In literature, for proving the existence of Green function for linear systems, it is argued that if for a linear differential equation like

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

$y(0)=y_0, \hspace{1cm} y'(0)=y_1, \hspace{1cm} ... \hspace{1cm}y^{N-1}(0)=y_{N-1}$

if we know the response of delta input $\delta(t-t_0)$ we can construct the answer for an arbitrary time-dependent input and here is the sketch of the proof:

Assuming that we know a function $g(t,t')$ that is the answer of equation

$\mathcal{D}[g] = \sum_{n=0}^N {a_n g^{(n)}} = \delta(t-t')$

and satisfies all the boundary conditions, then, for any arbitrary input, since $f(t) = \int{f(t')\delta(t-t')dt'}$ we can say use the linearity of $\mathcal{D}$ and say

$\mathcal{D}\big[\int{f(t')g(t,t')dt'}\big] = \mathcal{D}\big[y(t)] =\int{f(t')\delta(t-t')dt'} =f(t)$

Up to this, it's all clear, but I don't see how the initial condition will affect the answer $y(t)$. In other words, I think for a complete answer we must state the following expression:

$y(0) = \int_{0}^{\infty}{g(0,t')f(t')dt'}=y_0 $

$y'(0) = \int_{0}^{\infty}{g'(0,t')f(t')dt'}=y_1 $

$y''(0) = \int_{0}^{\infty}{g''(0,t')f(t')dt'}=y_2$


These additional constraints, which are forced by initial conditions, never discussed if the initial conditions are non-zero (if all of them are zero, these equations will be satisfied automatically).

Now, I'm totally confused, because these constraints, impose some sort of normalization over function $f$ and $g$ which is none sense. I appreciate any help that explains the last expressions, their intuition and the way people deal with that.

  • $\begingroup$ These are not constraints to $f(t)$, but to uniquely defining $g$ so that not only the equation is satisfied, but also the initial conditions are too. $\endgroup$ – Vladimir Kalitvianski Jun 12 '18 at 9:16
  • $\begingroup$ You're right because the $f(t)$ is indeed arbitrary. Yet my problem still holds @Vladimir $\endgroup$ – arash Jun 12 '18 at 9:19
  • 1
    $\begingroup$ Without imposing some initial conditions to $g$, the Green's function will have $N$ arbitrary constants of integration due to the differential equation order being $N$. It will lead to arbitrary initial conditions for $y$ and its derivatives, if $y$ is expressed via $f$ and $g$. Thus, in order to obtain a given set of the initial conditions, one has to impose them to $g$. Look at $g$ as at a solution of your equation. $\endgroup$ – Vladimir Kalitvianski Jun 12 '18 at 9:44
  • $\begingroup$ So are you suggesting that the initial conditions needed to be used solely for determining $g$ and not for the response of arbitrary input $y(t)$? $\endgroup$ – arash Jun 12 '18 at 9:56
  • $\begingroup$ Just the opposite: as soon as you express $y$ via $g$, it means specifying the initial conditions for $y$. Look, the differential equation may have many solutions, depending on the initial conditions. Or you construct $y$ directly, either via $g$, you will have the same arbitrariness without applying the initial conditions. $\endgroup$ – Vladimir Kalitvianski Jun 12 '18 at 10:04

Thanks to the hints and with a little bit research, I found what I was looking for in Sadri Hassani's Mathematical Physics book. I try to explain step by step.

First of all, since the ODE in non-homogeneous and also has non-homogeneous initial values, the convention is (thanks to the linearity) to decompose it to two functions $y_h$ and $y_i$ which are the homogeneous and non-homogeneous answer of the problem respectively. Regarding to this decomposition, we also have to take care of initial values.

Here comes the second convention that embraces the initial values. We request that the initial values for $y_i$ to be zero (homogeneous) but non-zero (non-homogeneous) for $y_h$.

So the original problem of

$\mathcal{D}[y(t)] = \delta(t-t_0)\ ;\hspace{2cm} \ y^{(n)}(0) = y_{n} \hspace{1cm} \forall n=0, ...,N-1$

is transformed to finding a $y_i$ and $y_h$ such that $y=y_h+y_h$ and

$\mathcal{D}[y_h(t)] = 0\ ;\hspace{3.3cm} \ y^{(n)}(0) = y_{n} \hspace{0.8cm} \forall n=0, ...,N-1$ $\mathcal{D}[y_i(t,t_0)] = \delta(t-t_0)\ ;\hspace{1.5cm} \ y^{(n)}(0) =0 \hspace{1cm} \forall n=0, ...,N-1$

The first problem is straightforward. One can solve the characteristic equation to obtain the powers of the exponential terms and solve a linear system for finding the coefficients that gives make the general solution of $y_h(t)$ fit the initial values. Note that the $y_h(t)$ is a function of $t$ only.

The second one can be solved thanks to the magic of Green function. i.e. we can look for a $g(t,t')$ that satisfies the second expression, which in practice is much simpler than what I'd mentioned in the question that would satisfy all initial conditions. Assuming we are able to find such $g$, we can write the $y_i$ as below:

$y_i (t) = \int_{0}^{\infty}{f(t') g(t,t') dt'}$

and since now $g(0,t'), g'(0,t'), g''(0,t'), ...$ are all zero, $y_i$ and all derivatives are zero, consistent with the equation above.

PS: I was more into a general routine that for any initial/boundary conditions, using Green function technique, produces the general answer for arbitrary input. This recipe, although works for this specific problem, is sort of nicely engineered and I think it may not be the most general case. The reason I didn't find this recipe general is that the decomposition I did at the first place, may not be useful for other cases.

And to show what I mean by a general case, I'd rather to bring up the following problem. Assume the same problem but this time with boundary conditions instead of initial values:

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

$y(x_0)=y_0, \hspace{1cm} y(x_1)=y_1, \hspace{1cm} ... \hspace{1cm}y(x_{N-1})=y_{N-1}$

Are we able to construct the answer using similar machinery? What if the BCs contain derivatives (whether first order or higher)?

PSS: I haven't thought about them yet, but appreciate if someone introduces me a reference about them in the comments.


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