Page 296 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

So the auxiliary equations are Notes
dx 1 dp 1 dx 2 dp 2 dx 3 dp 3
2x x 2p x 3x 2 2p p 0 p 2 2p x 6p x 2 ...(3)
1 3 1 3 2 2 3 2 1 1 2 3
of which integrals are obtained by integrating the equations

dx 1 dp 1
x 1 = p 1
dp = 0
2
or

F = x p = a ...(4)
1 1 1 1
F = p = a ...(5)
2 2 2
Now consider
F 1 F 2 F 1 F 2 F 1 F 2 F 1 F 2 F 1 F 2 F 1 F 2
F 1 ,F 2 = x 1 p 1 p 1 x 1 x 2 p 2 p 2 x 2 x 3 p 3 p 3 x 3

= p (0)  x (0) + 0 + 0 + 0 +0 = 0
1 1
So equations (4) and (5) can be taken as the two additional equations required. So
a
p = 1 , p a
1 x 1 2 2

And from equation (1) we have

p = 2x a 3a x 2 a 2 2a x 3a 2 3 x a 2
3 3 1 2 3 2 1 3 2 2
Hence
dz = p dx + p dx + p dx
1 1 2 2 3 3
a dx 2 dx
= 1 1 a dx 2 2a x 3a x 3
2 3
1 3
2
x 1 a 2 2
So on integration we get
1 2 3
z = a 1 log x 1 a x 2 a x a x a 3
2 2
2 3
1 3
a 2
as the complete integral.
2. Solve
2
x 2 x 3 p 2 p 3 zp 1 0 ...(1)
Solution:
This equation is not of Jacobi’s type as it involves z. But put
z = x
4
z x u u
so p = 4 p /p …(say)
1 x 1 x 1 x 1 x 4 1 4

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