2.1.31 \(u_x-u_y=1\) with \(u(x,0)=x^2\) Example 3.5.6 in Lokenath Debnath

problem number 31

Added June 2, 2019.

From example 3.5.6, page 214 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\) \begin {align*} u_x-u_y&=1 \end {align*}

with \(u(x,0)=x^2\)

Mathematica

ClearAll["Global`*"]; 
pde =  D[u[x, y], x] - D[u[x, y], y] ==1; 
ic  = u[x,0]==x^2; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde,ic} ,u[x, y], {x, y}], 60*10]];
 

\[\left \{\left \{u(x,y)\to x^2+2 x y+(y-1) y\right \}\right \}\]

Maple

restart; 
pde := diff(u(x,y),x)-diff(u(x,y),y)=1; 
ic  := u(x,2)=x^2; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,y))),output='realtime'));
 

\[u \left (x , y\right ) = -y +\left (x +y -2\right )^{2}+2\]

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