# equivalence of two linear PDE

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• Jul 20th 2011, 03:43 AM
Sogan
equivalence of two linear PDE
Hello,

i try to understand some basic theory about linear PDE with constant coefficients.
But my knowledge about PDEs and the notation used in books is not so good.(because i have just started to study this topic).

i don't understand this statement or perhaps just the notation :

"for any $\displaystyle a \in \mathbb{R}^n$, solving $\displaystyle P(D)u=f$ on$\displaystyle B_{R} = \{ x \in \mathbb{R}^n : \left|x\right| < R\}$ is equivalent to solving$\displaystyle P(D+a)v=g$, where $\displaystyle v=exp(-iax)*u$and $\displaystyle g=exp(-iax)f$."

Ok i don't understand, why this is true. I think the author mean by P(D) some general linear PDE with constant coefficients, that is $\displaystyle \sum_{\left| \alpha \right|\le k} b_{\alpha}*D^{\alpha}u=f$ with$\displaystyle b_{\alpha}$ constant coefficients.

So but what i don't know how to understand P(D+a)?
And why are both PDEs equivalent? Do you have an idea?

Regards