Page 294 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
,
,
F x 1 ,x x p p p 3 =  2 ...(3)
,
,
2
2
3
1
2
Here  and  are arbitrary constants. These equations are such that p , p , p can be found from
1 2 1 2 3
(1), (2), (3) as functions of x , x , x that make the equation
1 2 3
dz = p dx 1 p dx 2 p dx 3 ...(4)
2
1
3
integrable, for which the conditions are
p 2 = 2 z p 1 , p 3 2 z p 1 , p 3 p 2 ...(5)
x 1 x x 2 x 2 x 1 x x 3 x 1 x 2 x 3
1
1
Now by differentiating (1) partially with respect to x , keeping x , x constant, but regarding p ,
1 2 3 1
p , p , as dependent functions of x , x , x , we get
2 3 1 2 3
F F p 1 F p 2 F p 3 = 0 ...(6)
x 1 p 1 x 1 p 2 x 1 p 3 x 1
Similarly
F F p F p F p
1 1 1 1 2 1 3 = 0 ...(7)
x 1 p 1 x 1 p 2 x 1 p 3 x 1
F F
Multiplying equation (6) by 1 and equation (7) by , and subtracting we get
p 1 p 1
, F F 1 , F F 1 p 2 , F F 1 p 3
x ,p p ,p x p ,p x = 0 ...(8)
1 1 2 1 1 3 1 1
where
, F F 1 denotes "Jacobian" F F 1 F F 1 .
x 1 ,p 1 x 1 p 1 p 1 x 1

Similarly, like (8) we get

, F F , F F p , F F p
1 1 1 1 3
x ,p p ,p x p ,p x = 0 ...(9)
2 2 1 2 2 3 2 2
and

, F F 1 , F F 1 p 1 , F F 1 p 2
x 3 ,p 3 p 1 ,p 3 x 3 p 2 ,p 3 x 3 = 0 ...(10)

Add equation (8), (9) and (10) and noting that two pairs of terms are:

, F F p , F F p 2 z , F F , F F
1 2 1 1 1 1 0
p ,p x p ,p x x x p ,p p ,p
2 1 1 1 2 2 1 2 2 1 1 2
Similarly two other pairs of terms also vanish, leaving

, F F 1 , F F 1 , F F 1 = 0 ...(11)
x 1 ,p 1 x 2 ,p 2 x 3 ,p 3

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