3
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.