The document discusses finite difference methods for solving differential equations. It begins by introducing finite difference methods as alternatives to shooting methods for solving differential equations numerically. It then provides details on using finite difference methods to transform differential equations into algebraic equations that can be solved. This includes deriving finite difference approximations for derivatives, setting up the finite difference equations at interior points, and assembling the equations in matrix form. The document also provides an example of applying a finite difference method to solve a linear boundary value problem and a nonlinear boundary value problem.

1. Finite Difference Methods
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Gaurav Mallik
SAU/AM(M)/2014/22
South Asian University
Rupali Sharma
SAU/AM(M)/2014/27
South Asian University
Divyansh Verma
SAU/AM(M)/2014/14
South Asian University

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Finite Difference Methods
• The most common alternatives to the shooting
method are finite-difference approaches.
• In these techniques, finite differences are
substituted for the derivatives in the original
equation, transforming a linear differential
equation into a set of simultaneous algebraic
equations.

3. Finite Difference Method for Linear Problem
The finite difference method for the linear second-order BVP
y‘’ = p(x)y’ + q(x)y + r(x) for a ≤ x ≤ b with y(a) = α and y(b) = β
we select an integer N > 0 and divide the interval [a, b] into
(N+1) equal subintervals whose endpoints are the mesh points
xi = a + ih for i = 0, 1, . . . , N+1 where h = (b−a)/(N+1)
xi are called collocation points, we find the solution at these points.
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x1 xN+1x2 x3 . . .
h h
xN

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Finite Difference Method for Linear Problem
At the interior mesh points, xi, for i = 1, 2, . . . , N, the differential
equation to be approximated is y’’(xi) = p(xi)y’(xi) + q(xi)y(xi) + r(xi)
Expanding ‘y’ in a third Taylor polynomial about xi evaluated
at xi+1 and xi−1, assuming that y Є C4[xi-1,xi], we have,
y(xi+1) = y(xi + h) = y(xi ) + h.y(1)(xi) + (h2/2).y(2) (xi) + (h3/6).y(3) (xi)
+ (h4/24).y(4)(ξ i
+) where ξi
+ Є (xi,xi+1) (I)
y(xi-1) = y(xi - h) = y(xi ) - h.y(1)(xi) + (h2/2).y(2) (xi) - (h3/6).y(3) (xi)
+ (h4/24).y(4)(ξ i
-) where ξi
- Є (xi,xi+1) (II)

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Finite Difference Method for Linear Problem
Adding I and II , we get
y(xi + h) + y(xi - h) = 2y(xi) + h2.y(2) (xi) + (h4/24).[y(4)(ξ i
+) + y(4)(ξ i
-)]
(By intermediate value theorem, there exists ξi Є (ξ i
+,ξi
-) such that
y(4)(ξ i) = [y(4)(ξ i
+) + y(4)(ξ i
-)] / 2 or 2y(4)(ξ i) = [y(4)(ξ i
+) + y(4)(ξ i
-)] )
y(xi + h) + y(xi - h) = 2y(xi) + h2.y(2) (xi) + (h4/24).[2y(4)(ξ i)]
y(2) (xi) = [ [y(xi + h) + y(xi - h) - 2y(xi) ] / h2 ] - (h2/12).[y(4)(ξ i)]
Subtracting II from I , we get
y(xi + h) - y(xi - h) = 2hy(1) (xi) + (2h3/6).y(3) (xi)
y(1) (xi) = [ [y(xi + h) - y(xi - h)] / 2h ] - (h2/6).y(3) (xi)

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Finite Difference Method for Linear Problem
Now substituting the value of y(2) (xi) and y(1) (xi) in original
differential equation, we get
[ [y(xi + h) + y(xi - h) - 2y(xi) ] / h2 ] - (h2/12).[y(4)(ξ i)]
= p(xi) [ [y(xi + h) - y(xi - h)] / 2h ] - (h2/6).y(3) (xi) + q(xi)y + r(xi)
Simplifying the above equation, we get
-(1+h.p(xi)/2)yi+1 + (2+h2.q(xi))yi – (1- h.p(xi)/2)yi = -h2ri
For i=1, -(1+h.p(x1)/2)y2 + (2+h2.q(x1))y1 – (1- h.p(x1)/2)y1 = -h2r1
For i=2, -(1+h.p(x2)/2)y3 + (2+h2.q(x2))y2 – (1- h.p(x2)/2)y2 = -h2r2
.
.
.
For i=N, -(1+h.p(xN)/2)yN+1 + (2+h2.q(xN))yN – (1- h.p(xN)/2)yN = -h2rN

