Partial Differential Equation with no solution - Transversality condition? I have the following equation:
$$
x u_x + y u_y = \frac{2e^u }{xy } , x>0,y>0
$$
with the initial condition (corresponding to $t=0$ ):
$$
\Gamma =\{ (s,s,0) | 0<s<\infty  \}
$$
By using the transversality condition, we get that for this case, there is no solution at all. But, I can't see why (without using the transversality condition) . I know the general solution is:
$$
u(s,t) = \ln\left(\frac{c_1 (s) c_2 (s) }{e^{-2t } +c_3 (s) }\right)
$$
and in our specific case of $\Gamma$ : 
$$
x(s,t)=y(s,t)=se^t , 
u(s,t) = \ln\left(\frac{s^2 }{e^{-2t } +s^2 -1  }\right)
$$
Is there any way to see (without using the transversality condition) that this cannot be a solution to the equation ? 
Hope I made myself clear enough
Any help will be gratfully acknowledged ! 
 A: Using the method of characteristics to solve $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2e^u }{xy }$
The change of variables and function is : $u=u\left(x(t),y(t)\right)$
$$dt=\frac{dx}{x}=\frac{dy}{y}=\frac{du}{\frac{2e^u }{xy }}$$
Solving $dt=\frac{dx}{x}$ leads to : 
$$x=c_1(s)e^t$$
Solving $dt=\frac{dy}{y}$ leads to : 
$$y=c_2(s)e^t$$
Solving $dt=\frac{du}{\frac{2e^u }{xy }}$ :
$2dt=e^{-u}xy du=c_1(s)c_2(s)e^{2t}e^{-u}du$
$$u=-\ln\left(\frac{e^{-2t}}{c_1(s)c_2(s)}+c_3(s)\right)=\ln\left(\frac{c_1(s)c_2(s)}{e^{-2t}+c_1(s)c_2(s)c_3(s)}\right)$$
$c_1(s)c_(s)=xe^{-t}ye^{-t}=xye^{-2t}$
$$u=\ln\left(\frac{xye^{-2t}}{e^{-2t}+c_3(s)xye^{-2t}}\right)=\ln\left(\frac{1}{\frac{1}{xy}+c_3(s)}\right)$$
$\frac{x}{y}=\frac{c_1(s)e^t}{c_2(s)e^t}=\frac{c_1(s)}{c_2(s)}=$ any function of $s$
in inverse $s=$any function of $\frac{x}{y}$ and $c_3(s)=$any function of $\frac{x}{y}$ hense $c_3(s)=f\left(\frac{x}{y}\right)$ any function $f$
$$u=\ln\frac{1}{\frac{1}{xy}+f\left(\frac{x}{y}\right)}$$
Up to now no boundary condition is involved.
BOUNDARY CONDITIONS :
Apparently, it is the key point to clarify. As far as I can understand, the wording of the boundary conditions is :
$$
\Gamma =\{ (s,s,0) | 0<s<\infty  \}
$$
So, if I well understand the symbols, at $t=0$ we have $x=y=s$ and, as a consequence $c_1(s)e^0=c_2(s)e^0=s$ which determines those functions. 
$$u(s,t=0)=\ln\frac{1}{\frac{1}{s^2}+f(1)}=\ln\frac{s^2}{1+f(1)s^2}$$
I don't see in the given boundary conditions that $u(s,0)=0$ which could determine $f(1)=0$. 
At this point, $f(s)$ is not yet determined. Accordingly, I cannot understand why $c_3(s)=s^2-1$ with the symbol of $c_3$ used by "great Matematician" ( which is different from the symbol $c_3$ used in my answer).
A: $$
x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2e^u }{xy } , x>0,y>0
$$
Let $u=-\ln(v)$
$$
-\frac{x}{v} \frac{\partial v}{\partial x} - \frac{y}{v} \frac{\partial v}{\partial y} = \frac{2 }{xy\:v }$$
$$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = -\frac{2 }{xy }$$
Let $x=e^X $ and $y=e^Y$
$$\frac{\partial v}{\partial X} + \frac{\partial v}{\partial Y} = -2e^{-X-Y}$$
The solution of the homogeneous PDE $\frac{\partial V}{\partial X} + \frac{\partial V}{\partial Y} =0$ is $V=f(X-Y)$, any derivable function $f$.
An obvious particular solution of $\frac{\partial v_p}{\partial X} + \frac{\partial v_p}{\partial Y} = -2e^{-X-Y}$ is $v_p=e^{-X-Y}$.
The general solution of $\frac{\partial v}{\partial X} + \frac{\partial v}{\partial Y} = -2e^{-X-Y}$ is :
$$v=f(X-Y)+e^{-X-Y}$$
$$v=f\left(\ln(x)-\ln(y)\right)+e^{-\ln(x)-\ln(y)}=f\left(\frac{x}{y}\right)+\frac{1}{xy}$$
And, coming back to $u(x,y)=-\ln(v)$ we obtain the general solution of the initial ODE $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2e^u }{xy }$ :
$$u(x,y)=-\ln\left( f\left(\frac{x}{y}\right)+\frac{1}{xy}\right)$$
where $f$ is any derivable function.
It remains to take account to the boundary conditions in order to determine the function $f$.
Unfortunately, I do not understand anything of the wording about the boundary conditions. So, I cannot go further. 
Supposing that the boundary condition is $x(s,t)=y(s,t)=s e^t$ and nothing else, then
$$u(s,t)=-\ln\left( f\left(\frac{s e^t}{s e^t}\right)+\frac{1}{s^2 e^{2t}}\right)$$
$f\left(\frac{s e^t}{s e^t}\right)=f(1)=$constant
$$u(s,t)=-\ln\left( C+\frac{1}{s^2 e^{2t}}\right)=\ln\frac{s^2}{C s^2+ e^{-2t}}$$
This equation is different from the equation given in the wording of the question. That is why I do not understand correctly the boundary conditions as they are defined.
