- #1
Hoplite
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I have been working on a derivation in which the following simultateous ordinary differential equations have appeared:
f^{(4)}(x)-2 a^2 f''(x)+a^4 f(x)+b(g''(x)-a^2 g(x))=0,
g^{(4)}(x)-2 a^2 g''(x)+a^4 g(x)-b(f''(x)-a^2 f(x))=0,
where a and b are constants. I figured that I could solve this using Fourier transforms. First, I transform the above, using \nu as the transform space analogue of x, into the following:
(\nu^2 +a^2 )\hat f(\nu ) -b \hat g(\nu )=0,
(\nu^2 +a^2 )\hat g(\nu ) +b \hat f(\nu )=0.
I then rearrange these equations via substitution into
[(\nu^2 +a^2 )^2 +b^2 ] \hat f(\nu ) =0,
[(\nu^2 +a^2 )^2 +b^2 ] \hat g(\nu )=0.
Taking the inverse Fourier transform then results in
f^{(4)}(x)-2a^2 f''(x)+(a^4 +b^2 )f(x)=0,
g^{(4)}(x)-2a^2 g''(x)+(a^4 +b^2 )g(x)=0.
However, after I solve for these two ODEs using the boundary conditions, I find that the resulting answer is erroneous. I can only conclude that the above methodoly doesn't work, although I can't see why.
Could anyone point out the mathematical error in the above steps?
f^{(4)}(x)-2 a^2 f''(x)+a^4 f(x)+b(g''(x)-a^2 g(x))=0,
g^{(4)}(x)-2 a^2 g''(x)+a^4 g(x)-b(f''(x)-a^2 f(x))=0,
where a and b are constants. I figured that I could solve this using Fourier transforms. First, I transform the above, using \nu as the transform space analogue of x, into the following:
(\nu^2 +a^2 )\hat f(\nu ) -b \hat g(\nu )=0,
(\nu^2 +a^2 )\hat g(\nu ) +b \hat f(\nu )=0.
I then rearrange these equations via substitution into
[(\nu^2 +a^2 )^2 +b^2 ] \hat f(\nu ) =0,
[(\nu^2 +a^2 )^2 +b^2 ] \hat g(\nu )=0.
Taking the inverse Fourier transform then results in
f^{(4)}(x)-2a^2 f''(x)+(a^4 +b^2 )f(x)=0,
g^{(4)}(x)-2a^2 g''(x)+(a^4 +b^2 )g(x)=0.
However, after I solve for these two ODEs using the boundary conditions, I find that the resulting answer is erroneous. I can only conclude that the above methodoly doesn't work, although I can't see why.
Could anyone point out the mathematical error in the above steps?
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