PDE Problem, the solutions of a square drum

[jamesmcm]

Homework Statement

Question: Show that the solutions of the wave equation for a square drum head of side L can be written as:

u(x,y,t)=\sum_{k_x , k_y} A_{k_x , k_y} e^{-ik_x x - ik_y y}e^{i\omega t}

where:

\omega = a \sqrt{{k_x}^2 + {k_y}^2}

Where a is the wave-velocity and A_{k_x , k_y} is a constant independent of x, y and t.

Hint:

The PDE is:
\frac{1}{a^2} \frac{\partial^2 u}{\partial t^2} = \nabla ^2 u(x,y,t)
Take u=X(x)Y(y)T(t).

Homework Equations

The actual PDE to solve is: \frac{1}{a^2} \frac{\partial^2 u}{\partial t^2} = \nabla ^2 u(x,y,t)

The PDE is separable so u=X(x)Y(y)T(t).

The Attempt at a Solution

By assuming u is separable we see:

YTX'' + XTY'' = \frac{1}{a^2} XYT''

\therefore \frac{1}{a^2} \left ( \frac{X''}{X} + \frac{Y''}{Y} \right ) = \frac{T''}{T} = -\omega^2
Solve ODE for T:

T'' + \omega^2 T = 0 ~ \therefore ~ m^2 = - \omega^2 ~ \therefore ~ m=\pm i\omega ~ \therefore ~ T=Ae^{+i\omega t} + Be^{-i\omega t}

Let \frac{1}{a^2} \frac{X''}{X} = -{k_x}^2, giving us the condition -a^2{k_x}^2 -a^2{k_y}^2 =-\omega^2:

\therefore m^2 + a^2 k_x =0 \therefore m^2 = -a^2 {k_x}^2 \therefore m=\pm a{k_x}i \therefore X=Ce^{+ia{k_x}x} + De^{-ia{k_x}x}

Similarly:

Y=Ee^{+i{ak_y}y} + Fe^{-i{ak_y}y}

And so:

u=XYT=\left ( Ae^{+i\omega t} + Be^{-i\omega t} \right ) \left (Ce^{+ia{k_x}x} + De^{-ia{k_x}x} \right ) \left(Ee^{+ia{k_y}y} + Fe^{-ia{k_y}y} \right )

How do we discount the positive roots for X and Y and the negative root for T, to obtain the necessary answer?

Mathbin link, in case I screwed up this tex: http://mathbin.net/55696"

 

[LCKurtz]

I haven't checked all your work, but the answer to your question is generally that you use the boundary conditions at your separated variables step. Presumably your drum head is clamped down on the boundaries so you must have conditions like

u(0,y,t) = u(L,y,t) = 0
u(x,0,t) = u(x,L,t) = 0

Translate these to boundary conditions for your X and Y and use them.

 

[maves]

Hello,

I am trying to solve a problem, that looks something like this, with an addition of (the same) boundary conditions on all of the four sides of the square. I need to find the eigenvalues and eigenfrequencies (and find out how the spectrum of eigenfrequencies depends on the coupling constant \alpha in the boundary condition).
The boundary conditions are of 3rd order, which means sth like u+\alpha \frac{\partial u}{\partial x}\bigg |_{x=0}=0.

So, as I understand it, I can separate the equation in the way that jamesmcm did in his original post, so this also means that because the directions x and y are independent, I can also write the boundary condition for every direction separately, am I right? What does this condition even mean (physically) for my drum?

Later on, I inserted this expressions (the ones similar that jamesmsm wrote in the end of his post, although I do believe he had one unneded a in every exponent in X and Y) into the 4 boundary conditions - I got 2 equations for X and 2 for Y (here b is the side of the square):
Ce^{ik_x x}+De^{-ik_x x}+C \alpha ik_x e^{ik_x x}-D\alpha ik_x e^{-ik_x x} \bigg|_{x=0, x=b}=0 and
Ee^{ik_y y}+Fe^{-ik_y y}+E \alpha ik_y e^{ik_y y}-F\alpha ik_y e^{-ik_y y} \bigg|_{y=0, y=b}=0 , so if C/D=\tilde C and E/F=\tilde E:

