How to Solve a Heat Equation Using FFCT?

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Homework Statement


solve the following heat problem using FFCT:
A metal bar of length L, is at constant temperature of ## U_0 ## , at ##t=0## the end ##x=L## is suddenly given the constant temperature of ##U_1## and the end x=0 is insulated. Assuming that the surface of the bar is insulated, find the temperature at any point x of the bar at any time ##t>0## , assume ##k=1##

X5usGki.png

Homework Equations


heat eq.
## \frac {\partial^2 u} {\partial x^2} = \frac 1 k \frac {\partial u} {\partial t} ##
with the following additional equations:
GGjbJ4J.jpg


The Attempt at a Solution


my attempt goes like this:
$$ \frac {\partial^2 u} {\partial x^2} = \frac 1 k \frac {\partial u} {\partial t} $$
$$ \mathcal{F}_{fc} \left[ \frac {\partial u} {\partial t} \right] = \mathcal{F}_{fc} \frac {\partial^2 u} {\partial x^2} $$
$$ \frac {dU} {dt} = {-\left( \frac {{n} {\pi}} L \right)}ˆ{2} * F(x,t) + \left( {-1} \right)ˆn \frac {\partial{f(L,t)}} {\partial x} - \frac {\partial{f(0,t)}} {\partial x} $$
$$ \frac {dU} {dt} = - \left( \frac {{n} {\pi}} L \right)ˆ(2) * F(x,t) + \left( {-1} \right)ˆn \frac {\partial{f(L,t)}} {\partial x} $$

and i don't know how to continue...
 
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@Aows, I have not worked much with the FFCT, but it seems like the method is much like that of other transforms.
After you have rewritten the derivatives, you should separate the variables, setting F(x,t) = X(x)T(t) and use standard ODE methods to solve for T(t).
What is unclear to me it that your problem does not explicitly give
## \frac{\partial f}{\partial x} (L,t)##
It feels like the sudden change from ##U_0 ## to ##U_1## represents a discontinuity. Have you dealt with similar problems before? I know in Laplace transforms, there is a standard method for handling things that switch on. I am unfamiliar with the analogous method for the Cosine transform.
 
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RUber said:
@Aows, I have not worked much with the FFCT, but it seems like the method is much like that of other transforms.
After you have rewritten the derivatives, you should separate the variables, setting F(x,t) = X(x)T(t) and use standard ODE methods to solve for T(t).
What is unclear to me it that your problem does not explicitly give
## \frac{\partial f}{\partial x} (L,t)##
It feels like the sudden change from ##U_0 ## to ##U_1## represents a discontinuity. Have you dealt with similar problems before? I know in Laplace transforms, there is a standard method for handling things that switch on. I am unfamiliar with the analogous method for the Cosine transform.
Dr.RUber @RUber ,
1. separate the variables, you mean in my last step? if so can you tell me how ?
2. for the ## \frac{\partial f}{\partial x} (L,t) ## , this is my problem too, do you think that some information should be given or all the infos are available now ?
 
To separate the variables, you assume that your function ##F## is a product of two functions, one dependent only on x, ##X(x)## and one dependent only on t ##T(t)##.
If you let ## T(0) = 1##, then your initial conditions will fully describe ##X(x)##.
In the case that ## \frac{\partial f}{\partial x} (L,t) = 0## then you have a simple differential equation to solve for each n.

I am not sure using the Cosine series is the best choice, since the data given are Dirichlet. The source I saw online said that Sine series are more appropriate for Dirichlet data. Are you able to use the sine transform instead?
 
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Dr. @RUber ,
how to apply the separation of variable after my last step??

and regarding the using FFCT instead of FFST is because the problem says in the end ## x=0 ## is insulated which means that du/dx=0,
 
Oh, I see. I read that information as saying that ##f(0,t) = U_0, \quad f(L,t) = U_1##. In this case, you seem to have mixed boundary conditions then.
##\frac{\partial u}{\partial x} (0,t) = 0, u(L,t) = U_1 ##
To use the separation of variables after your last step, you will find an appropriate function of time that would satisfy your differential equation.
 
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RUber said:
Oh, I see. I read that information as saying that ##f(0,t) = U_0, \quad f(L,t) = U_1##. In this case, you seem to have mixed boundary conditions then.
##\frac{\partial u}{\partial x} (0,t) = 0, u(L,t) = U_1 ##
To use the separation of variables after your last step, you will find an appropriate function of time that would satisfy your differential equation.
Dr. @RUber , can you elaborate more please?

many thanks to all of your contributions...
 
any updates related to this problem ?
thank you
 
Are you required to solve it using the FFCT? Like @RUber said, this doesn't seem too appropriate because of the mixed boundary conditions. ( I think I see a trick way of solving it using the FFST.)

