Setting the limit while finding convolution

Question

I was going through an example related to application of convolution theorem in Nonhomogeneous Linear ODEs

The example question was:

And the solution was given as follows:

1/2

2/2

I am unable to understand the part where it is stated "Now comes an important point in handling convolution"

My take on the statements stated: If t<1 then r(t)=0,so the integral is zero,that's OK,then if 12 the limit is 1 to 2 because it covers all the values of t between 1 and 2 and hence the limit.But I feel what I think about limits is too vague and it doesn't have any purpose because when we are asked to solve differential equation we aren't told what the value of 't' is.We are suppose to find the solution for possible values of t.

Correct me if I am wrong and please tell me some concrete way so know the limits

Answer 1

You have indeed a solution for every value of $t$. $y(t)$ is this:

$$y(t)=\begin{cases} 0 &\text{if}\; t\le 1\\ {1\over 2}-e^{-(t-1)}+{1\over 2} e^{2(t-1)} &\text{if}\;1<t\le 2\\ e^{-(t-2)}-{1\over 2} e^{2(t-2)}-\left(e^{-(t-1)}-{1\over 2} e^{2(t-1)}\right) &\text{if}\;2<t \end{cases}$$

The solution comes straight from this integral

$$y(t)=\int_0^t q(t-\tau)r(t)\mathrm d\tau\;\;\text{with}\;r(t)=\begin{cases} 0 &\text{if}\; t\le 1\\ 1 &\text{if}\;1<t\le 2\\ 0 &\text{if}\;2<t \end{cases}$$

To calculate the integral you have to split it in two parts or three parts when $1<t\le 2$ and $2<t$ respectively because the function $r$ is defined piecewise. The word "handling" refers to how manage the integral limits in the convolution when the function is defined piecewise. The solution $y$ at some $t$ comes integrating from $0$ to $t$, so you calculate separately for the three pieces, adding to one piece, the result from the previous (you add zero to the second and zero and all the way from $1$ to $2$ to the third).

$$y(t)=\begin{cases} \int_0^t q(t-\tau)·0\mathrm d\tau &\text{if}\; t\le 1\\ \int_0^1 q(t-\tau)·0\mathrm d\tau+\int_1^t q(t-\tau)·1\mathrm d\tau&\text{if}\;1<t\le 2\\ \int_0^1 q(t-\tau)·0\mathrm d\tau+\int_1^2 q(t-\tau)·1\mathrm d\tau+\int_2^t q(t-\tau)·0\mathrm d\tau&\text{if}\;2<t \end{cases}$$

You can see the integrals are all from $0$ to $t$, but separated by intervals. Finally,

$$y(t)=\begin{cases} 0 &\text{if}\; t\le 1\\ 0+\int_1^t q(t-\tau)·1\mathrm d\tau&\text{if}\;1<t\le 2\\ 0+\int_1^2 q(t-\tau)·1\mathrm d\tau+0&\text{if}\;2<t \end{cases}$$