(1)\; \tilde C + 1+\alpha i k_x \tilde C -\alpha ik_x=0 \hspace{1cm}\longrightarrow <br /> \hspace{0.5cm} \tilde C (1+\alpha ik_x)=\alpha ik_x-1 \\ <br /> <br /> (2)\; \tilde C e^{i2k_x b} + 1+\alpha i k_x \tilde C e^{i2k_x b} -\alpha ik_x=0 \\<br /> <br /> (3)\; \tilde E + 1+\alpha i k_y \tilde E -\alpha ik_y=0 \hspace{1cm}\longrightarrow <br /> \hspace{0.5cm} \tilde E (1+\alpha ik_y)=\alpha ik_y-1 \\ <br /> <br /> (4)\; \tilde E e^{i2k_y b} + 1+\alpha i k_y \tilde E e^{i2k_y b} -\alpha ik_y=0\, .\\<br />
When I inserted equations for \tilde C and \tilde E from (1) and (3) into (2) and (4), I got:
(5)\; \frac{\alpha ik_x -1}{\alpha ik_x +1}(e^{2ik_x b}+\alpha i k_x e^{2ik_x b})=\alpha ik_x -1\, \\<br /> (6)\; \frac{\alpha ik_y -1}{\alpha ik_y +1}(e^{2ik_y b}+\alpha i k_y e^{2ik_y b})=\alpha ik_y -1\,<br />
and therefore
e^{2ik_x b}=1 \hspace{0.5cm} \text{and}\hspace{0.5cm} e^{2ik_y b}=1\;,
which gives me the requirement
k_x=\frac{\pi m}{b} \hspace{0.5cm} \text{and}\hspace{0.5cm} k_y=\frac{\pi n}{b}\;, where m and n are integers. This gives us k^2=k_x^2+k_y^2=(\frac{\pi}{b})^2(m^2+n^2) \,.

But isn't this the solution for a system with 1st order boundary conditions (such as u|_{x=0}=0 )?? I think so - also because the introductions also tells me to find out how the spectrum of eigenfrequencies depends on the coupling constant \alpha and I have "lost" \alpha somewhere during this calculation. Could anybody please tell me, where I am getting this wrong?

 

[LCKurtz]

@maves: When you separate variables you get$$
\frac{X''}{X} + \frac{Y''}{Y} = \frac{T''}{\alpha^2 T}$$If you set$$
\frac{X''}{X}=\lambda,\, \frac {Y''}{Y}=\mu$$then you have$$
\frac{T''}{\alpha^2 T}= \lambda + \mu$$or$$
T'' -\alpha^2(\lambda + \mu)T = 0$$ $$
T'' + \alpha^2(\frac{n^2\pi^2}{b^2}+\frac{m^2\pi^2}{b^2})T=0$$
That shows where your $\alpha$ went. As for the rest of the problem, I don't see where you have solved your $T(t)$ equation, constructed your solution $u(x,y,t)$, and applied your boundary conditions.

As an aside, I can't for the life of me figure out why you express everything in terms of complex exponentials when sines and cosines are more appropriate.

 

[maves]

LCKUrtz, first of all I would like to thank you for your answer. Only now I can see I didn't make some of my assumptions clear, so you couldn't really understand my question.
What I wanted to say is that I followed all of the steps that the original poster (jamesmcm) did in his first post of this thread, as I found them correct, and only after that I wanted to use my special boundary conditions. I believe this order of steps should be OK? (So to first find the general solution and only after that put in the boundary conditions, so that you get the ratio of the coefficients or sth like that.)

