Proving general solution of Helmholtz equation

AI Thread Summary
The discussion focuses on proving that F(k•r - ωt) is a solution to the Helmholtz equation under the condition ω/k = 1/(µε)^(1/2). Participants express confusion regarding the steps to show that ∇²F(xkx + yky + zkz) = 0 and how k²F(xkx + yky + zkz) = 0 leads to a trivial solution. There is a suggestion that the problem may actually require demonstrating that F(k•r - ωt) is a solution to the wave equation first, before confirming it satisfies the Helmholtz equation. The complexity of the problem is acknowledged, particularly because the Helmholtz equation does not involve time dependence. Clarification on the problem's requirements is deemed necessary for accurate resolution.
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Homework Statement



Prove that F(k•r -ωt) is a solution of the Helmholtz equation, provided that ω/k = 1/(µε)1/2, where k = (kx, ky, kz) is the wave-vector and r is the position vector. In F(k•r -ωt), “k•r –ωt” is the argument and F is any vector function.

Homework Equations


Helmholtz Equation: ∇2A+k2A=0

The Attempt at a Solution


I first completed the dot of k and r thus,
F(xkx+ yky+zkz)I am a little confused at this point, if I plug it into the equation I don't see how it would be 0
 
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It's not clear what you've done, since you haven't shown your work.
 
SteamKing said:
It's not clear what you've done, since you haven't shown your work.

I completed the dot between k and r within F.

k = (kx, ky, kz)
TA said to use r=(x,y,z)

thus the function becomes
F(xkx+ yky+zkz)

2F(xkx+ yky+zkz)=0, right?but then how would k2F(xkx+ yky+zkz)=0?
 
2F(xkx+ yky+zkz)=0 is not right. Because then, as you say, you would need k2F=0, which is the trivial solution.

It is a kind of weird problem in the first place, since the Helmholtz equation does not care about time dependence. Are you sure the problem wasn't to show that ##F(k\cdot r -\omega t)## is a solution of the wave equation, and to go on to prove that it is also a solution of the Helmholtz equation?
 

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