Using Fourier Bessel series to solve the annular cylinder heat problem

[shreddinglicks]

Hello, I see many examples for solutions of the heat problem in cylindrical coordinates. I notice one thing in common. They all apply to a solid cylinder. In this case the solution after separating variables gives,

separation equation
u(r,t) = RT

gives

R(r) = A*J(0,L*r)+B*Y(0,L*r)
and
T(t) = C*e^(-alpha*L^2*t)

In the typical examples it states B = 0 due to the radius r = 0 of the solid cylinder.


therefore,

R(r) = A*J(0,L*r)

and I can the obtain my lambdas written here as L based on the condition at the outside cylinder surface.

Then the typical textbook mentions the three typical boundary conditions and obtains the coefficients for them

J'=0
hJ + alpha*b*J' = 0
and J = 0




If I have a annular cylinder where a < r < b

How would I solve this? I have attached a page I found during my search. The problem I have is how do I solve for A and B?

 

[mario99]

If you have a Fourier Bessel series expansion like this:

[imath]\displaystyle f(x) = \sum_{k=1}^{\infty}a_kJ_m(z_{mk}x)[/imath]

And you want to find the coefficient \(\displaystyle a_k\), you multiply both sides of the equation by \(\displaystyle xJ_m(z_{mk}x)\), and integrate with respect to [imath]x[/imath].

 

[shreddinglicks]

If you have a Fourier Bessel series expansion like this:

[imath]\displaystyle f(x) = \sum_{k=1}^{\infty}a_kJ_m(z_{mk}x)[/imath]

And you want to find the coefficient \(\displaystyle a_k\), you multiply both sides of the equation by \(\displaystyle xJ_m(z_{mk}x)\), and integrate with respect to [imath]x[/imath].

Yes, but what of the situation where, a <r <b?

In this case Bessel Y does not get removed. I have two coefficients in this case. Where the function is,

 

[mario99]

If the boundary is: [imath]a < r < b[/imath], the solution will be more complicated. The boundary and initial conditions of most problems will be like this:

[imath]u(a,t) = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ t > 0[/imath]
[imath]u(b,t) = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ t > 0[/imath]
[imath]u(r,0) = f(r), \ \ \ \ \ a < r < b[/imath]

After solving the Bessel equation of order zero, you will get:

[imath]R(r) = AJ_{0}(\lambda r) + BY_{0}(\lambda r)[/imath]

The boundary conditions will be changed to:
[imath]R(a) = 0[/imath]
[imath]R(b) = 0[/imath]

Applying the boundary conditions will give us:

[imath]R(a) = AJ_{0}(\lambda a) + BY_{0}(\lambda a) = 0[/imath]
[imath]R(b) = AJ_{0}(\lambda b) + BY_{0}(\lambda b) = 0[/imath]

Use the first equation and solve for [imath]B[/imath]:

[imath]\displaystyle B = -A\frac{J_{0}(\lambda a)}{Y_{0}(\lambda a)}[/imath]

Go back to the solution [imath]R(r)[/imath], and plug the value of [imath]B[/imath], you should get:

[imath]\displaystyle R(r) = AJ_{0}(\lambda r) - A\frac{J_{0}(\lambda a)}{Y_{0}(\lambda a)}Y_{0}(\lambda r) = A\left(J_{0}(\lambda r) - \frac{J_{0}(\lambda a)}{Y_{0}(\lambda a)}Y_{0}(\lambda r)\right)[/imath]

Multiply and divide the first term by [imath]Y_{0}(\lambda a)[/imath]

[imath]\displaystyle R(r) = A\left(\frac{Y_{0}(\lambda a)J_{0}(\lambda r)}{Y_{0}(\lambda a)} - \frac{J_{0}(\lambda a)}{Y_{0}(\lambda a)}Y_{0}(\lambda r)\right) = \frac{A}{Y_{0}(\lambda a)}\left[Y_{0}(\lambda a)J_{0}(\lambda r) - J_{0}(\lambda a)Y_{0}(\lambda r)\right][/imath]

Everything inside the brackets is a function of [imath]r[/imath], so we can write:

[imath]g(r) = Y_{0}(\lambda a)J_{0}(\lambda r) - J_{0}(\lambda a)Y_{0}(\lambda r)[/imath]

[imath]\displaystyle R(r) = \frac{A}{Y_{0}(\lambda a)}g(r)[/imath]

If we solve the time equation, we will get something like this:

[imath]T(t) = Ke^{-\lambda^2 t}[/imath]

The solution to the original differential equation will be:

[imath]\displaystyle u(r,t) = R(r)T(t) = \frac{AK}{Y_{0}(\lambda a)}g(r)e^{-\lambda^2 t}[/imath]

We can combine the constants at the beginning as [imath]\displaystyle C = \frac{AK}{Y_{0}(\lambda a)}[/imath]

[imath]\displaystyle u(r,t) = Cg(r)e^{-\lambda^2 t}[/imath]

Apply the superposition principle:

[imath]\displaystyle u(r,t) = \sum_{n=1}^{\infty}C_ng_n(r)e^{-\lambda_n^2 t}[/imath]

Believe it or not, after all this hard work, we still did not find the constant [imath]C_n.[/imath] This will be the job of the initial condition and the Fourier Bessel series.

Remember that [imath]g_n(r)[/imath] is a Bessel function, and all Fourier Bessel series properties apply to it.

Apply the initial condition:

[imath]\displaystyle u(r, 0) = f(r) = \sum_{n=1}^{\infty}C_ng_n(r)[/imath]

Now you can find the constant [imath]C_n[/imath], by this hint below:

If you have a Fourier Bessel series expansion like this:

[imath]\displaystyle f(x) = \sum_{k=1}^{\infty}a_kJ_m(z_{mk}x)[/imath]

And you want to find the coefficient \(\displaystyle a_k\), you multiply both sides of the equation by \(\displaystyle xJ_m(z_{mk}x)\), and integrate with respect to [imath]x[/imath].

 

[shreddinglicks]

Wow, this is great!

OK, so to continue what you did I have,


Now to finish this, are my lambdas the zeroes of the respective Bessel J and Bessel Y divided by my integration limit b? Where The x variable represents the zeros of Bessel J and Bessel Y respectively.


So the first lambda values are plugged into the Bessel J function and the 2nd lambda is plugged into the Bessel Y function?

 

[mario99]

Wow, this is great!

OK, so to continue what you did I have,

View attachment 37913
Now to finish this, are my lambdas the zeroes of the respective Bessel J and Bessel Y divided by my integration limit b? Where The x variable represents the zeros of Bessel J and Bessel Y respectively.

View attachment 37914
So the first lambda values are plugged into the Bessel J function and the 2nd lambda is plugged into the Bessel Y function?

You have a little mistakes regarding the constant [imath]A_n[/imath].

[imath]\displaystyle A_n = \frac{\int_{a}^{b}rg_n(r)f(r) \ dr}{\int_{a}^{b}rg^2_n(r) \ dr}Y_0({\lambda a})[/imath]


Also you should write [imath]\displaystyle e^{-\lambda^2 t}[/imath], not [imath]e^{-\lambda t}[/imath]

The roots are wrong. If [imath]\displaystyle b\lambda_Y[/imath] is a root for [imath]Y_0[/imath], then [imath]Y_0({\lambda b}) = 0[/imath] which means [imath]Y_0({\lambda a}) = 0[/imath] which means
[imath]A_n = 0[/imath], which is not true.

We don't want [imath]J_0[/imath] or [imath]Y_0[/imath] to be zero, but we want [imath]g_n(r)[/imath] to be zero. Therefore, the roots or zeros are simply [imath]\lambda_n[/imath] where [imath]n = 1, 2, 3, \cdot\cdot\cdot\cdot\cdot[/imath]

Can you see why?

 

[shreddinglicks]

I'm trying to understand. You say I need g(r) to be equal to 0

 

[shreddinglicks]

The other thing I don't understand is for example

Say one of my zeroes for Bessel Y is .893
Then lambda is .893/b

Then BesselY(0,lambda*b) =0

which would be different than

BesselY(0,Lambda*a)

 

[shreddinglicks]

 

[mario99]

I'm trying to understand. You say I need g(r) to be equal to 0

View attachment 37918

But we already know the value of [imath]r[/imath] that makes the expression zero. If we pick [imath]r = a[/imath] or [imath]r = b[/imath] and change [imath]\lambda[/imath] to any [imath]\lambda_n[/imath], we still get zero, then [imath]\lambda_n[/imath], [imath]n = 1,2,3,\cdot\cdot\cdot[/imath] are the roots of this equation.

