Solving for non moving points of a 1-D wave

[jianxu]

Homework Statement

Hi!

I'm suppose to find the points x on the "string" 1-D wave which are not moving during the vibrations, i.e., 0<x<1 such that u(x,t) = 0 for all times t >0

Homework Equations

$$\left.u(x,t) = sin( \pi x)cos(\pi t) + \frac{1}{2}sin(3\pi x)cos(3\pi t) + 3sin(7\pi x) cos(7\pi t)$$

$$\left.u(x,0) = sin( \pi x) + \frac{1}{2}sin(3\pi x) + 3sin(7\pi x)$$

$$\left.\frac{du}{dx}(x,0) = \pi cos( \pi x) + \frac{3}{2}\pi sin(3\pi x) + 21\pisin(7\pi x)$$

The Attempt at a Solution

Here's what I have so far,
I reasoned that if there are stationary points, it doesn't matter at what t I'm at, I'll still achieve the same x values and therefore, I simplified the problem by using t = 0.

Now the next thing I said was for a x value to be non moving throughout the vibrations, this means that the velocity would be zero therefore, I took the derivative of the function.

so, to find the x value I simply set the derivative to zero so:

$$\left.\pi cos( \pi x) + \frac{3}{2}\pi sin(3\pi x) + 21\pi sin(7\pi x) = 0$$

I divided out the $$\pi$$ to simplify things:

$$\left. cos( \pi x) + \frac{3}{2} sin(3\pi x) + 21sin(7\pi x) = 0$$

At this point, I'm lost on how to approach this problem in order to find all the non moving points.

I asked my professor whether I can approximate the points by looking at the graph but he wants to points to be solved in a systematic fashion so...any type of suggestion would be greatly appreciated! Thank you!

 

[gabbagabbahey]

You seem to be very confused about what $u(x,t)$ actually represents. It represents the amplitude/height of the string at each point $x$ at a time $t$. For example, at time $t=0$, your string would look like this:

http://img526.imageshack.us/img526/6533/tzero.th.jpg

While at $t=1$, your string would look like this:

http://img197.imageshack.us/img197/3382/tone.th.jpg

The points that aren't moving with time are the points that are the same at all values of $t$. Just by comparing the two graphs above you should see that there are at most 8 such points (including the endpoints of the string).

To represent this mathematically, you expect both the x values and the amplitudes of any non-moving points to be constant in time. The string is not moving horizontally, so the x-values are all independent of time and so you want to find which points the amplitude is constant in time.

In other words, you want to find the values of $x$ for which $u(x,t)$ does not depend on $t$...what would you expect $\frac{\partial u}{\partial t}$ to be for those points?

 

[jianxu]

zero right because u(x,t) at those times would just be some kind of constant?

 

[gabbagabbahey]

Yes, so find the points where it is zero...

 

[jianxu]

So if I take the derivative of the function in respect to time, then

$$\left.\frac{du}{du}(u,t) = -\pi sin(\pi x)sin(\pi t) - \frac{3}{2}\pi sin(3\pi x)sin(3\pi t) - 21\pi sin(7\pi x)sin(7\pi t)$$

so now I set it equal to zero:
$$\left. -\pi sin(\pi x)sin(\pi t) - \frac{3}{2}\pi sin(3\pi x)sin(3\pi t) - 21\pi sin(7\pi x)sin(7\pi t) = 0$$

am I suppose to solve this assuming t is just some constant integer? Also, how would I approach this in order to solve the problem? Do I need to use trig identity to combine the terms and then solve for x that way? I'm not very good at solving for x with so many trig terms...so sorry for all the questions. Thanks

 

[gabbagabbahey]

I would use the multiple angle trig identities to write it as a polynomial in $\sin(\pi x)$ and $\sin(\pi t)$...then try to factor the polynomial to get it into the form $f(x)g(t)$...then, you simply find where $f(x)=0$

 

[jianxu]

so I'm kind of stuck on converting my function using multiple angle formula. This is the formula I found:
$$\left.sin(nx) = \sum^{\frac{n-1}{2}}_{k = 0}(-1)^{k}\left(\stackrel{n}{2k+1}\right)sin^{2k+1}xcos^{n-2k-1}x$$

so if my n is $$\left.7\pi$$, how does the summation work since $$\left.\frac{n-1}{2} = \frac{7\pi - 1}{2}$$ which is not an integer value?

