Solving ODE using Fourier Transform


by absolute76
Tags: fourier, solving, transform
absolute76
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#19
Apr11-10, 09:38 PM
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Quote Quote by gabbagabbahey View Post
Actually, I made an error in my previous post:

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex]

You will need to use the product rule to carry out the derivative.
error???

error in which part??

yes i got the answer of 2k-4k-k/(-2k+1) using....(u'v-uv')/v am i right??
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#20
Apr11-10, 10:00 PM
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Quote Quote by absolute76 View Post
error???

error in which part??
[tex]\frac{d}{dk}\left(k\tilde{u}\right)\neq k\frac{d\tilde{u}}{dk}[/tex]
absolute76
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#21
Apr11-10, 10:04 PM
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Quote Quote by gabbagabbahey View Post
Actually, I made an error in my previous post:

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex]

You will need to use the product rule to carry out the derivative.
[tex]
i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]
[/tex]

owk i think i got it, is it correct....i(iu~(k)-ku~(k))

that will give me -u~(k)+iku~(k)---->u~(k)[-1+ik]??

owk how do i separate u~(k) so that it don't cancel each other(left and right side)??
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#22
Apr11-10, 10:17 PM
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Quote Quote by absolute76 View Post
[tex]
i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]
[/tex]

owk i think i got it, is it correct....i(iu~(k)-ku~(k))
No.

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}\right]=-\frac{d}{dk}\left[k\tilde{u}\right]=-\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=-\tilde{u}-k\frac{d\tilde{u}}{dk}[/tex]

This is a basic application of the product rule.
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#23
Apr11-10, 10:32 PM
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Quote Quote by gabbagabbahey View Post
No.

[tex]i\frac{d}{dk}\left[(ik)\tilde{u}\right]=-\frac{d}{dk}\left[k\tilde{u}\right]=-\left(\tilde{u}\frac{dk}{dk}+k\frac{d\tilde{u}}{dk}\right)=-\tilde{u}-k\frac{d\tilde{u}}{dk}[/tex]

This is a basic application of the product rule.
Ok my mistake again,

so now the equation will be u~(k)[-2k+2]=-k du~/dk

so u~(k)= [-k/(-2k+2) du~/dk

Is it correct if i diffrentiate towards k on the right side???
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#24
Apr11-10, 10:40 PM
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Quote Quote by absolute76 View Post
so now the equation will be u~(k)[-2k+2]=-k du~/dk

so u~(k)= [-k/(-2k+2) du~/dk
That's correct. So now you have a 1st order ODE to solve instead of the 2nd order one you started with. This one is separable:

[tex]\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk[/tex]
absolute76
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#25
Apr11-10, 11:12 PM
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Quote Quote by gabbagabbahey View Post
That's correct. So now you have a 1st order ODE to solve instead of the 2nd order one you started with. This one is separable:

[tex]\frac{d\tilde{u}}{\tilde{u}}=2\frac{k^2-1}{k}dk[/tex]
Owk now is this correct?

u~(k)=e^[(k)-(2 ln k)]

....u~(k)=e^(k)/e^(2 ln k)

then i transform it using the table?? right?
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#26
Apr11-10, 11:31 PM
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Quote Quote by absolute76 View Post
Owk now is this correct?

u~(k)=e^[(k)-(2 ln k)]

....u~(k)=e^(k)/e^(2 ln k)

then i transform it using the table?? right?
Right. Although, I doubt your table will give you the inverse FT of this function, so you will probably want to use the convolution theorem with [itex]\tilde{g}(k)=\frac{1}{k^2}[/itex]
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#27
Apr11-10, 11:52 PM
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Quote Quote by gabbagabbahey View Post
Right. Although, I doubt your table will give you the inverse FT of this function, so you will probably want to use the convolution theorem with [itex]\tilde{g}(k)=\frac{1}{k^2}[/itex]
ok, so if i want to find the general solution,

should i just let it be in terms of convolution plus the homogenous equation right?

that should be my final general solution.
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#28
Apr12-10, 09:23 AM
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Right, although in this case, I think your second solution comes from solving [itex]u''(x)=0[/itex] and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".
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#29
Apr12-10, 09:57 PM
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Quote Quote by gabbagabbahey View Post
Right, although in this case, I think your second solution comes from solving [itex]u''(x)=0[/itex] and [itex]xu'-u=0[/tex]. which isn't really "the homogeneous equation".
owh ok..i think i get it...
anyway thanks a lot for your help!
i really appreciate it..


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