# Unsolvable PDE?

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?

$$(x+y)u_{x} + yu_{y} = 0 [/itex] [tex] u(1,y) = \frac{1}{y} + ln(y) [/itex] ## Homework Equations ... ## The Attempt at a Solution Here was my approach via method of characteristics: [tex] \frac{dy}{dt} = y \to y(t) = Se^{t}$$
$$y(t=0) = S$$

$$\frac{dx}{dt} = x+y$$
$$x'-x = Se^{t}$$
Via integrating factor we arrive at:
$$x(t) = Ste^{t} + g(s)e^{t}$$
$$x(t=0) = 1 \to 1 = g(s)$$

Thus we are left with:

$$x(t) = Ste^{t} + e^{t}$$
and
$$y(t) = Se^{t}$$

You can check for yourself that the following is true:

$$\frac{dU(x(t),y(t))}{dt} = (x+y)u_{x} + yu_{y} = 0$$

Thus our general solution is:

$$U(t) = f(s)$$

A general function of s.

However, we have found that it is impossible to explicitly solve for s. Am I missing something?

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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".