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It's said in a book I am reading that:

The form $X_1\ dx_1+X_2\ dx_2+X_3\ dx_3$ has an integrating factor, i.e: $$ X_1\left(\frac{\partial X_3}{\partial x_2}-\frac{\partial X_2}{\partial x_3}\right) +X_2\left(\frac{\partial X_1}{\partial x_3}-\frac{\partial X_3}{\partial x_1}\right) +X_3\left(\frac{\partial X_2}{\partial x_1}-\frac{\partial X_1}{\partial x_2}\right)=0$$ The above should be derived from the following condition: "If there exists an integrating factor, $\lambda(x_1,x_2,x_3)$, such that $\lambda(X_1\ dx_1+X_2\ dx_2+X_3\ dx_3)=d\phi(x_1,x_2,x_3)$, then we have: $$\frac{\partial \phi}{\partial x_1} = \lambda X_1 , \ \frac{\partial \phi}{\partial x_2} = \lambda X_2 , \ \frac{\partial \phi}{\partial x_3} = \lambda X_3$$ By setting $\displaystyle \frac{\partial^2 \phi}{\partial x_1 \partial x_2} = \frac{\partial^2 \phi}{\partial x_2 \partial x_1}$, etc. and eliminating $\lambda$ from them, the above equation is derived as the condition of integrability.

Here's what I tried so far: I got after equating the mixed partial derivatives the following condition: $$\lambda \left(\frac{\partial X_2}{\partial x_1}-\frac{\partial X_1}{\partial x_2}\right) = \frac{\partial \lambda}{\partial x_2}X_1-\frac{\partial \lambda}{\partial x_1} X_2$$ $$\lambda \left(\frac{\partial X_3}{\partial x_2}-\frac{\partial X_2}{\partial x_3}\right) = \frac{\partial \lambda}{\partial x_3}X_2-\frac{\partial\lambda}{\partial x_2} X_3$$ $$\lambda \left(\frac{\partial X_1}{\partial x_3}-\frac{\partial X_3}{\partial x_1}\right) = \frac{\partial \lambda}{\partial x_1}X_3-\frac{\partial\lambda}{\partial x_3} X_1$$

How to procceed from here?

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  • $\begingroup$ Are $X_1, X_2, X_3$ all known functions of $x_1,x_2,x_3$? $\endgroup$ – Dylan Nov 21 '17 at 22:51
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For ease of notation, I'm going to write $(x_1,x_2,x_3) = (x,y,z)$, $(X_1,X_2,X_3) = (X,Y,Z)$ and the partials as subscripts. The system has the form

$$ A\lambda = Y\lambda_z - Z\lambda_y \tag{1} $$ $$ B\lambda = Z\lambda_x - X\lambda_z \tag{2} $$ $$ C\lambda = X\lambda_y - Y\lambda_x \tag{3} $$

where $(A,B,C) = (Z_y - Y_z,X_z - Z_x, Y_x - X_Y) = \nabla \times (X,Y,Z)$

Cancel out $\lambda_z$ from $(1)$ and $(2)$ to get

$$ (AX + BY)\lambda = -Z(X\lambda_y -Y\lambda_z) = -CZ\lambda $$

Thus we have

$$ (AX + BY + CZ)\lambda = 0 $$

In order for a solution to exist, we need $$ AX + BY + CZ = (X,Y,Z) \cdot (A,B,C) = 0 $$

If this condition is satisfied, we maybe able to find a solution in two variables if $A = A(y,z)$, $B = B(z,x)$ or $C = C(x,y)$

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