Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
If the question is silly for you, you should improve it. –  Vobo Jan 3 '13 at 11:39
    
1. because $\delta|_{\mathbb{R}^n\setminus\{0\}} = 0$ ... 2. $0\neq k\delta$. –  Willie Wong Jan 3 '13 at 11:40
1  
@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 '13 at 11:42

1 Answer 1

up vote 2 down vote accepted

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 $0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.