2.1.77 \(u_x u_y = 1\)

problem number 77

Added May 27, 2019.

From UMN Math 5587 HW2, Fall 2016, problem 5(a).

Solve for \(u(x,y)\)

\begin{align*} u_x u_y = 1 \end{align*}

Mathematica ✓

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

\[\left \{\left \{u(x,y)\to \frac {x}{c_2}+c_2 y+c_1\right \}\right \}\]

Maple ✓

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

\[u \left (x , y\right ) = \frac {x}{\textit {\_c}_{2}}+c_{1} +\textit {\_c}_{2} y +c_{2}\]

Hand solution

Solve \(u_{x}u_{y}=1\). Let \(u=f\left ( x\right ) +g\left ( y\right ) \). The pde becomes \(f^{\prime }\left ( x\right ) g^{\prime }\left ( y\right ) =1\) or \(f^{\prime }\left ( x\right ) =\frac {1}{g^{\prime }\left ( y\right ) }\). Hence both are constant, say \(\lambda \). Therefore \(f^{\prime }\left ( x\right ) =\lambda \) or \(f\left ( x\right ) =\lambda x+C_{1}\) and similarly, \(\frac {1}{g^{\prime }\left ( y\right ) }=\lambda \) or \(g^{\prime }\left ( y\right ) =\frac {1}{\lambda }\) or \(g\left ( y\right ) =\frac {y}{\lambda }+C_{2}\). Therefore the solution becomes

\begin{align*} u\left ( x,y\right ) & =f\left ( x\right ) +g\left ( y\right ) \\ & =\lambda x+C_{1}+\frac {y}{\lambda }+C_{2}\end{align*}

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