Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Point Collocation Method used in the solving of Differential Equations, particularly in Finite Element Methods

4,263 views

Published on

Methods of Solving Differential Equations as Boundary Value Problems using the Point Collocation Method.

Published in: Engineering
  • There is a useful site for you that will help you to write a perfect and valuable essay and so on. Check out, please ⇒ www.HelpWriting.net ⇐
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • I can definitely recommend a website that really helped me with my essay. I found out it was due the day before I had to submit it. Went into full-on panic mode. Worst experience of my senior year by far. It’s called HelpWriting.net. The quality of the writing is passable but the completion rate is super quick. You get to pick your own writer to do your stuff and that’s also a big bonus.
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here

Point Collocation Method used in the solving of Differential Equations, particularly in Finite Element Methods

  1. 1. Point Collocation Method FEM - Introduction - Methods of Solving Differential Equations Suddhasheel Ghosh, PhD Department of Civil Engineering Jawaharlal Nehru Engineering College N-6 CIDCO, 431003 Advanced Numerical Methods Series shudh (JNEC) PCM MEStru2k1617 1 / 12
  2. 2. DiffEq1 Introduction to terminology Given a differential equation Ψ d2y dx2 , dy dx , y, x = 0, (1) and the initial conditions, F1 dy dx , y, x = a = 0 F2 dy dx , y, x = b = 0 So, given the points x1 = a, x2, x3, . . . , xi, . . . , xn, xn+1 = b, it is desired to find the solution of the differential equation at the points xj, ∀j = 2, . . . , n. The points xj, j = 2, . . . , n are known as the points of collocation. shudh (JNEC) PCM MEStru2k1617 2 / 12
  3. 3. DiffEq1 A second-order Boundary Value Problem A boundary value problem is given as follows: d2y dx2 + P(x) dx dy + Q(x)y = R(x) along with the conditions y(x1) = A, y(xn+1) = B shudh (JNEC) PCM MEStru2k1617 3 / 12
  4. 4. Collocation Method Point collocation Method Derivative calculation Assume that y = n i=0 αixi . Therefore, we will have dy dx = n i=0 i · αixi−1 , and d2y dx2 = n i=0 i(i − 1)αixi−2 shudh (JNEC) PCM MEStru2k1617 4 / 12
  5. 5. Collocation Method Point collocation Method I Substitution and formulation Substituting these in the differential equation, we have n i=0 i(i − 1)αixi−2 + P(x) n i=0 i · αixi−1 + Q(x) n i=0 αixi = R(x). Thus giving, n i=0 αi i(i − 1)xi−2 + ixi−1 P(x) + xi Q(x) = R(x) The aim of the interpolation method is to “agree” at the node points, and therefore, we shall have: n i=0 αi i(i − 1)xi−2 j + ixi−1 j P(xj) + xi jQ(xj) = R(xj), ∀j = 2, . . . , n shudh (JNEC) PCM MEStru2k1617 5 / 12
  6. 6. Collocation Method Point collocation Method II Substitution and formulation For the nodes x1 and xn+1, we have the following conditions: n i=0 αixi 1 = A n i=0 αixi n+1 = B shudh (JNEC) PCM MEStru2k1617 6 / 12
  7. 7. Collocation Method Point collocation Method Matrix formulation of the problem   1 x1 x2 1 . . . xn 1 Q(x2) P(x2) (2 + 2x2P(x2) + x2 2 Q(x2)) . . . [n(n − 1)xn−2 2 + nP(x2)xn−1 2 + xn 2 Q(x2)] ... ... . . . . . . ... Q(xn) P(xn) (2 + 2xnP(xn) + x2 nQ(xn)) . . . [n(n − 1)xn−2 n + nP(xn)xn−1 n + xn nQ(xn)] 1 xn+1 x2 n+1 . . . xn n+1     α0 α1 ... αn−1 αn   =   A R(x2) ... R(xn) B   The solution can then be achieved by any of the standard methods like Gauss-Siedel, Gaussian Elimination or Gauss-Jordan Elimination. shudh (JNEC) PCM MEStru2k1617 7 / 12
  8. 8. Collocation Method Point collocation method I Example Use the point collocation method to solve the following differential equation: d2y dx2 − y = x Use the boundary conditions y(x = 0) = 0 and y(x = 1) = 0. Choose x = 0.25 and x = 0.5 as collocation points. (Desai, Eldho, Shah) Solution: There are four points where we are considering the solution for, x = 0, 0.25, 0.5, 1. We label them as x1, x2, x3, x4. Since there are four points, we will consider a cubic polynomial. y = α0 + α1x + α2x2 + α3x3 shudh (JNEC) PCM MEStru2k1617 8 / 12
  9. 9. Collocation Method Point collocation method II Example We have dy dx = α1 + 2α2x + 3α3x2 d2y dx2 = 2α2 + 6α3x Substituting these in the given differential equation, we have 2α2 + 6α3x − α0 − α1x − α2x2 − α3x3 = x −α0 − α1x + (2 − x2 )α2 + (6x − x3 )α3 = x From the first boundary condition y(x = 0) = 0, we have α0 + α1(0) + α2(02 ) + α3(03 ) = 0 =⇒ α0 = 0 (2) shudh (JNEC) PCM MEStru2k1617 9 / 12
  10. 10. Collocation Method Point collocation method III Example From the second boundary condition y(x = 1) = 0, we have α0 + α1(1) + α2(12 ) + α3(13 ) = 0 =⇒ α1 + α2 + α3 = 0 (3) At the collocation points, we have the following equations: For x = 0.25, we have −α1(0.25) + (2 − (0.25)2 )α2 + (6 × 0.25 − (0.25)3 ) = 0.25 −0.25α1 + 1.9375α2 + 1.4844α3 = 0.25 (4) For x = 0.5, we have −0.5α1 + (2 − 0.52 )α2 + (6 × 0.5 − 0.53 ) = 0.5 −0.5α1 + 1.75α2 + 2.875α3 = 0.5 (5) shudh (JNEC) PCM MEStru2k1617 10 / 12
  11. 11. Collocation Method Point collocation method IV Example Using the equations above, we have the following matrix based arrangement   1 1 1 −0.25 1.9375 1.4844 −0.5 1.75 2.875     α1 α2 α3   =   0 0.25 0.5   (6) which gives on the inverse operation, α1 = −0.1459, α2 = −0.006738, α3 = 0.1526 Therefore the polynomial approximation for y is y = 0.1459x − 0.006738x2 + 0.1526x3 (7) shudh (JNEC) PCM MEStru2k1617 11 / 12
  12. 12. Collocation Method Thank you! shudh (JNEC) PCM MEStru2k1617 12 / 12

×