Solution of $y'+2y=g(t)$, $y(0)=0$, for a given $g(t)$ having a jump discontinuity

Question

The differential equations book that I'm reading states the following:

Linear differential equations sometimes occur in which one or both of the functions $p$ and $g$ have jump discontinuities. If $t_0$ is such a point of discontinuity, then it is necessary to solve the equation separately for $t < t_0$ and $t > t_0$. Afterward, the two solutions are matched so that $y$ is continuous at $t_0$; this is accomplished by a proper choice of the arbitrary constants.

Then it asks to solve the following initial value problem:

$y' + 2y = g(t),\ y(0) = 0$

where

$g(t)=\begin{cases} 1 & \text{ if $0 \leq t \leq 1$} \\ 0 & \text{ if $t > 1$}\end{cases}$

For $0 \leq t \leq 1$:

$$y' + 2y = 1$$

The integrating factor is $e^{2t}$:

$$(ye^{2t})' = e^{2t}$$

$$ye^{2t} = \dfrac{e^{2t}}{2} + C_1$$

Since the interval $0 \leq t \leq 1$ contains the initial point $t = 0$, the constant $C_1$ is determined by the initial condition $y(0) = 0$:

$0e^{0} = \dfrac{e^{0}}{2} + C_1$. So, $C=-\dfrac{1}{2}$

Therefore, the solution for this interval is $y = \dfrac{1}{2}\left(1-e^{-2t}\right)$.

For $t > 1$:

$$y' + 2y = 0$$

The integrating factor is $e^{2t}$:

$$(ye^{2t})' = 0$$

$$ye^{2t} = C_2$$

$$y = C_2e^{-2t}$$

The above solution should be "matched" with the previous one at $t=1$. So, setting $t = 1$ and equalling both solutions gives the value of $C_2$:

$$\dfrac{1}{2}\left(1-e^{-2}\right) = C_2e^{-2}$$

$$C_2 = \dfrac{1}{2}(e^2-1)$$

So, the second solution becomes $y = \dfrac{1}{2}(e^2-1)e^{-2t}$

Now, a continuous function can be created, joining these two solutions gives the general solution:

$$y(t)=\begin{cases} \dfrac{1}{2}(1-e^{-2t}) & \text{ if $0 \leq t \leq 1$} \\ \dfrac{1}{2}(e^2-1)e^{-2t} & \text{ if $t > 1$}\end{cases}$$

This attempt is correct, because this is the answer given by the book. But, although the function above is continuous at $t=1$, it is not differentiable, since the derivative of $y(t)$ is not continuous at $t=1$ (there is a "kink" at $y(1)$). So, why is the function above a solution for $t\geq 0$? Shouldn't a solution be differentiable everywhere?

Answer 1

Great attempt! The function $y(t)$ is a solution because it satisfies the OE in two parts. $g$ is discontinuous at $t=1$ so, as you pointed, we solve the OE in two parts. This lets us to have the desired probable solutions for $t>1$ and another interval separately. In some practical cases such as Electrical Circuits, we always want the particular solution a continuous function and that is why you checked $$\lim_{t\to 1^+}y(x)=y(1)$$ This later solution is not differentiable everywhere because it is not continuous at $t=1$ (A Cusp!) and indeed you don't have an OE in this point, but two parts of it are differentiable everywhere. There is no need to have such this strong condition everywhere. What can I add for additional points here is that we use the Picard's Existence of a unique solution separately in the way the final solution be continuous everywhere.