Question of damping vibrating strings

[yungman]

This is not homework, I need someone to verify (9) and (10) below whether I am correct because I don't have the answer. I need to find the solution of the following:

Question
Find the solution of the equation for damping vibration string stretched from x=0 to x=L where

\frac{\partial ^2 u}{\partial t^2} \;+\; 2k\frac{\partial u}{\partial t} \;=\; c^2\frac{\partial ^2 u}{\partial x^2} \;\;\;\;\; (1)

With boundary and initial condition given:

u(0,t) = u(L,t)= 0 \;\;\; (2)
u(x,0) = f(x)\; ,\; \frac{\partial u}{\partial t}(x,0) = g(x) \;\;\;\;(3)

We assume u(x,t)=X_{(x)}T_{(t)} \;\Rightarrow\; X''+\mu^2X=0 \;\;and\;\; T'' \;+\; 2kT' \;+\; (\mu c)^2 T \;=\; 0

Attempted steps:
1)\;\; X''+\mu^2X=0 \Rightarrow \mu=\mu_n = \frac{n\pi}{L} \;\;\;\Rightarrow\;\;\; X \;=\; X_n \;=\; sin(\frac{n\pi}{L})x \;\;\;\;n=1,2,3...(4)


2)\;\; T'' \;+\; 2kT' \;+\; (\frac{n\pi}{L}c)T \;=\; 0\;\; \;\;(5) Using constant coeficients ODE, m=\frac{-2k^+_-\sqrt{4k^2-4(\frac{n\pi}{L}c)^2}}{2}

Three cases of k^2-(\frac{n\pi}{L}c)^2 \;\;\;\;Let\;\lambda_n =\sqrt{|k^2-(\frac{n\pi}{L}c)^2|}

Case \;1\;\;\;\;k^2-(\frac{n\pi}{L}c)^2\;< 0 \Rightarrow\; n > (\frac{kL}{\pi c}) \;\;\;\Rightarrow\;\; \; T= e^{-kt}[c_n cos(\lambda_n t) \;+\; d_n sin(\lambda_n t)]\;\;(6)

Case \;2\;\;\;k^2-(\frac{n\pi}{L}c)^2\;= 0 \Rightarrow\; n = (\frac{kL}{\pi c}) \;\;\;\Rightarrow\;\; \; T= h_{n=\frac{kL}{\pi c}}e^{-kt} \;+\; j_{n=\frac{kL}{\pi c}}te^{-kt} \;\;\;\;(7)

Case \;3\;\;\;\;k^2-(\frac{n\pi}{L}c)^2\;> 0 \Rightarrow\; n < (\frac{kL}{\pi c}) \;\;\;\Rightarrow\;\; \; T= e^{-kt}[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)] \;\;\;(8)
For n=1,2,3...



Solution of wave equation:
3)For (\frac{kL}{\pi c}) is not a possitive integer:
\Rightarrow u(x,t) = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L})x[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)] \;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L})x[c_n cos(\lambda_n t) \;+\; d_n sin(\lambda_n t)] \;\;\;(9)




4)For (\frac{kL}{\pi c}) is a possitive integer:

We add the term wheren= (\frac{kL}{\pi c}) \;\Rightarrow\; sin(\frac{k}{c}x)[ h_{n=\frac{kL}{\pi c}}e^{-kt} \;+\; j_{n=\frac{kL}{\pi c}}te^{-kt} ] to (9) above.

\Rightarrow u(x,t) = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L})x[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)] \;+\; sin(\frac{k}{c}x)[ h_{n=\frac{kL}{\pi c}}e^{-kt} \;+\; j_{n=\frac{kL}{\pi c}}te^{-kt} ] \;
.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L})x[c_n cos(\lambda_n t) \;+\; d_n sin(\lambda_n t)] \;\;\;(10)


My question:
1) k is the attenuation constant?

2) Bottom line the solution depend on (\frac{kL}{\pi c}) and solution can be combination of all three cases shown above?

3) Is it true solution require separation constant -\mu ^2\;<0\;? Because if not, we only get trivial solution?

