I am working on a problem where I have the following ODE. $$m\dot{v}+bv=\delta_I(t)$$
where
$$\delta_I(t)=\begin{cases}0, & \text{for}&t\ne0\\ I, & \text{for} &t=0\end{cases}.$$
The solution $v(t)$ was derived using Laplace transforms, the ODE in the Laplace domain is (with $0$ initial conditions)$$(ms+b)V(s)=I$$ giving $$v(t)=\frac{I}{m}e^{-bt/m}.$$
How does this solution satisfy the original ODE though? At $t\ne0$ everything is good,$$-\frac{Ib}{m}e^{-bt/m}+\frac{Ib}{m}e^{-bt/m}=0,$$ while at $t=0$, $$-\frac{Ib}{m}+\frac{Ib}{m}=I$$ the result seems to be saying $0=1$ which is obviously false. What am I missing here?
[Math] Solution to ODE with Dirac Delta satisfies ODE
dirac deltalaplace transformordinary differential equations
Related Solutions
Rewrite $\hat{y}$ as $\hat{y}(s) = \frac{e^{-\pi s}+(s+1)+1}{(s+1)^2+1} = \frac{e^{-\pi s}}{(s+1)^2+1} + \frac{(s+1)}{(s+1)^2+1} + \frac{1}{(s+1)^2+1}$.
I am assuming that $t \geq 0$ in the following:
If $x_1(t) = e^{-t} \cos t$, then $\hat{x}_1(s) = \frac{(s+1)}{(s+1)^2+1}$.
If $x_2(t) = e^{-t} \sin t$, then $\hat{x}_2(s) = \frac{1}{(s+1)^2+1}$.
Using time-shifting, if $x_3(t) = x_2(t-\pi)u(t-\pi)$, then $\hat{x}_3(s) = \frac{e^{-\pi s}}{(s+1)^2+1}$.
Consequently, you have $y(t) = x_1(t)+x_2(t)+x_3(t)$, or explicitly,
$$y(t) = e^{-t} \cos t + e^{-t} \sin t + e^{-(t-\pi)} \sin (t-\pi) u(t-\pi)$$
1. Consider the homogeneous case, i.e., $h=0$.
The two linearly independent solutions are $e^{\pm \sqrt{ir} t}$, where we take the principle branch cut of $\sqrt\cdot$. In order to make the solution $L^2[0,\infty)$, you need to discard the exponentially growing solution $e^{\sqrt{ir} t}$ and keep only the decaying one $e^{-\sqrt{ir} t}$. So $\sinh$ and $\cosh$ can not be used alone but only in equal weight pair in each of your terms. Explicitly $$w(t)=w(0)e^{-\sqrt{ir} t},$$ and $w'(0)=-w(0)\sqrt{ir}$. Substitute in the linear form of $w(0)$ and $w'(0)$, we can easily obtain the condition on $a$ which is $a=-\frac{c_1\sqrt{ir}+c_3}{c_2\sqrt{ir}+c_4}$ if the denominator does not vanish, and any $a\in\Bbb C$ if both the denominator and numerator vanish, and does not exist if only the denominator vanishes.
2. Now consider the inhomogeneous case but with homogeneous boundary condition, i.e., $h\not\equiv0$ and $w(0)=w'(0)=0$.
There are two interesting cases for two different Green's functions. The difference of these two Green's function is a homogeneous solution.
2.1 The first Green's function is $$G_1(t)=\frac1{k}\sinh(kt)\Theta(t)$$
where $k=\sqrt{ir}$ and $\Theta$ is the Heaviside step function. $$w(t)=-\int_0^t h(u)G_1(t-u)du.$$ There are many $h$'s (counterexamples saught by the OP) making $w\notin L^2[0,\infty)$ for $G_1$: (1) $h(t)=e^{-at}\in L^2[0,\infty)$ with some positive $a$; (2) any $h\in C[0,\infty)$ compactly supported and positive on $(0,T)$ for some positive $T$.
For $G_1$, to produce $w\in L^2[0,\infty)$ with a nonzero $h\in L^2[0,\infty)\cap C[0,\infty)$, both the real part and imaginary part of $h$ have to alternate their signs over $t\in[0,\infty)$.
2.2 The second Green's function is $$G_2(t)=-\frac1{2k}e^{-k|t|}.$$ Then $$w(t)=-\int_0^\infty h(u)G_2(t-u)du+\int_0^\infty h(u)G_2(-u)du=-h*G_2+(h*G_2)(t=0),$$ where $*$ stands for convolution. The second term is to ensure the null initial condition. $G_2\in L^1[0,\infty)$. By Young's convolution inequality, $$\|h*G\|_2\le \|h\|_2\|G_2\|_1,$$ and $w\in L^2[0,\infty),\, \forall h\in L^2[0,\infty)$.
Note: As indicated before, $G_1-G_2=\frac{e^{kt}}k$ a solution of the homogeneous equation.
Separating out the "explosive" exponentially growing term $e^{kt}$ from $\sinh(kt)$ of $G_1(t)$, we have the following proposition regarding the square integrability of $w$.
