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Pauly Man
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I have an assignment due at the end of the week, and I was wondering if someone could check my working for me, as I am prone to making errors. Also, in Step Five I am unsure how to solve for T(t), can anyone point me in the right direction?
∂u/∂t = (c/r)*(r(∂u/∂r)) + Q(r,t)
0 < r < a ; c > 0
u(0,t) is finite
u(a,t) = 0
u(r,0) = 0
Q(r,t) = qJ0((μ1r)/a)
Where μ1 is the first crossing of J0(x), which is a Bessel Function of the first kind of order zero, q is a constant.
Solve for u(r,t).
Step One-
Use separation of variables on the homogenous equation, (ie. Set Q(r,t) = 0):
u(r,t) = R(r)T(t)
⇒ ∂u/∂t = RT'
⇒ ∂u/∂r = R'T
⇒ ∂2u/∂r2 = R"T
⇒ (1/c)*(T'/T) = (R"/R) + (1/r)*(R'/R) = -λ
Step Two-
Solve for R(r) first:
r2R" + rR' + λr2R = 0
let λ = k2
and ρ = kr
⇒ dR/r = (dR/dρ)*(dρ/dr) = k(dR/dρ)
⇒ d2R/dr2 = k2(d2R/dρ2)
⇒ ρ2R"(ρ) + ρR'(ρ) + ρ2R(ρ) = 0
which is Bessels equation, with ν = 0.
⇒ R(ρ) = AJ0(ρ) + BY0(ρ)
⇒ R(r) = AJ0(kr) + BY0(kr)
where Y(x) is a Bessel Function of the second kind.
Now Y(x) approaches infinity as x approaches zero, therefore if R(r) is to satisfy the first boundaryb condition;
u(0,t) finite;
then B must equal zero.
⇒ R(r) = AJ0(kr)
The second boundary condition is;
u(a,t) = R(a)T(t) = 0;
⇒ R(a) = 0 and/or T(t) = 0
Now T(t) = 0 is a trivial solution, as this implies u(r,t) = 0. So we look at R(a) = 0;
⇒ R(a) = AJ0(ka) = 0
A = 0 is another trivial solution, so we look at;
J0(ka) = 0
We define μn to be the nth zero of J0, then
kn = μn/a
Setting A = 1 we get;
⇒ R(r) = J0((μnr)/a)
Step Three-
Expand Q(r,t) as a Fourier series of Rn(r):
Let Q(r,t) = ∑ bn(t)Rn(r)
Use the orthogonality condition:
∫ rJ0(knr)J0(kmr)dr = 0 ; n ≠ m
∫ rJ0(knr)J0(kmr)dr = (a2/2)*J12(kma) ; n = m
⇒ bm(t) = 0 ; m ≠ 1
⇒ bm(t) = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr ; m = 1
(So far so good, I think. I have a solution for R(r) with no unknowns, and have expanded Q(r,t) out in terms of the eigenfunctions R(r), and have "solved" for the coefficient bm).
Step Four-
Substitute the solution;
u(r,t) = ∑Rn(r)Tn(t);
into the original PDE;
∑RnT'n = c∑(R"n + (1/r)R'n)Tn + ∑bnRn
Now R' + (1/r)R' = -λR
⇒ ∑(T'n + c&lambdanTn)Rn = ∑bnRn
Use the orthogonality condition for Rn to get;
∑T'n + c&lambdanTn = ∑bn
⇒ ∑T'n + c&lambdanTn = 0 ; n ≠ 1
⇒ ∑T'n + c&lambdanTn = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr ; n = 1
Step Five-
Now I have to solve for T(t) using the equations above. For n > 1 the solution is easy to find;
T'n + cλnTn = 0
⇒ Tn(t) = Cncos((√c)(μn/a)) + Dnsin((√c)(μn/a))
For n = 1 the I am unsure how to find the solution. The DE to solve is given below, and I've never had to solve a DE like it before:
T1' + c&lambdanTn = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr
I can find the homogenous solution, however I can't find the particular solution. HELP!
