One-Dimensional Wave Equation & Steady-State Temperature Distribution

[Athenian]

Homework Statement

The steady-state temperature distribution, $T(x,y)$, in a flat metal sheet obeys the partial differential equation

$\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0$.

Separate the variables in this equation just as we did in the one-dimensional wave equation and find $T$ everywhere on a square flat plate of side length $S$, with the boundary conditions $T(0,y) = T(S,y) = T(x,0) = 0$, and $T(x,S) = T_0$.

In your write-up, be very explicit in how you apply the Orthogonal Function/Fourier Series methods from the OFFS tutorial when finding expressions for the unknown coefficients. Which orthogonality relationships do you use? Why?

Use the below equation as a guide for how to state your final answer:
$$y(x,t) = 8A \sum_{n \: odd} (-1)^{\frac{n-1}{2}} \bigg(\frac{1}{n \pi} \bigg)^2 sin \bigg(\frac{n \pi x}{L} \bigg) cos \bigg( \frac{n \pi v t}{L} \bigg)$$

Relevant Equations

Refer Below $\longrightarrow$

To begin with, I can first let $T(x,y) = X(x) Y(y)$ to be the given solution. With this, I can then continue by writing:

$$Y \frac{\partial^2 X}{\partial x^2} + X \frac{\partial^2 Y}{\partial y^2} = 0$$

$$\Longrightarrow \frac{1}{X} \frac{\partial ^2 X}{\partial x^2} + \frac{1}{Y} \frac{\partial ^2 Y}{\partial y^2} = 0$$

Continuing on, I could then introduce the linking constant $-k^2$ by writing:

$$\frac{1}{X} \frac{\partial ^2 X}{\partial x^2} = - \frac{1}{Y} \frac{\partial ^2 Y}{\partial y^2} = -k^2$$

With this equation, I can conclude that:

$\frac{\partial ^2 X}{\partial x^2} = -k^2 X$ and $\frac{\partial ^2 Y}{\partial y^2} = k^2 Y$

Calculating for both $X(x)$ and $Y(y)$ respectively:

$$X(x) = A \, sin(kx) + B \, cos(kx)$$

$$Y(y) = C \, e^{ky} + D \, e^{-ky}$$
Therefore, $T(x,y)$ can now be expressed as below:

$$T(x,y) = X(x) \, Y(y) = \begin{Bmatrix} A \, sin(kx) \\ B \, cos(kx) \end{Bmatrix} \begin{Bmatrix} C \, e^{ky} \\ D \, e^{-ky} \end{Bmatrix} \Longrightarrow \begin{Bmatrix} A \, \frac{e^{ikx} – e^{-ikx}}{2i} \\ B \, \frac{e^{ikx} + e^{-ikx}}{2} \end{Bmatrix} \begin{Bmatrix} C \, e^{ky} \\ D \, e^{-ky} \end{Bmatrix}$$

Next, I need to begin applying my boundary conditions to the equation.

However, this is where my problem begins to set in.

I did use the multiplication (i.e. FOIL method) to get a moderately long chain of algebraic expressions upon multiplying $X(x)$ and $Y(y)$ together. When taking the condition $T(0,y) =0$ and plugging 0 into all my $x$’s in the equation, I got a relatively elegant equation of $T(0,y) = Be^{-ky} (Ce^{2ky} + D) = 0$. However, when calculating for $T(x,0) =0$, I got a really complicated equation in its place – which, to me, seems to be a telltale sign that I am in the wrong direction.

Therefore, any help and assistance to help me know how to properly apply my boundary condition to the given equation would be greatly appreciated! Of course, if I made any errors in my above calculations, I would sincerely appreciate it if you would point it out. Thank you very much for reading through this as well as presenting your assistance!

 

[BvU]

Hi,

I am surprised by your boundary condition application: $$T(0,y) = Be^{-ky} (Ce^{2ky} + D) = 0$$ can't be right: for $x=0$ you want $X=0$ so $B = 0$ immediately follows. Next to apply is $x=S$ which gives a set of allowed $k_n$.

