Respuesta :
Answer:
Φ (ρ ,ϕ , z ) = V a∫In (0 ∞) J₁ (ka) J₀ (k ρ)e ∧ −k z dk
Explanation:
(a)
By recalling the Laplace equation, we know that the general solution to the Laplace equation in cylindrical coordinates is:
Φ (ρ ,ϕ , z )= ( A₀,₀+B₀,₀ l n ρ) (C₀,₀ + D₀,₀ ϕ) ( F₀, ₀+ G₀, ₀ z )
+ Σ In(ν≠0) (Aν , 0ρν + Bν , 0 ρ ∧ −ν) (Cν ,0 e∧i ν ϕ + Dν ,0 e∧−i νϕ) (Fν ,0 + Gν ,0 z )
+Σ In(k ≠0) (A₀, k J₀(k ρ) + B₀, k N₀(k ρ))(C₀,k+D₀, k ϕ)(F₀,k e k z+G₀,k e−k z)
+Σ In(ν≠0) Σ In(k ≠0) ( Aν , k J ν (k ρ) + Bν ,k N ν(k ρ)) (Cν ,k e ∧ i νϕ+ Dν , k e−i νϕ) (Fν ,k e k z+Gν, k e−k z)
Therefore,
Consequentially, the problem has azimuthal symmetry. Hence, immediately dictating that ν = 0 and the function be single valued in the azimuthal angle, leading to:
Φ (ρ ,ϕ , z ) = ( A₀+B₀ ln ρ) ( F₀+ G₀ z ) + ΣIn( k ≠0) ( Ak J ₀ (k ρ) + Bk N₀ (k ρ))( Fk e k z+Gk e−k z)
Therefore,
The solution must then be finite as z approaches infinity, telling us that Fk = 0 and G0 =0.
Also, the solution must be finite along the z axis (ρ = 0), telling us that Bk = 0 and B₀ = 0.
Resolving further, we now have our solution to be:
Φ(ρ ,ϕ , z )=Σk A ∨k J₀ (k ρ) e ∧−k z
So, typically at this point, we would apply a boundary condition at a finite radius, but there is not one for this problem. With no constraint on the radius, k therefore becomes a continuous spectrum and not a discrete set.
So, Φ (ρ ,ϕ , z ) = ∫In(0 ∞) A (k ) J₀ (k ρ)e ∧−k z dk
Now, if we apply the last boundary condition,
We have:
Φ ( z= 0 )= V (ρ)
V (ρ)=∫ In (0 ∞) A(k ) J₀ (k ρ) dk
We can then take the advantage of the orthogonality statement to solve this:
∫In(0 ∞ ) x J₀ (k x) J₀ (k ' x)dx = 1 / k δ(k '−k )
Therefore,
A(k ) = k ∫ In(0 ∞) V (ρ) ρ J₀ (k ρ)d ρ
If we now insert the actual boundary condition in this problem,
We have,
A(k ) = V k ∫In(0 a) J₀ (k ρ)ρd ρ
A(k )= V a J₁ (ka)
Hence, the final solution therefore is
Φ (ρ ,ϕ , z ) = V a∫In (0 ∞) J₁ (ka) J₀ (k ρ)e ∧ −k z dk
