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

problem number 78

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

Failed

Maple

$u \left (x , y\right ) = x y$

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$u\left ( x,y\right ) =xy$