Question

solve f(x, u_x,u_z)=g(y,u_y,u_z) by Jacobi method.

Answer 1

Let $f(x,y,z,u_x,u_y,u_z) = 0$ (1)

The Fundamental idea of the Jacobi method is to introduce two first order PDE involving two arbitrary constants a and b of the following form,

$h_1(x,y,z,u_x,u_y,u_z,a) = 0$ (2)

$h_2(x,y,z,u_x,u_y,u_z,b) = 0$ (3)

such that $\frac{\partial(f,h_1,h_2)}{\partial(u_x,u_y,u_z)} \neq 0$ (4)

and

Equations (1),(2) and (3) can be solved for $u_x,u_y,u_z$

and equation $du = u_xdx + u_ydy+u_zdz$ is integrable

If $h_1 = 0$ and $h_2=0$ are compitable with$f =0$ then $h_1,h_2$ satisfy,

$\frac{\partial(f,h)}{\partial(x,u_x)}+\frac{\partial(f,h)}{\partial(y,u_y)}+\frac{\partial(f,h)}{\partial(z,u_z)} =0$ for $h = h_i$, i =1,2

Last equation will lead to the PDE of the form,

$f_{u_x}\frac{\partial h}{\partial x}+f_{u_y}\frac{\partial h}{\partial y}+f_{u_z}\frac{\partial h}{\partial z} - f_x\frac{\partial h}{\partial u_x}- f_y\frac{\partial h}{\partial u_y}- f_z\frac{\partial h}{\partial u_z} = 0$

It's auxiliary equation will be,

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

After getting this we need to those fractions which will lead to the solution.

Since given equation is

$f(x,u_x,u_z) = g(y,u_y,u_z)$

it can be simplified as

$f(x,u_x,u_z) - g(y,u_y,u_z) = 0$

or it can be written as

$F(x,y,u_x,u_y,u_z) = 0$

Then auxiliary equation will be

$\frac{dx}{F_{u_x}}=\frac{dy}{F_{u_y}} =\frac{dx}{F_{u_z}}= \frac{du_x}{-F_x}= \frac{du_y}{-F_y}$

Solve it to obtain the solution.