 

[elibj123]

1) Yes, the bigger it is the bigger the damping is. (The exponential decay is faster)

2) Yes. For some values of n the differential equation for T would describe over damping, for others it will describe under damping and there might be one n that will describe a critical damping.

As you can see from the conditions of the different cases, there exists such n, that for any other bigger indices, the solutions will be under-damped. So most of the vibration consists of decaying oscillations, and only some of them are over-damped.

3) First of all, no. \mu is real, so it will never satisfies \mu^{2}<0

What is true, is that usually, the PDE is associated with it's Sturm-Liouville problem:

X''(x)=-\lambda X
X(0)=X(L)=0

(This is the specific SL problem of this PDE)
(The sign of the lambda is a convention)

Then it is found due to boundary conditions, that lambda must be a positive number to provide oscillatory motion, and then denoted \lambda=\mu^{2}
The set of lambdas that satisfies the SL problem, are called eigenvalues, and the set of X that are solutions with a given eigenvalue, are the corresponding eigenfunctions.

 

[yungman]

1) Yes, the bigger it is the bigger the damping is. (The exponential decay is faster)

2) Yes. For some values of n the differential equation for T would describe over damping, for others it will describe under damping and there might be one n that will describe a critical damping.

As you can see from the conditions of the different cases, there exists such n, that for any other bigger indices, the solutions will be under-damped. So most of the vibration consists of decaying oscillations, and only some of them are over-damped.

3) First of all, no. \mu is real, so it will never satisfies \mu^{2}<0

What is true, is that usually, the PDE is associated with it's Sturm-Liouville problem:

X''(x)=-\lambda X
X(0)=X(L)=0

(This is the specific SL problem of this PDE)
(The sign of the lambda is a convention)

Then it is found due to boundary conditions, that lambda must be a positive number to provide oscillatory motion, and then denoted \lambda=\mu^{2}
The set of lambdas that satisfies the SL problem, are called eigenvalues, and the set of X that are solutions with a given eigenvalue, are the corresponding eigenfunctions.

Thanks for you help again.
I forgot the sign, separation constant equal to -\mu^2.

I am working of finding the coeficients. My additional questions are:

1) So far I assumed the coeficients of the three cases are different representing bya_n,\; b_n,\; c_n,\;d_n,\;h_,\;j_n But the book equate .a_n=c_n\;,\; b_n=d_n. Are they the same? I don't understand how.

2)\;\;\;Are\;cosh(\lambda_n t) \;and\; sinh(\lambda_n t)\; orthogonal\; functions \;that \;\;(cosh(\lambda_n t),cosh(\lambda_m t))=0 \; for\; n\;not\;equal\;m\;?

3) Do I need to study complex Fourier series to solve this problem?

 

[elibj123]

1) It is strange and I don't see how it is possible. Especially where a_{n},b_{n} are define for finite numbers of n, wile c_{n},d_{n} are infinite sequences.
I think what the book meant is that a & c, and b& d (I ignore h,j since they're a special case, and mostly do not occure) can be both represented by the same closed formula (one for a&c, and one for b&c). Where a (b) are the first terms, which continued by c (d). I hope you get it.

And it is true, because if you plug t=0 you get:

u(x,0)=\sum^{\infty}_{n=1}\alpha_{n} sin(\frac{n\pi x}{L})=f(x) (1)
Where alpha is constructed from a continued by c:

\alpha_{n}=(a_{1},a_{2},...,c_{1},c_{2},...)

If you take the derivitive at t=0 you'll get:

u_{t}(x,0)=-k\sum^{\infty}_{n=1}\alpha_{n}sin(\frac{n\pi x}{L})+\sum^{\infty}_{n=1}\lambda_{n}\beta_{n}sin(\frac{n\pi x}{L})=g(x)

Where beta is the same as apha only with b & d.

So you rearrange to get:

\sum^{\infty}_{n=1}\lambda_{n}\beta_{n}sin(\frac{n\pi x}{L})=g(x)+kf(x) (2)

Now you have it all laid out for you. Break f(x) into its sinusoidals components to get the alphas. And break the combination g(x)+kf(x) into its sinusoidals components to get the beta's.