 

[gabbagabbahey]

so I'm kind of stuck on converting my function using multiple angle formula. This is the formula I found:
$$\left.sin(nx) = \sum^{\frac{n-1}{2}}_{k = 0}(-1)^{k}\left(\stackrel{n}{2k+1}\right)sin^{2k+1}xcos^{n-2k-1}x$$

so if my n is $$\left.7\pi$$, how does the summation work since $$\left.\frac{n-1}{2} = \frac{7\pi - 1}{2}$$ which is not an integer value?

It is easier if you use a different variable in your trig identity:

$$\left.sin(n\theta) = \sum^{\frac{n-1}{2}}_{k = 0}(-1)^{k}\left(\stackrel{n}{2k+1}\right)\sin^{2k+1}\theta \cos^{n-2k-1}\theta$$

Then, use this with $\theta=\pi x$

 

[jianxu]

Thanks, so after expanding a bit I get this:

$$\left. -\pi sin(\pi x)sin(\pi t) + \frac{3}{2}\pi (3cos^{2}(\pi x)sin(\pi x) - sin^{3}(\pi x))(3cos^{2}(\pi t)sin(\pi t) - sin^{3}(\pi t)) - 21\pi (3sin(\pi x)cos^{6}(\pi x)$$
$$\left. - 21sin^{3}(\pi x)cos^{4}(\pi x) + 35sin^{5}(\pi x)cos(\pi x))(3sin(\pi t)cos^{6}(\pi t) - 21sin^{3}(\pi t)cos^{4}(\pi t) + 35sin^{5}(\pi t)cos(\pi t))$$

So I mean this term is very long already so should I be trying to use more trig identities on the polynomials or should I multiply the terms and hope there are like-terms that I can factor out?(I've never worked with so many trig terms before heh) Thanks!

 

[gabbagabbahey]

That doesn't look right, you shouldn't have any odd powers of cosine in there...this gets a little ugly, so you'll want to use a program like Mathematica or Maple (something with a built in function for factorizing).
After you fix up your expression use the fact that $\cos^2\theta=1-\sin^2\theta$ to write everything in terms of powers of $\sin(\pi x)$ and $\sin(\pi t)$...then define two new variables $\alpha\equiv\sin(\pi x)$ and $\beta\equiv\sin(\pi t)$ and write your expression as a polynomial in them, then factor it and find any values of $\alpha$ for which the expression is zero

 

[jianxu]

I've only started using maple until recently but I entered my original derivative, and I right clicked on it and told maple to expand it?

I got the following:
$$\left.-\pi sin(\pi x)sin(\pi t) - 6\pi sin(3\pi x)sin(\pi t)cos^{2}(\pi t) + \frac{3}{2}\pi sin(3\pi x)sin(\pi t)$$
$$\left.- 1344\pi sin(7\pi x)sin(\pi t)cos^{6}(\pi t) + 1680\pis in(7\pi x)sin(\pi t)cos^{4}(\pi t)$$
$$\left. - 504\pi sin(7\pi x)sin(\pi t) cos^{2}(\pi t) + 21\pi sin(7\pi x)sin(\pi t)$$

So now you said to use the trig identity of $\cos^2\theta=1-\sin^2\theta$

but so what do I do about the other powers of cosine? Do I need to apply trig identities to bring them down to powers of 2 as well? Thanks!

 

[gabbagabbahey]

$$\cos^6 (\pi t)=(\cos^2 (\pi t))^3=(1-\sin^2 (\pi t))^3$$

You will also need to use the same multiple angle trig identity to write the $\sin(7\pi x)$ and $\sin(3\pi x)$ terms as powers of $\sin(\pi x)$

 

[jianxu]

Hm alright thanks, do you know why maple only applied multiple angle rule to the t variable and not the x? Is there a way to specify that in maple?

