How to solve this first order PDE using method of characteristics, cannot find any resources anywhere online.

Question

The PDE in question is $u_{x}u_{y}=1, u(x,0)=\sqrt{x}$. The process for solving a first order PDE using method of characteristics when the terms are summed is relatively well-established, but I cannot figure out how to generalize this to a first order PDE with the terms multiplied together.

Answer 1

Let me present a solution using the method of separation of variables. Let's assume that $u(x,y)=f(x)g(y)$; then \begin{align} u_xu_y=1 &\implies f'(x)g(y)f(x)g'(y)=1 \\ &\implies f(x)f'(x)=\lambda=\frac{1}{g(y)g'(y)}, \tag{1} \end{align} where $\lambda$ is a nonzero constant. Solving for $f(x)$ we obtain $$ \frac{1}{2}(f^2(x))'=\lambda\implies f^2(x)=2\lambda x+ a\implies f(x)=\pm\sqrt{2\lambda x+a}\,. \tag{2} $$ Similarly, $$ g(y)=\pm\sqrt{\frac{2y}{\lambda}+b}\,. \tag{3} $$ Combining $(2)$ and $(3)$ we obtain $$ u(x,y)=\pm\sqrt{(2\lambda x + a)\left(\frac{2y}{\lambda}+b\right)}\,. \tag{4} $$ The condition $u(x,0)=\sqrt{x}$ is satisfied by the positive solution, provided $a=0$ and $b=\frac{1}{2\lambda}$; hence, we finally obtain $$ u(x,y)=\sqrt{2\lambda x\left(\frac{2y}{\lambda}+\frac{1}{2\lambda}\right)}=\sqrt{4xy+x}\,. \tag{5} $$