Proving general solution of Helmholtz equation

[physicslove2]

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 = (k_x, k_y, k_z) 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(xk_x+ yk_y+zk_z)I am a little confused at this point, if I plug it into the equation I don't see how it would be 0

 

[SteamKing]

It's not clear what you've done, since you haven't shown your work.

 

[physicslove2]

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 = (k_x, k_y, k_z)
TA said to use r=(x,y,z)

thus the function becomes
F(xk_x+ yk_y+zk_z)

∇²F(xk_x+ yk_y+zk_z)=0, right?but then how would k²F(xk_x+ yk_y+zk_z)=0?

 

[BruceW]

∇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?

 

[rezeew]

.

To prove that F(k•r -ωt) is a solution of the Helmholtz equation, we need to show that it satisfies the equation ∇^2A + k^2A = 0.

Let's start by taking the gradient of F(k•r -ωt):
∇F(k•r -ωt) = F'(k•r -ωt) * ∇(k•r -ωt)
= F'(k•r -ωt) * (k - 0)
= kF'(k•r -ωt)

Next, we can take the Laplacian of F(k•r -ωt):
∇^2F(k•r -ωt) = ∇•(∇F(k•r -ωt))
= ∇•(kF'(k•r -ωt))
= ∇k•F'(k•r -ωt) + k•∇F'(k•r -ωt)
= k•∇F'(k•r -ωt) + k•∇F'(k•r -ωt)
= 2k•∇F'(k•r -ωt)

Now, we can substitute these expressions into the Helmholtz equation:
∇^2F(k•r -ωt) + k^2F(k•r -ωt) = 0
2k•∇F'(k•r -ωt) + k^2F(k•r -ωt) = 0

Since F'(k•r -ωt) is any vector function and k is a constant vector, we can rewrite this as:
2k•∇F'(k•r -ωt) = -k^2F(k•r -ωt)

Dividing both sides by k^2:
2∇F'(k•r -ωt) = -F(k•r -ωt)

Now, we can use the given relationship between ω and k to simplify this equation:
2∇F'(k•r -ωt) = -F(k•r -ωt)
2∇F'(k•r -ωt) = -F(k•r -ωt)
2∇F'(k•r -ωt) = -F(k•r -ωt)
2∇F