3. Proposition 1.: $\forall k\in\Bbb C\,\ni \mathbf{Re}(k)>0 \implies \exists h\in L^2[0,\infty) \ni \big(w[h,a](t):=\int_0^t h(u)e^{k(t-u)}\,du-ae^{kt}\notin L^2[0,\infty),\, \forall a\in\Bbb C-\{0\} \big).$
Proof: Take the Laplace transformation of $w[h,a]$ \begin{align} \mathscr L[w]&=\mathscr L[h]\mathscr L[e^{kt}]-a\mathscr L[e^{kt}], \tag1\\ \mathscr L[e^{kt}](s)&=\frac1{s-k}. \end{align} $\mathscr L[h]\in H^{2+}$ the Hardy function space on the right half complex plane iff $h\in L^2[0,\infty)$. Let $$\mathscr L[h](s)=\frac{s-k}{(s+\alpha)(s+\beta)},$$ or equivalently $$h(t)=\frac{(\beta+k)e^{-\beta t}-(\alpha+k)e^{-\alpha t}}{\beta-\alpha},\tag2$$ for some $\alpha,\beta\in\Bbb C$ where $\mathbf{Re}(\alpha)>0,\mathbf{Re}(\beta)>0$. $\mathscr L[h]\in H^{2+}$ since it is a proper fractional function and its poles are all in the left half complex plane. By Eq. (1), there is always a simple pole at $k$ in the right half complex plane so long as $a\ne0$. Then $\mathscr L\big[w[h,a]\big]\notin H^{2+}$ and $w[h,a]\notin L^2[0,\infty),\,\forall a\ne0$. This can also be verified by direct computation of $w$ with the chosen $h$ with the explicit expression Eq. (2). $\quad\square$
4. Proposition 2: $$k\in\Bbb C\,\ni \mathbf{Re}(k)>0, h\in L^2[0,\infty) \implies \exists !a\in\Bbb C \ni \big(w[a](t):=\int_0^t h(u)e^{k(t-u)}\,du-ae^{kt}\in L^2[0,\infty)\big).$$
Proof: Take the Laplace transformation of $w[h,a]$ \begin{align} \mathscr L[w]&=\mathscr L[h]\mathscr L[e^{kt}]-a\mathscr L[e^{kt}], \tag1\\ \mathscr L[e^{kt}](s)&=\frac1{s-k}. \end{align} $\mathscr L[h]\in H^{2+}$ the Hardy function space on the right half complex plane iff $h\in L^2[0,\infty)$. Let $D(\omega;R)$ be the closed disk centered at $\omega$ and radius $0<R<\mathbf{Re}(k)$. For an arbitrary $x\ge 0$, let $\Omega_x:=\{x+iy\}\setminus D(\omega-k,R)$. We consider two cases.
(1)$k$ is not one of the zeros.
$$\frac{\mathscr L[h]}{s-k}=\frac{\mathscr L[h](s)-\mathscr L[h](k)}{s-k}+\frac{\mathscr L[h](k)}{s-k}$$ $\phi(s):=\frac{\mathscr L[h](s)-\mathscr L[h](k)}{s-k}$. $\phi(s)$ is holomorphic on the closed right half complex plane.
We will prove that $\phi(s)$ is square integrable and uniformly bounded along all vertical line on the right hand side complex plane. $|\phi(s)|$ has a maximum on the compact $D(k,R)$.
$$|\phi(s)|^2\le \frac{|\mathscr L[h](s)|^2}{R^2}+\frac{|\mathscr L[h](k)|^2}{|s-k|^2},\ \forall |s-k|\ge R.$$ We have $$\int_{\Omega_x} |\phi(x+iy)|^2 dy\le \frac1{R^2}\int_{\Omega_x} |\mathscr L[h](x+iy)|^2dy+\frac{\pi |\mathscr L[h](k)|^2}{2\,\max(x,R)}.$$ The integral on the right hand side is bounded uniformly over all $x\ge0$ as $\mathscr L[h]\in H^{2+}$. So now too is the left hand side. We conclude $\phi\in H^{2+}$.
Then setting and only setting $a=\mathscr L[h](k)$ leads to the desired result.
(2) $k$ is one of the zeros.
$\phi(s):=\frac{\mathscr L[h](s)}{s-k}$. $\frac{\mathscr L[h]}{s-k}\in H^{2+}$. Again $\phi$ is holomorphic on $B(k;R)$.
We will prove that $\phi(s)$ is square integrable and uniformly bounded along all vertical line on the right hand side complex plane. $|\phi(s)|$ has a maximum on the compact $D(k,R)$. $$\int_{\Omega_x} |\phi(x+iy)|^2 dy\le \frac1{R^2}\int_{\Omega_x} |\mathscr L[h](x+iy)|^2dy.$$ The integral on the right hand side is bounded uniformly over all $x\ge0$ as $\mathscr L[h]\in H^{2+}$. Then so is the left hand side. We conclude $\phi\in H^{2+}$.
Setting and only setting $a=0$ leads to the desired result.
$\quad\square$
Best Answer
So I start by find the solution to the homogeneous eqn: $$my'+by=0 \to y_h=Ce^{-bt/m}$$
Then, we can determine the inhomogeneous part using a Fourier transform of the original eqn: $$my'+by=\delta(t) \to i\omega m \tilde{y}+b\tilde{y}=1 \to \tilde{y}=\frac{1}{m}\frac{1}{b/m+i\omega}$$
From a table of Fourier transforms $$\frac{1}{b/m+i\omega} \to \theta(t)e^{-bt/m}$$ where $\theta(t)$ is the heaviside step function.
So, $f_i= \theta(t)e^{-bt/m}/m$, which when combined with the homogeneous solution will give the general solution for this differential equation.