The final solution-
The final solution is:
u(r,t) = J0((μ0r)/a)[C0cos((√c)(μ0/a)) + D0sin((√c)(μ0/a))] + (2 to infinity)∑J0((μnr)/a)[Cncos((√c)(μn/a)) + Dnsin((√c)(μn/a))] + J0((μ1r)/a)[(Particular Solution of T1(t))
∂u/∂t = (c/r)*(r(∂u/∂r)) + Q(r,t)
0 < r < a ; c > 0
u(0,t) is finite
u(a,t) = 0
u(r,0) = 0
Q(r,t) = qJ0((μ1r)/a)
Where μ1 is the first crossing of J0(x), which is a Bessel Function of the first kind of order zero, q is a constant.
Solve for u(r,t).
Step One-
Use separation of variables on the homogenous equation, (ie. Set Q(r,t) = 0):
u(r,t) = R(r)T(t)
⇒ ∂u/∂t = RT'
⇒ ∂u/∂r = R'T
⇒ ∂2u/∂r2 = R"T
⇒ (1/c)*(T'/T) = (R"/R) + (1/r)*(R'/R) = -λ
Step Two-
Solve for R(r) first:
r2R" + rR' + λr2R = 0
let λ = k2
and ρ = kr
⇒ dR/r = (dR/dρ)*(dρ/dr) = k(dR/dρ)
⇒ d2R/dr2 = k2(d2R/dρ2)
⇒ ρ2R"(ρ) + ρR'(ρ) + ρ2R(ρ) = 0
which is Bessels equation, with ν = 0.
⇒ R(ρ) = AJ0(ρ) + BY0(ρ)
⇒ R(r) = AJ0(kr) + BY0(kr)
where Y(x) is a Bessel Function of the second kind.
Now Y(x) approaches infinity as x approaches zero, therefore if R(r) is to satisfy the first boundaryb condition;
u(0,t) finite;
then B must equal zero.
⇒ R(r) = AJ0(kr)
The second boundary condition is;
u(a,t) = R(a)T(t) = 0;
⇒ R(a) = 0 and/or T(t) = 0
Now T(t) = 0 is a trivial solution, as this implies u(r,t) = 0. So we look at R(a) = 0;
⇒ R(a) = AJ0(ka) = 0
A = 0 is another trivial solution, so we look at;
J0(ka) = 0
We define μn to be the nth zero of J0, then
kn = μn/a
Setting A = 1 we get;
⇒ R(r) = J0((μnr)/a)
Step Three-
Expand Q(r,t) as a Fourier series of Rn(r):
Let Q(r,t) = ∑ bn(t)Rn(r)
Use the orthogonality condition:
∫ rJ0(knr)J0(kmr)dr = 0 ; n ≠ m
∫ rJ0(knr)J0(kmr)dr = (a2/2)*J12(kma) ; n = m
⇒ bm(t) = 0 ; m ≠ 1
⇒ bm(t) = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr ; m = 1
(So far so good, I think. I have a solution for R(r) with no unknowns, and have expanded Q(r,t) out in terms of the eigenfunctions R(r), and have "solved" for the coefficient bm).
Step Four-
Substitute the solution;
u(r,t) = ∑Rn(r)Tn(t);
into the original PDE;
∑RnT'n = c∑(R"n + (1/r)R'n)Tn + ∑bnRn
Now R' + (1/r)R' = -λR
⇒ ∑(T'n + c&lambdanTn)Rn = ∑bnRn
Use the orthogonality condition for Rn to get;
∑T'n + c&lambdanTn = ∑bn
⇒ ∑T'n + c&lambdanTn = 0 ; n ≠ 1
⇒ ∑T'n + c&lambdanTn = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr ; n = 1
Step Five-
Now I have to solve for T(t) using the equations above. For n > 1 the solution is easy to find;
T'n + cλnTn = 0
⇒ Tn(t) = Cncos((√c)(μn/a)) + Dnsin((√c)(μn/a))
For n = 1 the I am unsure how to find the solution. The DE to solve is given below, and I've never had to solve a DE like it before:
T1' + c&lambdanTn = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr
I can find the homogenous solution, however I can't find the particular solution. HELP!
The final solution-
The final solution is:
u(r,t) = J0((μ0r)/a)[C0cos((√c)(μ0/a)) + D0sin((√c)(μ0/a))] + (2 to infinity)∑J0((μnr)/a)[Cncos((√c)(μn/a)) + Dnsin((√c)(μn/a))] + J0((μ1r)/a)[(Particular Solution of T1(t))
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