explicit answer of and initial condition ODE with delta input

Question

Assume this initial value problem ODE with constant coefficient:

$\mathcal{D}[u] = \sum_{n=0}^N {a_n u^{(n)}}=0$

$u(0)=u_0\hspace{0.2cm} ;\hspace{0.2cm} u'(0)=u_1\hspace{0.2cm} ;\hspace{0.2cm} ... \hspace{0.2cm} ;\hspace{0.2cm} u^{(n-1)}(0)=u_{(n-1)} $

I want to explicitly find the answer of the a delta function input given that I know the answer of homogeneous input that satisfies the initial conditions (that I will call $g$). i.e. I want to find the answer of the following equation in a piece-wise manner of $t<t_0$ and $t>t_0$ :

$\mathcal{D}[u] = \sum_{n=0}^N {a_n u^{(n)}}= f_0 \delta(t-t_0)$

$u(0)=u_0\hspace{0.2cm} ;\hspace{0.2cm} u'(0)=u_1\hspace{0.2cm} ;\hspace{0.2cm} ... \hspace{0.2cm} ;\hspace{0.2cm} u^{(n-1)}(0)=u_{n-1} $

Knowing that the input is active only at $t=t_0$, it's just fine to express the first part of answer($t<t_0$) exactly as $g(t<t_0)$. For the second part, I want to use again the same $g$ only with different initial conditions (now at $t=t_0$ instead of 0).

To do so, I can integrate the ODE over and infinitesimal interval $(t_0-\epsilon,t_0+\epsilon)$ which yields the following constraints on the new initial conditions:

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

where $c_i$s are the discontinuities (shifts) on each derivatives at $t=t_0$. and since $u^{(n-1)} (t_0^-)=g^{(n-1)}(t_0)$, then

$ \sum_{n=1}^N a_n u^{(n-1)} (t_0^+) = C$

where C is a linear combination of the same derivatives at $t=t_0^-$ and also $f_0$ and now I can rewrite my new initial conditions as follows:

$u(t_0^+)=g(t_0^+)\hspace{0.3cm};\hspace{0.3cm} u'(t_0^+)=g'(t_0^+)+c_1\hspace{0.3cm} ;\hspace{0.3cm} ... \hspace{0.3cm} ;\hspace{0.3cm} u^{(n-1)}(t_0^+)=g^{(n-1)}(t_0^+)+c_{n-1} $

Now, I'm wondering if this additional constraint prevents me to find a unique answer for the second part since now, it seems that I've got to chose the amount of shift each derivative undergoes.

Answer 1

One can obtain an explicit answer using the Laplace transform: $$U=F/P+Q/P,$$ where $U$ is the Laplace transform of $u$, $P=\sum_{k=0}^Na_kz^k$, $F=e^{-t_0z}$ is the Laplace transform of the delta-dunction, and $$Q(z)=u_0z^{N-1}+(u_1+a_{N-1}u_0)z^{N-2}+\ldots +(u_{N-1}+a_{N-1}u_{N-2}+\ldots+a_2u_1+a_1u_0).$$ This is an "explicit form" because the inverse Laplace transform of a rational function can be computed by partial fraction decomposition. To do this, you need to factor $P$. But this is unavoidable with any form of the solution, even for the homogeneous equation. But if you cannot factor $P$, you still obtain an "explicit answer", in the form of an integral.