# Second order ODE with dirac delta and fourier tranform

I am struggling with the following equation $$\tau^2 \ddot{x}=x-m^2-Dm\tau^2 \delta(t)$$ This equation can be easy solved by applying the Fourier transform on both sides $$(1+\omega \tau^2) \widehat{x} =m^2 \delta(\omega)+2D\tau$$ which gives $$\widehat{x} = \frac{m^2 \delta(\omega)+2 D\tau}{1+\omega \tau^2}$$ and by applying the inverse transform one gets $$x(t) = m^2 + D e^{-|t|/\tau}$$

One the other side one could consider the first equation and multiply both sides by $\dot{x}$ (using the same idea explained here Second-order nonlinear ODE with Dirac Delta ). After few passages one gets $$\frac{d}{dt} \left( \tau^2\frac{\dot{x}^2}{2} + V(x) \right) = - 2 D \delta(t) \dot{C}(0)$$ where $$V(x) = -\frac{x^2}{2} + m^2 x$$ Following the same procedure used here Second-order nonlinear ODE with Dirac Delta , one has $$\frac{d}{dt}[\tau^2\frac{\dot{x}^2}{2} + V(x)] = 0 \ \, t>0$$ where $$\tau^2\frac{\dot{x}(0)^2}{2} + V(x(0)) = -2 D \dot{x}(0)$$ fixes the initial condition. One can realize that $m^2+K e^{-t/\tau}$ is a solution of the first order differential equation and $\tau^2\frac{\dot{x}^2}{2} + V(x) = m^4/2$

The problem is that if I try to fix $K$ using the last equation, I do not get $K=D$ as obtained applying the Fourier transform, but $\tau m^4 /(4 D)$. I do not understand where the second method fails and why they give different results.

## 1 Answer

Substitutions \begin{equation*} s=t/\tau ,\;x(t)=m^{2}y(s),\;u(s)=y(s)-1,\;u(s)=\frac{D\tau }{m}v(s) \end{equation*} Then \begin{equation*} \partial _{s}^{2}v(s)=v(s)-\delta (s) \end{equation*} Fourier transform \begin{equation*} f(s)=\int d\omega \exp [-i\omega s]\tilde{f}(\omega ),\;\tilde{f}(\omega )=% \frac{1}{2\pi }\int ds\exp [+i\omega s]f(t) \end{equation*} Now \begin{eqnarray*} v(s) &=&\frac{1}{2}\exp [-|s|] \\ \partial _{s}v(s) &=&\frac{1}{2}\{-\exp [-s]\theta (s)+\exp [+s]\theta (-s)\} \\ \partial _{s}^{2}v(s) &=&v(s)-\delta (s) \end{eqnarray*} Note that, given the solution obtained by Fourier transforming, \begin{equation*} \{\partial _{s}v(s)\}\delta (s),\;\frac{1}{\partial _{s}v(s)}\delta (s) \end{equation*} are ill-defined. Thus the second method becomes problematic if the integral over $[-\varepsilon ,+\varepsilon ]$ is contemplated. Of course you can proceed assuming that everything is all right and maybe end up with an answer but you cannot expect it to coincide with the solution already obtained.

• I found out how to proceed. There might be an error in math.stackexchange.com/questions/383750/… Starting from $$\tau^2 \ddot{x}=x-m^2-2Dm\tau^2 \delta(t)$$ One can integrate both side between $-a$ and $a$. Taking the limit $a \to 0$ one obtains $$\tau( \lim_{t \to 0^+} \dot{x} - \lim_{t \to 0^-} \dot{x} ) = 2Dm\tau^2$$ Dec 7 '15 at 16:14
• I found out how to proceed. There might be an error in math.stackexchange.com/questions/383750/… Starting from $$\tau^2 \ddot{x}=x-m^2-2Dm\tau^2 \delta(t)$$ By integrating both side in $[-a,a]$. taking the limit $a \to 0$ one obtains $$\tau( \lim_{t \to 0^+} \dot{x} - \lim_{t \to 0^-} \dot{x} ) = 2Dm\tau^2$$ Since $\dot{x}$ is not continuous the limits do not coincides. If one assumes that $x$ is an even function, one can use $\lim_{t \to 0^+} \dot{x} = - \lim_{t \to 0^-} \dot{x}$ to recover the result obtained with the Fourier transform Dec 7 '15 at 16:20