Are the derivatives of eigenfunctions orthogonal?

  • #1
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Main Question or Discussion Point

We know that modes of vibration of an Euler-Bernoulli beam are given by eigenfunctions, with the natural frequency of each mode being given by its eigenvalue. Thus these modes are all mutually orthogonal.


Can anything be said of the derivatives of these eigenfunctions? For example, I have the general solution

[tex]
w(x,t)=\sum_{n=1}^{\infty}q_{n}(t)\Big[ \left(C_{1}cosh(s_{1}x)+C_{2}sinh(s_{1}x)+C_{3}cos(s_{2}x)+C_{4}sin(s_{2}x)\right)\Big]

[/tex]

where [itex] q_{n}(t) [/itex] is the dynamic component of the solution, [itex]s_{1,2}[/itex] are known constants and [itex] C_{1-4} [/itex] are known coefficients.

Can we say that its second and fourth derivatives are also orthogonal, such that

[tex]
\int_{0}^{L} \left(\frac{\partial^{2}w_{n}(x,t)}{\partial x^{2}}\right) \left(\frac{\partial^{2}w_{m}(x,t)}{\partial x^{2}}\right) dx = 0 [/tex]
and
[tex]
\int_{0}^{L} \left(\frac{\partial^{4}w_{n}(x,t)}{\partial x^{4}}\right) \left(\frac{\partial^{4}w_{m}(x,t)}{\partial x^{4}}\right) dx = 0 \text{?} [/tex]

And what about the combination of the two:

[tex]\int_{0}^{L} \left(\frac{\partial^{2}w_{n}(x,t)}{\partial x^{2}}\right) \left(\frac{\partial^{2}w_{n}(x,t)}{\partial x^{2}}\right) dx = 0 \text{?} [/tex]


Thanks in advance for the help, it would be much appreciated.
 
Last edited:

Answers and Replies

  • #2
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What operator are these orthogonal eigenfunctions of? If it's the second partial derivative with respect to x, then taking the second derivative will just give you an eigenvalue times the original function...in which case the integral of the product over the region would still be zero...and zero times any multiplicative constant is still zero.
 
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  • #3
Ssnow
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Yes, according with @Megaquark when you take the second derivative (or even) respect to ##x## you have the same form of the original function multiplied by a eigenvalue, so the orthogonality corresponds to the orthogonality of original functions ...
 
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  • #4
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Thanks very much for the replies. What you both say makes a lot of sense, and is definitely the case with a simple solution such as ##
w(x,t)=\sum_{n=1}^{\infty}q_{n}(t)C_{n}sin(s_{2}x)\text{.}## Here, as you say, any even derivative will be a multiple of the original function.

In the function I quoted in my original question, however, there are two different arguments, ##s_{1}## and ##s_{2}##. The even derivatives are hence not a multiple of the general solution, for example: $$X^{(4)}(x)=s_{1}^{4}\left(C_{1}cosh(s_{1}x)+C_{2}sinh(s_{1}x)\right) +s_{2}^{4}\left(C_{3}cos(s_{2}x)+C_{4}sin(s_{2}x)\right)\text{.}$$ This means the eigenvalue cannot be 'taken outside the integral sign' when the inner product is calculated to check for orthogonality: $$\lambda_{n}\int_{0}^{L}w_{n}(x,t)w_{m}(x,t)dx=0\text{.}$$ Does this affect your answers?
 
  • #5
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Where do the equations come from? They are eigenfunctions - of what?
 
  • #6
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I'm solving for the vibrations of an Euler-Bernoulli beam subject to a tensile axial load ##P.## The governing equation that I've solved is $$\rho A\frac{\partial^{2}w(x,t)}{\partial t^{2}}=-EI\frac{\partial^{4}w(x,t)}{\partial x^{4}}+P\frac{\partial^{2}w(x,t)}{\partial x^{2}}-\rho Ag\text{, }$$ where ##w(x,t)=\sum_{n=1}^{\infty}q_{n}(t)X(x)\text{.}##

The simple case I referred to in my previous reply (##
w(x,t)=\sum_{n=1}^{\infty}q_{n}(t)C_{n}sin(s_{n}x)##)
is the solution for a simply supported beam, but I'm really interested in the clamped-clamped case, which yields the solution I gave in the original question. When the simply supported solution is substituted into the above governing equation we can use orthogonality to simplify it a great deal; multiplying through by ##\int_{0}^{L}sin(s_{m}x)dx## causes the first and second term on the RHS to disappear (as long as ##m\neq n##). My question is really whether it is possible to substitute in the more complex clamped-clamped solution and multiply through in the same way, but by ##\int_{0}^{L}X''(x)dx## or ##\int_{0}^{L}X''''(x)dx## instead.

Would orthogonality cause the ##X''(x)## and ##X''''(x)## terms to disappear in this case? Thanks once again for your help, I really do appreciate it.
 
  • #7
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I think s1 equals s2 in the above general solution...especially since we're dealing with something physical like a vibrating beam. Wouldn't sn be the related to the resonance frequencies of the vibration? I don't know how one solution could have two different resonance frequencies happening at once.
 
  • #8
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Thanks again for the reply. It's not that simple unfortunately, due to the axial load term. There is still only one natural frequency as ##s_{1}## and ##s_{2}## are functions of ##\omega^{2}##: $$s_{1}=\sqrt{\gamma-\sqrt{\left[\gamma^{2}+\xi\omega^{2}\right]}}$$ and $$s_{2}=\sqrt{\gamma+\sqrt{\left[\gamma^{2}+\xi\omega^{2}\right]}}\text{,}$$where ##\gamma=\frac{P}{2EI}## and ##\xi=\frac{\rho A}{EI}##. This is verified by all the books and papers on the topic of axially loaded beams, such as Rao S., "Vibration of Continuous Systems," John Wiley & Sons, 2007 and Gatti, P. & Ferrari, V., "Applied Structural and Mechanical Vibrations," Taylor & Francis Group, 2003.

So you can see that having the ##s_{1}## and ##s_{2}## terms as arguments in the solution above is less than convenient! But does affect the orthogonality of even derivatives?
 

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