When dealing with 2-D Laplace equation how is Differential quadrature method applied to get the solution. How are the four boundary conditions incorporated into the formulation?
Hi, Raj
I had the same problem as you had for 2D vibration problems. Unfortunately, the best experts did not help me either. I figured out myself. you know that we need a matrix like A *u= B for solution. You have to number the nodes first and relates them to the matrix elements A. In other words for node (i,j) specify a node number that number is used for referring to matrix A.
Consider we have 5 * 5 nodes along x and y =25 nodes in total. So the matrix A should be 25*25. So, you have to relate (i,j) of the node to a number like
(1,1) =1
(1,2)=2
.
(5,5)=25
Then , you know your boundaries i,j and their node numbers. Now, you can easily modify the matrix A elements of the boundaries.
In most of the references, the matrix A is built in such a way that the boundaries and inner nodes equations are separate which is useless in my opinion. You can have the Matrix A with mixed equations in between and still have the same answer.
In Differential quadrature method the unknown u and its partial derivatives are expressed in terms of node points where boundary points are also included. Use them at the boundary to form as many equations as unknowns.
Thank you sir.
When dealing with one dimensional problems we can get a set of linear equations directly. But in the case of 2-D problems am not able to form linear equations. Am using the method proposed by "Mingle" to substitute for the partial derivatives. I have the weighting coefficients corresponding to the second order derivatives but am having trouble in getting the system of equations. Please help with this.
Could you please some of our work in the area .Still you face problem, we will help you.
Sir I am trying to solve a 2-D Laplace equation using Differential Quadrature method,
d2u/dx2+d2u/dy2=0 x,y runs from 0 to 1
BC`s are
u(x,0)=0
u(x,1)=1
u(0,y)=0
u(1,y)=0
let my grid is x and y divided as 0,1/3,2/3,1.
Initially before applying BC my coefficient matrix would be 16x16.
Then after applying boundary it will become a 4x4 matrix.
1)My question is when applying the second BC(x=0,1/3,2/3,1 and y=1) how will my RHS change?
Kindly see the attached answer for your question. For more clarification our papers on 2 dimension problems where we have solved time dependent problems but treatment at the boundary is the same.
Sir I got the solution for the 2-D Poisson equation and is matching with the analytic solution.
Now I am trying to solve the free vibration of a cantilever beam. I am attaching a word document in which I have given the progress of my work. I am not able to converge to fundamental frequency. I have tried increasing the sample points but still I am not getting the result. Can you please refer my work and advise me where I have gone wrong.
I have referred the review paper of C.W.Bert and Moinuddin Malik (page No 10).
Hi, Raj
I had the same problem as you had for 2D vibration problems. Unfortunately, the best experts did not help me either. I figured out myself. you know that we need a matrix like A *u= B for solution. You have to number the nodes first and relates them to the matrix elements A. In other words for node (i,j) specify a node number that number is used for referring to matrix A.
Consider we have 5 * 5 nodes along x and y =25 nodes in total. So the matrix A should be 25*25. So, you have to relate (i,j) of the node to a number like
(1,1) =1
(1,2)=2
.
(5,5)=25
Then , you know your boundaries i,j and their node numbers. Now, you can easily modify the matrix A elements of the boundaries.
In most of the references, the matrix A is built in such a way that the boundaries and inner nodes equations are separate which is useless in my opinion. You can have the Matrix A with mixed equations in between and still have the same answer.
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