Then $T(x,0)=0$ gives $C + D = 0$ which is anything but complicated.

The complication sets in with looking at $T(x,S)=T_0$ !

I don't know what's in your OFFS tutorial, but I see a solution of the form $\sum \sin(...x)\sinh(...y)$ emerging, so I can't help you with the book answer.

 

[Athenian]

Thank you so much for the help, @BvU ! And, judging by the solution form you got, I am fairly certain you are correct. Not to mention, another person working on the project stated that $sinh$ should appear in the Fourier series as well.

Anyway, with that said, I went ahead and redid my calculations.

Starting where I left off and calculating for $T(0,y) = 0$, I got:
$$T(0,y) = X(0) \, Y(y) = B(Ce^{ky} + De^{-ky}) = 0$$

Since $X(0)$ needs to be equal to zero for the entire product to ultimately go to zero, I thus inferred that $B$ must equal to zero in my equation, i.e. $B=0$.

With that said, my equation now looks like below:
$$T(x,y) = A \, sin(kx) \cdot (Ce^{ky} + De^{-ky})$$

Next, I have the boundary condition $T(S,y) = 0$:
$$T(S,y) = A \, sin(kS) \cdot (Ce^{ky} + De^{-ky}) = 0$$

Assuming $A \neq 0$ (otherwise, the amplitude will be zero and ultimately not satisfy the problem), I can state that $sin(kS) = 0$ in order for the entire product to go to zero. Therefore:

$$sin(kS) = 0$$
$$\Rightarrow kS = n \pi $$
$$\Rightarrow k = \frac{n \pi}{S}$$

From now on, I will refer to this $k$ that has a value of $\frac{n \pi}{S}$ as $k_n$.

Therefore, the "updated" equation is as seen below:
$$T(x,y) = A \, sin(k_n x) \cdot (Ce^{k_n y} + De^{-k_n y})$$

Next, I have the boundary condition $T(x,0) = 0$ to apply to my equation:
$$T(x,0) = A \, sin(k_n x) \cdot (C+ D) = 0$$

To satisfy the above condition, I can say that $C + D = 0$.

Finally, I have the boundary condition $T(x,S) = T_0$:
$$T(x,S) = A \, sin(k_n x) \cdot (Ce^{k_n S} + De^{-k_n S}) = T_0 $$

Now, from here, I am completely confused on how to properly continue. You were right when you said this was when the complication begins to set in.

Provided that I am on the right solution-finding track, how should I go about continuing my mathematical calculations?

Thank you so much for the awesome help! I really appreciate it!

 

[BvU]

Provided that I am on the right solution-finding track

I think you are. Note that the solution so far is an infinite sum (over n) and the last boundary condition is going to assign weights to each of the terms, so that you end up with a unique solution.

how to properly continue

The next step is above my pay grade, so I will have to leave you in other hands:

Please consult
http://www.eng.auburn.edu/~dmckwski/mech7210/condbook.pdf page 95 - 98

 

[LCKurtz]

@Athenian : One thing that will simplify you work at this point is to use sinh and cosh functions instead of exponentials. So your $Y_n(y)$ would be $C_n\cosh(k_ny) + D_n\sinh(k_ny)$. Then when you apply the boundary condition $Y(0) = 0$ you get $C_n = 0$. So your solution will be a something like

$$T(x,y) = \sum_{n=1}^\infty a_n\sin(k_nx)\sinh(k_ny)$$
and you still have your boundary condition on $T(x,S)$ to work. Thats where your Fourier Series comes in.

 

[Athenian]

I think you are. Note that the solution so far is an infinite sum (over n) and the last boundary condition is going to assign weights to each of the terms, so that you end up with a unique solution.

Thank you for letting me be aware of this. I'll be sure to keep this in mind as I continue my calculations. Thank you for all your helpful comments thus far and I really appreciate the assistance!