So, my constructed solution is - written a little bit differently as in the first post of this thread, because \omega^2=c^2(k_x^2+k_y^2), moreover I am going to use c for the wave velocity, because I can see from your answer, there was some confusion between a (wave velocity in the first post) and \alpha (the coupling coefficient in my boundary condition):

u(x,y,z)=TXY=(Ae^{ict\sqrt{k_x^2+k_y^2}}+Be^{-ict\sqrt{k_x^2+k_y^2}})(Ce^{ik_x x}+De^{-ik_x x})(Ee^{ik_y y}+Fe^{-ik_y y}),
and only then I used the boundary conditions, which only have an impact on X and Y - this means in my case

X+\alpha \frac{\partial X}{\partial x} \bigg|_{x=0}=0 \\<br /> X+\alpha \frac{\partial X}{\partial x} \bigg|_{x=b}=0 \\<br /> Y+\alpha \frac{\partial Y}{\partial y} \bigg|_{y=0}=0 \\<br /> Y+\alpha \frac{\partial Y}{\partial y} \bigg|_{y=b}=0\, ,
where (b is the side of the square, the origin of the coordinate system is positioned in the bottom left corner of the square).

So when I inserted my X and Y into those equations, I got eq. (1)-(4) (if you look at my earlier post) and so on.

It's a shame I can no longer edit my previous post, as to put everything together nicely.

So my question, again, is, where did I lose the coupling coefficient of the boundary conditions, which means \alpha. It would be very nice if anyone could help me out.

PS: I somehow prefer the complex exponentials to sines&cosines, as I believe it is easier to compute with them.

 

[LCKurtz]

The boundary conditions are of 3rd order, which means sth like u+\alpha \frac{\partial u}{\partial x}\bigg |_{x=0}=0.

LCKUrtz, first of all I would like to thank you for your answer. Only now I can see I didn't make some of my assumptions clear, so you couldn't really understand my question.
What I wanted to say is that I followed all of the steps that the original poster (jamesmcm) did in his first post of this thread, as I found them correct, and only after that I wanted to use my special boundary conditions. I believe this order of steps should be OK? (So to first find the general solution and only after that put in the boundary conditions, so that you get the ratio of the coefficients or sth like that.)

So, my constructed solution is - written a little bit differently as in the first post of this thread, because \omega^2=c^2(k_x^2+k_y^2), moreover I am going to use c for the wave velocity, because I can see from your answer, there was some confusion between a (wave velocity in the first post) and \alpha (the coupling coefficient in my boundary condition):

u(x,y,z)=TXY=(Ae^{ict\sqrt{k_x^2+k_y^2}}+Be^{-ict\sqrt{k_x^2+k_y^2}})(Ce^{ik_x x}+De^{-ik_x x})(Ee^{ik_y y}+Fe^{-ik_y y}),
and only then I used the boundary conditions, which only have an impact on X and Y - this means in my case

X+\alpha \frac{\partial X}{\partial x} \bigg|_{x=0}=0 \\<br /> X+\alpha \frac{\partial X}{\partial x} \bigg|_{x=b}=0 \\<br /> Y+\alpha \frac{\partial Y}{\partial y} \bigg|_{y=0}=0 \\<br /> Y+\alpha \frac{\partial Y}{\partial y} \bigg|_{y=b}=0\, ,
where (b is the side of the square, the origin of the coordinate system is positioned in the bottom left corner of the square).

So, does this mean your original boundary conditions in u were
$$
u(0,y,t) + \alpha u_y(0,y,t)=0$$ $$
u(b,y,t) + \alpha u_y(b,y,t)=0$$ $$
u(x,0,t) + \alpha u_x(x,0,t)=0$$ $$
u(x,b,t) + \alpha u_x(x,b,t)=0$$

instead of what the OP had, not in addition to his BC's? I am going to assume the answer to that is yes. This means that when you separate your variables getting$$
X''-\lambda X = 0\quad Y''-\mu Y = 0$$you have corresponding BC's. For $X$ you would have$$
X(0)+\alpha X'(0)=0\,\quad X(b)+\alpha X'(b) = 0$$and for $Y$:$$
Y(0)+\alpha Y'(0)=0\,\quad Y(b)+\alpha Y'(b) = 0$$
This is where you use the $\alpha$ and those boundary conditions, and it will affect everything. You don't do the u = XYT thing before you have done this.

Generally a problem like this will have an initial condition $u(x,y,0) = f(x,y)$ which is where the Fourier Series comes into play although I don't know whether that is relevant to what you are trying to do.

 

[maves]

Hello,
this is me again, I unfortunately still haven't managed to finish my work yet.

