The Green's function of the beam deflection equation

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.

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• 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. – Chris Mueller Oct 1 '15 at 12:00
• 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 – Olorun Oct 1 '15 at 12:06

Since the deflection $u(x)$ satisfies the above differential equation, the Green's function satisfies $$\frac{\partial^4 g}{\partial x^4}+\alpha^4 g = \delta(x - \xi),$$ 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}