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

(x+y)u_{x} + yu_{y} = 0 [/itex]<br /> <br /> u(1,y) = \frac{1}{y} + ln(y) [/itex]&lt;h2&gt;Homework Equations&lt;/h2&gt;&lt;br /&gt; ...&lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;&lt;br /&gt; Here was my approach via method of characteristics:&lt;br /&gt; &lt;br /&gt; \frac{dy}{dt} = y \to y(t) = Se^{t}&lt;br /&gt; y(t=0) = S&lt;br /&gt; &lt;br /&gt; \frac{dx}{dt} = x+y&lt;br /&gt; x&amp;amp;#039;-x = Se^{t}&lt;br /&gt; Via integrating factor we arrive at:&lt;br /&gt; x(t) = Ste^{t} + g(s)e^{t}&lt;br /&gt; x(t=0) = 1 \to 1 = g(s)&lt;br /&gt; &lt;br /&gt; Thus we are left with:&lt;br /&gt; &lt;br /&gt; x(t) = Ste^{t} + e^{t}&lt;br /&gt; and&lt;br /&gt; y(t) = Se^{t}&lt;br /&gt; &lt;br /&gt; You can check for yourself that the following is true:&lt;br /&gt; &lt;br /&gt; \frac{dU(x(t),y(t))}{dt} = (x+y)u_{x} + yu_{y} = 0&lt;br /&gt; &lt;br /&gt; Thus our general solution is:&lt;br /&gt; &lt;br /&gt; U(t) = f(s)&lt;br /&gt; &lt;br /&gt; A general function of s.&lt;br /&gt; &lt;br /&gt; However, we have found that it is impossible to explicitly solve for s. Am I missing something?

 

[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".