Let $m \in \mathbb{N}$. For $k=1,...,m$ let $a_k : \mathbb{R} \rightarrow \mathbb{C}$ be a $C^{\infty}$ function. And suppose that:

$a_m(x) \neq 0 \; \forall x \in [x_0, \infty[$

And let P be the differential operator:

$P(\frac{d}{dx})=\displaystyle{\sum_{k=0}^m} a_k (x) \frac{d^k}{dx^k}$

Given $f \in C(\mathbb{R},\mathbb{C})$ how do I show that the solution to the following differential equation in $D^{'}([x_0,\infty])$, the space of Schwartz distribuitions is unique?

$PU(x)=f(x)$, $\; \; \;$if $x > x_0$, $\; \; \;$ $U^{(k)}(x_0)=0, \; \; k=0,...,m-1$

I've tried proving that the only solution to

$PU(x)=0$, $\; \; \;$if $x > x_0$, $\; \; \;$ $U^{(k)}(x_0)=0, \; \; k=0,...,m-1$

Is the null distribuition but I get stuck. Any hint would be appreciated.


What does it mean for $PU$ to be zero? It means that for every test function $\phi$ $$\left\langle \sum_{k=0}^m a_k U^{(k)},\phi \right\rangle = \left\langle U, \sum_{k=0}^m (-1)^k (a_k \phi)^{(k)} \right\rangle =0 \tag1$$ where the middle part is the definition of the left. If one can prove that every test function $\psi$ can be written as $\sum_{k=0}^m (-1)^k (a_k \phi)^{(k)}$ for some test function $\phi$, then (1) implies $U\equiv 0$. Thus, the problem of uniqueness turns into a problem of existence and regularity for the adjoint equation $$\sum_{k=0}^m (-1)^k (a_k \phi)^{(k)} = \psi \tag2$$ The assumption $a_m\ne 0$ ensures that (2) is an equation of order $m$ with nonvanishing coefficients of the highest derivative: this allows the standard existence/regularity results to apply. To ensure that $\phi$ vanishes for all sufficiently large $x$, solve (2) with homogeneous initial conditions $\phi^{(k)}(b)=0$ for $k=0,\dots,m-1$ where $b$ is such that $\psi = 0 $ on $[b,\infty)$.

  • $\begingroup$ Cool trick, thanks for the help. $\endgroup$ – grizzlyjoker Jun 4 '13 at 8:53
  • $\begingroup$ Just one thing: where do the initial conditions for $U$ come into play? It seems to me that if this result were true then any $U$ such that $PU=0$ would be the null distribution, regardless of the initial conditions. A result that is false ($U=x$ comes to mind, with $P=\frac{d^2}{dx^2}$). I am probably not understanding some detail of your proof, sorry. $\endgroup$ – grizzlyjoker Jun 4 '13 at 16:50
  • $\begingroup$ @grizzlyjoker Integration by parts (1) requires them, otherwise boundary terms would appear. $\endgroup$ – ˈjuː.zɚ79365 Jun 4 '13 at 23:10
  • $\begingroup$ Ah, ok I see it now $\endgroup$ – grizzlyjoker Jun 5 '13 at 9:34

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