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)


$\displaystyle-\int_{-\infty}^{\infty} \phi'(x)u(x)dx = -\int_{-\infty}^{0} \phi'(x) * 0 dx - \int_{0}^{\infty} \phi'(x) * 1 dx$

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?


  • $\begingroup$ I don't know that I should give you a hint or not? because I often give this example to my students, lol ^^ $\endgroup$
    – Black Mild
    Oct 15 at 14:31
  • $\begingroup$ $u'+u$ is the sort of thing that should remind you of integrating factors: if $v= e^x u$ then $v'=e^x(u'+u)$. Then you can directly integrate for the guessing $\endgroup$ Oct 15 at 14:57

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 $$ (e^xu(x))'=H_0(x)e^x\implies e^xu(x)=c+H_0(x)(e^x-e^0) $$

  • $\begingroup$ This perfectly explained what I missed! I would have never found that on my own, I wasn't even thinking about it. That literally slipped my mind. Thanks for clarifying! $\endgroup$
    – Nolan P
    Oct 15 at 22:54

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} $


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