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Tough differential equation

  • Thread starter Troels
  • Start date
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1. Homework Statement

The problem concerns the free undamped oscialtions of an elactic beam, clamped at one end.
The system is governed by a partial differential equation (one spatial dimension + time) which are to be solved by separation of the variables.

Constants:

E Young's Modulus, Real, and positive
I: Moment of inertia, real and positive
L: Lenght of the beam, real and positive

2. Homework Equations
after some preparations, one ends with the following eigenvalue equation for the spatial component:

[tex] EI\frac{d^4}{dz^4}Z(z)=\lambda Z(z)[/tex]
Where lambda is the seperation constant and eigenvalue of the problem
With the boundary conditions:

  1. [tex]Z(0)=0[/tex]
  2. [tex]\left.\frac{d}{dz}Z(z)\right|_{z=0}=0[/tex]
  3. [tex]\left.EI\frac{d^2}{dz^2}Z(z)\right|_{z=L}=0[/tex]
  4. [tex]\left.EI\frac{d^3}{dz^3}Z(z)\right|_{z=L}=0[/tex]


3. The Attempt at a Solution

I have now tried every trick 4 years of physics education have taught me, and I still cannot find any nontrivial solution to this equation. I only seem to get an insane misture of exponential functions, real and complex, with no obvious way to fit the BC's. Any thought or suggestions on how to tackle this one will be most helpfull.
 

Answers and Replies

What makes you suspect there exists a function Z(z) satisfying the ODE and the boundary conditions...?

Exponentials seem right, if you can't make them satify the boundary conditions, there is not much more you can try.

Maybe you made same mistake when seperating the original PDE.
 
125
2
What makes you suspect there exists a function Z(z) satisfying the ODE and the boundary conditions...?
Nothing, except my professors persistent claim that there does and that we are to find it without his help for a graded hand-it problem due for monday the 14th.

Maybe you made same mistake when seperating the original PDE
That is possible, but unlikely, as I have seperatet quite a number of PDEs over the years. But if you would like to check, the original equation is

[tex]\rho A \frac{\partial^2}{\partial t^2}x(z,t)+EI\frac{\partial^4}{\partial z^4}x(z,t)=0[/tex]

Then assume that [tex]x(z,t)=Z(z)T(t)[/tex]

Note that there is given no initial contitions on the temporal part, as we are only asked to find the periods of the oscilations (for which we need the eigenvalue) and their shape (for wich we need a solution to the spatial part)
 
tiny-tim
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Homework Helper
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[tex] EI\frac{d^4}{dz^4}Z(z)=\lambda Z(z)[/tex]

I only seem to get an insane misture of exponential functions, real and complex …
Put µ^4 = λ/EI.

Then (d^4/dz^4)Z - (µ^4)Z

= [d^2/dz^2 + µ^2][d^2/dz^2 - µ^2]Z

= … ? :smile:
 
I assume that's what the OP did, getting

[tex]
Z(z)=A_\pm e^{\pm i \mu z}\text{ and } Z(z)=B_\pm e^{\pm \mu z}
[/tex],

the "insane mixture of exponentials." :smile:
 
tiny-tim
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Homework Helper
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I assume that's what the OP did, getting

[tex]
Z(z)=A_\pm e^{\pm i \mu z}\text{ and } Z(z)=B_\pm e^{\pm \mu z}
[/tex],

the "insane mixture of exponentials." :smile:
Well, I get a sum of cos sin cosh and sinh … I wouldn't call that insane … I end up with a formula for sinhµL/sinµL (… I think! :rolleyes:). :smile:
 
HallsofIvy
Science Advisor
Homework Helper
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Are you assuming that [itex]\lambda[/itex] is positive or negative? The whole question was, for what values of [itex]\lambda[/itex] does this exist a non-trivial solution to the original equation that satisfies all the boundary conditions.
 
125
2
Okay, I think I got it now.

First, use a laplace transform method, as this can take care of BC 1 and 2 straight away. The solution is then:

[tex]Z(z)=c_1(\cosh(\omega z)-\cos(\omega z))+c_2(\sinh(\omega z)-\sin(\omega z)), \quad \omega = \sqrt[4]{\frac{\lambda}{EI}}[/tex]

(this can also be done using the exponential approach, but is is much more crumblesome)
Now, do a backdifferentiation, insert the remanings BCs and do about one page of closely packed calculations to eliminate the two constants and bring the equation on this form:

[tex]-\frac{2(\cosh(\omega L)\cos(\omega L)-1)}{\sinh(\omega L) - \sin(\omega L)} = 0[/tex]

The roots of this trancedental equation, which most of all resembles a Bessel function of the second kind determines the allowed (possitive) values of [tex]\omega[/tex] and thus [tex]\lambda[/tex]
 

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