#### 2.1.29 $$(y-z) u_x + (z-x) u_y + (x-y) u_z = 0$$ (Example 3.5.4 in Lokenath Debnath)

problem number 29

From example 3.5.4, page 212 nonlinear pde’s by Lokenath Debnath, 3rd edition.

First order PDE of three unknowns. Solve for $$u(x,y,z)$$ \begin {align*} (y-z) u_x + (z-x) u_y + (x-y) u_z &= 0 \end {align*}

Mathematica

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Maple

$u \left (x , y , z\right ) = c_{3} c_{4} c_{5} {\mathrm e}^{c_{1} x} {\mathrm e}^{c_{1} y} {\mathrm e}^{c_{1} z} {\mathrm e}^{\frac {c_{2} x^{2}}{2}} {\mathrm e}^{\frac {c_{2} y^{2}}{2}} {\mathrm e}^{\frac {c_{2} z^{2}}{2}}$

Hand solution

Solve $\left ( y-z\right ) u_{x}+\left ( z-x\right ) u_{y}+\left ( x-y\right ) u_{z}=0$ The following method works only when the sum of the coeﬃcients of $$u_{x},u_{y},u_{z}$$ is zero. This is the case here. Hence we write\begin {align} du & =0\tag {1}\\ dx+dy+dz & =\left ( y-z\right ) +\left ( z-x\right ) +\left ( x-y\right ) =0\tag {2} \end {align}

We need one more equation which is\begin {equation} xdx+ydy+ydz=x\left ( y-z\right ) +y\left ( z-x\right ) +z\left ( x-y\right ) =0\tag {3} \end {equation} Integrating (1,2,3) gives\begin {align*} u & =C_{1}\\ x+y+z & =C_{2}\\ x^{2}+y^{2}+z^{2} & =C_{3} \end {align*}

And since we know that $$C_{1}=F\left ( C_{2},C_{3}\right )$$ where $$F$$ is arbitrary function, then the solution is$u\left ( x,y,z\right ) =F\left ( x+y+z,x^{2}+y^{2}+z^{2}\right )$ Again, the above method only worked because of the special value of the coeﬃcients. If the PDE was $$\left ( 2y-z\right ) u_{x}+\left ( z-x\right ) u_{y}+\left ( x-y\right ) u_{z}=0$$ for example, then the above method will not work and we have to use the Lagrange-Charpit equations. But since the equations will be coupled, this would make the solution harder.

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