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 again something from Stein-Shakarchi I would really appreciate some help with. Any references are also welcome!

Suppose $L$ is a linear partial differential operator with constant coefficients. Show that when $d \geq 2$, the linear space of solutions $u$ of $L(u)=0$ with $ u \in C^{\infty}(\mathbb{R}^{d})$ is not finitely dimensional.

Thanks in advance!

EDIT: $L$ takes the form $$L= \sum_{|\alpha| \leq n}{a_{\alpha}\left(\frac{\partial{}}{\partial{x}}\right)^{\alpha}}$$ with $a_{\alpha} \in \mathbb{C}$ constants.

share|improve this question
    
How is a partial differential operator defined? –  Davide Giraudo May 2 '12 at 8:02
    
Edited. Sorry for not being thorough. –  Anna May 2 '12 at 23:58
add comment

1 Answer

For $p=(p_1,p_2)\in\Bbb C^2$, define $f_p(x)=\exp\left(p_1x_1+p_2x_2\right)$. If $(p_1^{(k)},p_2^{(k)})\in\Bbb C^2$ are distinct (i.e. $(p_1^{(k)},p_2^{(k)})\neq (p_1^{(j)},p_2^{(j)})$ if $j\neq k$, then the set of maps $\{f_{p^{(j)}},1\leq j\leq N\}$ is linearly independent. Indeed, if $\sum_{j=1}^Nc_jf_{p^{(j)}}$, let $J_1,\ldots,J_l$ a partition of $\{1,\ldots,N\}$ such that for all $j_1,j_2\in J_k,p_1^{(j_1)}=p_1^{(j_2)}$. Then we have $$\forall x_2\in\Bbb R,\forall 1\leq k\leq l,\quad \sum_{j\in J_k}c_je^{p_2^{(j)}x_2}=0.$$ Since $p_2^{(j)}$ are distinct, we get that $c_j=0$ for all $j\in J_k$ hence $c_j=0$ for all $j$.

Now, we just use the fact that a polynomial of two variables has infinitely many zeros.


It's a huge contrast with $d=1$ (in this case the space is necessary finite dimensional).

share|improve this answer
add comment

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.