- #1
mikeleeds
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Homework Statement
Find analytic solution of some kind:
[tex]0&=Y''(y)-\frac{\alpha^2 [u(y) + U]}{\epsilon}Y(y)[/tex]
U eigenvalue, u(y) known, epsilon & alpha paramertres,and& Y thing to be found
Homework Equations
u(y) is a parallel flow of some kind and laplace transform given by:
[tex] \mathcal{L}[Y(y)]=\bar{Y}(s)=\int_0^{\infty} e^{-sy} Y(y) dy[/tex]
The Attempt at a Solution
Do the transform:
[tex]
-sY(0)+s^2 \bar{Y}(s) &= \frac{\alpha^2}{\epsilon} \left( \mathcal{L}[u(y)Y(y)] + U \bar{Y}(s) \right)
[/tex]
Do the u(y)Y(y) transform
[tex]
\mathcal{L}[u(y)Y(y)]&=\int_0^{\infty} u(y)Y(y) e^{-sy}dy
[/tex]
by parts
[tex]
&=\bar{Y}(s)u(y)|_0^{\infty}-\int_0^{\infty} u'(y)\mathcal{L}[Y(y)] dy
[/tex]
by parts again and again...
[tex]
&=\bar{Y}(s)u(y)|_0^{\infty}-\frac{\bar{Y}(s) u'(y)|_0^{\infty}}{s} + \frac{\bar{Y}(s) u''(y)|_0^{\infty}}{s^2} - ...
[/tex]
this series might converge so we can truncate
Put it back together again and get a solutn. in Laplace space.
4. My problem:
I don't seem to get the right kind of answers, when I numerically undo my laplace transform they are either the wrong way round, or completely messed up (oscillating like a nutcase)! I don't know is you are allowed to do that kind of thing (the whole by parts business) or the validity of my series truncation... Could anyone clarify for me?
And if anyone is feeling REALLY helpful, they might tell me how to get the eigenvalue out and find what it is?!
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