@Athenian : One thing that will simplify you work at this point is to use sinh and cosh functions instead of exponentials. So your $Y_n(y)$ would be $C_n\cosh(k_ny) + D_n\sinh(k_ny)$. Then when you apply the boundary condition $Y(0) = 0$ you get $C_n = 0$.

Thank you for your help, @LCKurtz ! While I understand the given logic of using sinh and cosh functions instead of exponentials, am I allowed to "legally" do this - mathematically speaking? As I am assuming this is the case, why is that? Would I need to revert the solution back to its original form once I find the solution to the problem?

I ask this because isn't $\frac{\partial^2 Y}{\partial y^2} = k^2 Y$ supposed to be only $Y(y) = C e^{ky} + D e^{-ky}$ rather than hyperbolic trignometrical functions?

Now, I did notice that you added a subscript $n$ to your $Y(y)$, [i.e. $Y_n (y)$]. Is this why all this is "transformation" is allowed?

Finally, looking over your solutions, would my old solution here below be technically illegal since I added a subscript $n$ for all my $k's$ upon finding $k= \frac{n \pi}{S}$?

Therefore, the "updated" equation is as seen below:

$$T(x,y) = A \, sin(k_n x) \cdot (Ce^{k_n y} + De^{-k_n y})$$

Once again, thank you so much for the help. However, before moving on, I would just like to make sure I properly understand what is going on. Thank you for all your assistance and I sincerely appreciate your insightful previous reply!

 

[LCKurtz]

This reply is about using $\{\sinh(kx), \cosh(kx)\}$ pair instead of the $\{e^{kx}, e^{-kx}\}$ pair. When are solving $y'' - k^2 y = 0$, with characteristic equation $r^2 - k^2 = 0$ and roots $r = \pm k$ that gives you an independent solution pair $\{ e^{kx}, e^{-kx}\}$ and you would write the general solution as $y =C_1e^{kx} + C_2e^{-kx}$. Nothing wrong with that. But you don't have to use those two functions. Any pair of linearly independent functions in the same space will do. As you get more experience you will discover that, depending on the boundary conditions, a different pair might be simpler to work with. To keep it simple, let's take $y''-y=0$ with $k=1$. Suppose your boundary conditions are $y(0) =0,~y'(0) = 2$. If you are using the exponentials you have $y = C_1e^{x} + C_2e^{-x}$. Then $y'=C_1e^{x} - C_2e^{-x}$. Putting the boundary conditions in those gives $C_1+C_2 =0,~C_1-C_2=2$ giving $C_1=1,~C_2 = -1$ and solution $y = e^{x} - e^{-x}$.
Now consider the same boundary conditions but using the hyperbolic functions. Here $y = A\cosh(x) + B\sinh(x)$ and $y' = A\sinh(x) + B\cosh(x)$. Putting the boundary conditions in these gives $0 = A \cdot 1 + B\cdot 0$ and $2 = A\cdot 0 +B\cdot 1$ for a solution $y = 2\sinh(x)$. Same solution, and even in this easy example, simpler.
You get the same thing with $y'' + y = 0$. Here your characteristic equation gives $r^2 + 1 = 0, ~ r = \pm i$. So you can choose between the pair $\{e^{ix},~e^{-ix}\}$ or $\{\sin x,~\cos x\}$. Since equations like this with real coefficients and real boundary conditions have real solutions, nobody in his right mind would use the complex pair. (I'm old, so I can say stuff like that).

 

[Athenian]

@LCKurtz , wow, thank you very much for the informative and - frankly - amazing explanation. That really cleared things up for me and I really appreciate the detailed example as well.

Anyway, just to make sure I am properly understanding everything thus far, I am going to backtrack a bit in my calculations and continue from there before I hit a bump.

