Solution of Euler-Bernolli differential equation for a simple beam

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 $n$?

• 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. – Novice C Mar 2 '17 at 16:04
• 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? – Alessandro Zunino Mar 2 '17 at 16:07
• 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. – Novice C Mar 2 '17 at 16:20
• 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. – Alessandro Zunino Mar 2 '17 at 16:23