Have you tried RUber's suggestion of approaching it using the standard separation of variables technique? What differential equations do you get for ##X(x)## and ##T(t)##?
 
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  • #10
TSny said:
Are you required to solve it using the FFCT? Like RUber said, this doesn't seem too appropriate because of the mixed boundary conditions. ( I think I see a trick way of solving it using the FFST.)

Have you tried RUber's suggestion of approaching it using the standard separation of variables technique? What differential equations do you get for ##X(x)## and ##T(t)##?
Hello Mr. @TSny ,
yes am required to use the FFCT to solve this problem. it is said that if there is an insulated side $dp/dx = 0 $ , then we are required to use the FFCT as a method of solution.
can you help with how to solve it using FFCT? @TSny
here is the solution of the problem using laplace transform:

F2dduhK.jpg


nqwOE4R.jpg
 
  • #11
The LT solution looks good except for a couple of minor errors that cancel each other:

In equation (6), there should be a minus sign in front of C2.

cosh is defined with a positive sign in the numerator, not a negative sign.

Using the FFCT, I don't see how to get past the snag of having U specified at x = L rather than having Ux specified at x = L.
 
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  • #12
TSny said:
The LT solution looks good except for a couple of minor errors that cancel each other:

In equation (6), there should be a minus sign in front of C2.

cosh is defined with a positive sign in the numerator, not a negative sign.

Using the FFCT, I don't see how to get past the snag of having U specified at x = L rather than having Ux specified at x = L.
thanks indeed for your notes on the problem, I also noticed those errors too.
regarding FFCT solution, this is my problem too, @TSny do you think that there are some missing information (i mean more should be given) or those info are enough but i need to learn more ??
 
  • #13
I don't think there is any missing information. The problem is well-posed.

Note that the FFCT over the interval ##0<x<L## leads to an expansion of ##U## in terms of the set of functions ##\cos \left(\frac{n \pi x}{L} \right)##. But, the LT solution is in terms of the set ##\cos \left(\frac{(2n-1) \pi x}{2L} \right)##. The members of the second set are not contained in the first set. So, I don't see how the FFCT is going to lead to the solution.
 
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  • #14
TSny said:
I don't think there is any missing information. The problem is well-posed.

Note that the FFCT over the interval ##0<x<L## leads to an expansion of ##U## in terms of the set of functions ##\cos \left(\frac{n \pi x}{L} \right)##. But, the LT solution is in terms of the set ##\cos \left(\frac{(2n-1) \pi x}{2L} \right)##. The members of the second set are not contained in the first set. So, I don't see how the FFCT is going to lead to the solution.
Mr. @TSny , what do you mean by this ((The members of the second set are not contained in the first set.)),
in my last reply, i meant: in order to use FFCT, do you think there are some more infos should be given ?
 
  • #15
For example, in the set ##\cos \left(\frac{(2n-1) \pi x}{2L} \right)##, you have ##\cos \left(\frac{3 \pi x}{2L} \right)## when ##n = 2##. But this function is not contained in the set ##\cos \left(\frac{n \pi x}{L} \right)## for any value of the integer ##n##.

I think there might be a "trick" way to solve the problem using the FFCT. Consider a related problem where you have a rod of length 2L which is insulated at both ends as well as the sides. Suppose the initial temperature distribution is ##U(x, 0) = U_0## for ##0 < x < L## and ##U(x, 0) = -U_0## for ##L < x <2L##. Also, suppose that ##U(L, 0) = 0##. You can solve this using the FFCT since the boundary condition at both ends is ##U_x = 0##. I think the solution for this problem relates in a simple way to the solution for the original problem.

upload_2017-9-3_12-58-37.png


Here, the "original problem" is your problem with the right end set at 0 rather than ##U_1## for convenience. But, that's easily taken care of.

I have not actually carried out the calculation, but I think it should work.
 
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  • #16
Mr. @TSny , that's exactly my problem, if i substitute ##U_1## , i don't know what to do after that ??
 
  • #17
I suggest first getting rid of the inhomogeneous boundary condition at x = L by shifting the solution.

Once you have done that, you can solve the problem using the base functions given in #13 or make an odd extension around x=L to end up with a domain 0<x<2L with homogeneous Neumann conditions at both boundaries - a problem you can directly apply FFCT on.

Edit: oops, that is exactly what #19 said. Should reload before posting after some time ...
 
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  • #18
Orodruin said:
I suggest first getting rid of the inhomogeneous boundary condition at x = L by shifting the solution.