You can see this better if you go back to the step where we applied the boundary conditions.

[imath]R(a) = AJ_{0}(\lambda a) + BY_{0}(\lambda a) = 0[/imath]
[imath]R(b) = AJ_{0}(\lambda b) + BY_{0}(\lambda b) = 0[/imath]

Use the first equation and solve for [imath]B[/imath]:

[imath]\displaystyle B = -A\frac{J_{0}(\lambda a)}{Y_{0}(\lambda a)}[/imath]


Plug the value of [imath]B[/imath] in the second equation.


[imath]\displaystyle AJ_{0}(\lambda b) - A\frac{J_{0}(\lambda a)Y_{0}(\lambda b)}{Y_{0}(\lambda a)} = 0[/imath]


[imath]A\left(\displaystyle J_{0}(\lambda b) - \frac{J_{0}(\lambda a)Y_{0}(\lambda b)}{Y_{0}(\lambda a)}\right) = 0[/imath]


Clearly [imath]A \neq 0[/imath], so we must have:


[imath]\displaystyle J_{0}(\lambda b) - \frac{J_{0}(\lambda a)Y_{0}(\lambda b)}{Y_{0}(\lambda a)} = 0[/imath]


[imath]\displaystyle Y_{0}(\lambda a)J_{0}(\lambda b) - J_{0}(\lambda a)Y_{0}(\lambda b) = 0[/imath]

Change [imath]\lambda[/imath] to [imath]\lambda_n[/imath]

[imath]\displaystyle Y_{0}(\lambda_n a)J_{0}(\lambda_n b) - J_{0}(\lambda_n a)Y_{0}(\lambda_n b) = 0[/imath]

The above equation means only one thing: the roots of this equation are [imath]\lambda_n[/imath] where [imath]n = 1,2,3,\cdot\cdot\cdot[/imath]

Treat the above equation like if it were [imath]f(ax_n) - f(bx_n) = 0[/imath]. If you plug any value in [imath]\displaystyle x_n[/imath], you get zero, this means [imath]\displaystyle x_n[/imath] are the roots of this equation.

The other thing I don't understand is for example

Say one of my zeroes for Bessel Y is .893
Then lambda is .893/b

Then BesselY(0,lambda*b) =0

which would be different than

BesselY(0,Lambda*a)

I understand why you are confused. Because you are thinking in the same way as if you were solving

[imath]\displaystyle J_{0}(\lambda_n a) = 0[/imath]

In this case above, yes, [imath]\displaystyle \lambda_n a = z_n[/imath] where [imath]z_n[/imath] is the nth zero of [imath]\displaystyle J_{0}[/imath] and [imath]\displaystyle \lambda_n = \frac{z_n}{a}[/imath].

 

[mario99]

The other thing I don't understand is for example

Say one of my zeroes for Bessel Y is .893
Then lambda is .893/b

Then BesselY(0,lambda*b) =0

which would be different than

BesselY(0,Lambda*a)

I will explain to you this in a different way and I hope this time you can see what I mean.

We have already seen how to get this equation.

[imath]\displaystyle Y_{0}(\lambda_n a)J_{0}(\lambda_n b) - J_{0}(\lambda_n a)Y_{0}(\lambda_n b) = 0[/imath]

Look at the equation above carefully. Our goal is to let the left side equal to the right side.

Let us assume [imath]\displaystyle Y_{0}(\lambda_n b) = 0[/imath].

