2nd order nonlinear imaginary partial dif eqn

  • Thread starter FD2010
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Hi all,

I am having a hard time solving a partial second order differential equation with an imaginary part. I basically took a much bigger function with real and imaginary parts and simplified it down to this. I also know the solution to a similar equation (shown in image). Any help would be appreciated!

Nick
 

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  • #2
tiny-tim
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Welcome to PF!

Hi FD2010! Welcome to PF! :wink:

Can't you just replace Ω by Ω+iK (and use polar form to find the square-root) ? :smile:
 
  • #3
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I am terribly sorry but I wound up writing the original expression wrong. I was so hasty for help I had an error in the equation (I left out a u!). Here is what I am trying to solve (new image).

Any help is sincerely appreciated!

Nick
 

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  • #4
tiny-tim
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Hi Nick! :wink:

in that case, you need a particular solution to add to the general solution you already have …

try u = a constant :smile:
 
  • #5
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I am terribly sorry but I wound up writing the original expression wrong. I was so hasty for help I had an error in the equation (I left out a u!). Here is what I am trying to solve (new image).

Any help is sincerely appreciated!

Nick
I don't see how you're getting all that unless the constants are eliminating the complex components of the solution. For starters, it's just one derivative so you could replace it with just an ODE.

[tex]u''-ku=0[/tex]

and keep in mind the i is just a constant and as long as you use correct complex-value arithmetic, it's just like real variables. So the solution to that is:

[tex]u(y)=c_1 e^{\sqrt{k}y}+c_2 e^{-\sqrt{k}y}[/tex]

however, that expression is already (implicitly) taking out the two values of the multi-valued square root. Now in the case of [itex]\sqrt{i}[/itex], it's best to explicitly evaluate the square root of i:

[tex]\sqrt{\frac{i\omega}{\nu}}=\sqrt{\omega/\nu}e^{\pi i/4}{, \sqrt{\omega/\nu}e^{-3\pi i/4}}[/tex]

then the solution with the i would be:

[tex]u(y)=c_1\text{exp}\left(y\sqrt{\omega/\nu}e^{\pi i/4}\right)+c_2\text{exp}\left(y\sqrt{\omega/\nu}e^{-3\pi i/4}\right)[/tex]

And if it was indeed a partial of say u(y,t), then the constants of integration would be functions of t and the solutions would be:

[tex]u(y,t)=c_1(t)\text{exp}\left(y\sqrt{\omega/\nu}e^{\pi i/4}\right)+c_2(t)\text{exp}\left(y \sqrt{\omega/\nu}e^{-3\pi i/4}\right)[/tex]
 
  • #6
tiny-tim
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no, forget what i said about Ω + iK, that was for the (original) wrong problem …

just put u = A, then ∂2u/dy2 - (iΩ/ν)u + κ/ν = (-iΩA + κ)/ν :wink:
 
  • #7
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Sincere thanks for the help. I would go into detail about how I arrived at that very simple equation but it would take awhile.

I wound up having to solve the original equation I was working with using non-dimensionalization. u (a velocity in this case) was unsteady, and depended on a oscillating pressure gradient. Due to this, I couldn't just let du/dt go to 0 because u is not a constant. The result wound up being quite a mess, but I believe I'm good now.

Thanks again!
 
  • #8
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Hi all,

I am having a hard time solving a partial second order differential equation with an imaginary part. I basically took a much bigger function with real and imaginary parts and simplified it down to this. I also know the solution to a similar equation (shown in image). Any help would be appreciated!

Nick

Hi you can Take (i*omg-K)/v as one whole part(is omg).
 

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