2.1.34 \(x u_x+y u_y=x e^{-u}\) with \(u=0\) on \(y=x^2\) Example 3.5.10 in Lokenath Debnath

problem number 34

Added June 2, 2019.

From example 3.5.10, page 218 nonlinear pde’s by Lokenath Debnath, 3rd edition.

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

with \(u=0\) on \(y=x^2\)

Mathematica ✓

ClearAll["Global`*"]; 
pde =  x*D[u[x, y], x] +y*D[u[x, y], y] ==x*Exp[-u[x,y]]; 
ic  = u[x,x^2]==0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to \log \left (-\frac {y}{x}+x+1\right )\right \}\right \}\]

Maple ✓

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

\[u \left ( x,y \right ) =\ln \left ( x+{\it \_F1} \left ( {\frac {y}{x}} \right ) \right ) \] Maple does not accept this form of Cauchy data

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