Green's Function for a Fourth-Order Differential Equation I have just begun to look at Green's functions and am studying a fourth-order equation
$\frac{d^4y}{dx^4}=f(x) \:\:\:\:\:\:y(0)=y'(0)=0\:\:\:\:\:\:y(1)=y'(1)=0$
In particular I have to show that the Green's function $G(x,u)$ for this equation satisfies a condition
$\lim_{\epsilon\to0}\bigg[\frac{\partial^3G}{\partial x^3}\bigg]_{x=u-\epsilon}^{u+\epsilon}=1$
and must demonstrate the continuity of the Green's function and its first and second partial derivatives with respect to $x$ at $x=u$.  I am fairly new to Green's functions so would appreciate some pointers.
 A: Based on your previous question on the site, I'll assume you're an applied mathematician/physicist, and you're not looking to be too rigorous. If so, then you can think of the Green's function $G(x,u)$ as the solution to the differential equation,
$$ \frac{\partial^4 G(x,u)}{\partial x^4}  = \delta(x - u),$$
with $G(0,u)=\partial_x G(0,u) = G(1,u) = \partial_x G(1, u) = 0$. You can think of the $\delta (x - u)$ as a "sharp impulse" at $x = u$, and you can think of $G(x,u)$ as your "system's response to this impulse".
To analyse the behaviour of $G(x,u)$ and its derivatives at $x = u$, consider integrating both sides of our differential eequation over the interval $u-\epsilon \leq x \leq u + \epsilon$, where $\epsilon $ is a small positive number. By the fundamental theorem of calculus, the integral of the left-hand side is
$$ \int_{u-\epsilon}^{u + \epsilon} dx \frac{\partial^4 G(x,u)}{\partial x^4} = \left[ \frac{\partial^3 G(x,u)}{\partial x^3}\right]_{u-\epsilon}^{u + \epsilon}$$
Meanwhile, regardless of what value we choose for $\epsilon$, the integral of the right-hand side is
$$ \int_{u - \epsilon}^{u + \epsilon} dx \delta (x - u) = 1$$
Thus
$$ \left[ \frac{\partial^3 G(x,u)}{\partial x^3}\right]_{u-\epsilon}^{u + \epsilon} =  1$$
which is to say that  $\partial_x^3 G(x,u)$ has a discontinuity of size $1$ at $x = u$.
For the next part, simply observe that for any $\epsilon > 0$,  $\partial_x^3 G(x,u)$ is piecewise continuous, hence bounded, on the closed interval $u - \epsilon \leq x \leq u + \epsilon$, so
$$ \lim_{\epsilon \to 0}\left[ \frac{\partial^2 G(x,u)}{\partial x^2}\right]_{u-\epsilon}^{u + \epsilon} =  \lim_{\epsilon \to 0} \int_{u - \epsilon}^{u + \epsilon} dx \frac{\partial^3 G(x,u)}{\partial x^3}= 0,$$
i.e. $\partial^2_x G(x,u)$ is continuous at $x = u$. By similar arguments, $\partial_x G(x,u)$ and $G(x,u)$ are also continuous at $x = u$.
