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To let us consider our first multidimensional example of a differential equation, and let's see what

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is different to the one dimensional case.

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So first of all, we must load all of these modules as previously and especially we have to load also

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from PSI Pi and the integrated module because we can use the same command that we used previously for

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the one dimensional examples.

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So the first example will be a ball in a bowl.

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So physically or better to the same?

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Mathematically, this is nothing different to two uncoupled harmonic oscillators because this ball in

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a ball which is two dimensional, can be described by the following differential equation.

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So we have here to force.

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So this is the mass times, the acceleration.

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So the double of R.

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So the position vector and into second time derivative, which is the acceleration, which is also a

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vector, of course.

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And then we have here several terms.

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First of all, we have a damping term, which we also had in the previous case of the one dimension.

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And then we have here an external force, which will be a driving force later on.

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But in the beginning, we will neglect this.

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And then we have here a force that is generated by some potential and the potential will model the bowl.

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So in a second, you will see what I mean by that.

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And basically, we have now two coordinates for our bold X and Y, and then the Z component is basically

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indirectly models by the potential.

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So in a sense, you could say this is like the potential energy of the Z component and to be able to

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model this.

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We must consider some bowl shape for the bowl or for the potential better to see.

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And in this case, I want to consider a quadratic function.

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Of course, this will strongly depend on the precise profile and the precise shape of the bowl.

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But to make it a bit more simple, I use here does quadratic dependence.

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And this is also why we have these two uncoupled harmonic oscillators.

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Because if we take the radius square, then this is just a square root of x squared plus y squared and

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then we square everything.

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So it's basically just a sum of two terms.

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U zero times x square and use zero times y square.

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And then to calculate the corresponding force, we have to calculate minus gradient of you and the gradient

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of you in this case is just two times use zero times X for the X component and the same thing with Y

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for the Y components.

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So it's basically two times use zero times to position Vector.

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So this we will later on include in this equation of motion that we will then solve with our method

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that we have established previously.

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But first of all, I want to show you the bowl.

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So I prepared already here some plotting commands where I have prepared an X and Y list using the Mesh

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Grit Command, where we will mash two of these grids.

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This is what we have also encountered in the crash course, and now we only have to specify what is

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the value of the Z component.

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And this is, of course, use zero times x squared plus y square.

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So just for simplicity, I neglect to use zero.

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Or you could say it's equal to one.

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So we just have this one x squared plus y square.

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And when I run it, then this will be the output.

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So you see, we could now cut this profile in a circular way, and then we would have a nice ball.

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But it's not really that necessary.

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We could just say we have an infinitely large ball that extends along the X and Y direction until plus

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and minus infinity.

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And it's just important that it has to scratch or take shape.

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So in a few minutes or in the next lecture, we will consider how we can solve this differential equation.

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But first of all, let me show you what we have encountered previously.

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So when I delete all of these cells and show you what I have prepared previously, then I want to revisit

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with you the one dimensional harmonic oscillator.

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And this is just the code that I have copied from our previous lectures.

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So here we have just defined the yeah, the differential equations.

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And I say equations because they are actually two differential equations.

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This is because the harmonic oscillator is described by a second order differential equation, and we

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have learned that we can transform this problem by introducing a new variable called Z.

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So this I have written down here again, this is what we have discussed previously.

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We introduced new variable C zero and Z one, which describe why and the first derivative of Y.

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And so we can replace the second order differential equation by two first order differential equations.

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So here you see them, and this is basically what we have implemented and in the mathematical way I've

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written it down here.

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This already is a two dimensional example because we have now these two variables.

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We have Y and we have the first derivative of Y.

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And our unseats was here that these two variables are independent.

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So in our case, they are not really independent, but they could be independent mathematically.

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So really, we already learned how to solve multidimensional examples.

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And of course, this are normal.

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Harmonic oscillator was one dimensional in the physical sense, but mathematically it was two dimensional

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because we have these two equations.

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So you can already guess what happens next.

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We will now consider this two dimensional differential, but this two dimensional harmonic oscillator

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or these two uncoupled oscillators, and they will give rise to four equations.

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And the way we solve this is very similar to the one dimensional case.

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We just define these this function and calculate the solution by using integrate to solve, underscore

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IBP.

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And then we plot the solution as we did previously.

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And as I said now we just have to take into account four different functions here, and this one will

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do and we will discuss this in more detail in the next lecture.