Find a weak solution of the ODE Find a weak solution to the following ODE:
$u' + u = H_0(x)$ where $H_0(x) = \begin{cases}
0 & x < 0 \\
1 & x \geq 0 \end{cases}$
My professor advised that we try to guess the solution and then verify it. My first guess was naive because I did not know the "derivative" (I put quotes here because this isn't really a derivative) of $H_0(x)$ was $\delta_0(x)$. I thought it was $0$. I still state this because something consistent is happening.
If you do guess $u = H_0$, we can go ahead and attempt to find the weak derivative. Consider $\phi \in C_{c}^{\infty}$ (continuous functions with compact support)
Then,
$\displaystyle-\int_{-\infty}^{\infty} \phi'(x)u(x)dx = -\int_{-\infty}^{0} \phi'(x) * 0 dx - \int_{0}^{\infty} \phi'(x) * 1 dx$
Using integration by parts,
$\displaystyle-[\phi(x) * 0 |_{-\infty}^{0} + \int_{-\infty}^{0} \phi(x) * 0 dx - [\phi(x) * 1 |_{0}^{\infty} + \int_{0}^{\infty} \phi(x) * 0 dx$
The first term is $0$ due to the multiplication. The third term only leaves the lower limit because $\phi$ is compactly supported. Therefore, I am left with
$\displaystyle\boxed{\phi(0)} + \int_{-\infty}^{\infty} \phi(x) * 0 dx$.
This is very close to what I wanted, but I have an extra $\phi(0)$.
After our next lecture, I found out that the weak derivative of $H_0(x)$ does not exist and we need the distribution derivative to make it $\delta_0(x)$. Therefore, I knew my initial guess was wrong.
My next guess was to solve the ODE for both "components." What I mean is solve $u'+u = 0$ and $u' + u = 1$. Just to see if this worked, I first plugged these into wolfram alpha and got $c e^{-x}$ and $c e^{-x} + 1$ respectively. Therefore, my guess was $u = \begin{cases}
ce^{-x} & x < 0 \\
ce^{-x} + 1 & x \geq 1 \end{cases}$.
Now, I attempted to find $u'$
\begin{align}&-\int_{-\infty}^{\infty} \phi'(x) u(x) dx = -\int_{-\infty}^{0} \phi'(x) ce^{-x} dx - \int_{0}^{\infty} \phi'(x) (ce^{-x}+1) dx \\&= -\int_{-\infty}^{0} \phi'(x) ce^{-x} dx - \int_{0}^{\infty} \phi'(x) ce^{-x} dx - \int_{0}^{\infty} \phi'(x) dx\\ 
&= -\int_{-\infty}^{\infty} \phi'(x) ce^{-x} dx - [\phi(x) |_{0}^{\infty} \\ &= -[\phi(x) ce^{-x} |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \phi(x) (-ce^{-x} dx) + \phi(0)\\ 
&= \boxed{\phi(0)} + \int_{-\infty}^{\infty} \phi(x) (-ce^{-x})dx\\ \end{align}.
Again. $\phi(0)$ is there, At this point, I asked my professor if he could give me a hint. He told me my initial guess should be the solution to the ODE involving each component, which is exactly what I did. Since I did get the solutions through wolfram alpha,I went ahead and solved both ODEs by hand just to make sure something weird didn't happen. I ended up with $ce^{-x}$ and $1 - ce^{-x}$ respectively.
While the second one is slightly different, it shouldn't make a difference because $c$ is a constant, so it could "absorb" the $-$ sign. I won't go through the details again, but one will end up with
$\displaystyle\boxed{\phi(0)} + \int_{-\infty}^{0} \phi(x) (-ce^{-x}) dx + \int_{0}^{\infty} \phi(x) (ce^{-x}) = \boxed{\phi(0)} + \int_{-\infty}^{\infty} \phi(x) u'(x) dx$ where $u'(x) = \begin{cases} 
-ce^{-x} & x < 0 \\
ce^{-x} & x \geq 1 \end{cases}$.
Again, the $\phi(0)$ is still there and I'm not sure how to get rid of it! Does anyone see what I'm doing wrong? Is my initial guess still wrong?
Thanks!
 A: In your attempt for the piecewise solution you get different constants for both regions and have to ensure the continuity of the solution
$$
u(x)=\begin{cases}c_1e^{-x}&x<0\\1+c_2e^{-x}&x\ge 0\end{cases}
$$
with
$$
c_1e^0=1+c_2e^0
$$
so that
$$
u(x)=\begin{cases}ce^{-x}&x<0\\1+(c-1)e^{-x}&x\ge 0\end{cases}
\\=ce^{-x}+H_0(x)(1-e^{-x})
$$

You could also work with an integrating factor
$$
(e^xu(x))'=H_0(x)e^x\implies e^xu(x)=c+H_0(x)(e^x-e^0)
$$
A: This is what my knowledge allows me:
Given differential system $\dot{x} = A \, x + B \, u$,  and output $y = C \, x$, its solution is below.
$x(t, \tau) = e^{A(t - \tau)} x(\tau) + \int^{\tau}_{t} \, e^{A \, (t - s)} \, B \, u(s) \, ds$
Suppose $u(s) = H_0(s)$ as you state. Hence, the solution for it is the equality below. THem matrix $X$ corresponds to the integral $\int^{\tau}_{t} \, e^{- A \, s} ds$. Finally, the output is $y(t, \tau) = C \, e^{A(t - \tau)} x(\tau) + C \, e^{A \, t} \, X(t, \tau) \, B$
$
\begin{aligned}
    x(t, \tau) & = e^{A(t - \tau)} x(\tau) + \int^{\tau}_{t} \, e^{A \, (t - s)} \, B \, H_0(s) \, ds \\
    & = e^{A(t - \tau)} x(\tau) + e^{A \, t} \, \left( \int^{\tau}_{t} \, e^{- A \, s} ds \right) B \\
    & = e^{A(t - \tau)} x(\tau) + e^{A \, t} \, X(t, \tau) \, B 
\end{aligned}
$
For your case, matrix $A$ is a scalar $-1$, B is the scalar $1$ and C is the scalar $1$. Therefore, the solution is
$
\begin{aligned}
u(t, 0) = u(t) & = e^{-t} x(0) + e^{-t} \, X(t, 0) \\
& = e^{-t} x(0) + e^{-t} \, (e^t - 1) \\ 
& = e^{-t} x(0) + e^{-t} \, (e^t - 1) \\
& = e^{-t} x(0) - e^{-t} + 1
\end{aligned}
$
