And then we have here an external force, which will be a driving force later on.
But in the beginning, we will neglect this.
And then we have here a force that is generated by some potential and the potential will model the bowl.
So in a second, you will see what I mean by that.
And basically, we have now two coordinates for our bold X and Y, and then the Z component is basically
indirectly models by the potential.
So in a sense, you could say this is like the potential energy of the Z component and to be able to
model this.
We must consider some bowl shape for the bowl or for the potential better to see.
And in this case, I want to consider a quadratic function.
Of course, this will strongly depend on the precise profile and the precise shape of the bowl.
But to make it a bit more simple, I use here does quadratic dependence.
And this is also why we have these two uncoupled harmonic oscillators.
Because if we take the radius square, then this is just a square root of x squared plus y squared and
then we square everything.
So it's basically just a sum of two terms.
U zero times x square and use zero times y square.
And then to calculate the corresponding force, we have to calculate minus gradient of you and the gradient
of you in this case is just two times use zero times X for the X component and the same thing with Y
for the Y components.
So it's basically two times use zero times to position Vector.
So this we will later on include in this equation of motion that we will then solve with our method
that we have established previously.
But first of all, I want to show you the bowl.
So I prepared already here some plotting commands where I have prepared an X and Y list using the Mesh
Grit Command, where we will mash two of these grids.
This is what we have also encountered in the crash course, and now we only have to specify what is
the value of the Z component.
And this is, of course, use zero times x squared plus y square.
So just for simplicity, I neglect to use zero.
Or you could say it's equal to one.
So we just have this one x squared plus y square.
And when I run it, then this will be the output.
So you see, we could now cut this profile in a circular way, and then we would have a nice ball.
But it's not really that necessary.
We could just say we have an infinitely large ball that extends along the X and Y direction until plus
and minus infinity.
And it's just important that it has to scratch or take shape.
So in a few minutes or in the next lecture, we will consider how we can solve this differential equation.
But first of all, let me show you what we have encountered previously.
So when I delete all of these cells and show you what I have prepared previously, then I want to revisit
with you the one dimensional harmonic oscillator.
And this is just the code that I have copied from our previous lectures.
So here we have just defined the yeah, the differential equations.
And I say equations because they are actually two differential equations.
This is because the harmonic oscillator is described by a second order differential equation, and we
have learned that we can transform this problem by introducing a new variable called Z.
So this I have written down here again, this is what we have discussed previously.
We introduced new variable C zero and Z one, which describe why and the first derivative of Y.
And so we can replace the second order differential equation by two first order differential equations.
So here you see them, and this is basically what we have implemented and in the mathematical way I've
written it down here.
This already is a two dimensional example because we have now these two variables.
We have Y and we have the first derivative of Y.
And our unseats was here that these two variables are independent.
So in our case, they are not really independent, but they could be independent mathematically.
So really, we already learned how to solve multidimensional examples.
And of course, this are normal.
Harmonic oscillator was one dimensional in the physical sense, but mathematically it was two dimensional
because we have these two equations.
So you can already guess what happens next.
We will now consider this two dimensional differential, but this two dimensional harmonic oscillator
or these two uncoupled oscillators, and they will give rise to four equations.
And the way we solve this is very similar to the one dimensional case.
We just define these this function and calculate the solution by using integrate to solve, underscore
IBP.
And then we plot the solution as we did previously.
And as I said now we just have to take into account four different functions here, and this one will
do and we will discuss this in more detail in the next lecture.