Can anyone help me solve a Non-Homogenous PDE for my assignment?

Pauly Man · Apr 27, 2003

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 μ_n to be the nth zero of J₀, then

k_n = μ_n/a

Setting A = 1 we get;
⇒ R(r) = J₀((μ_nr)/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:

∫ rJ₀(k_nr)J₀(k_mr)dr = 0 ; n ≠ m

∫ rJ₀(k_nr)J₀(k_mr)dr = (a²/2)*J₁²(k_ma) ; n = m

⇒ b_m(t) = 0 ; m ≠ 1

⇒ b_m(t) = (2q/a²J₁²(μ_m))*∫ re^rtJ₀²((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 b_m).

Step Four-

Substitute the solution;

u(r,t) = ∑R_n(r)T_n(t);

into the original PDE;

∑R_nT'_n = c∑(R"_n + (1/r)R'_n)T_n + ∑b_nR_n

Now R' + (1/r)R' = -λR

⇒ ∑(T'_n + c&lambda_nT_n)R_n = ∑b_nR_n

Use the orthogonality condition for R_n to get;

∑T'_n + c&lambda_nT_n = ∑b_n

⇒ ∑T'_n + c&lambda_nT_n = 0 ; n ≠ 1

⇒ ∑T'_n + c&lambda_nT_n = (2q/a²J₁²(μ_m))*∫ re^rtJ₀²((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 + cλ_nT_n = 0

⇒ T_n(t) = C_ncos((√c)(μ_n/a)) + D_nsin((√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:

T₁' + c&lambda_nT_n = (2q/a²J₁²(μ_m))*∫ re^rtJ₀²((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((μ_nr)/a)[C_ncos((√c)(μ_n/a)) + D_nsin((√c)(μ_n/a))] + J₀((μ₁r)/a)[(Particular Solution of T₁(t))

quantumdude · Apr 27, 2003

Sit tight--I'm printing this out and taking it home.

Till tomorrow...

Pauly Man · Apr 28, 2003

I realized last nite that I for some reason solved the equation for T(t) as a second order DE, when it is clearly a first order DE. So ignore the final solution and part of step five.

For n ≥ 1-

T'_n + cλ_nT_n = 0

T(t) = e^{-(cλ_nt)}

quantumdude · Apr 28, 2003

OK.

I had some trouble getting through your initial post because of what I am convinced are typos.

For instance, think that this:

Originally posted by Pauly Man
∂u/∂t = (c/r[/color])*(r(∂u/∂r)) + Q(r,t)

Should be this:

∂u/∂t=(c∂/∂r[/color])*(r(∂u/∂r))+Q(r,t)

At least, that would be consistent with what you write later.

Also this:

⇒ dR/r[/color] = (dR/dρ)*(dρ/dr) = k(dR/dρ)

Should presumably be this:

⇒ dR/dr[/color] = (dR/dρ)*(dρ/dr) = k(dR/dρ)

Also, I don't know what

⇒

is supposed to be. It shows up as a rectangle.

If I'm right about those typos, then you are OK through Steps 1 and 2. I am still checking Step 3.

Pauly Man · Apr 29, 2003

Originally posted by Tom

∂u/∂t=(c∂/∂r[/color])*(r(∂u/∂r))+Q(r,t)

At least, that would be consistent with what you write later.

Yep, there was a typo there. Although it should still be c/r, and not c:

∂u/∂t=((c/r)∂/∂r[/color])*(r(∂u/∂r))+Q(r,t)

Also this:

Originally posted by Tom

Should presumably be this:

⇒ dR/dr[/color] = (dR/dρ)*(dρ/dr) = k(dR/dρ)

Yep, that is correct.

Originally posted by Tom

Also, I don't know what

⇒

is supposed to be. It shows up as a rectangle.

That is supposed to be a "implies" symbol, it doesn't seem to work on all broswers and OS's. Sorry.

Originally posted by Tom

If I'm right about those typos, then you are OK through Steps 1 and 2. I am still checking Step 3.

Cool.

Pauly Man · Apr 30, 2003

I think I've solved it!

I found out today that the reason I was having so much difficulty with the problem was that the assignment had a "typo". Q(r,t) is not given by:

Q(r,t) = qJ₀((μ₁r)/a)e^rt

But by:

Q(r,t) = qJ₀((μ₁r)/a)e^{ν[/color]t}

This makes the problem a heck of a lot easier to solve.

Step one and step two remain unchanged, step three changes only with the final answer:

b_m(t) = 0 ; m ≠ 1
b_m(t) = (qe^νtJ₀²(μ₁))/(J₀²(μ_m)) ; m =1

Step four reamins unchanged until the end result:

where T'_n + ck_n²T_n = b_n(t) , where b_n is given above.

Step five becomes:

Step Five-

For n ≥ 1-

T_n(t) = D_ne^-ck_n²t

For n = 1-

Let α = ck_n²
Let β = q(J₀²(μ₁)/J₀²(μ_n))

then;

T'₁ + αT₁ = βe^νt

The integrating factor is:

I.F- e^αt

So;

e^αtT₁(t) - T₁(0) = (β/(α + ν))(e^{(α + ν)t} - 1)

Now, T_n(0) = 0 from the initial condition. SO the above simplifies. Now we add the T(t)'s together:

T_n(t) = (β/(α + ν))(e^νt - e^-αt) + (2 to infinity)∑D_ne^-ck_n²t

Now let's look at the initial condition to determine D_n-

T_n(0) = (β/(α + ν))(1 - 1) + (2 to infinity)∑D_n = 0

T_n(0) = (2 to infinity)∑D_n = 0

So D_n = 0

Therefore T_n(t) is given by:

T_n(t) = T₁(t) = (q(J₀²(μ₁)/J₀²(μ₁))
/(ck_n² + ν))(e^νt - e^-ck_n²t)

Which upon simplification becomes:

T_n(t) = T₁(t) = (q/(ck_n² + ν))(e^νt - e^-ck_n²t)

Therefore the final solution is given by:

u(r,t) = ∑R_n(r)T_n(t) = R₁(r)T₁(t)

u(r,t) = J₀((μ₁r)/a)(q/(c(μ₁/a)² + ν))(e^νt - e^{-c(μ₁/a)²t})

This solution satifsies the boundary and initial conditions, and so I hopw that it is right.

Another God · Apr 30, 2003

sweet jesus!

quantumdude · Apr 30, 2003

I'll check it and get back to you tomorrow.

Pauly Man · Apr 30, 2003

Thanx for all the help Tom, I really appreciate it.

Most of the class has the same result, so I'm crossing fingers.

Pauly Man · Apr 30, 2003

Originally posted by Another God
sweet jesus!

LOL. Looks daunting I agree.

quantumdude · May 1, 2003

Looks good to me.

Pauly Man · May 4, 2003

Thanx for all the help Tom.

Can anyone help me solve a Non-Homogenous PDE for my assignment?

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