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[tex]

\frac{d^{2}u}{dx^{2}} = \frac{1}{c^{2}} \frac{d^{2}u}{dt^{2}}

[/tex]

and;

[tex]

u(x,0) = \phi (x)

[/tex]

[tex]

\frac{d^{2}u(x,0)}{dt^{2}} = \theta(x)

[/tex]

Show that the Fourier Transform of the u(x,t) w.r.t. to x is;

[tex]

\tilde{u}(k,t) = \tilde{\phi} (k) cos(ckt) + \frac{\tilde{\theta(k)}}{ck} sin(ckt)

[/tex]

My attempt at the question

Well I know I can FT both sides in the following way;

[tex]

\tilde{F}[ \frac{d^{2}u}{dx^{2}} ] = \frac{1}{c^{2}} \tilde{F}[ \frac{d^{2}u}{dt^{2}} ]

[/tex]

L.H.S.

[tex]

\tilde{F}[ \frac{d}{dx} \frac{du}{dx} ] = ik \tilde{F}[ \frac{du}{dt} ]

[/tex]

[tex]

\therefore

[/tex]

[tex]

\tilde{F}[ \frac{d}{dx} \frac{du}{dx} ] = -k^{2} \tilde{F}[ u ]

[/tex]

Where i was the imaginary number i. Thus leaving me with;

[tex]

-k^{2} \tilde{F}[ u ] = \frac{1}{c^{2}} \tilde{F}[ \frac{d^{2}u}{dt^{2}} ]

[/tex]

From here I'm not sure what to do (to be frank I'm not even sure I know precisely what the questuon is asking so bear with me). Oh the d's should be partial derivatives, hence I reason that because t is just a parameter value I can take the second derivative out of the R.H.S. of the last eq. thus leaving me with an F;

[tex]

-k^{2} \tilde{F}[ u ] = \frac{1}{c^{2}} \frac{d^{2}}{dt^{2}} \tilde{F}[ u ]

[/tex]

[tex]

-k^{2}c^{2} = \frac{d^{2}}{dt^{2}}

[/tex]

Which looks so tantilisingly close to the wave equation it appears be on the right track. Of course I want the wave equation because I know that the general solution is given in terms of a sum of a sin and cos wave;

[tex]

\tilde{u}(k,t) = A cos(kct) + B sin(kct)

[/tex]

Now because k is the my fourier coeffecient. I guess that makes the second time derivative 1 equation back on the R.H.S. one with respect to k. Hence why I can write the above equation. Yes or no?

If Yes, how do I present a mathamatical argument that get's me from my A and B constants to the functions of theta and phi?

If no, why is this wrong, and how do I go about getting to the solution?

Thanks.

PS: Why aren't the formula coming up quite right?

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# Homework Help: Fourier Transform of the Wave Eq.

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