2.1.29 (yz)ux+(zx)uy+(xy)uz=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) (yz)ux+(zx)uy+(xy)uz=0

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(x,y,z)=c3c4c5ec1xec1yec1zec2x22ec2y22ec2z22

Hand solution

Solve (yz)ux+(zx)uy+(xy)uz=0 The following method works only when the sum of the coefficients of ux,uy,uz is zero. This is the case here. Hence we write(1)du=0(2)dx+dy+dz=(yz)+(zx)+(xy)=0

We need one more equation which is(3)xdx+ydy+ydz=x(yz)+y(zx)+z(xy)=0 Integrating (1,2,3) givesu=C1x+y+z=C2x2+y2+z2=C3

And since we know that C1=F(C2,C3) where F is arbitrary function, then the solution isu(x,y,z)=F(x+y+z,x2+y2+z2) Again, the above method only worked because of the special value of the coefficients. If the PDE was (2yz)ux+(zx)uy+(xy)uz=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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