Solving PDE heat problem with FFCT

[Aows]

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 $k=1$

Homework Equations

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

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...

 

[RUber]

@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.

 

[Aows]

@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 ?

 

[RUber]

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?

 

[Aows]

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,

 

[RUber]

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.

 

[Aows]

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...

 

[Aows]

any updates related to this problem ?
thank you

 

[TSny]

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 $T(t)$?

 

[Aows]

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:

 

[TSny]

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 U_x specified at x = L.

 

[Aows]

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 U_x 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 ??

 

[TSny]

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.

 

[Aows]

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 ?

 

[TSny]

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.

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.

 

[Aows]

Mr. @TSny , that's exactly my problem, if i substitute $U_1$ , i don't know what to do after that ??

 

[Orodruin]

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 ...

 

[Aows]

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.

 

[Orodruin]

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.

 

[Aows]

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

 

[Orodruin]

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.

 

[Aows]

t

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:

$$ \frac {dU} {dt} = - \left( \frac {{n} {\pi}} L \right)ˆ(2) * U(x,t) + \left( {-1} \right)ˆn * U_1 $$

 

[Aows]

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

 

[Orodruin]

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 $x=L$?

 

[Aows]

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 ?

 

[Aows]

I didn't use this anstaz before $u(x,t) = v(x,t) + h(x)$ @Orodruin

 

[Aows]

Hello Mr. @Orodruin , any updates regarding the problem ?

very thankful,
Aows K.

 

[RUber]

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
$U_n(x,t) = a_n \cos (n \omega x) f(t) + b_n \cos (n \omega x) .$

 

[Orodruin]

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!

 

[Aows]

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 ?

 

[Orodruin]

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.

 

[Aows]

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

 

[Orodruin]

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?

 

[Aows]

I don't know how to apply it, because i didn't use it before... @Orodruin

 

[Aows]

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.