How to derive the jump condition at $x=0$ for a fourth-order ordinary differential equation involving a derivative of the Dirac delta function?

Question

Let's consider the fourth-order ordinary differential equation (ODE): $$ f^{(4)}(x) - 2 f^{(2)}(x) + f(x) = \delta'(x) , \tag{1} $$ where $f^{(n)}$ denotes the $n$th derivative with respect to $x$.

Given that the Dirac delta function is related to its derivative via $x\delta'(x) = -\delta(x)$, we can rewrite Eq. (1) as [Wikipedia]: $$ x \left[ f^{(4)}(x) - 2 f^{(2)}(x) + f(x) \right] = -\delta(x) . \tag{2} $$

Our objective is to determine the jump condition at the interface $x=0$.

To obtain the jump condition, we start by integrating both sides of Eq. (2) by parts. This leads to the following expression: $$ \left[ x \left( f^{(3)}-2f^{(1)}+f^{(-1)} \right) + f^{(2)}-2f+f^{(-2)} \right]_{x=-\epsilon}^\epsilon = 1\, , $$ where $\epsilon$ is a small positive value, approaching zero.

How can we rigorously obtain the jump condition at $x=0$?

It seems that the discontinuity in the second derivative might play a crucial role, but I'm struggling to establish a concrete connection. Any assistance or guidance on this matter would be greatly appreciated.

Thank you very much!

Answer 1

Transform the equation into a system of second order equations, using the freedom to chose the new component to combine the highest derivatives on both sides by setting $$f''=g(x)+\theta,$$ $\theta$ the unit jump function, thus $\theta'=\delta$. Then $$ g''+\delta'-2(g+\theta)+f=\delta' \\~\\ g''-2g+f=2\theta $$ The linear system for $(f,g)$ now has piecewise constant right sides. Thus its solutions are $C^1$ and piecewise smooth.

In consequence, $f$ and $f'$ are continuous at $x=0$, $f''$ has a jump of size 1, $f'''=g'+\delta$ has no jump in the values left and right from $x=0$.

Answer 2

Let $\partial$ denote the ordinary derivative where it is defined and let us use $'$ for the distributional derivative.

Assume that $f$ is smooth everywhere except at $0$ where the function itself and its derivatives might have steps.

Then, $$ \begin{align} f' &= \partial f + (f(0^+)-f(0^-))\,\delta \\ f'' &= \partial^2 f + (f(0^+)-f(0^-))\,\delta' + (\partial f(0^+)-\partial f(0^-))\,\delta \\ f''' &= \partial^3 f + (f(0^+)-f(0^-))\,\delta'' + (\partial f(0^+)-\partial f(0^-))\,\delta' + (\partial^2 f(0^+)-\partial^2 f(0^-)) \delta \\ f'''' &= \partial^4 f + (f(0^+)-f(0^-))\,\delta''' + (\partial f(0^+)-\partial f(0^-))\,\delta'' + (\partial^2 f(0^+)-\partial^2 f(0^-)) \delta' \\ &\ \ \ \ \ \ \ \ \ \ \ + (\partial^3 f(0^+)-\partial^3 f(0^-)) \delta \\ \end{align} $$

Inserting these into the differential equation we find that $$ \begin{align} f(0^+)-f(0^-) &= 0 \\ \partial f(0^+)-\partial f(0^-) &= 0 \\ \partial^2 f(0^+)-\partial^2 f(0^-) &= 1 \\ \partial^3 f(0^+)-\partial^3 f(0^-) &= 0 \\ \end{align} $$

Answer 3

For what its worth, OP's ODE can in fact be solved exactly via factorization:

$$\begin{align} \delta^{\prime}(x) ~\stackrel{(1)}{=}~&\left(\frac{d^2}{dx^2}-1\right)^2f(x)\cr ~=~&\left(\frac{d}{dx}+1\right)^2\left(\frac{d}{dx}-1\right)^2f(x)\cr ~=~&e^{-x}\frac{d^2}{dx^2}e^{2x}\frac{d^2}{dx^2}e^{-x}f(x), \end{align}$$ or $$\begin{align} \frac{d^2}{dx^2}e^{2x}\frac{d^2}{dx^2}e^{-x}f(x) ~=~&e^x\delta^{\prime}(x)\cr ~=~&\delta^{\prime}(x)+(e^x-1)\delta^{\prime}(x)\cr ~=~&\delta^{\prime}(x)-e^x\delta(x)\cr ~=~&\delta^{\prime}(x)-\delta(x). \end{align}$$ Integrating twice yields $$ \frac{d^2}{dx^2}e^{-x}f(x) ~=~e^{-2x}\left(\frac{{\rm sgn}(x)-|x|}{2}+ax+b\right). $$ Integrating a third time yields $$ \frac{d}{dx}e^{-x}f(x) ~=~e^{-2x}\left(\frac{2|x|-{\rm sgn}(x)}{8}+Ax+B\right)+\frac{{\rm sgn}(x)}{8}+C. $$ Integrating a fourth time yields the complete solution $$ f(x) ~=~e^{-x}\left(-\frac{|x|}{8}+\alpha x+\beta\right)+e^x\left(\frac{|x|}{8}+\gamma x+\delta\right). $$ A particular solution is $$f(x)~=~\frac{|x|}{4}\sinh x. $$ It is easy to check that $f^{\prime\prime}$ has a jump of size 1 at $x=0$.