Page 298 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 19: Jacobi’s Method for Solving Partial Differential Equations

as F , F do not contain x , x , x and x . Notes
1 2 1 2 3 4
From (3) and (5) we have
a
P a , P 3
1 1 4
x
4
From (4) we have
P = P + a ...(7)
2 3 2
Substituting in (2) we have
2
x 2 x 3 2P 3 a 2 a a = 0
1 3
a a
P P 2P a = 1 3 ...(8)
2 3 3 2
x 2 x 3
a a
1 3
 2P = a ...(9)
2 2 x 2 x 3
a a
2P = a 1 3 ...(10)
3 2
x 2 x 3
du = P dx + P dx + P dx + P dx
1 1 2 2 3 3 4 4
a dx 4 a 2 1 a a
1 3
3
= a dx dx dx dx dx
1 1 2 3 2 3
x 2 2 x x
4 2 3
on integration we get
a 1 1/2
u a x a log x 2 x x a a x x a
1 1 3 4 2 3 1 3 2 3 4
2 2
so u = 0 gives, replacing x by z, and dividing by a we have
4 3
a 1 x log z a 2 x x a 1 x x 1/2 a 4 0
a 1 2a 2 3 a 2 3 a
3 3 3 3
a a a
Let 1 A 1 , 2 A 2 , 4 A we have the required equation:
3
a 2a a
3 3 3
1/2
log z A x A x 2 x 3 A x 2 a 3 A 3 0 ...(11)
2
1
1
3. Solve
2
2
p x + q x = z ...(1)
1 2
Solution:
Let z = x ; let u x ,x ,x 0 be the solution.
3 1 2 3
z x 3 p 1 u u
p , where P 1 ,P 3
x 1 x 1 P 3 x 1 x 3
z x P u
q 3 2 where P 2
x 2 x 2 P 3 x 2

LOVELY PROFESSIONAL UNIVERSITY 291