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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) = qJ

Where μ

Solve for u(r,t).

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

⇒ ∂

⇒ (1/c)*(T'/T) = (R"/R) + (1/r)*(R'/R) = -λ

Solve for R(r) first:

r

let λ = k

and ρ = kr

⇒ dR/r = (dR/dρ)*(dρ/dr) = k(dR/dρ)

⇒ d

⇒ ρ

which is Bessels equation, with ν = 0.

⇒ R(ρ) = AJ

⇒ R(r) = AJ

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) = AJ

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) = AJ

A = 0 is another trivial solution, so we look at;

J

We define μ

k

Setting A = 1 we get;

⇒ R(r) = J

Expand Q(r,t) as a Fourier series of R

Let Q(r,t) = ∑ b

Use the orthogonality condition:

∫ rJ

∫ rJ

⇒ b

⇒ b

(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 b

Substitute the solution;

u(r,t) = ∑R

into the original PDE;

∑R

Now R' + (1/r)R' = -λR

⇒ ∑(T'

Use the orthogonality condition for R

∑T'

⇒ ∑T'

⇒ ∑T'

Now I have to solve for T(t) using the equations above. For n > 1 the solution is easy to find;

T'

⇒ T

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:

T

I can find the homogenous solution, however I can't find the particular solution. HELP!

The final solution is:

u(r,t) = J

∂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) = qJ

_{0}((μ_{1}r)/a)Where μ

_{1}is the first crossing of J_{0}(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

⇒ ∂

^{2}u/∂r^{2}= R"T⇒ (1/c)*(T'/T) = (R"/R) + (1/r)*(R'/R) = -λ

__Step Two-__Solve for R(r) first:

r

^{2}R" + rR' + λr^{2}R = 0let λ = k

^{2}and ρ = kr

⇒ dR/r = (dR/dρ)*(dρ/dr) = k(dR/dρ)

⇒ d

^{2}R/dr^{2}= k^{2}(d^{2}R/dρ^{2})⇒ ρ

^{2}R"(ρ) + ρR'(ρ) + ρ^{2}R(ρ) = 0which is Bessels equation, with ν = 0.

⇒ R(ρ) = AJ

_{0}(ρ) + BY_{0}(ρ)⇒ R(r) = AJ

_{0}(kr) + BY_{0}(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) = AJ

_{0}(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) = AJ

_{0}(ka) = 0A = 0 is another trivial solution, so we look at;

J

_{0}(ka) = 0We define μ

_{n}to be the nth zero of J_{0}, thenk

_{n}= μ_{n}/aSetting A = 1 we get;

⇒ R(r) = J

_{0}((μ_{n}r)/a)__Step Three-__Expand Q(r,t) as a Fourier series of R

_{n}(r):Let Q(r,t) = ∑ b

_{n}(t)R_{n}(r)Use the orthogonality condition:

∫ rJ

_{0}(k_{n}r)J_{0}(k_{m}r)dr = 0 ; n ≠ m∫ rJ

_{0}(k_{n}r)J_{0}(k_{m}r)dr = (a^{2}/2)*J_{1}^{2}(k_{m}a) ; n = m⇒ b

_{m}(t) = 0 ; m ≠ 1⇒ b

_{m}(t) = (2q/a^{2}J_{1}^{2}(μ_{m}))*∫ re^{rt}J_{0}^{2}((μ_{1}r)/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 b

_{m}).__Step Four-__Substitute the solution;

u(r,t) = ∑R

_{n}(r)T_{n}(t);into the original PDE;

∑R

_{n}T'_{n}= c∑(R"_{n}+ (1/r)R'_{n})T_{n}+ ∑b_{n}R_{n}Now R' + (1/r)R' = -λR

⇒ ∑(T'

_{n}+ c&lambda_{n}T_{n})R_{n}= ∑b_{n}R_{n}Use the orthogonality condition for R

_{n}to get;∑T'

_{n}+ c&lambda_{n}T_{n}= ∑b_{n}⇒ ∑T'

_{n}+ c&lambda_{n}T_{n}= 0 ; n ≠ 1⇒ ∑T'

_{n}+ c&lambda_{n}T_{n}= (2q/a^{2}J_{1}^{2}(μ_{m}))*∫ re^{rt}J_{0}^{2}((μ_{1}r)/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}+ cλ_{n}T_{n}= 0⇒ T

_{n}(t) = C_{n}cos((√c)(μ_{n}/a)) + D_{n}sin((√c)(μ_{n}/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:

T

_{1}' + c&lambda_{n}T_{n}= (2q/a^{2}J_{1}^{2}(μ_{m}))*∫ re^{rt}J_{0}^{2}((μ_{1}r)/a)drI can find the homogenous solution, however I can't find the particular solution. HELP!

__The final solution-__The final solution is:

u(r,t) = J

_{0}((μ_{0}r)/a)[C_{0}cos((√c)(μ_{0}/a)) + D_{0}sin((√c)(μ_{0}/a))] + (2 to infinity)∑J_{0}((μ_{n}r)/a)[C_{n}cos((√c)(μ_{n}/a)) + D_{n}sin((√c)(μ_{n}/a))] + J_{0}((μ_{1}r)/a)[(Particular Solution of T_{1}(t))
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