Continuity of $f(t)$ from $\frac{df}{dt}+kf(t)=\delta (t)$

Question

Consider the differential equation $$\frac{df}{dt}+kf=\delta (t)$$ where $k$ is constant. I'm asked to give statement about the continuity of $f(t)$ and $f'(t)$ at $t=0$.

I tried solving the equation using Laplace transform seeing the $\delta $ function, it seems much convenient. $$\mathcal{L}\left(\frac{df}{dt}+kf\right)=\mathcal{L}(\delta(t))$$ $$s\mathcal{L}(f)-f(0^+)+k\mathcal{L}(f)=1$$ $$\mathcal{L}(f)=\frac{1+f(0^+)}{s+k}$$ $$f(t)=(1+f(0^+))\mathcal{L^{-1}}\left(\frac{1}{s+k}\right)=(1+f(0^+))e^{-kt}$$ But looking on the function, I don't How can I say about it anything. For example $$\lim_{t\rightarrow 0^+}f(t)=f(0^+)=(1+f(0^+))e^0\Rightarrow 1=0$$ which is complete non-sense. I don't understand what I'm missing. Can any help me figure it out?

Answer 1

On $\Bbb R\setminus\{0\}$, we have to deal with $$ f'(t)+kf(t)=0,$$ i.e., $$ f(t)=Ce^{-kt}$$ where $C$ is locally constant, i.e., ptoentially different constants $C_+$ and $C_-$ for $(-\infty)$ and $(0,\infty)$. Now $f$ is continuous at $0$ only when $C_+=C_-$ (and $=f(0)$). You want to check what happens when $C_+\ne C_-$ and $f'(0)$ only exists in the distributional sense.

Answer 2

The direct answer without solving anything is that the primary source for the singularity on the right side is the highest derivative on the left. So for $f'$ to give $\delta$ then $f$ must have a jump of size $1$ at $t=0$ and consequently $f'$ a jump of $-k$.

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To quickly solve this, apply the integrating factor and sifting property of delta to get $$ (e^{kt}f(t))'=e^{k·0}δ(t) $$ Now use that delta integrates to the unit jump to get $$ e^{kt}f(t)-f(0^+)=u(t)-u(0^+),\\ f(t)=(u(t)-1+f(0^+))e^{-kt}. $$ To ensure that $f(t)=0$ for $t<0$ one needs that $f(0^+)=1$.

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The Laplace transform is mostly concerned with things that happen at $t>0$. Using singularities like jumps or Dirac deltas at $t=0$ is just another form of encoding initial conditions. The usual treatment is to integrate a short way formally $$ f(ε)-f(-ε)+k\int_{-ε}^εf(s)\,ds=1\implies f(0^+)=1+f(0^-). $$ One could also just shift the Dirac delta some small distance $ε$. If shifting to the left, it leaves the scope of the Laplace transform and one is left with a transformed equation $$ sF(s)-f(0^+)+kF(s)=0. $$ Shifting the Dirac pulse to the right to some position $ε$ results in the transform $$ sF(s)-f(0^-)+kF(s)=e^{-εs},~~~ε\to 0 $$ Both interpretations lead to consistent results.