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 $f \in \map {C^\infty} \R$ 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 $F \in \map {C^\infty} \R$ is a classical solution to $D F = f$.
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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:
- $\paren {DT = T_f, f \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R}$
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:
- $\paren {DT = T_f, f \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R}$
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$
- $\paren {\paren {\dfrac \d {\d x} - \lambda_{n + 1} } D_1 T = T_f, f \in \map {C^\infty} \R} \implies \paren {D_1 T = T_g, g \in \map {C^\infty} \R : \paren {\dfrac \d {\d x} - \lambda_{n + 1} } g = f}$
By the induction hypothesis:
- $\paren {D_1 T = T_g, g \in \map {C^\infty} \R} \implies \paren {T = T_F, F \in \map {C^\infty} \R : D_1 F = g}$
Just to make sure:
\(\ds D F\) | \(=\) | \(\ds \paren {\dfrac \d {\d x} - \lambda_{n + 1} } D_1 F\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\dfrac \d {\d x} - \lambda_{n + 1} } g\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f\) |
Hence, in the distributional sense we have that $D T = T_f$ is solved by $T = T_F$ where $D F = f$ and $F \in \map {C^\infty} \R$.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.4$: A glimpse of distribution theory. Multiplication by $C^\infty$ functions