We have now:

[imath]\displaystyle Y_{0}(\lambda_n a)J_{0}(\lambda_n b) - J_{0}(\lambda_n a)*0 = Y_{0}(\lambda_n a)J_{0}(\lambda_n b) = 0[/imath]

To satisfy the equation above, it's either [imath]\displaystyle Y_{0}(\lambda_n a) = 0[/imath] or [imath]\displaystyle J_{0}(\lambda_n b) = 0[/imath]

If [imath]\displaystyle Y_{0}(\lambda_n b) = 0[/imath], it is impossible [imath]\displaystyle J_{0}(\lambda_n b) = 0[/imath] because Bessel function order zero of first kind doesn't have the same roots as Bessel function order zero of second kind.

Then, we must have [imath]\displaystyle Y_{0}(\lambda_n a) = 0[/imath] which also impossible because if [imath]\displaystyle Y_{0}(\lambda_n a) = 0[/imath], then [imath]\displaystyle A_n = 0[/imath].

This leave us with only one option. To satisfy the equation, [imath]\displaystyle \lambda_n[/imath] must be the roots of the equation for [imath]\displaystyle n = 1,2,3, \cdot \cdot \cdot \cdot [/imath]

This is why we are not looking for the zeros of [imath]\displaystyle J_{0}[/imath] or [imath]\displaystyle Y_{0}[/imath].

 

[shreddinglicks]

I will explain to you this in a different way and I hope this time you can see what I mean.

We have already seen how to get this equation.

[imath]\displaystyle Y_{0}(\lambda_n a)J_{0}(\lambda_n b) - J_{0}(\lambda_n a)Y_{0}(\lambda_n b) = 0[/imath]

Look at the equation above carefully. Our goal is to let the left side equal to the right side.

Let us assume [imath]\displaystyle Y_{0}(\lambda_n b) = 0[/imath].

We have now:

[imath]\displaystyle Y_{0}(\lambda_n a)J_{0}(\lambda_n b) - J_{0}(\lambda_n a)*0 = Y_{0}(\lambda_n a)J_{0}(\lambda_n b) = 0[/imath]

To satisfy the equation above, it's either [imath]\displaystyle Y_{0}(\lambda_n a) = 0[/imath] or [imath]\displaystyle J_{0}(\lambda_n b) = 0[/imath]

If [imath]\displaystyle Y_{0}(\lambda_n b) = 0[/imath], it is impossible [imath]\displaystyle J_{0}(\lambda_n b) = 0[/imath] because Bessel function order zero of first kind doesn't have the same roots as Bessel function order zero of second kind.

Then, we must have [imath]\displaystyle Y_{0}(\lambda_n a) = 0[/imath] which also impossible because if [imath]\displaystyle Y_{0}(\lambda_n a) = 0[/imath], then [imath]\displaystyle A_n = 0[/imath].

This leave us with only one option. To satisfy the equation, [imath]\displaystyle \lambda_n[/imath] must be the roots of the equation for [imath]\displaystyle n = 1,2,3, \cdot \cdot \cdot \cdot [/imath]

This is why we are not looking for the zeros of [imath]\displaystyle J_{0}[/imath] or [imath]\displaystyle Y_{0}[/imath].

I see.

I will attempt this problem tomorrow and see what results I get. I will post it here.

Thanks for your help so far. This has been a good learning experience.

 

[mario99]

I will be so glad to see your solution.

 

[shreddinglicks]

 

[mario99]

I like your approach. This is one way of how may someone get the benefit of the PDE's solution. The more terms you add, the approximation of the temperature value gets better. With using only 5 eigenfunctions, you got very close to the exact result.

Bravo shreddinglicks



You should by now know how to solve the heat equation easily when the boundary is [imath]a < r < b[/imath], and this is the most important point of this thread. Other things that you were confused of would be more clear every time you solve a PDE.

 

[shreddinglicks]

Yes, thank you. Your assistance has been invaluable in my self study.

I also attempted to create the solution to the other common boundary conditions.

Can you recommend a good textbook on this subject?

 

[shreddinglicks]

I forgot to write,

u(r,o) = f(r) in both cases.

 

[mario99]

Can you recommend a good textbook on this subject?

For me, I don't read from a specific book, but from multiple resources. But my advice for you is to warm up with any differential equation book that explains how to solve ordinary differential equations as well as PDEs. Then, if you feel that you have a solid foundation on the subject, read a book that only solve PDEs. In this manner, you will deal with almost every type of linear PDE.