LCKurtz, I think you were right, it is beter to use the combination of sines and cosines in my solution, so now I have:
<br /> X=C\cos{k_x x}+D\sin{k_x x} \\<br /> Y= E\cos{k_y y}+F\sin{k_y y}<br />

I figured out, that the original boundary condition is actually (u+ \alpha \frac{\partial u}{\partial n})\bigg|_{\partial \Omega}=0, (where \partial \Omega is the boundary of the area, where we want to find the solution for u)so we have a normal derivative in it. I believe that this results in change of sign before \alpha where the normal is negatively directed:
(XY-\alpha\frac{\partial X}{\partial x}Y)\bigg|_{x=0} =0 \\<br /> (XY+\alpha\frac{\partial X}{\partial x}Y)\bigg|_{x=a} =0 \\<br /> (XY-\alpha X\frac{\partial Y}{\partial y})\bigg|_{y=0} =0 \\<br /> (XY +\alpha X\frac{\partial Y}{\partial y})\bigg|_{y=a} =0

I deliberately left "Y" in first two eq. and "X" in the last two, as the BC works on the whole product (XY), so I wonder, whether this way of writing tells me sth new, or is Y=0 or X=0 only a trivial solution?

The 1st and the 3rd eq. gave me relation between coefficient: C=\alpha D k_x and E=\alpha F k_y, so the only constants left are the ones in T.

The other 2 BCs gave me 2 transcendental equations, each in 1 direction:
\tan{k_i b}=\frac{-2\alpha b (k_i b)}{b^2-\alpha^2k_i^2b^2},
for i=x,y (b is a side of the square, which I am going to keep constant).

So what I can do now is to find the roots of this transc.eq. for various coupling constants in BCs (\alpha), which gives me k_{in}b=... for the n-th root. But this k does of course not simplify the term for u, as it usually did with the BCs of 1st kind.
So I have

u=\sum\limits_{m=1}^\infty \sum\limits_{n=1}^\infty (\alpha k_{xm}\cos{k_{xm} x}+\sin{k_{xm }x}) (\alpha k_{yn}\cos{k_{yn} y}+\sin{k_{yn} y})(A_{mn}\cos{\omega t}+B_{mn}\sin{\omega t}),
where j in k_{ij} denotes the "number of the root" of the transc.eq. and i is the direction (x,y).

I can write the solution as
u=\sum\limits_{m=1}^\infty \sum\limits_{n=1}^\infty (\alpha^2 k_{xm} k_{yn}\cos{k_{xm} x}\cos{k_{yn}y} + \alpha k_{xm}\cos{k_{xm} x}\sin{k_{yn} y} + \\ + \alpha k_{yn} \sin{k_{xm} x}\cos{k_{yn} y} + \sin{k_{xm}x}\sin{k_{yn} y})(A_{mn}\cos{\omega t}+B_{mn}\sin{\omega t}),
but my task is to find the eigenfrequencies of this expression (and their dependency on \alpha). How should I do that, as u is a sum of functions, not a product?

As a hint, I was told that the solution had sth to do with the parity - that I get 4 different eigenfunctions for 4 different combinations (odd-odd, even-odd, odd-even, even-even), but I don't understand, the parities of what should I compare - of functions (sin/cos), of solutions (m=1,2,...) ?

PS. I don't have any given initial conditions, which is why A and B remain undefined.

 

[LCKurtz]

For $X$ you would have$$
X(0)+\alpha X'(0)=0\,\quad X(b)+\alpha X'(b) = 0$$and for $Y$:$$
Y(0)+\alpha Y'(0)=0\,\quad Y(b)+\alpha Y'(b) = 0$$

I figured out, that the original boundary condition is actually (u+ \alpha \frac{\partial u}{\partial n})\bigg|_{\partial \Omega}=0, (where \partial \Omega is the boundary of the area, where we want to find the solution for u)so we have a normal derivative in it. I believe that this results in change of sign before \alpha where the normal is negatively directed:
(XY-\alpha\frac{\partial X}{\partial x}Y)\bigg|_{x=0} =0 \\<br /> (XY+\alpha\frac{\partial X}{\partial x}Y)\bigg|_{x=a} =0 \\<br /> (XY-\alpha X\frac{\partial Y}{\partial y})\bigg|_{y=0} =0 \\<br /> (XY +\alpha X\frac{\partial Y}{\partial y})\bigg|_{y=a} =0

I deliberately left "Y" in first two eq. and "X" in the last two, as the BC works on the whole product (XY), so I wonder, whether this way of writing tells me sth new, or is Y=0 or X=0 only a trivial solution?

