Very nice, simple, and lucid lecture about the point collocation method. I became interested in finding out how the method of collocation works because I recently learned to use the solve_bvp function of SciPy for Python and it uses collocation to solve the BVP. The collocation method isn't usually mentioned in numerical methods books. The shooting and finite difference methods are the ones usually mentioned. The collocation method is mentioned in only one of the books I have and only in passing without detailing how it works and no examples.
You can choose any point for the collocation. What matters is that those collocation points satisfy the governing differential equation, which you do by substituting those points into the ODE.
Here, the limit is 0 to 1. We need two gauss points. So, we need to divide 0 to 1 into three equal parts to get two gauss points (shown at 9.07 minutes in the video). So, points (1/3) & (2/3) divide limit into three equal parts and hence two gauss points are obtained.
The ode for this problem is a 2nd-order, linear, non-homogenous ODE with constant coefficients. For a non-homogenous ODE, you can use the method of undetermined coefficients (MUC) or method of variational parameters (MVP). Since the we have constant coefficients for this one, MUC is applicable. The forcing function here is just a constant, -1, so your steady state solution is of the form yss = k. k = -1 for this problem. The total solution is composed of the homogenous solution and the non-homogenous solution (y = -1). For the homogenous solution, you just equate the sum of y and its derivatives to zero and substitute y=e^(at) in the ODE and solve for a.
Here, the limit is 0 to 1. We need two gauss points. So, we need to divide 0 to 1 into three equal parts to get two gauss points (shown at 9.07 minutes in the video). So, points (1/3) & (2/3) divide limit into three equal parts and hence two gauss points are obtained.
Here, the limit is 0 to 1. We need two gauss points. So, we need to divide 0 to 1 into three equal parts to get two gauss points (shown at 9.07 minutes in the video). So, points (1/3) & (2/3) divide limit into three equal parts and hence two gauss points are obtained.
The way you explain is so good, thanks so much
Glad it was helpful!
This is so good. After breaking my head on reading theory, this simple explanation made more sense. Thank you so Shubham :)
Very nice, simple, and lucid lecture about the point collocation method. I became interested in finding out how the method of collocation works because I recently learned to use the solve_bvp function of SciPy for Python and it uses collocation to solve the BVP. The collocation method isn't usually mentioned in numerical methods books. The shooting and finite difference methods are the ones usually mentioned. The collocation method is mentioned in only one of the books I have and only in passing without detailing how it works and no examples.
Good Luck!
Commendable job dear sir
Thank You Sir!
Nice video to grab the method...
Thank you sir
Great teaching style
How we are choosing Gauss points here?
same doubt i had as well
You can choose any point for the collocation. What matters is that those collocation points satisfy the governing differential equation, which you do by substituting those points into the ODE.
Here, the limit is 0 to 1. We need two gauss points. So, we need to divide 0 to 1 into three equal parts to get two gauss points (shown at 9.07 minutes in the video). So, points (1/3) & (2/3) divide limit into three equal parts and hence two gauss points are obtained.
Thank you!
How find the exact soulution
The ode for this problem is a 2nd-order, linear, non-homogenous ODE with constant coefficients. For a non-homogenous ODE, you can use the method of undetermined coefficients (MUC) or method of variational parameters (MVP). Since the we have constant coefficients for this one, MUC is applicable. The forcing function here is just a constant, -1, so your steady state solution is of the form yss = k. k = -1 for this problem.
The total solution is composed of the homogenous solution and the non-homogenous solution (y = -1). For the homogenous solution, you just equate the sum of y and its derivatives to zero and substitute y=e^(at) in the ODE and solve for a.
Sir, can we choose any point as gauss point?
Here, the limit is 0 to 1. We need two gauss points. So, we need to divide 0 to 1 into three equal parts to get two gauss points (shown at 9.07 minutes in the video). So, points (1/3) & (2/3) divide limit into three equal parts and hence two gauss points are obtained.
Sir can you explain how you found that gauss points,??? 1/3 and 2/3?
Here, the limit is 0 to 1. We need two gauss points. So, we need to divide 0 to 1 into three equal parts to get two gauss points (shown at 9.07 minutes in the video). So, points (1/3) & (2/3) divide limit into three equal parts and hence two gauss points are obtained.
@@shubhamvdeshmukh7625 Ohk....Thank You So Much Sir🙏😊
How did we understand that we need 2 gauss points?
please provide me a reference book
Your name is shoom dishoom ? 😅