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Finite Difference Method for Linear Problem
The system of equations can be expressed in Tri-diagonal nXn
matrix form Aw=b, where
=

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Finite Difference Method for Linear Problem
let yi=wi

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Example for Linear BVP
Solve (d2y/dx2) = xy with y(0)+y’(0)=1 and y(1)=1
such that 0≤x≤1
Solution : Here let h= 1/3
So by the formula discussed earlier we have,
(yi-1 -2yi +yi+1)/h2 = xi yi
(yi-1 -2yi +yi+1) = h2 xi yi
(yi-1 -2yi +yi+1) = (1/9)xi yi
Now, for i=0 we have,
(y-1 -2y0 +y1) = (1/9)x0 y0
(y-1 -2y0 +y1) = 0 1

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for i=1 we have, (y0 -2y1 +y2) = (1/9)x1 y1
(y0 -2y1 +y2) = (1/9)(1/3)y1
(y0 -2y1 +y2) = (1/27) y1
for i=2 we have, (y1 -2y2 +y3) = (1/9)x2y2
(y1 -2y2 +y3) = (1/9)(2/3)y2
(y1 -2y2 +y3) = (2/27)y2
The unknowns are y-1 , y1 , y2 and y0
Using (yi )‘ = (yi+1 – yi-1 +o(h3))/2h we get,
y’(0)= (y1 - y-1)/2h
1+ y0 = (y1 - y-1)/2h
y-1 = y1 –(2/3)(1- y0)
putting Eqn (4) in (1), we get, -2 y0 + 3y1 =1
Example for Linear BVP
2
3
4

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So,
-2 y0 + 3y1 =1
(y0 -2y1 +y2) = (1/27) y1
(y1 -2y2 +y3) = (2/27)y2
The matrix will be :
The Soln. is y1 = - 0.9879518, y2 = -0.3253012, y3 = 0.3253012
Example for Linear BVP
-2 3 0
1 -2-(1/27) 1
0 1 -2-(2/27)
y0
y1
y2
1
0
-1
=

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Finite Difference Method for Non- Linear Problem
General form of Non linear BVP:
y”= f(x,y,y’) for a≤x≤b such that y(a)=α , y(b)=β
ie
(yi+1 -2yi + yi-1)/h2 = f(xi ,yi , (yi+1- yi-1 )/2h –(h2)/6 y”(η))-
(h2)/12 y(n)(ξ i)
y0 = GIVEN and yN+1 = GIVEN
For i=1 y2 -2y1 = h2 * f(x1 ,y1 , (y2 – α )/2h) – α
i=2 y3 -2y2 + y1 = h2 * f(x2 ,y2 , (y3- y2 )/2h)
i=N -2yN + yN-1)/h2 = f(xN ,yN , (β- yN-1 )/2h )-β

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Example for Non-Linear BVP
Solve y”=(3/2)y2 with y(0)=4, y(1)=1 such that 0≤x≤1
using Newton Method
Solution :
yi+1 -2yi + yi-1 = (3/2) h2 (yi )2 = (3/2)(1/9)(yi )2
for i=1 we have,
y2 -2y1 + y0 =(1/6)(y1)2
for i=2 we have,
y3 -2y2 + y1 =(1/6)(y2 )2
So we get,
(y1)2 +12y1 -6y2 -24=0 ≡ F(y1,y2)
(y2 )2 -6y1 + 12y2 -6 =0 ≡ F(y1,y2)

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Example for Non-Linear BVP
Now, Jacobian J =
=
J-1 = 1/[(2y1+12)(2y2+12)-36]
∂F1 /∂y1 ∂F1 /∂y2
∂F2 /∂y1 ∂F2 /∂y2
2y1 +12 -6
-6 2y2 +12
2y1 +12 -6
-6 2y2 +12

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Example for Non-Linear BVP
Method : = - J-1 ((y1)N , (y2)N ) F ((y1)N , (y2)N )
for N=0
= - J-1 ((y1)0 , (y2)0 ) F ((y1)0, (y2)0 )
Now choose,
=
So we have, J
-1
= [1/(144-36)]
(y1)N+1
(y2)N+1
(y1)N
(y2)N
(y1)1
(y2)1
(y1)0
(y2)0
(y1)0
(y2)0
0
0
12 6
6 12

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Example for Non-Linear BVP
= (1/108)
And F((y1)0, (y2)0 ) =
= - (1/108) =
12 6
6 12
(y1)1
(y2)1
0
0
12 6
6 12
-24
-6
3
2
-24
-6

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References
Numerical Analysis (9th Edition) Richard L. Burden, J. Douglas Faires, 2010

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THANK
YOU