First Order PDE Cauchy problem Using Method of Characteristics

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Homework Statement


Ok, so I can get through most of this but I can't seem to get the last part... Here is the problem

[tex]xU_x + (y^2+1)U_y = U-1; U(x,x) = e^x[/tex]


Homework Equations





The Attempt at a Solution



Characteristic equations are:

[tex]\frac{dx}{x} = \frac{dy}{y^2+1} = \frac{dU}{U-1}[/tex]

Solving the first and third gives:

[tex]\frac{U-1}{x} = c_1[/tex]

The first and second equation yield:

[tex]tan^{-1}(y) - lnx = c_2[/tex]

Put the two together in the form

[tex]c_1 = f(c_2)[/tex]

[tex]\frac{U-1}{x} = f(tan^{-1}(y) - lnx)[/tex]

Sub in the Cauchy data and you get

[tex]\frac{e^x-1}{x} = f(tan^{-1}(x) - lnx)[/tex]

Now how do I find what my arbitrary function f is? I have spent hours on this. Is there something that relates inverse tan to natural log? Arrggghhhh!

Thanks for any help.
 

Answers and Replies

  • #2
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Where did the function f come from? The way I see it, you end up with three equations:

[tex]\begin{cases}
x=C_1 e^{\tan^{-1}y}\\
U=C_2 x+1\\
U=C_3 e^{\tan^{-1}y}+1
\end{cases}[/tex]

Reconciling those, you get
[tex]U=Cxe^{\tan^{-1}y}+1[/tex]



Oh, and there is a relation between inverse tan and natural log: http://functions.wolfram.com/ElementaryFunctions/ArcTan/02/0001/

[tex]\tan^{-1}z={i \over 2}\ln{1-iz \over 1+iz}[/tex]
 

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