This is a problem in a textbook used in my class:

Suppose we have an infinite elastic beam, where the deflection $u(x)$ satisfies the differential equation $$\frac{d^4 u}{dx^4}+k^4 u = > f(x),$$ where $k^4$ is a positive constant regarded as known, and $f(x)$ is a load.

For part (a) of the question, we assume that the load is a unit concentrated load at $x = \xi$, so that it satisfies we equation $$\frac{d^4 u}{dx^4}+k^4 u = \delta(x - \xi),$$ where $\delta$ is the Dirac-delta equation.

So, for this problem I would like to find the deflection (which is equal to the free space Green's function). I've looked in several mechanics textbooks to find a detailed approach as to how to calculate this, but only found solutions for different problems.


migrated from engineering.stackexchange.com Oct 1 '15 at 17:28

This question came from our site for professionals and students of engineering.

  • 1
    $\begingroup$ Finding the Green's function of a fourth order equation isn't an easy task, and honestly seems a bit beyond the scope of a course on beam deflection; are you sure that is what the problem is asking for? At any rate, you may find this journal article helpful: Green’s function for the deflection of non-prismatic simply supported beams by an analytical approach. $\endgroup$ – Chris Mueller Oct 1 '15 at 12:00
  • $\begingroup$ The course is taught in the (graduate) mathematics department and named "Applied Mathematics" and is more about Green's functions and the theory of distributions so far - this question was just used as an example during the course. Thanks for the link $\endgroup$ – Olorun Oct 1 '15 at 12:06

Finally found a way to do it using Fourier transform, but a friend of mine did it just using calculations (much more tedious).

Since the deflection $u(x)$ satisfies the above differential equation, the Green's function satisfies \begin{equation} \frac{\partial^4 g}{\partial x^4}+\alpha^4 g = \delta(x - \xi), \end{equation} and assuming that $g = \frac{\partial g}{\partial x} = \frac{\partial^2 g}{\partial x^2} = \frac{\partial g^3}{\partial x^3} = 0$ at $|x| = \infty$, we have the solution $$u(x) = \int_{-\infty}^{\infty}g(x,\xi)f(\xi)d\xi.$$

Next, define the Fourier transforms $$\hat{u}(k) = \int_{-\infty}^{\infty}u(\xi) e^{ik\xi}d\xi; \; u(\xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ik\xi}\hat{u}(k)dk$$ and $$\hat{u'} = \int_{-\infty}^{\infty}u'(\xi)e^{ik\xi}d\xi = u(\xi)e^{ik\xi}|_{-\infty}^{\infty} - ik \int_{-\infty}^{\infty}u(\xi)e^{ik\xi}d\xi = (-ik)\hat{u}(k).$$

Applying the Fourier transform to our solution, we get $(k^4+\alpha^4)\hat{g} = e^{ik\xi},$ so \begin{equation*} \hat{g} = \frac{e^{ik\xi}}{k^4 + \alpha^4}. \end{equation*} Using the Fourier inversion formula, we obtain \begin{align} g(x,\xi)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{ikx}e^{ik\xi}}{k^4 + \alpha^4}dk\\ & = \frac{1}{\pi}\int_0^\infty \frac{\cos k(x-\xi)}{k^4+\alpha^4}dk\\ &=\frac{e^{-\alpha|x-\xi|/\sqrt{2}}}{2\alpha^3}\sin\left(\frac{\alpha|x-\xi|}{\sqrt{2}}+\frac{\pi}{4}\right). \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.