While finding green function, how to define deflection $u(x)$

Question

In the PDE below, $u$ describes the deflection of a stiff beam with length density $\rho_l$, elastic modulus $E$, moment of inertia $I$ and external force density $f(x,t)$.

The question is to define the green function to the beams stationary deflection when $f(x,t) = f(x)$, that is, when the external force density is independent of time.

PDE: $$\rho_l u_{tt} - EIu_{xxxx} = f(x,t)$$ Boundary conditions: $$ u(0,t) = u_x(0,t) = u(l,t) = u_x(l,t) = 0$$

To my problem: I got help from a teacher who told me that I should start by defining

$$u(x) = \int_0^l G(x,s)\frac{f(s)}{EI}ds$$

From here on, I have an idea of how to proceed. But I have no idea how he knew what to put as $u(x)$, and would love it if I could get an explanation of this. For example, how did he know that $f(s)$ should be divided by $EI$? Thank you!

Answer 1

When the beam is stationary, your PDE becomes \begin{equation} -E I u_{xxxx} = f(x), \end{equation} or \begin{equation} -u_{xxxx} = \frac{f(x)}{EI}. \end{equation} So that's where that division comes from.

Now, looking at the meaning of a Green's function, let's assume that it satisfies your differential equation, with $\delta(x-s)$ in place of the source term. In this case the source term is $\frac{f(x)}{EI}$, so we assume that \begin{equation} -G_{xxxx} = \delta(x-s). \end{equation} (To get a solution for any particular problem, you'll usually need to find a particular Green's function satisfying your equation and the boundary conditions. But for now, all we need is to just assume you already have one.)

Then, the reason Green's functions are useful is that we can use a solution for $G$ to find $u$ by expressing the source term as an integral involving this $\delta$-function, and then replace $\delta$ with that Green's function quantity: \begin{align} -\partial_x^4 u &= \frac{f(x)}{EI} \\ &= \int \delta(x-s) \frac{f(s)}{EI} ds \\ &= \int -\partial_x^4 G(x,s) \frac{f(s)}{EI} ds \\ &= -\partial_x^4 \int G(x,s) \frac{f(s)}{EI} ds. \end{align} Now, comparing the first line with the last, we can see that one solution for $u$ is simply \begin{equation} u(x) = \int G(x, s) \frac{f(s)}{EI} ds. \end{equation} Assuming $G$ also satisfies your boundary conditions, this is the solution you're looking for.

This is a very general idea, and is basically always what you do when using Green's functions. You should just read that Wikipedia page I linked above to understand this idea more generally. You will almost certainly use it again, but in slightly different forms, so you really will need to understand it on a more abstract level.

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I'll note that you don't actually have to divide by $EI$. You could have treated your source as just $f(x)$, so that the PDE was just \begin{equation} -E I u_{xxxx} = f(x). \end{equation} Remember that you replace the source [here that's $f(x)$] by $\delta$, and then assume that $G$ satisfies that equation, so you would have to have \begin{equation} -E I G_{xxxx} = \delta(x-s). \end{equation} The result would have been the same, except that the thing you call "the Green's function" would have been rescaled by $EI$.

You also could have divided by $-EI$, in which case \begin{equation} u_{xxxx} = -\frac{f(x)}{EI} \end{equation} says that $-\frac{f(x)}{EI}$ is your source, so your Green's function would have to satisfy \begin{equation} G_{xxxx} = \delta(x-s). \end{equation} It's all just a matter of when you divide by any given constant.