 

[gabbagabbahey]

Sorry, I haven't used Maple in many years...thee should be some documentation that will help you though.

 

[jianxu]

Ok so for the x terms I ended up double checking my previous expansion(fixed the mistakes) and then substituted for the x terms. I also used $\cos^2\theta=1-\sin^2\theta$

and combined some like terms after multiplying stuff through and got the following:

$$\left.-\pi sin(\pi x)sin(\pi t)$$
$$\left.-\frac{3}{2}\pi (3sin(\pi x)-4sin^{3}(\pi x))(3sin(\pi t)-4sin^{3}(\pi t))$$
$$\left.-21\pi (7sin(\pi x)-21sin^{3}(\pi x)+35sin^{5}(\pi x)-21sin^{7}(\pi x))(7sin(\pi x)-21sin^{3}(\pi t)+35sin^{5}(\pi t)-21sin^{7}(\pi t))$$

So am I suppose to find the roots of the x terms? Thanks!

 

[gabbagabbahey]

That's closer, but still a little off:

$$\sin(7\theta)=7 \cos^6(\theta) \sin(\theta)-35 \cos^4(\theta) \sin^3(\theta)+21 \cos^2(\theta)\sin^5(\theta)-\sin^7(\theta)$$
$$=7 [1-\sin^2(\theta)]^3 \sin(\theta)-35 [1-\sin^2(\theta)]^2 \sin^3(\theta)+21 [1-\sin^2(\theta)]\sin^5(\theta)-\sin^7(\theta)$$
$$=7\sin(\theta)-56\sin^3(\theta)+112\sin^5(\theta)-64\sin^7(\theta)$$

After you correct your above expression, substute $\alpha=\sin(\pi x)$ and $\beta=\sin(\pi t)$...then use Maple to factorize the resulting polynomial...are there any factors that are polynomials in $\alpha$ only?

 

[jianxu]

Is there a thread where I can ask questions about maple because I've been searching the internet trying to find out how to factor the polynomial and was unable to find an answer. Apparently the Factors command is suppose to factor multivariate polynomials but when I type in the command to factor the expression, it just placed a bracket around the expression and then wrote Factors next to it...

But I assume there would be factors that are only in alpha which means they are independent of time? Would I take simply those factors, set them to zero and solve for x? By solving for x, I should get the six roots(excluding the ends)? Thanks

 

[gabbagabbahey]

You can start a thread in the appropriate place yourself, but for now I'll just quote Mathematica's output for the factored form:

$-\frac{1}{2} \pi \alpha \beta \left(2087-16500 \alpha ^2+32928 \alpha ^4-18816 \alpha ^6-16500 \beta ^2+131760 \alpha ^2 \beta ^2-263424 \alpha ^4 \beta ^2+150528 \alpha ^6 \beta ^2+32928 \beta ^4-263424 \alpha ^2 \beta ^4$
$+526848 \alpha ^4 \beta ^4-301056 \alpha ^6 \beta ^4-18816 \beta ^6+150528 \alpha ^2 \beta ^6-301056 \alpha ^4 \beta ^6+172032 \alpha ^6 \beta ^6\right)$

Now you want to find values of $x$ where this expression is zero for all $t$...$\alpha$ is a function of $x$ only, and $\beta$ is a function of $t$ only; so you are looking for values of $\alpha$ for the above expression is zero, for all values of $\beta$...what are those value(s)?

 

[jianxu]

I understand the reasoning but I am having troubles translating the meaning into math. I mean I set the expression to zero but mathematically, how does "for all values of t" fit in.

This is what I'm thinking currently:

$$\left. 0 = 2087-16500\alpha^{2}+32928\alpha^{4}-18816\alpha^{6}$$

This is because all the other terms contain beta while these do not. In addition, the non moving x values are independent of t and so these are the only terms that matter?

