Explicitly Solving a PDE: Is There a Solution?

[Clever-Name]

I just had a 'quiz' in my PDE class today and there was a problem my friends and I are convinced has no explicit solution. I want to know if maybe we are doing something wrong?

Homework Statement

[tex](x+y)u_{x} + yu_{y} = 0 [/itex]<br /> <br /> [tex]u(1,y) = \frac{1}{y} + ln(y) [/itex]<h2>Homework Equations</h2><br /> ...<h2>The Attempt at a Solution</h2><br /> Here was my approach via method of characteristics:<br /> <br /> $$\frac{dy}{dt} = y \to y(t) = Se^{t}$$<br /> $$y(t=0) = S$$<br /> <br /> $$\frac{dx}{dt} = x+y$$<br /> $$x'-x = Se^{t}$$<br /> Via integrating factor we arrive at:<br /> $$x(t) = Ste^{t} + g(s)e^{t}$$<br /> $$x(t=0) = 1 \to 1 = g(s)$$<br /> <br /> Thus we are left with:<br /> <br /> $$x(t) = Ste^{t} + e^{t}$$<br /> and<br /> $$y(t) = Se^{t}$$<br /> <br /> You can check for yourself that the following is true:<br /> <br /> $$\frac{dU(x(t),y(t))}{dt} = (x+y)u_{x} + yu_{y} = 0$$<br /> <br /> Thus our general solution is:<br /> <br /> $$U(t) = f(s)$$<br /> <br /> A general function of s.<br /> <br /> However, we have found that it is impossible to explicitly solve for s. Am I missing something?[/tex][/tex]

 

[Coelum]

I suspect the problem cannot be solved. Along the initial line $y=1$ you can differentiate the initial conditions and get

$u_y(1,y)=-1/y^2+1/y=(y-1)/y^2=0$

that you can substitute back into the PDE in order to obtain $u_x$

$(x+1)u_x=0 → u_x=0$

which means $u(x,y)=u(y)$. Hence, $u(x,y)$ should be equal to the initial conditions everywhere. But this is not true, as one can check substituting
$u_x=0$ into the PDE:

$yu_y=y(1-y)/y^2=0$

true only when $y=0$ or $y=1$.

I am not an expert of the method of characteristics, but I believe the curve on which the initial conditions are given is "noncharacteristic".