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 ✗
ClearAll["Global`*"]; pde = (y - z)*D[u[x, y, z], x] + (z - x)*D[u[x, y, z], y] + (x - y)*D[u[x, y, z], z] == 0; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y, z], {x, y, z}], 60*10]];
$Aborted
Maple ✓
restart; pde :=(y-z)*diff(u(x,y,z),x)+(z-x)*diff(u(x,y,z),y)+(x-y)*diff(u(x,y,z),z)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y,z),'build')),output='realtime'));
\[u \left ( x,y,z \right ) ={{\rm e}^{{\frac {{\it \_C2}\,{x}^{2}}{2}}}}{{\rm e}^{{\it \_C1}\,x}}{{\rm e}^{{\frac {{\it \_C2}\,{y}^{2}}{2}}}}{{\rm e}^{{\it \_C1}\,y}}{\it \_C3}\,{\it \_C5}\,{\it \_C4}\,{{\rm e}^{{\frac {{\it \_C2}\,{z}^{2}}{2}}}}{{\rm e}^{{\it \_C1}\,z}}\]
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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