2.1.78 \(u_x u_y = u\) with \(u(x,0)=0,u(0,y)=0\)

problem number 78

Added May 27, 2019.

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

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

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

With \(u(x,0)=0,u(0,y)=0\).

Mathematica ✓

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

\[\{\}\]

Maple ✓

restart; 
pde := diff(u(x,y),x)*diff(u(x,y),y)=u(x,y); 
bc  := u(x,0)=0,u(0,y)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y),HINT=`*`)),output='realtime'));
 

\[u \left (x , y\right ) = x y\]

Hand solution

Solve \(u_{x}u_{y}=u\) with \(u\left ( x,0\right ) =0,u\left ( 0,y\right ) =0\).

Let \(u=X\left ( x\right ) Y\left ( y\right ) \). The pde becomes \(\left ( X^{\prime }Y\right ) \left ( Y^{\prime }X\right ) =XY\) or \(X^{\prime }Y^{\prime }=1\) or \(X^{\prime }=\frac {1}{Y^{\prime }}\). Hence both sides are constant, say \(\lambda \). Therefore we have 2 differential equations to solve

\[ X^{\prime }=\lambda \]

And

\[ \frac {1}{Y^{\prime }}=\lambda \]

The solution to the first one is \(X\left ( x\right ) =\lambda x+C_{1}\). The solution to the second equation is \(Y\left ( y\right ) =\frac {1}{\lambda }y+C_{2}\).

Since \(u=X\left ( x\right ) Y\left ( y\right ) \) then the solution is

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

Using the condition \(u\left ( x,0\right ) =0\) gives

\[ 0=C_{2}\lambda x+C_{1}C_{2}\]

Differentiating gives

\[ 0=C_{2}\lambda \]

But \(\lambda \neq 0\) for nontrivial solution, hence \(C_{2}=0\). Therefore (1) reduces to

\[ u\left ( x,y\right ) =xy+\frac {C_{1}}{\lambda }y \]

Using the condition \(u\left ( 0,y\right ) =0\) gives

\[ 0=\frac {C_{1}}{\lambda }y \]

Hence \(C_{1}=0\). Therefore the solution becomes

\[ u\left ( x,y\right ) =xy \]