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Consider a hypothetical model of an extended mechanical system (in which a derivatives of higher order than acceleration may exist d) as bellow:

$$\sum_{n=0}^N {a_n x^{(n)}}= f_0 \delta(t-t_0)$$

which is imposed into a non-resting initial values:

$$x(t_0^-)=x_0\hspace{0.2cm} ;\hspace{0.2cm} x'(t_0^-)=x_1\hspace{0.2cm} ;\hspace{0.2cm} ... \hspace{0.2cm} ;\hspace{0.2cm} x^{(n-1)}(t_0^-)=x_{(n-1)} .$$

How can I write the correct initial values for $t=t_0^+$? This is ambiguous since integrating the differential equation above, yields:

$$ \sum_{n=1}^N a_n \big[x^{(n-1)} (t_0^+)-x^{(n-1)} (t_0^-)\big] = \sum_{n=1}^N a_n c_{n-1} = f_0$$

where $c_i$s are the jumps in initial conditions. In literature, it is said that only the highest order will undergo a discontinuity, yet it's not clear for me why.

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  • $\begingroup$ Quick hint : If suppose $\frac{d_{}^{k}}{dt_{}^{k}}x$ for $k<N$ has a jump discontinuity, what kind of singular behavior you expect for $\sum_{k=0}^{N}a_{n}^{}(t)\frac{d_{}^{k}}{dt_{}^{k}}x$ (supposing $a_{k}^{}(t)$ are all smooth functions of $t$). $\endgroup$ – Sunyam Jun 12 '18 at 9:31
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Thanks Sunyam for the hint, I revisited this problem with another point of view which gives the answer right away. Since the $x$ is at least $N-1$ times differentiable, it's derivatives (up to order $N-2$) and itself must be continuous. So no shift until the last derivative. Using the my question's notation,

$c_n =0\hspace{1cm} \forall n<N-1$

$c_{N-1} =f_0/a_N$

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