Starting from here:
$$T(S,y) = A \, sin(kS) \cdot (Ce^{ky}+De^{−ky})=0 $$
$$\Rightarrow T(x,y) = A \, sin(k_n x) \cdot (Ce^{ky}+De^{−ky})$$
Note that I just skipped the steps of finding for $k_n$ here as it was already calculated in another reply above.

However, noting that $Y(y) = (Ce^{ky}+De^{−ky})$, I could state that $Y_n (y) = C_n \, cosh(k_n y) + D_n \, sinh(k_n y)$ can be used instead of the former equation (i.e. $Y(y)$) as not only does it make the calculation process easier but also they are linearly independent functions at the same given space from the beginning part of the problem.

Therefore, the equation would look like below:
$$T(x,y) = A \, sin(k_n x) \cdot (C_n \, cosh(k_n y) + D_n \, sinh(k_n y))$$

From there, applying the condition $T(x,0) = 0$:
$$T(x,0) = A \, sin(k_n x) \cdot (C_n \, cosh(0) + D_n \, sinh(0)) = 0$$
$$\Rightarrow T(x,0) = A \, sin(k_n x) \cdot (C_n) = 0$$

Since $Y_n(y)$ needs to be equal to zero, I can safely say that the constant $C_n=0$.

Now, the equation looks like:
$$T(x,y) = \sum^{\infty}_{n=1} A_n \, sin(k_n x) \, sinh(k_n x)$$

Note that $D_n$ got absorbed into $A$, resulting in $A_n$.

Finally, applying the given boundary condition $T(x,S) = T_0$:
$$T(x,S) = \sum^{\infty}_{n=1} A_n \, sin(k_n x) \, sinh(k_n S) = T_0$$

However, noting that $k_n = \frac{n \pi}{S}$, I can then write $sinh(k_n S) \Rightarrow sinh(\frac{n \pi}{S} S) \Rightarrow sinh(n \pi)$.

Therefore, the equation is now:
$$T(x,S) = \sum^{\infty}_{n=1} A_n \, sin(k_n x) \, sinh(n \pi) = T_0$$

From, here, I am lost.

I want to find for $A_n$, however, I am not sure how to go about accomplishing it. I thought about making the above equation into Fourier Series form and solve for $A_n$. But, that isn't going too well. Not to mention, I am not sure how $T_0$ fits into all this so I can finally find for $T(x,y)$.

Once again, thank you for your stellar explanation earlier and I really appreciate the help. Thank you and I look forward to your insightful comments!

 

[LCKurtz]

I didn't check the details, but it looks pretty good to here and the below equation is what I would expect.

Therefore, the equation is now:
$$T(x,S) = \sum^{\infty}_{n=1} A_n \, \sin(k_n x) \, \sinh(n \pi) = T_0$$
From, here, I am lost.

Not too lost I think. You are almost done. Let's call $A_n\sinh(n\pi) = a_n$. So your equation is $$
\sum^{\infty}_{n=1} a_n sin(\frac{n\pi} S x)= T_0$$Hopefully, if you were given that equation out of the blue, you would recognize that is indeed a FS problem and to get the $a_n$ you would use the formula for the half-range sine expansion of the constant $T_0$ on $(0,S)$. Once you have $a_n$ you have $A_n$ to put in your answer. It may get a bit complicated, but that's life in the FS domain.

 

[LCKurtz]

Now, the equation looks like:
$$T(x,y) = \sum^{\infty}_{n=1} A_n \, sin(k_n x) \, sinh(k_n \color{red}{x})$$

Probably just a typo, but that $x$ should be a $y$.

 

[Athenian]

@LCKurtz , thank you so much for the help as well as pointing out my typo. But, yes, that should be a $y$ there in $sinh(k_n y)$.

