Núcleo del operador diferencial

Question

Supongamos que tenemos una ecuación diferencial lineal: %#% $ de #% donde $$\left[D^{n}+a_{1}D^{n-1}+\dots+a_{n}D^{0}\right]y(t)=0$ es una función analítica.

¿Cómo podemos demostrar que el núcleo del operador diferencial $y$ $ tiene dimensión $$\left[D^{n}+a_{1}D^{n-1}+\dots+a_{n}D^{0}\right]$?

Muchas gracias.

Answer 1

Que llamamos la expresión $$a_nD^n+a_{n-1}D^{n-1}+...+a_2D^2+a_1D+a_0~~~(*)$$ a linear $ n$th-order differential operator. Since $(*)$ is a polynomial in symbol $D$, somone can denote it as $P(D)$. It seems you know that when all $a_i$ are constant, $P(D)$ can be factored into differential operators of lower order, so that makes us to treat $P(D)$ as an ordinary polynomial.However, these extracted factors can commute with each other. Now remind The Fundamental Theorem of Algebra about $P(D)$. For example, if $$(D^2+5D+6)y=0$$ then $$(D+3)(D+2)y=0$$ Or when $$[D^2-2\alpha D+(\alpha^2+\beta^2)]^n$$ annihilate $y(x)$, then $y(x) $ contains a linear combination of the following independent functions: $% $ $\{\text{e}^{\alpha x}\cos(\beta x),x\text{e}^{\alpha x}\cos(\beta x),...x^{n-1}\text{e}^{\alpha x}\cos(\beta x),\text{e}^{\alpha x}\sin(\beta x),x\text{e}^{\alpha x}\sin(\beta x),...x^{n-1}\text{e}^{\alpha x}\sin(\beta x)\}$