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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?

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

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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 $L(D)(Hu)=δ_0$.

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  • $\begingroup$ Brilliant! Thank you very much for the answer. Still, I would like to know how to solve this using the 'hint' (although I like your method more). Do you happen to have an idea? $\endgroup$
    – BigbearZzz
    May 30, 2017 at 9:50

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