Once you have done that, you can solve the problem using the base functions given in #13 or make an odd extension around x=L to end up with a domain 0<x<2L with homogeneous Neumann conditions at both boundaries - a problem you can directly apply FFCT on.

Edit: oops, that is exactly what #19 said. Should reload before posting after some time ...
thanks indeed for your contribution Mr. @Orodruin , can you write the full answer in details ?
exams are on the doors and you know there are a lot of subjects along with it.
your help will be highly appreciated.
 
  • #19
Aows said:
thanks indeed for your contribution Mr. @Orodruin , can you write the full answer in details ?
your help will be highly appreciated.
No, this would be against the forum rules and posting the solution would just generate a warning for me. You need to solve the problem yourself. Feel free to ask further questions about things you still find unclear.
 
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  • #20
Orodruin said:
No, this would be against the forum rules and posting the solution would just generate a warning for me. You need to solve the problem yourself. Feel free to ask further questions about things you still find unclear.
appreciate your kind reply a lot,
but i really don't know how to apply your idea of shifting the boundary condition after substituting the ##U_1## . @Orodruin
 
  • #21
What exactly do you mean by "substituting the ##U_1##"? Please write it out explicitly, it helps for understanding your thought process and whether you understood previous replies or not.
 
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  • #22
t
Orodruin said:
What exactly do you mean by "substituting the ##U_1##"? Please write it out explicitly, it helps for understanding your thought process and whether you understood previous replies or not.
this is my last step:

$$ \frac {dU} {dt} = - \left( \frac {{n} {\pi}} L \right)ˆ(2) * U(x,t) + \left( {-1} \right)ˆn * U_1 $$
 
  • #23
after my last step, i should separate the variables but if i have ##U_1##, i don't know hot to perform the separation with the ##U_1## exist ... @Orodruin
 
  • #24
You should get rid of the inhomogeneous boundary condition before you attempt the transform. Essentially you can do this by the ansatz ##u(x,t) = v(x,t) + h(x)## where ##h(x)## is the stationary solution for your inhomogeneous boundary conditions. You will then get an ODE for ##v(x,t)## that you can solve using either the eigenfunctions proposed in #13 or by the extension proposed in #19.

So first step: What is the stationary solution?

If you have problems with this, you can also start by solving the problem for ##U_1=0## and deal with this in the end. In that case, do you understand the odd extension around ##x=L##?
 
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  • #25
Orodruin said:
You should get rid of the inhomogeneous boundary condition before you attempt the transform. Essentially you can do this by the ansatz ##u(x,t) = v(x,t) + h(x)## where ##h(x)## is the stationary solution for your inhomogeneous boundary conditions. You will then get an ODE for ##v(x,t)## that you can solve using either the eigenfunctions proposed in #13 or by the extension proposed in #19.

So first step: What is the stationary solution?

If you have problems with this, you can also start by solving the problem for ##U_1=0## and deal with this in the end. In that case, do you understand the odd extension around ##x=L##?
Mr. @Orodruin , I really didn't understand this, why not write the answer after my last step ?
 
  • #26
I didn't use this anstaz before ##u(x,t) = v(x,t) + h(x)## @Orodruin
 
  • #27
Hello Mr. @Orodruin , any updates regarding the problem ?

very thankful,
Aows K.
 
  • #28
Aows said:
I didn't use this anstaz before ##u(x,t) = v(x,t) + h(x)## @Orodruin
You should try to do this.
At time = 0, you get
## u(x,0)=v(x,0) + h(x) ##
Your cosine transform is just in terms of x, right? And for each n, you have an ODE to solve in terms of a function of t, which should be of the form:
## f'(t) = g(t) + c ##
where your functions of x are treated as constants in terms of t. What methods do you know for solving ODEs of this type?

If you use separation of variables, the function might look like:
## u(x,t) = v(x)f(t) + h(x) ##
or in the transformed space
##U_n(x,t) = a_n \cos (n \omega x) f(t) + b_n \cos (n \omega x) . ##
 
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  • #29
Aows said:
t

this is my last step:

$$ \frac {dU} {dt} = - \left( \frac {{n} {\pi}} L \right)ˆ(2) * U(x,t) + \left( {-1} \right)ˆn * U_1 $$
I strongly suggest you follow my approach. A priori, you are only allowed to expand in the cosines if the function satisfies the appropriate homogeneous boundary conditions. Also, there cannot be any ##x## dependence left after the Fourier transform - you are transforming it away, it is the entire point, to get an ##x## independent differential equation for every Fourier mode!
 
