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Recently,I'm trying to solve this differential equation: $$ m\frac{d^2x(t)}{dt^2}=(V_1-V_2)\delta(x(t)) $$

The equation has a physical significance:a particle moves from one half-space with potential energy $V_1$ to the other half-space with potential energy $V_2$. So it can be solved by conversation of energy as follows. $$ m\frac{d^2x(t)}{dt^2}\frac{dx}{dt}=(V_1-V_2)\delta(x)\frac{dx}{dt} $$ $$ \frac{1}{2}m\frac{d}{dt}\left(\frac{dx}{dt}\right)^2 = (V_1-V_2)\delta(x)\frac{dx}{dt} $$ Then we can multipy $dt$ to both sides and integral,the equation will be solved easily. So the original equation does make sense.

However,it becomes different if we do not use the conversation of energy. First,we know $$ x(t) = \left\{ \begin{aligned} & at+b & t < -\frac{b}{a} \\ & (k + 1)at+(k + 1)b & t > -\frac{b}{a} \end{aligned} \right. $$ Then, $$ \frac{d^2x}{dt^2}=ka\delta(t + \frac{b}{a}) $$ But it seems that the relation between $\delta(x)$ and $\delta(t + \frac{b}{a})$ is not defined because of the difference of the left and right derivative of $x(t)$ at $t=-\frac{b}{a}$. So the original equation does not make sense.

Why are the two results different? What is the mistake? Thanks for your help.

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  • $\begingroup$ 1) The last equation is a bit strange in form: it somehow knows about the solution that you are plugging in. Because you know how solutions behave, isn't $\delta(t+\frac{b}{a}) = \delta (a x ) = \delta (x)$ ? (I might be mistaken here) 2) Note that the first equation also doesn't make sense from point of view of classic functions and you have to understand it at least in generalized functions or treat as a shorthand for some integral equation. $\endgroup$ – Evgeny Oct 1 '16 at 8:21
  • $\begingroup$ @Evgeny As far as I know,from the generalized function theory,we have $\delta(f(x)) = \sum_{f(x_i)=0} \frac{\delta(x-x_i)}{|f'(x_i)|}$. So $\delta(f(x))$ doesn't make sense at all if $f'(x_i)$ doesn't exist when $f(x_i) = 0$. I also tried a integral equation,but I found it depends on $H(0) = \int^{0}_{-\infty} \delta(t) dt$ ,which doesn't also make sense. $\endgroup$ – goodqt Oct 1 '16 at 11:11

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