On a side note, do you know which thread I should post into ask the question about maple? Would it be in general discussion?

 

[gabbagabbahey]

No, for example, $3\alpha^2+2-6\beta=0$ does not mean that $3\alpha^2+2=0$...it means that there are no values of $\alpha$ where $3\alpha^2+2-6\beta=0$ is true for all $\beta$...On the otherhand if you had $f(\alpha)(3\alpha^2+2-6\beta)=0$ then any values of $\alpha$ that made $f(\alpha)=0$ would make the entire expression $f(\alpha)(3\alpha^2+2-6\beta)=0$ true for any value of $\beta$...

 

[jianxu]

But in this case we don't know what f($$\alpha$$) is though? You're saying to find all the values of f($$\alpha$$) that gives us zero right? is f($$\alpha$$) u(x,0)? when solving for the equation we had learned to use u(x,0) = f(x) and the derivative of it to be the velocity so...other than that where does f($$\alpha$$) originate from?

 

[gabbagabbahey]

I used $f(\alpha)$ to represent any function of just $\alpha$.

The expression in post #18 is fully factored... are there any factors that are independent of $\beta$?

 

[jianxu]

well no because all the terms have$$\left.-\frac{1}{2}\pi\alpha\beta$$ in common...

 

[gabbagabbahey]

Well $\alpha$ is clearly a factor isn't it?

 

[jianxu]

by itself alpha is independent of beta

 

[gabbagabbahey]

If I gave you the function f(x,y)=3xy(x-7)(y+2)(x+2y), could you tell me which values of 'x' made that expression zero?

 

[jianxu]

I would say 0, 7, and -2y if we knew what y is

but-2y would mean there's some kind of dependency between the two variables?

 

[gabbagabbahey]

Yes, the roots x=0 and x=7 are independent of y, while the root x=-2y is not...the same ideas apply to the expression in post #18...

$\alpha=0$ is a root, and is independent of $\beta$ (and hence independent of $t$), and so values of $x$ where $\alpha=0$ will be stationary points (since the expression will be zero for all $t$ anytime $\alpha=0$).

There will also be six complex or real roots of the factor:

$(2087-16500 \alpha ^2+32928 \alpha ^4-18816 \alpha ^6-16500 \beta ^2+131760 \alpha ^2 \beta ^2$
$-263424 \alpha ^4 \beta ^2+150528 \alpha ^6 \beta ^2+32928 \beta ^4-263424 \alpha ^2 \beta ^4$
$+526848 \alpha ^4 \beta ^4-301056 \alpha ^6 \beta ^4-18816 \beta ^6+150528 \alpha ^2 \beta ^6-301056 \alpha ^4 \beta ^6+172032 \alpha ^6 \beta ^6)$

But they will all depend on $\beta$; and hence they also depend on $t$; and so they are not stationary points (stationary points are stationary for all $t$, not just specific values!).

So...the only stationary points are the points where $\alpha=0$

 

[jianxu]

so to solve for the roots then, would just plain old factoring be the best way?

 

[gabbagabbahey]

You mean solving for the 6 roots of that ugly expression in my last post?

 

[jianxu]

yes heh, considering I don't know how to get maple to do anything correctly, what are some alternatives to find the roots? Thanks

 

[gabbagabbahey]

Well, the expression In post #18 is fully factored...that means that the roots of the ugly factor are going to be difficult to find...BUT! you don't need to find them because they will all depend on $\beta$, whioch means they will only be roots for certain values of $t$, and as I said in post #28, that means that those roots are not stationary points...do you not understand this?

 

[jianxu]

yes but...so am I suppose to say that even though they seem to be stationary, that's not the case since they will be dependent on beta?

 

[gabbagabbahey]

Why do you say "they seem to be stationary"?

 

[jianxu]

That's because we were to graph 10 plots at various t, I decidedly went t=1..10 and from the graphs the roots looked stationary

Also I had the impression that the problem would've been simpler in terms of all the trig stuff...