When you go back and add the temporal solution, don't forget that some of the alphas and the betas go to the over damped sum, and the rest go to the oscillations (as long as \frac{KL}{\pi c} is non-integer you won't have to deal with the critical-damping)

2) I'm almost certain it's not (it can be easily checked with integration), and why would you need this property? As I've shown above, all of the orthogonality property is used only with the sinusoids which are the solution to the Sturm Liouville problem.

3) No. You'll need to have some background in Generalized Fourier Series, or at least know Sturm-Liouville Theorem (it will include the Generalized Fourier Series by itself).

But mostly series exapnsion method in PDE is technical:

Find the associated Sturm-Liouville problem (It will be a problem in the spatial coordinates involving the boundary conditions) and solve for eigenvalues and eigenfunctions.
Assume your solution u(x,t) is a series of these eigenfunctions with coefficients T_{n}(t).
Express every other given functions in the problem (Initial Conditions, Driving Forces) as a series of these eigenfunctions. (This will involve some integrations)
Plug everything into the equation, equate expressions component-wise, and solve an ODE for Tn(t) (f(x) and g(x) will provide IC for this ODE's).
Write u(x,t) as a sum of known Xn(x)Tn(t).

 

[yungman]

Thanks, I have no respond yet, I need time to digest what you wrote.

 

[elibj123]

Do you know generally about Fourier Series/orthogonal function basis?

 

[yungman]

Do you know generally about Fourier Series/orthogonal function basis?

Yes. I have spent quite a bit of time studying Fourier Series and Orthogonal function. Also I have studied Sturm Liouville problem. I was just mistakenly thinking of something like:

(f(x),cosh(\lambda_m t) )=(cosh(\lambda_m t),\sum a_n cosh(\lambda_n t))=\int_0^L a_m\;cosh^2(\lambda_m t)\;dt

But obvious this is not right because as we make t=0, the term disappeared. Obviously that wasn't right after I sit down and write it out.

I spent some time on what you wrote and I feel I understand what you wrote. I am working out the step and I'll post it later to verify with you. Actually the book is correct of using only a_n \; and \; b_n for all three cases because the coeficients do not overlap. Instead they form a continuous sequence of n=1,2,3,4...

I'll post my work sometime today.

Thanks for all your help. You are of big help on this.

Alan

 

[yungman]

This is what I have so far. Took a long time to type all these. There are two cases:

1)For\; \frac{kL}{\pi c} \;\; is\; not\; integer

u(x,t) = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x)[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)] \;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x)[a_n cos(\lambda_n t) \;+\; b_n sin(\lambda_n t)] \;\;\;

\Rightarrow \; u(x,t) = \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x)[a_n \; e^{-kt} \; cosh(\lambda_n t)] \;+\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x) [b_n\; e^{-kt}\; sinh(\lambda_n t)]

.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x)[a_n \; e^{-kt} \; cos(\lambda_n t)] \;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x) [b_n\; e^{-kt}\; sin(\lambda_n t)]

u(x,0) \;=\; f(x) \;=\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n\; sin(\frac{n\pi}{L}x) \;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n \; sin(\frac{n\pi}{L}x) \;=\; \sum_{n=1} ^{\infty} \; a_n\; sin(\frac{n\pi}{L}x)

\frac{du(x,t)}{dt} = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} a_n sin(\frac{n\pi}{L}x )[-k\;cosh(\lambda_n t) + \lambda_n sinh(\lambda_n t)] \;+ \;\;e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} b_n sin(\frac{n\pi}{L}x )[-k\;sinh(\lambda_n t) + \lambda_n cosh(\lambda_n t)]

.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;a_n sin(\frac{n\pi}{L}x )[-k\;cos(\lambda_n t) - \lambda_n sin(\lambda_n t)] \;+\;\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n sin(\frac{n\pi}{L}x )[-k\;sin(\lambda_n t) + \lambda_n cos(\lambda_n t)]

