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SFF and BVP present Warp Speed Whiteboard Space Tech Solving Problems Together

Tech and Space are two industriesexploring new frontiers and they both face similar challenges. Introducingquot;Warp Speed Whiteboard: Space and Tech Solving Problems Togetherquot;. Bessemer Venture Partners and The Space Frontier Foundation handselected 22 participantsincluding individuals from Google, Oculus Virgin Galactic, and more, to meet and collaborate on problem solving What if we apply Tech business to New Space technologies, are we destined to advance bothé We can't solve everything in a day but great things usually come from great conversation, and that's a great place to start.

Finite Difference Method for Solving ODEs Example Part 1 of 2

. . In this segment we're going to take an example of Finite Difference Method of solving boundary value ordinary differential equations.So let's go and see that how we can use the theory behind the finite difference methods to solve ordinary differential equation boundary value problems.And we are going to look at an example of doing that.So let's go and take a

simple, so this is the problem statement which is given to us for the example. You have a second order differential equation which is of this particular form. . So you are given this and you are given the boundary conditions here that the value of u at 2 is 0.008 and the value of u at 6.5 is 0.003.So what you want to be able to do is, you want to be able to find out what u is as a function of r

because u is the dependent variable and r is the independent variable and you want to be able to find what u is at the different points.In order to keep the problem simple, so somebody's asking so basically the problem statement is find u as a function of, function of r.But since we are given the numerical methods we cannot find u as a function of r at every data point, so somebody might say hey, use 4 nodes. Use 4 nodes to do the problem. Let me say 4, equidistant. Equidistant meaning that the nodes have the same distance between each other.

So what that implies is that we need to break up our interval which is going from 2 to 6.5 into 4 nodes to be able to solve the problem. So I am here, so let's suppose if I am here, at r equal to 2 and then here I am at r equal to 6.5. So if I am going to break it up into 4 nodes I'll have another node here and another node right here so that means that three segments here.So the segment width will be divided by 3, the difference between the two. So if I want to calculate delta r it's just (6.52)3 which is 1.5.And I have four nodes here so

this will become the first, second node will be at r + delta r which will be at 3.5 and the next node will be at 3.5 plus delta r which will be at 5.0 so that's where the next node is going to be at 5.0.And what I am going to do is I'm going to number these nodes.I am going to number this node as 1, number this node as 2, and number this node as 3, and number this node as 4 and that's how I'll be designating that.So I already know what the value of u at 1 is because it is already given to me. The value of u is already given to me as 0.008 and I already know what the value at 4 is

which is given to me as the value at r=6.5 is 0.003.So what that implies that I have to find out what the value of the u is, the dependent variables at the second node and the third node and if I am able to do that, that means that I have been able to solve the problem.That I can draw the profile of u, then I can do things like interpolation or spline interpolation to be able to find the value of u at some other point approximately.Even the values which I get at 2 and 3 are known approximately because I will be using approximations for my derivatives.So let's go and see that how do we go about doing this problem here by using Finite Difference Methods is

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