- #1
Haths
- 33
- 0
Given;
[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?
[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?