1 Chapter 9 NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS

2 CLASSIFICATION A general second order partial differential equation may be written as AU XX + BU XY + CU YY + DU X + EU Y + FU = 0 where A, B, C, D, E, F are in general functions of x and y. The above equation is classified as follows: ● Elliptic if B 2 – 4AC < 0 ● Parabolic if B 2 – 4AC = 0 ● Hyperbolic if B 2 – 4AC > 0

3 Standard Examples

4 ELLIPTIC EQUATIONS Finite Difference Method—In solving elliptic equations, we approximate the derivatives using finite differences.

5 Graphical Representation I = i = m j = n J = 2 J = 1 J = 0

6 Representation & Approximation

7 Solution of Elliptic Equations The most important type of elliptic equation is Laplace equation u xx + u yy = 0. Approximating the derivatives by difference expressions, we get when h = k This is called diagonal averaging.

8 Diagonal averaging (u i-1, j+1 ) (u i+1,j+1 ) (u i-1,j-1 ) (u i+1,j-1 )

9 Liebmann’s Method To solve u xx + u yy = 0, in a square region R whose boundary is C. The square region is sub-divided into small squares. The values on the boundary points are given. We have to find the values for the function u (x,y) in the interior mesh points. For this we use the approximation using cross-averaging wherever possible and diagonal averaging otherwise. This iterative process is continued until the values at each mesh point converge. While applying this method, symmetry about the horizontal, vertical and diagonal lines should be taken into consideration.

10 Example Find by Liebmann’s Method the values at the interior lattice points of a square plate of the harmonic function u whose boundary values are given in the figure. U1U1 U2U2 U3U3 U4U4 U5U5 U6U6 U7U7 U8U8 U9U9

11 Solution Using cross averaging and diagonal averaging, the successive iterations yield the following values:

12 PARABOLIC EQUATION Most important parabolic equation One dimensional heat equation =  2 where  2 = and c is the specific heat,  is the density and k is the thermal conductivity of the material. The above equation can be written as: u xx = au t where a =

13 Bender – Schmidt Method Consider the equation u xx = au t with boundary conditions u(0, t) = T 0 and u(1, t) = T 1 The initial condition is u (x, 0) = f(x). Let h be the spacing for x and k be the spacing for t. Using finite difference approximation for derivatives, and applying boundary conditions and considering the special case = = ½, we get

14 u i,j+1 = ½ (u i-1,j + u i+1,j ) This formula means that the value of u at x = x i and t = t j+1 is the arithmetic mean of the values of u at the surrounding points x i-1 and x i+1 at the previous time t j. u i,j+1 A u i+1,j u i,j u i-1,j BC Value at A = ½ (Value at B + Value at C) Bender – Schmidt Method

15 Example Find the value of the function u (x,t) satisfying the equation The boundary conditions are and the initial condition is at the points x = i, i = 0, 1, 2, 3, 4 and t = (1/8) j, j = 0, 1, 2, 3, 4, 5.

16 Solution

17 HYPERBOLIC EQUATION Most important hyperbolic equation One – dimensional wave equation

18 Solution of One–Dimensional Wave Equation a 2 u xx – u tt = 0 The boundary conditions are The initial conditions are Let h be the spacing for x and k be the spacing for t. Using finite difference approximation for derivatives, and applying boundary conditions and considering the special case = =, we get

19 Solution of One–Dimensional Wave Equation The boundary conditions u(0, t) = 0, u(1, t) = 0 can be written in the from u 0, j = 0 and u n,j = 0 when l = nh. The initial condition u (x, 0) = f (x) can be expressed as: u i,0 = f (ih) i = 1, 2,…, That is u i,0 = fi. For the initial condition u t (x, 0) = 0, we use Central Difference Approximation for the derivative and write

20 u i, j+1 = u i+1, j + u i-1,j – u i, j-1 u i-1, j-1 u i, j-1 u i+1, j u i-1, j u i, j u i+1, j u i, j+1 Note: u(x i,t j+1 ) = (sum of surrounding values at previous time) – u(x i,t j-1 ) Solution of One – Dimensional Wave Equation

21 Example Tabulate the pivotal values for the equation Given that u (0, t) = 0, u (5, t) = 0, u (x, 0) = x 2 (5 – x) and u t (x, 0) = 0 Assume h = 1.

22 Solution u i,j+1 = u i-1, j + u i+1, j – u i, j-1

23 The End…… Wish you all the best…