There is no point in leaving the other variable in these equations. For example, the first one is$X(0)Y(y)-\alpha X'(0)Y(y)=0$. If there is any point where $Y(y)\ne 0$, which I assume there is, this implies $X(0)-\alpha X'(0)=0$. So you get the same equations I had above except for the sign changes in $\alpha$.

The 1st and the 3rd eq. gave me relation between coefficient: C=\alpha D k_x and E=\alpha F k_y, so the only constants left are the ones in T.

The other 2 BCs gave me 2 transcendental equations, each in 1 direction:
\tan{k_i b}=\frac{-2\alpha b (k_i b)}{b^2-\alpha^2k_i^2b^2},
for i=x,y (b is a side of the square, which I am going to keep constant).

So what I can do now is to find the roots of this transc.eq. for various coupling constants in BCs (\alpha), which gives me k_{in}b=... for the n-th root. But this k does of course not simplify the term for u, as it usually did with the BCs of 1st kind.
So I have

u=\sum\limits_{m=1}^\infty \sum\limits_{n=1}^\infty (\alpha k_{xm}\cos{k_{xm} x}+\sin{k_{xm }x}) (\alpha k_{yn}\cos{k_{yn} y}+\sin{k_{yn} y})(A_{mn}\cos{\omega t}+B_{mn}\sin{\omega t}),
where j in k_{ij} denotes the "number of the root" of the transc.eq. and i is the direction (x,y).

I can write the solution as
u=\sum\limits_{m=1}^\infty \sum\limits_{n=1}^\infty (\alpha^2 k_{xm} k_{yn}\cos{k_{xm} x}\cos{k_{yn}y} + \alpha k_{xm}\cos{k_{xm} x}\sin{k_{yn} y} + \\ + \alpha k_{yn} \sin{k_{xm} x}\cos{k_{yn} y} + \sin{k_{xm}x}\sin{k_{yn} y})(A_{mn}\cos{\omega t}+B_{mn}\sin{\omega t}),
but my task is to find the eigenfrequencies of this expression (and their dependency on \alpha). How should I do that, as u is a sum of functions, not a product?

As a hint, I was told that the solution had sth to do with the parity - that I get 4 different eigenfunctions for 4 different combinations (odd-odd, even-odd, odd-even, even-even), but I don't understand, the parities of what should I compare - of functions (sin/cos), of solutions (m=1,2,...) ?

PS. I don't have any given initial conditions, which is why A and B remain undefined.

I'm not sure how much more help I can give you. I will show you how I would have proceeded with the equations, which may just duplicate what you did but didn't show.
You have the equations$$
X''-\lambda X = 0$$ $$
X(0)-\alpha X'(0) = 0$$ $$
X(b) +\alpha X'(b) = 0$$
Taking the case $\lambda = -c^2 < 0$ gives$$
X = A\sin(cx) + B\cos(cx)$$ $$
X' = Ac\cos(cx) -Bc\sin(cx)$$Applying the boundary conditions gives$$
X(0)-\alpha X'(0) = B -\alpha Ac = 0$$ $$
X(b)+\alpha X'(b) = A\sin(cb) + B\cos(cb) +\alpha(Ac\cos(cb)-Bc\sin(cb)=0$$So the equations are$$
-\alpha c A + B =0$$ $$
(\sin(cb) +\alpha c\cos(cb))A + (\cos(cb)-\alpha c \sin(cb)B = 0$$
This requires the determinant of coefficients to be $0$.

