# Fundamental solution of a linear differential operator.

Let $L$ be a linear ordinary differential operator $$L(D)=a_nD^n + a_{n-1}D^{n-1}+\dots + a_1 D + a_0.$$ A fundamental solution is a distribution $E$ satisfying $L(D)E=\delta_0$. We want to show that $E$ is of the form $$E=Hu,$$ where $H$ is the Heaviside function and $u$ satisfies $L(D)u=0$.

The hint is to consider the expression $$L(D)\varphi=\frac 1{2\pi}\int \frac{\hat\varphi(-x-ic)}{L(x+ic)}dx$$ from Malgrange-Ehrenpreis construction. I am not good at complex analysis so could anyone please guide me the way?

• Just a remark, there must be a condition somewhere that $u$ is not identically zero. May 30, 2017 at 8:29

Let $u$ be the solution of $L(D)u=0$ with $$u(0)=u'(0)= … =u^{(n-2)}(0)=0, \quad u^{(n-1)}(0)=\frac1{a_n}.$$ The multiple root at zero can also be expressed in the factorization $u(x)=x^{n-1}v(x)$ with a smooth function $v$, $v(0)=\frac{1}{(n-1)!\,a_n}$.
Then $Hu$ has the correct jump in the $(n-1)$th derivative at $x=0$ with all lower derivatives continuous and zero so that $L(D)(Hu)=δ_0$.