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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 !

Alex
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Great Mathematician
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2 Answers2

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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).

JJacquelin
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0

$$ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2e^u }{xy } , x>0,y>0 $$ 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.

JJacquelin
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  • thanks a lot for your answer, but I can't understand how does it help me. Are you familiar with the method of characteristics? – Great Mathematician May 08 '15 at 13:43
  • The general solution of a PDE with variables $x$ and $y$ is function of $x$ and $y$, not of a variable $s$ which is not in the PDE. That was what I intended to point out. Of course, later on, taking account of the boundary conditions and if the boundary conditions are function of another variable $s$, then the related particular solution of the PDE (not the general solution) becomes function of $s$ as you write it. – JJacquelin May 08 '15 at 14:34
  • But does this help us to see that together with the initial condition I pointed , the equation has no solution ? – Great Mathematician May 08 '15 at 14:39
  • See the addition at the end of my first answer. Please, comment it. – JJacquelin May 08 '15 at 15:49
  • JJacquelin, are you familiar with the method of characteristics ? I will be glad to detail my solution so that you will see how I arrived to the mentioned solution. Thanks a lot for your patience – Great Mathematician May 08 '15 at 16:43
  • I also used the method of characteristics to solve the PDE $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2e^u }{xy }$ and I obtained the same result than with the above direct method of solving : $$u(x,y)=-\ln\left( f\left(\frac{x}{y}\right)+\frac{1}{xy}\right)$$ where $f$ is any derivable function. Would you mind detail your solution in order to see where is the difference compared to my own calculus. – JJacquelin May 09 '15 at 08:11
  • OK. Great! The characteristic equations are: $x_t = x , y_t = y , u_t = \frac{2e^u}{xy}$ which implies: $x=c_1(s) e^t , y=c_2(s) e^t $ and $u=ln(\frac{c_1 (s) c_2 (s)}{e^{-2t}+c_3(s) } )$ and this is our parameterization of the characteristic curves. Now, after substituting: $x(s,0)=y(s,0)=s, u(s,0)=0$ we obtain: $$ x(s,t)=y(s,t)=se^t , u(s,t) = ln(\frac{s^2}{e^{-2t}+s^2-1} ) $$ i.e. - $c_1(s)=c_2(s)=s$ , $c_3(s)=s^2-1$ . What do you think? I can't find any error. Thanks a lot ! – Great Mathematician May 09 '15 at 09:07
  • I posted another answer, using the method of characteristics. So that, one can compare with your results. – JJacquelin May 09 '15 at 09:24
  • At this point I see no contradiction between my and your answers, because both are not comparable. My answer doesn't take account of boundary condition and your answer is on the form of $u(s,t)$ instead of $u(x,y)$. – JJacquelin May 09 '15 at 09:46
  • Apparently, the key point is the definition of the boundary conditions and the consistensy of the boundary conditions between the system of variables $(x,y)$ and $(s,t)$. I made an addition to my previous answer; but I continue to not understand exactly what are the boundary conditions as I said at the first beginning. – JJacquelin May 09 '15 at 21:40
  • Why do you suppose $u(x=s,y=s,t=0)=0$ ? Is there something about in the context of the problem ? By the way, what is the context, where the PDE is comming from, eventually with what boundary conditions ? – JJacquelin May 10 '15 at 10:30