$$\left |
\begin{array}{cc}
-\alpha c & 1\\
\sin(cb) + \alpha c\cos(cb) & \cos(cb)-\alpha c \sin(cb)
\end{array} \right |=0$$So$$
-\alpha c \cos(cb) + \alpha^2c^2\sin(cb)-\sin(cb)-\alpha c\cos(cb)=0$$ $$
\sin(cb)(\alpha^2c^2-1)=2\alpha c \cos(cb)$$ $$
\tan(cb) = \frac{2\alpha c}{\alpha^2c^2-1}$$
I think that pretty much agrees with yours although I'm not sure because of the notation. Of course, you get a similar equation for Y. Note that with $c$ chosen to satisfy this equation, you have $B=\alpha C$ in your solution.

I am aware that I may not be telling you anything you don't already know here, but I have written it so, to me, it is a bit easier to understand than your notation. Anyway, I'm afraid you are on your own from here because I don't think I have anything further helpful to add.

 

[maves]

Yes, LCKurtz, you understood me correctly, thank you. So I was wondering, if maybe anyone else could help me out here:

The expression for the eigenmodes of the square drum, that I got, taking homogenous boundary conditions of 3rd kind, into account, is:

u=\sum\limits_{m=1}^\infty \sum\limits_{n=1}^\infty (\alpha k_{xm}\cos{k_{xm} x}+\sin{k_{xm }x}) (\alpha k_{yn}\cos{k_{yn} y}+\sin{k_{yn} y})(A_{mn}\cos{\omega t}+B_{mn}\sin{\omega t}),

where k_{xm}, k_{yn}, \alpha are constants.

Or written in a different way,

u=\sum\limits_{m=1}^\infty \sum\limits_{n=1}^\infty (\alpha^2 k_{xm} k_{yn}\cos{k_{xm} x}\cos{k_{yn}y} + \alpha k_{xm}\cos{k_{xm} x}\sin{k_{yn} y} + \\ + \alpha k_{yn} \sin{k_{xm} x}\cos{k_{yn} y} + \sin{k_{xm}x}\sin{k_{yn} y})(A_{mn}\cos{\omega t}+B_{mn}\sin{\omega t}),

Talking about the spatial part: In all the cases I found on the web, there was always at least one BC of 1st kind, which made a solution much simpler. But I still have a sum of sin and cos in every direction. So, my question is: what is the eigenfrequency, that belongs to every mode? I don't think I can simply write k^2=k_x^2+k_y^2 \hspace{0.3cm} \text{and}\hspace{0.3cm} \omega=kc \hspace{0.3cm}\longrightarrow \hspace{0.3cm} \omega_{mn}=c\sqrt{k_{xm}^2+k_{yn}^2},
as my solution u includes a sum. In other words - how can you get the eigenfrequency of a function, which is a sum of expressions, not just a product.

Can I make 4 "subcases" for every sumand that I have (that is to say, if we only have sinus in x direction and cosinus in y etc.), or is this restricted, as only the whole expression is the eigenfunction?This would be in agreement with the hint that I got - to have 4 different eigenfunctions for each case, but I am not sure, whether I can separate the expression on my own, if the BCs didn't give me a result like that...

Please, this is getting really frustrating, as I believe, that this homework is one of the easiests :(

 

[LCKurtz]

Yes, LCKurtz, you understood me correctly, thank you. So I was wondering, if maybe anyone else could help me out here:

(

I will see if I can get another Homework Helper who knows more about PDE's to drop by.

 

[vela]

So, my question is: what is the eigenfrequency, that belongs to every mode? I don't think I can simply write k^2=k_x^2+k_y^2 \hspace{0.3cm} \text{and}\hspace{0.3cm} \omega=kc \hspace{0.3cm}\longrightarrow \hspace{0.3cm} \omega_{mn}=c\sqrt{k_{xm}^2+k_{yn}^2},
as my solution u includes a sum. In other words - how can you get the eigenfrequency of a function, which is a sum of expressions, not just a product.

Your expression for u(x,y,t) is the general solution to the wave equation satisfying the boundary conditions expressed as a sum of modes. Each term in the sum is one mode, and your expression for $\omega_{mn}$ is the frequency of that mode.