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

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

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### 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