State the definitions of the space $𝒟(ℝ)$ of test functions and of the space $𝒟'(ℝ)$ of distributions. For a distribution $u ∈ 𝒟'(ℝ)$ define its distributional derivative $u'$ and explain how it is an extension of the usual concept for $\mathrm{C}^1$ functions.
For a $\mathrm{C}^{∞}$ function $a: ℝ → ℝ$ and a distribution $u ∈ 𝒟'(ℝ)$ define the product $a u$, and prove Leibniz' rule: $(a u)'=a u'+a' u$.
Assume that $u ∈ 𝒟'(ℝ)$ satisfies $u'=0$ in $𝒟'(ℝ)$. Prove that $u=c$ for some constant $c$. [You may assume that there exists a non-negative $θ ∈ 𝒟(ℝ)$ with $∫_{ℝ} θ \mathrm{d} x=1$]

Let $f ∈ \mathrm{L}_\text{loc}^1(ℝ)$ with $f(x)=0$ for almost all $x<0$ and define \[ F(x)=∫_{-∞}^x f(t) \mathrm{d} t,   x ∈ ℝ . \] Show that $F$ is continuous and that $F'=f$ in $𝒟'(ℝ)$.

To show that $F$ is continuous it suffices to show that $F(x_j) → F(x)$ whenever $x_j → x$. For $x ≦ 0$ we have $F(x)=0$ and so there is nothing to prove for $x<0$. We have \begin{aligned} & 𝟏_{(0, x_j)} f → 𝟏_{(0, x)} f \text{ a.e., } \\ & 𝟏_{(0, x_j)} |f| ≦ 𝟏_{(0, \sup _j x_j)}|f| ∈ L^1(ℝ), \end{aligned} and so $F(x_j) → F(x)$ by DCT. Thus $F$ is continuous and so in particular $F ∈ 𝒟'(ℝ)$.
For $φ ∈ 𝒟(ℝ)$ compute: \begin{aligned} ⟨F,-φ'⟩&= -∫_{-∞}^{∞} ∫_{-∞}^x f(t)\mathrm d t\ φ'(x)\mathrm d x \stackrel{\text{ Fubini}}{=}-∫_{-∞}^{∞} ∫_t^{∞} φ'(x)\mathrm d x\ f(t)\mathrm d t \\ & \stackrel{\text{FTC}}{=} ∫_{-∞}^{∞} φ(t) f(t)\mathrm d t=⟨f, φ⟩ . □ \\ & \end{aligned}

Denote by $δ_0$ Dirac's delta function at 0 and by $g: ℝ → ℝ$ a fixed bounded and continuous function. Find the general solutions to each of the following ordinary differential equations in $𝒟'(ℝ)$ : \begin{align} v'+v=δ_0 \\ u''+2 u'+u=δ_0 \\ w''+2 w'+w=g \end{align} What would the general solution to (2) be if we instead considered the equation in $𝒮'(ℝ)$ of tempered distributions? [No detailed justification is required for the general solution of (2) in $𝒮'(ℝ)$.]

  1. From Leibniz we get \[ δ_0=e^x δ_0=e^x v'+e^x v=(e^x v)' , \] and so $e^x v=H+c$ by (a) and since \[ H'=δ_0  (H=\text{Heaviside's function}) \] general solution is therefore $v=c e^{-x}+H e^{-x}$, where $c ∈ ℝ$.
  2. Note first that $(\frac{d}{d x}+I)^2=\frac{d^2}{d x^2}+2 \frac{d}{d x}+I$ so that from (1):$$u'+u=c_1 e^{-x}+H(x) e^{-x}$$ or$$(e^x u)'=c_1+H$$ whereby, from (b), $x^{+}=∫_{-∞}^x H(t) d t$ satisfies $(x^{+})'=H$, so$$(e^xu-x^{+})'=c_1$$ so$$(e^x u-x^{+}-c_1x)'=0$$ by (a)$$e^x u-x^{+}-c_1 x=c_2, c_2 ∈ ℝ$$ But then general solution is $u=(c_1 x+c_2) e^{-x}+x^{+} e^{-x}$, where $c_1, c_2 ∈ ℝ$.
  3. Put $E=x^{+} e^{-x}$. Then $E$ solves (2) and if $w=w(x)=(E * g)(x)=∫_0^{∞} g(x-y) y e^{-y} d y,\ x ∈ ℝ$, then $w$ is continuous and \[ w''+2 w'+w=(E''+2 E'+E) * g=δ_0 * g=g \] so that $E * g$ is a particular solution to (3). From calculations in (2) we get general solution $w=(c_1 x+c_2) e^{-x}+∫_0^{∞} g(x-y) y e^{-y}\mathrm d y$ where $c_1, c_2 ∈ ℝ$.
Finally if we seek general solution for (2) in $𝒮'(ℝ)$ we only get $x^{+} e^{-x}$ since $(c_1 x+c_2) e^{-x} ∈𝒮'(ℝ)$ iff $c_1=c_2=0$.