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I have two questions about solving PDEs. $L$ is an linear differntial operator

  1. In the complement of the origin, the equation $LE =\delta$ reduces to $LE = 0$. Why?
  2. What can say about solutions of $ LE=k\delta$, where $k$ is a constant, are them solutions of $LE=0$?

Thanks in advance

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If the question is silly for you, you should improve it. – Vobo Jan 3 at 11:39
1. because $\delta|_{\mathbb{R}^n\setminus\{0\}} = 0$ ... 2. $0\neq k\delta$. – Willie Wong♦ Jan 3 at 11:40
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@MAK: If you study PDEs, you should know about "weak solutions". To your first question: For any test function $\varphi$ with support away from the origin, you will have $\delta(\varphi) = 0$, hence $(LE)(\varphi) = 0$. – Vobo Jan 3 at 11:42

1 Answer

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You should not forget the meaning of $\delta$. It is a linear functional on (for instance) the compactly supported smooth functions on $\mathbf R^d$. That is, $LE = \delta$ means for all such $\psi$ that $$\langle LE, \psi \rangle = \langle \delta, \psi \rangle = \psi(0).$$ Also, the support of $\delta$ is $\{0\}$ as can be seen above. Hence, if we test against $\psi$ which vanishes in $0$ we get $$\langle LE, \psi \rangle = 0.$$ Obtaining that $LE = 0$ outside the support of $\delta$, that is, outside $\{0\}$.

Concerning your second question, the question reduces to "which class of functions do you test again", as recall that, distributions are functionals that have a certain domain. If $k \neq 0$, this only works on classes of test functions vanishing in $0$.

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