In that case, starting once again where I left off and following your helpful instructions.
Making $a_n = A_n \, sinh(n \pi)$, I have the below equation (while applying the boundary condition):

$$\sum_{n=1}^{\infty} a_n sin \bigg( \frac{n \pi}{S} x \bigg) = T_0$$

Now, solving for $a_n$ and noticing that it's a half range sine series:
$$a_n = \frac{2}{S} \int_0^S T_0 \, sin \bigg( \frac{n \pi}{S} x \bigg) dx$$

Avoiding writing out all the math of solving the above integration over here, I went ahead and solved it using an online calculator. Rest assured, I'll solve it by hand later.

Anyway, the answer the online calculator showed is:
$$a_n = \frac{2T_0(-(-1)^n +1)}{\pi n}$$

Note that if $n$ is even, then $a_n = 0$. However, if $n$ is odd, then $a_n = \frac{4T_0}{\pi n}$.

Next, noting that $a_n = A_n \, sinh(n \pi)$, I can also say that $A_n = \frac{a_n}{sinh(n \pi)}$.

Therefore, when $n$ is even, $A_n = 0$. And when $n$ is odd, $A_n = \frac{4T_0}{\pi n \, sinh(\pi n)}$.

To wrap up, the previous equation,
$$T(x,y) = \sum_{n=1}^{\infty} A_n \, sin(k_n x) \, sinh(k_n y)$$

can now turn into the below one-dimensional wave equation:
$$T(x, y) = \sum_{n=1}^{\infty} \frac{4T_0}{\pi n \, sinh(\pi n)} \, sin(k_n x) \, sinh(k_n y)$$

where $k_n = \frac{n \pi}{S}$.

Is the above solution attempt correct and is this how it is supposed to look like? Of course, if I made any mistakes, please do let me know. Thank you for your kind assistance!

 

[LCKurtz]

I don't want to slog through all the details to look for typos etc., I'm sure you can find them yourself. One more point I would give you. When you separated variables and had the equation for $Y(y)$ in the form $Y''(y) = k^2Y$ and $X(x)Y(0) = 0$ that gives you $Y(0) = 0$. At that point you (now) would take Y(y) = $C\cosh(ky) + D\sinh(ky)$ and your BC would have given you $Y_n(y) =D_n \sinh(k_ny)$ and you wouldn't have had those two annoying coefficients all the way through.
Also, another note, in your tex if you put a \ before the trig and hyperbolic functions it gives the appropriate typeface. This will be my last post here, at least for a while, because I have a busy afternoon starting about now.

 

[vela]

Next, I have the boundary condition $T(x,0) = 0$ to apply to my equation:
$$T(x,0) = A \, sin(k_n x) \cdot (C+ D) = 0$$
To satisfy the above condition, I can say that $C + D = 0$.

Finally, I have the boundary condition $T(x,S) = T_0$:
$$T(x,S) = A \, sin(k_n x) \cdot (Ce^{k_n S} + De^{-k_n S}) = T_0 $$

Is there any reason you didn't use the fact that $C+D=0$ to eliminate $D$ and write your solution as
\begin{align*}
T(x,S) &= \sum_n A_n \, \sin(k_n x) \cdot (C_n e^{k_n S} - C_n e^{-k_n S}) \\
&= \sum_n A_n \, \sin(k_n x) \cdot \sinh k_n S
\end{align*} where $C_n$ and the factor of 2 are absorbed into $A_n$ in the last step?

Is the above solution attempt correct and is this how it is supposed to look like? Of course, if I made any mistakes, please do let me know. Thank you for your kind assistance!

One way to check your answer would be to plot your solution and see if it satisfies all the boundary conditions.

 

[BvU]

Therefore, when $n$ is even, $A_n = 0$. And when $n$ is odd, $A_n = \frac{4T_0}{\pi n \, sinh(\pi n)}$.

Looks good. Link in post #4 (4.20) then says: Therefore, the index can be replaced by $2n-1\ , \quad n = 1,2,...$

Is the above solution attempt correct and is this how it is supposed to look like? Of course, if I made any mistakes, please do let me know. Thank you for your kind assistance!

