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I'm trying to solve the Euler-Bernolli differential equation for an homogeneous rectangular beam without load:

$$ EI{\frac {\partial ^{4}w}{\partial x^{4}}}+\mu {\frac {\partial ^{2}w}{\partial t^{2}}}=0 $$

A way to solve it is to separate the variables: $$ w(x,t)=\phi(x)Y(T)$$ and then to divide the differential equation by $\phi(x)Y(T)$, obtaining:

$$ \frac {\partial ^{4}_x \phi}{\phi}+\frac{\mu}{EI} \frac {\partial ^{2}_t Y}{Y}=0. \tag{1}$$

Clough and Penzien's Dynamics of Structures textbook claims that since the first term in this equation is a function of $x$ only and the second term is a function of $t$ only, the entire equation can be satisfied for arbitrary values of $x$ and $t$ only if each term is a constant in accordance with

$$ \frac {\partial ^{4}_x \phi}{\phi}=-\frac{\mu}{EI} \frac {\partial ^{2}_t Y}{Y}=a^4. \tag{2}$$

Why can I separate the variables? i.e. why can I assume that my real system will follow the separated solution I built and neglect all the remaining solutions? Why can I assume that the single terms in (1) are necessarily equal to a constant?

Later, using the boundary conditions for the cantilevered beam, one finds the solution:

$$\phi_n(x) = A\left[(\cosh \beta _nx-\cos \beta _nx) + \frac {\cos \beta _{n}L+\cosh \beta _{n}L}{\sin \beta _{n}L+\sinh \beta _{n}L}(\sin \beta _nx-\sinh \beta _nx)\right]$$

Is the constant $A$ real or complex? Is it the same for every mode or it depends on $n$?

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  • $\begingroup$ My understanding is that you are just making an assumption when you separate the variables. For a lot of difficult mathematics you start by assuming something about the solution. You are without a doubt however solving this equation for all solutions that CAN be separated, and hopefully that is enough physically. You set them equal to a constant because that equation holds true for all x and all t. So hold t constant and vary x, that equation must still hold true. The only possible way for that to work if they are both constant, and it's the equal and opposite since they equal 0 together. $\endgroup$
    – Novice C
    Mar 2, 2017 at 16:04
  • $\begingroup$ So, I am indeed excluding all the solutions that cannot be separated. How can I be sure that my real system will use the separated solution instead of the not separable ones? $\endgroup$
    – DrManhattan
    Mar 2, 2017 at 16:07
  • $\begingroup$ Consider: $$w(x,t) = \cos(\frac{x}{EI^{1/4}} + \frac{t}{\sqrt{\mu}}).$$ This solves the equation and I believe cannot be separated. Does this describe the physics? Perhaps. Most the time I think the philosophy is we hope it doesn't. There might be reasons to be motivated into using it physically outside being a crutch for our ignorance to a true general solution, but I unfortunately cannot help on that front. To me it's always been about doing the best we can do, even if it is complete. $\endgroup$
    – Novice C
    Mar 2, 2017 at 16:20
  • $\begingroup$ Yes, my question is indeed: "why can I assume that my real system will follow the separated solution I built and neglect all the remaining solutions?". Thank you for your time and your help, anyway. $\endgroup$
    – DrManhattan
    Mar 2, 2017 at 16:23
  • $\begingroup$ Is there a reason why you don't consider a solution as a simple superposition of waves? Also what are the boundary conditions of your problem? $\endgroup$
    – user8736288
    Mar 27, 2020 at 22:50

1 Answer 1

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Separating variables works for a lot of physical based differential equations, so you give it a try. It leads to equation 1.

In equation 1, the first term depends on only x, and the second term depends only on t. The only way to get them to sum to zero for all values of x, and all values of t, is if they are both equal to the same constant, but with opposite signs.

The value of A is a real constant that may be different for each mode. It is the strength of the contribution of each particular mode to the overall solution.

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  • $\begingroup$ Yes, I know that it works, but how can I know that by the separation of the variables I'm not losing a piece of the solution? I.e. is the solution I find in this way unique? About the second issue: I know that if the two terms are equal to a constant then the equation is true, but how can I prove that there is no other solution that is not a constant? Thank you for your help. $\endgroup$
    – DrManhattan
    Feb 27, 2017 at 1:11

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