\Rightarrow \;\;\frac{d(u}{dt}(x,0) \;=\;g(x) \;=\; -\;k\sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n sin(\frac{n\pi}{L}x ) \;\;+\;\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-\;\;k \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n sin(\frac{n\pi}{L}x ) \;\;+\;\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n \lambda_n sin(\frac{n\pi}{L}x )

\Rightarrow \;\;\frac{d(u}{dt}(x,0) \;=\;g(x) \;=\; -\;k\sum_{n=1} ^{\infty} \; a_n sin(\frac{n\pi}{L}x ) \;\;+\;\; \sum_{n=1} ^{\infty} \; b_n \lambda_n sin(\frac{n\pi}{L}x )\;=\; -\; kf(x) \;\;+\;\; \sum_{n=1} ^{\infty} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

Therefore \;\; g(x)+kf(x) \;=\; \sum_{n=1}^{\infty}b_n \lambda_n sin(\frac{n \pi x}{L})\;\;. This agree with what you wrote.

((g(x)+kf(x)) \;,\; sin(\frac{m \pi x}{L}) ) \;=\; ( sin(\frac{m \pi x}{L}) \;,\; \sum_{n=1} ^{\infty} \; b_n \lambda_n sin(\frac{n\pi}{L}x ))

\Rightarrow\; \int_0^L [g(x)+kf(x)]sin(\frac{m\pi}{L}x ) dx \;=\; b_n\lambda_m \int_0^L sin^2(\frac{m \pi x}{L}) dx

a_n=\frac{2}{L}\int_0^L f(x)]sin(\frac{n\pi}{L}x) dx \;\;\;and\;\;\; b_n=\frac{2}{\lambda_n L}\int_0^L [g(x)+kf(x)]sin(\frac{n\pi}{L}x ) dx








2)For\; \frac{kL}{\pi c} \;\; is\; integer

u(x,t) = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; sin(\frac{n\pi}{L}x)[a_n cosh(\lambda_n t) \;+\; b_n sinh(\lambda_n t)] \;+\; sin(\frac{k}{c}x)[ a_{n=\frac{kL}{\pi c}}e^{-kt} \;+\; b_{n=\frac{kL}{\pi c}}te^{-kt} ] \;

.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; sin(\frac{n\pi}{L}x)[a_n cos(\lambda_n t) \;+\; b_n sin(\lambda_n t)]

u(x,0) \;=\; f(x) \;=\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n\; sin(\frac{n\pi}{L}x) \;+\; a_{n=\frac{kL}{\pi c}} sin(\frac{k x}{c}) \;+\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n \; sin(\frac{n\pi}{L}x)

u(x,0) \;=\; f(x) \;=\;\sum_{n=1} ^{\infty} \; a_n\; sin(\frac{n\pi}{L}x)

\frac{du(x,t)}{dt} = e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} a_n sin(\frac{n\pi}{L}x )[-k\;cosh(\lambda_n t) + \lambda_n sinh(\lambda_n t)] \;+ \;\;e^{-kt} \sum_{n=1} ^{n < (\frac{kL}{\pi c})} b_n sin(\frac{n\pi}{L}x )[-k\;sinh(\lambda_n t) + \lambda_n cosh(\lambda_n t)]

.\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;+\; sin(\frac{k}{c}x)[ -a_{n=\frac{kL}{\pi c}}ke^{-kt} \;+\; b_{n=\frac{kL}{\pi c}}(e^{-kt}-kte^{-kt}) ] \;

.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;a_n sin(\frac{n\pi}{L}x )[-k\;cos(\lambda_n t) - \lambda_n sin(\lambda_n t)] \;+\;\; e^{-kt} \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n sin(\frac{n\pi}{L}x )[-k\;sin(\lambda_n t) + \lambda_n cos(\lambda_n t)]


\Rightarrow \;\;\frac{du}{dt}(x,0) \;=\;g(x) \;=\; -k\sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; a_n sin(\frac{n\pi}{L}x ) \;+\; \sum_{n=1} ^{n < (\frac{kL}{\pi c})} \; b_n \lambda_n sin(\frac{n\pi}{L}x )

