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Finite Difference Methods
Gaurav MallikSAU/AM(M)/2014/22
South Asian University
Rupali SharmaSAU/AM(M)/2014/27
South Asian University
Divyansh VermaSAU/AM(M)/2014/14South 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.
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Finite Difference Method for Linear Problem
The finite difference method for the linear second-order BVPy‘’ = 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.
x1 xN+1 x2 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
+,ξ i-) 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)/2h1+ y0 = (y1 - y-1)/2hy-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 01 -2-(1/27) 10 1 -2-(2/27)
y0
y1
y2
10-1
=
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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≤1using 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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