Unsolvable PDE?

  • #1
380
1
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]

[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} [/tex]
[tex] y(t=0) = S [/tex]

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

Thus we are left with:

[tex] x(t) = Ste^{t} + e^{t} [/tex]
and
[tex] y(t) = Se^{t} [/tex]

You can check for yourself that the following is true:

[tex] \frac{dU(x(t),y(t))}{dt} = (x+y)u_{x} + yu_{y} = 0 [/tex]

Thus our general solution is:

[tex] U(t) = f(s) [/tex]

A general function of s.

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

Answers and Replies

  • #2
64
9
I suspect the problem cannot be solved. Along the initial line [itex]y=1[/itex] you can differentiate the initial conditions and get

[itex]u_y(1,y)=-1/y^2+1/y=(y-1)/y^2=0[/itex]

that you can substitute back into the PDE in order to obtain [itex]u_x[/itex]

[itex](x+1)u_x=0 → u_x=0 [/itex]

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

[itex]yu_y=y(1-y)/y^2=0[/itex]

true only when [itex]y=0[/itex] or [itex]y=1[/itex].

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

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