With the $2n-1$ modification your answer looks like his (4.21). A contour plot is shown at the top of the same page. The solution is then discussed and found to be slightly unrealistic.

I watched the last nine posts from the sidelines and now understand better what is done. Bravo.

 

[Athenian]

I don't want to slog through all the details to look for typos etc., I'm sure you can find them yourself. One more point I would give you. When you separated variables and had the equation for $Y(y)$ in the form $Y''(y) = k^2Y$ and $X(x)Y(0) = 0$ that gives you $Y(0) = 0$. At that point you (now) would take Y(y) = $C\cosh(ky) + D\sinh(ky)$ and your BC would have given you $Y_n(y) =D_n \sinh(k_ny)$ and you wouldn't have had those two annoying coefficients all the way through.
Also, another note, in your tex if you put a \ before the trig and hyperbolic functions it gives the appropriate typeface. This will be my last post here, at least for a while, because I have a busy afternoon starting about now.

Thank you for noting that, @LCKurtz . I'll be sure to keep that in mind. Also, I'll add a \ before the trig and hyperbolic functions in the future. Surprisingly, I didn't know this LaTeX technique. Beyond that, please don't worry about checking the typos for me. Like you said, I could very easily check it myself and find any unintentional mishaps in my work. Rather, what I was more concerned about was whether the solution process as well as the attained solution being correct or not. But, judging everyone's comments here, it looks like I got it correct. Thank you for the help despite your busy schedule! I really appreciate it and I have learned a lot through your instructions.

Is there any reason you didn't use the fact that $C+D=0$ to eliminate $D$ and write your solution as
\begin{align*}
T(x,S) &= \sum_n A_n \, \sin(k_n x) \cdot (C_n e^{k_n S} - C_n e^{-k_n S}) \\
&= \sum_n A_n \, \sin(k_n x) \cdot \sinh k_n S
\end{align*} where $C_n$ and the factor of 2 are absorbed into $A_n$ in the last step?

Admittingly, I overlooked that fact. I really should have known better there. Regardless, thank you so much for pointing that out, @vela !

One way to check your answer would be to plot your solution and see if it satisfies all the boundary conditions.

After trying to plot my final $T(x,y)$ equation on Mathematica for a while to no avail (as I am a novice at the program), a user from StackExchange kindly helped me out on that front. Below in the quote box is the URL link for it. Essentially, he created an interactive plot via Mathematica (available to view in the given URL) which allows the user to change the constants $T_0$ and $S$ as well as $n$ in the $\sum$ of the $T(x,y)$ equation.

What confused me above all is this statement from the user:

That is "steady state wave equation". Steady state wave PDE becomes Laplace PDE at steady state. since at time infinity, the time dependency goes away and we are left with only the Laplacian.

Source: https://mathematica.stackexchange.c...-one-dimensional-wave-equations/220424#220424

So, if I am interpreting this statement correctly, my solution is not a one-dimensional wave equation? I feel pretty confused right now. If anything, though, the given graph is able to fulfill all the given boundary conditions when I change the equation a bit. Therefore, it should be valid. However, like I said already, his statement does confuse me a bit since I have not yet learned anything about Laplacian (even though I have heard of it before).

Looks good. Link in post #4 (4.20) then says: Therefore, the index can be replaced by $2n-1\ , \quad n = 1,2,...$With the $2n-1$ modification your answer looks like his (4.21). A contour plot is shown at the top of the same page. The solution is then discussed and found to be slightly unrealistic.

I watched the last nine posts from the sidelines and now understand better what is done. Bravo.

Thank you so much for pointing that out, @BvU . Rather than simply writing when $n$ is odd under the $\sum$ notation, I will also include the $2n-1$ modification. Beyond that, I did check equation 4.21 in the PDF and my equation does look incredibly similar to the one written in the book. Quite surprising, I got to admit.

Once again, thank you very much everybody for all of your kind assistance through these couple of days. I have learned a lot just from this problem alone and I sincerely appreciate the helpful guidance!