.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; -\;\;k a_{n=\frac{kL}{\pi c}} sin(\frac{k}{c}x)\;+\; b_{n=\frac{kL}{\pi c}} sin(\frac{k}{c}x)\;\;\;-\;\;\;k \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \; a_n sin(\frac{n\pi}{L}x )\;+\;\; \sum_{(\frac{kL}{\pi c})<n} ^{\infty} \;b_n \lambda_n sin(\frac{n\pi}{L}x )

\int_0^L g(x)sin(\frac{n\pi}{L}x ) dx \;=\; (-ka_n +b_n\lambda_n)\int_0^L g(x)sin^2(\frac{n\pi}{L}x ) dx n=\frac{kL}{\pi c}

\Rightarrow\;\; -\;ka_n\;+\;b_n\lambda_n \;=\; \frac{2}{L}\int_0^L g(x)sin^2(\frac{n\pi x}{L}) dx =\frac{kL}{\pi c}

\Rightarrow\;\; b_n \;=\; \frac{2}{\lambda_n L}\int_0^L g(x) sin(\frac{n\pi x}{L}) dx \;+\; \frac{ka_n}{\lambda_n} \;\;except\;\; n=\frac{kL}{\pi c} \;\;\;where\;\;\; b_{n=\frac{kL}{\pi c}} \;=\; \frac{2}{L}\int_0^Lg(x)sin(\frac{k}{c}x) dx \;+\; ka_{n=\frac{kL}{\pi c}}


a_n\; =\; \frac{2}{L}\int_0^L f(x)sin(\frac{n\pi}{L}x) dx

b_n=\frac{2}{\lambda_n L}\int_0^L [g(x)+kf(x)]sin(\frac{n\pi}{L}x ) dx \;\;\;+\;\;\; \frac{2k}{\lambda_n L}\int_0^L f(x)sin(\frac{n\pi}{L}x) dx

With\;\;\; b_{n=\frac{kL}{\pi c}} \;=\; \frac{2}{L}\int_0^Lg(x)sin(\frac{k}{c}x) dx \;+\; ka_{n=\frac{kL}{\pi c}}

In conclusion, there is two answer of the coeficients as show above. Please take a look and see whether I got all these correct or not.

 

[yungman]

1)
u(x,0)=\sum^{\infty}_{n=1}\alpha_{n} sin(\frac{n\pi x}{L})=f(x) (1)
Where alpha is constructed from a continued by c:

\alpha_{n}=(a_{1},a_{2},...,c_{1},c_{2},...)

If you take the derivitive at t=0 you'll get:

u_{t}(x,0)=-k\sum^{\infty}_{n=1}\alpha_{n}sin(\frac{n\pi x}{L})+\sum^{\infty}_{n=1}\lambda_{n}\beta_{n}sin(\frac{n\pi x}{L})=g(x)


\sum^{\infty}_{n=1}\lambda_{n}\beta_{n}sin(\frac{n\pi x}{L})=g(x)+kf(x) (2)

Now you have it all laid out for you. Break f(x) into its sinusoidals components to get the alphas. And break the combination g(x)+kf(x) into its sinusoidals components to get the beta's.

When you go back and add the temporal solution, don't forget that some of the alphas and the betas go to the over damped sum, and the rest go to the oscillations (as long as \frac{KL}{\pi c} is non-integer you won't have to deal with the critical-damping)

As you can see, my derivation agree with you if : .\frac{kL}{\pi c}\;\;not\; an\; integer.

But the book give the equation without explaining:

-\;k a_n\;+\;\lambda_n b_n\;=\;\frac{2}{L}\int_0^L f(x)sin(\frac{n\pi x}{L} dx

I tried and I cannot arrive to this formula no matter what. It look very similar to the formula for b_{n=\frac{kL}{\pi c}}\;\; when \;\;\frac{kL}{\pi c}\;\;is \; an\; integer.

But the book seldom make mistake! What did I do wrong?

 

[yungman]

Anyone please.

Just look at the last question why my derivation don't agree with the book.

Many thanks

Alan

 

[yungman]

I since use my way to solve a few problems already with different f(x) and g(x). It work every time. So I guess I am not wrong. I did use the book's formula to compare, they both get the same answer every time I tried.