2.1.32 \((y-x)u_x+(u-x)u_y=x-y\) with \(u=0\) on \(x y=1\) Example 3.5.7 in Lokenath Debnath

problem number 32

Added June 2, 2019.

From example 3.5.7, page 215 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\) \begin {align*} (y-u)u_x+(u-x)u_y&=x-y \end {align*}

with \(u=0\) on \(x y=1\)

Mathematica ✗

ClearAll["Global`*"]; 
pde =  (y-u[x,y])*D[u[x, y], x] +(u[x,y]-x)*D[u[x, y], y] ==x-y; 
ic  = u[x,1/x]==0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic} ,u[x, y], {x, y}], 60*10]];
 

Failed Kernel error

Maple ✗

restart; 
pde :=(y-u(x,y))*diff(u(x,y),x)+(u(x,y)-x)*diff(u(x,y),y)=x-y; 
ic  := u(x,1/x)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
 

unexpected occurrence of the variables x in the 2nd operand of u(x,1/x) in the given initial conditions Maple does not accept this form of Cauchy data

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