General solution to a first-order partial differential equation

Question

$$ \begin{cases} \displaystyle u(x+u)\frac {\partial }{\partial } - y(y+u)\frac {\partial }{\partial } = 0 \\ u=\sqrt y ,x =1 \end{cases} $$

my idea: Can we solve by the method : $$\frac {x}{u(+u )} = \frac{y}{-y( +u)}$$

Answer 1

This problem is solvable by using the method of characteristics. Letting $t$ be the characteristic coordinate, we obtain this system of differential equations $$\frac{dx}{dt} = u(x+u), \quad\frac{dy}{dt} = -y(y+u), \quad\frac{du}{dt} = 0$$ We can easily solve for $u$ provided the initial condition that $u(x(0),y(0))=\sqrt{y_0}$, $$u = \sqrt{y_0}$$ Plugging this result into the differential equation for $x$ and integrating gives $$x(t)=ce^{\sqrt{y_0}t}-\sqrt{y_0}$$ Using the initial condition that $x(0)=1$ gives $$x(t) = \left(1+\sqrt{y_0}\right)e^{\sqrt{y_0}t}-\sqrt{y_0}$$ Upon solving the differential equation for $y$, we see after applying the initial condition $y(0)=y_0$ (and after some algebra) that $$y(t)=\frac{y_0}{\left(1+\sqrt{y_0}\right)e^{\sqrt{y_0}t}-\sqrt{y_0}}$$ We can rewrite the equation for $y$ simply in terms of $u$ and $x$ thereby inverting the equation and eliminating the characteristic coordinate $$y=\frac{u^2}{x}$$ Rearrangement gives the simple solution to the complicated-appearing PDE $$\boxed{u(x,y) = \sqrt{xy}}$$ Plugging this back into the original PDE verifies the result. May the Fourth be with you.

Answer 2

Comparing

$$ \left\{ \begin{array}{rcl} u(x+u)u_x-y(y+u)u_y & = & 0\\ u_x dx + u_y dy & = & du \end{array} \right. $$

we conclude $u(x,y) = C_0$ along the characteristic curves and

$$ \frac{dx}{x+C_0}= - \frac{C_0dy}{y(y+C_0)} $$

Integrating both sides

$$ \ln(C_0+x)+\ln(y)-\ln(y+C_0) = C_1 $$

From now on please follow the LutzL comments below.

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(LutzL copy from comment) $$ \frac{(x+C_0)y}{y+C_0}=\pm e^{C_1}=C_2=\phi(C_0), $$ as the solution is a one-parameter family of characteristic curves, that is there is only one degree of freedom between the two integration constants. Then insert the initial condition $C_0=u(1,y)=\sqrt{y}=t$, $y=t^2$, to get $$ \phi(t)=\frac{(1+t)t^2}{t^2+t}=t $$ so that (using again that $C_0=u(x,y)$) $$ (x+C_0)y=C_0(y+C_0)\iff xy=C_0^2=u(x,y)^2. $$