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 ✓

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]];
 

\[\{\{u(x,y,z)\to c_1(x+y+z,-x (y+z)-y z)\}\}\]

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'));
 

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

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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