I am considering the following Initial Value problem: $$ \begin{aligned} &u'+\alpha u = \cos \omega t\\ & u(0)=u_0 \end{aligned} $$ The solution is: $$u(t) ={\rm e}^{-\alpha t} \left(u_0-{\frac { \alpha}{{\alpha}^{2}+{\omega}^{2}}} \right)+\frac {\alpha\cos\omega t+\omega\sin\omega t}{\alpha^2+\omega^2} $$ I now want to solve the equation using a distribution of the form $U(t)=H(t)u(t)$ where $H(t)$ is the Heaviside distribution. Since $U'=u'+u_0\delta$, we have to solve $$U'+\alpha U=\cos \omega t + u_0\delta$$ in the sense of distributions, which means, using classical convolution formulation $$(\delta'+\alpha\delta)*U=\cos\omega t + u_0\delta$$ The convolution inverse of $\delta'+\alpha\delta$ is $G(t)=H(t)\exp(-\alpha t)$ and so, if I am correct, the sought solution is $U(t)=G(t)*(\cos\omega t+u_0\delta)$ that is: $$U(t)=H(t)\Bigl[u_0\exp(-\alpha t)+\frac{\alpha\cos \omega t +\alpha\sin\omega t}{\alpha^2+\omega^2}\Bigr]$$ Obviously there is something wrong in the derivations above, since the two solutions do not match. I cannot find where.


Based on md2perpe's comments and reply, below is a detailed solution. From $U'+\alpha U=H(t)\cos \omega t + u_0\delta$, we can say that $$U(t)=G(t)*(H(t)\cos\omega t+u_0\delta)$$ where $G(t)=H(t)\exp(-\alpha t)$. Expanding the convolution yields: $$U(t)=\int_0^{t}{\rm e}^{-\alpha s}\cos\omega (t-s)\mathrm{d}s+u_0{\rm e}^{-\alpha t}$$ that is $$U(t)=H(t)\Bigl({\rm e}^{-\alpha t} \left(u_0-{\frac { \alpha}{{\alpha}^{2}+{\omega}^{2}}} \right)+\frac {\alpha\cos\omega t+\omega\sin\omega t}{\alpha^2+\omega^2}\Bigr)$$ as expected.

  • $\begingroup$ Without having deep-dived, one question hits me: can there really exist a solution that vanishes for $t<0$ but not for $t>0$? $\endgroup$ – md2perpe Apr 20 at 13:46
  • $\begingroup$ @md2perpe We assume that the solution vanishes for negative $t$, or more exactly, we do not really care about what happens for negative $t$, so we have a look at a distributional solution in the $\mathcal D'_+$ algebra... $\endgroup$ – pluton Apr 20 at 15:56
  • $\begingroup$ There is an important difference between "assume that the solution vanishes for negative $t$" and "do not really care about what happens for negative $t$". The first one forces the solution to be zero for negative $t$ which might result in $u'$ having a $\delta(t)$-term. The second does not. $\endgroup$ – md2perpe Apr 20 at 16:05
  • $\begingroup$ If you want to force $u(t)=0$ for $t<0$ and solve it using distributions, try to change the problem to $$ u'(t) + \alpha u(t) = \cos\omega t \, H(t) + c \, \delta(t), \quad u(0) = u_0. $$ $\endgroup$ – md2perpe Apr 20 at 16:10
  • $\begingroup$ @md2perpe you are right, I thought about the difference right after my comment and we should assume that $u$ vanishes for negative $t$. What I can say is that the approach described in my post works well without the $\cos \omega t$ term. Things go differently with it and the initial condition seems to be a problem. I will think about your suggestion. $\endgroup$ – pluton Apr 20 at 16:41

The solutions to $u' + \alpha u = \cos \omega t$ on all of $\mathbb{R}$ have the form $$u(t) = A \cos \omega t + B \sin \omega t + C e^{-\alpha t}.$$

Let $v(t) = u(t)$ for $t>0$ and $v(t) \equiv 0$ for $t<0$. This can be written $v(t) = u(t) \, H(t).$

What happens when we insert this into the differential equation, treating $v$ as a distribution? We have $$v'(t) = u'(t) \, H(t) + u(t) \, H'(t) = u'(t) \, H(t) + u(0) \, \delta(t)$$ giving $$\begin{align} v'(t) + \alpha v(t) &= \left( u'(t) \, H(t) + u(0) \, \delta(t) \right) + \alpha \left( u(t) \, H(t) \right) \\ &= \left( u'(t) + \alpha u(t) \right) H(t) + u(0) \, \delta(t) \\ &= \cos \omega t \, H(t) + u(0) \, \delta(t). \end{align}$$ We can see here that you have missed the factor $H(t)$ after $\cos \omega t$.

  • $\begingroup$ Thanks. I thought about the missing $H(t)$ but if we follow the convolution procedure, the $H(t)$ in "my" $G(t)$ above will simplify, I think, with the $H(t)$ in $\cos\omega t H(t)$ so I am not whether it makes a difference? $\endgroup$ – pluton Apr 20 at 17:57
  • $\begingroup$ How do you mean that it will simplify? $\endgroup$ – md2perpe Apr 20 at 18:00
  • $\begingroup$ Your solution should be $$U(t)=G(t)*(\cos\omega t \, H(t) + u_0\delta).$$ Do you mean that the $H(t)$ in $G(t)$ and the $H(t)$ in the parenthesis should multiply to form $H(t)$? $\endgroup$ – md2perpe Apr 20 at 18:03
  • $\begingroup$ yes! I see that it becomes $H(t)H(s-t)$ in the convolution. I have to compute it. $\endgroup$ – pluton Apr 20 at 18:06
  • $\begingroup$ $H*H = t H$ but you also have the factor $e^{-\alpha t}$ which complicates matter a bit. $\endgroup$ – md2perpe Apr 20 at 18:10

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