Question

Find the complete integral of the p.d.e P2x+Q2x= Pz using Jacobi's method

Answer 1

Step 1: (Converting the given PDE into the form $f(x, y, z, u_x, u_y, u_z)$ = 0.

Set $p = −\dfrac{ux}{uz} and \text q = −\dfrac{u_y}{u_z}$ in the given PDE to obtain

$\dfrac{(u_x)^2}{(u_z)^2} x + \dfrac{u_y^2}{ u_z^2} y = z =⇒ xu_x^2 + yu_y^2 − zu_z^2 = 0$ .

Thus,

$f(x, y, z, u_x, u_y, u_z) = xu_x^2 + yu_y^2 − zu_z^2 = 0$ .

Step 2: Solving above PDE by Jacobi’s method

Compute $f_{u_x} , f_{u_y} , f_{u_z} , f_x, f_y, f_z$

$f_{u_x} = 2xu_x, f_{u_y} = 2yu_y, f_{u_z} = −2zu_z, f_x = u_x^2 , f_y = u_y^2 , f_z = −u_z^2$ .

.

Step 2(b): Writing auxiliary equation and solving for $u_x, u_y and u_z$ .

The auxiliary equations are given by

$\dfrac{dx}{ f_{u_x}} = \dfrac{dy}{ f_{u_y}} = \dfrac{dz}{ f_{u_z}} = \dfrac{du_x}{ −f_x} = \dfrac{du_y}{ −f_y} = \dfrac{du_z}{ −f_z}$

⇒ $\dfrac{dx}{ 2xu_x} = \dfrac{dy}{ 2yu_y} = \dfrac{dz}{ −2zu_z} = \dfrac{du_x}{ −u_x^2} = \dfrac{du_y}{ −u_y^2} = \dfrac{du_z}{ u_z^2}$

Now, $\dfrac{dx }{2xu_x} = \dfrac{du_x}{ −u_x^2} ⇒ \dfrac{u_xdx}{ 2xu_x^2} = −\dfrac{2xdu_x}{ 2xu_x^2}$

$⇒ u_xdx = −2xdu_x \\= ⇒ \dfrac{dx}{ x} = −2 \dfrac{du_x} {u_x} \\ ⇒ log x = −2 log(ux) + log(a) \\ ⇒ log x + log(u_x^2 ) = log(a)\\ ⇒ xu_x^2 = a \\ ⇒ u_x = (\dfrac{a}{ x} )^\frac{1}{2}$

. Similarly, we get

.$yu_y^2 = b ⇒ u_y = (\dfrac{ b}{ y} )^\frac{1}{2}$

and

$u_z = [ \dfrac{(a + b) }{2} ]^\frac{1}{2}$

Step 2(c): Solving the equation .$du = u_xdx + u_ydy + u_zdz$

$du = (\dfrac{a }{x} )^\frac{1}{2} dx + (\dfrac{ b}{ y} )^\frac{1}{2} dy + (\dfrac{ a + b}{ z} )^\frac{1}{2} dz\\ ⇒ u = 2(a_x)^ \frac{1}{2} + 2(b_y)^ \frac{1}{2} + 2((a + b)z)^ \frac{1}{2} + c.$