Solution to Distributional Ordinary Differential Equation with Constant Coefficients

Theorem
Let $D$ be an ordinary differential operator with constant complex coefficients:


 * $\ds D = \sum_{k \mathop = 0}^n a_k \paren {\dfrac \d {\d x}}^k$

Let $f \in \map {\CC^\infty} \R$ be a smooth real function.

Let $T \in \map {\DD'} \R$ be a distribution.

Let $T_f$ be a distribution associated with $f$.

Suppose $T$ is a distributional solution to $D T = T_f$.

Then $T = T_F$ where $F \in \map {\CC^\infty} \R$ is a classical solution to $D F = f$.

Proof
Let $\map P \xi$ be a polynomial over complex numbers such that:


 * $\ds \map P \xi = \sum_{k \mathop = 0}^n a_k \xi^k = a_n \prod_{k \mathop = 0}^n \paren {\xi - \lambda_k}$

where $a_n \ne 0$.

Then there exists a polynomial $\map Q \xi$ such that:


 * $\map P \xi = \paren {\xi - \lambda_n} \map Q \lambda$

Let:


 * $\ds D = \sum_{k \mathop = 0}^n a_k \paren {\dfrac \d {\d x}}^k$

Then:


 * $D = \map P {\dfrac \d {\d x} }$

Furthermore:


 * $D = \paren {\dfrac \d {\d x} - \lambda_n} D_1$

where:


 * $D_1 := \map Q {\dfrac \d {\d x} }$

Now we will use the principle of mathematical induction to show that:


 * $\paren {DT = T_f, f \in \map {\CC^\infty} \R} \implies \paren {T = T_F, F \in \map {\CC^\infty} \R}$

Basis for the Induction
$n = 1$ case is covered by Dx - k is Hypoelliptic.

Induction Hypothesis
Suppose for $D$ of the order $n \in \N$ it holds that:


 * $\paren {DT = T_f, f \in \map {\CC^\infty} \R} \implies \paren {T = T_F, F \in \map {\CC^\infty} \R}$

We will show that the same property holds for $D$ of the order $n + 1$.

Induction Step
Let $D$ be of the order $n + 1 \in \N$.

Let:


 * $D = \paren {\dfrac \d {\d x} - \lambda_{n + 1}} D_1$

where $D_1$ is of the order $n \in \N$.

Then:


 * $ \paren {\dfrac \d {\d x} - \lambda_{n + 1}} D_1 T = T_f$

By Dx - k is Hypoelliptic:


 * $\paren {\paren {\dfrac \d {\d x} - \lambda_{n + 1} } D_1 T = T_f, f \in \map {\CC^\infty} \R} \implies \paren {D_1 T = T_g, g \in \map {\CC^\infty} \R : \paren {\dfrac \d {\d x} - \lambda_{n + 1} } g = f}$

By the induction hypothesis:


 * $\paren {D_1 T = T_g, g \in \map {\CC^\infty} \R} \implies \paren {T = T_F, F \in \map {\CC^\infty} \R : D_1 F = g}$

Just to make sure:

Hence, in the distributional sense we have that $D T = T_f$ is solved by $T = T_F$ where $D F = f$ and $F \in \map {\CC^\infty} \R$.