Page 300 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

2. Find the complete integral for Notes
p x p p x x 0
3 3 1 2 1 2

19.2 Simultaneous Partial Differential Equations

In Jacobi’s method two additional equations are needed to solve the partial differential equation
by Jacobi’s method.
In this section the problem of finding the solution of the partial differential equation F = 0 with
some work of finding F is already done. The method can be illustrated by the following
1
examples:

Example 1: Find the complete integral for the partial differential equations.
F = p x p x p 2 0 ...(1)
1 1 2 2 3
F = p  p + p  1 = 0 ...(2)
1 1 2 3
Here

F F 1 F F 1 F F 1 F F 1 F F 1 F F 1
, F F 1 = 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
= p 1 1 x 1 (0) p 2 ( 1) x 2 (0) 0 (1) 2 (0) p 1 p 2 ...(3)
3
Now F ,F 1 0, now to make
, F F 1 = 0, we have p = p 2 ...(4)
1
From equation (2) p = 1 ...(5)
3
1
So From (1), p x 1 x 2 1 = 0, so p 1 ...(6)
1
x 1 x 2
dz = p dx + p dx + p dx
1 1 2 2 3 3
dx dx
= 1 2 1dx 3
x x x x
1 2 1 2
or

dx dx
dz = 1 2 dx 3 ...(7)
x x
1 2
on integrating (7) we have
z = log x 1 x 2 x 3 a ...(8)
which is the complete integral of (1).

Example 2: Find the complete integral for
F = 2x p p  x p = 0 ...(1)
3 1 3 4 4
F1 = 2p  p = 0 ...(2)
1 2
Now
F F F F F F F F
, F F 1 = 1 1 1 1
x p p x x p p x
1 1 1 1 2 2 2 2

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