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  • #30
Orodruin said:
I strongly suggest you follow my approach. A priori, you are only allowed to expand in the cosines if the function satisfies the appropriate homogeneous boundary conditions. Also, there cannot be any ##x## dependence left after the Fourier transform - you are transforming it away, it is the entire point, to get an ##x## independent differential equation for every Fourier mode!
I don't know how to use your approach, can you help with the rest of the solution ?
 
  • #31
I have already pointed out several mistakes you have made and told you that your approach will not work and why. I cannot help you more unless you specify exactly what it is that you are having trouble with in the proposed approach.
 
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  • #32
i have another example it is also using FFCT but the value of the first derivative is zero which makes it easy to solve, here is this example the value of the first derivative is #U_1# which i don't know how to deal with it, this is my problem...
add to that, i don't know how to use your approach, we never use it before ...
your help is highly appreciated,... @Orodruin
 
  • #33
Aows said:
i don't know how to use your approach, we never use it before ...
But I described to you how to use it. What in that description poses a problem?
 
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  • #34
I don't know how to apply it, because i didn't use it before... @Orodruin
 
  • #35
Hello dear gents, @RUber @Orodruin ,
my exam will take place on saturday, can you provide a full detailed answer for the problem or not ?

regards,
Aows K.
 
  • #36
Aows said:
Hello dear gents, @RUber @Orodruin ,
my exam will take place on saturday, can you provide a full detailed answer for the problem or not ?

regards,
Aows K.
What you are asking is agains the forum rules. You need to solve the problem yourself based on the hints that you have been given. If there are things you do not understand about those hints, ask about it specifically.
 
  • #37
dear Mr. @Orodruin ,
i read the forum rules since the first day that i signed up, and it says that you need to show your attempt so others can help with what you need, and i posted all my attempts.
so that's why am asking for the answer...
 
  • #38
i don't know how to solve it using FFCT @Orodruin
 
  • #39
Aows said:
dear Mr. @Orodruin ,
i read the forum rules since the first day that i signed up, and it says that you need to show your attempt so others can help with what you need, and i posted all my attempts.
so that's why am asking for the answer...
Yes, and you have been given help and guidance. That you are refusing to work with that guidance is up to you. Providing full answers is against the forum rules and you should not be expecting people to do so.
 
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  • #40
Ok, then Mr. @Orodruin , can you tell me what is the first step ?
 
  • #41
Aows said:
Ok, then Mr. @Orodruin , can you tell me what is the first step ?
I already did.
Orodruin said:
You should get rid of the inhomogeneous boundary condition before you attempt the transform. Essentially you can do this by the ansatz ##u(x,t) = v(x,t) + h(x)## where ##h(x)## is the stationary solution for your inhomogeneous boundary conditions. You will then get an ODE for ##v(x,t)## that you can solve using either the eigenfunctions proposed in #13 or by the extension proposed in #19.

So first step: What is the stationary solution?

If you have problems with this, you can also start by solving the problem for ##U_1=0## and deal with this in the end. In that case, do you understand the odd extension around ##x=L##?
 
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  • #42
Orodruin said:
I already did.
I will try my best to follow this step even though this is the first time using it, hopefully we reach to a solution.
minutes and i will show you my progress @Orodruin
 
  • #43
so, this first problem is to find the h(x) right? if so, how to find it ? Mr. @Orodruin
 
  • #44
  • #45
@Orodruin , any ideas on how to find the ## h(x) ## ?
 
  • #46
As I said already, it is the stationary solution to the problem. What is particular about the stationary solution?
 
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  • #47
what is the meaning of stationary solutions ? and how to find it ?? @Orodruin
 
  • #48
These are really basic questions that should be covered in your textbook. A stationary solution is a time-independent solution.
 
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  • #49
Mr. @Orodruin , please excuse my basic questions for the moment, just put yourself in my shoes, exam is two days away.
the definition of the stationary solutions is clear now, but how to find it so that i can then use it here ##u(x,t) = v(x,t) + h(x) ## ?
 
  • #50
Aows said:
just put yourself in my shoes, exam is two days away.
This has nothing to do with how we offer to help people here. You created this thread over two weeks ago and until now you have shown very little interest in putting in the effort necessary to actually solve the problem. We are volunteers and mostly help people who want to be helped. Stating that you need help immediately because your exam is two days away is likely to have the exact opposite effect as compared to what you are going for. In order to learn this properly, you need to sit down with the material and think about each step. I have already given you several hints that should be sufficient to at least find the stationary solution.

You are saying that you know understand what a stationary solution is, but I have my doubts because it seems that you put exactly zero effort into thinking about what this means and until you do so you will not be able to really learn the subject. Just seeing a lot of examples will not get you far, you need to sit down to think about and understand what the meaning behind what we are telling you is. Since it does not seem that we are going in that direction, I am done with this thread.
 
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