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Two part question: If I define a linear operator $L=\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$ and then use that operator on a variable, say $L\alpha$ then is it accurate to say that the $L$ maps $\alpha$ onto itself like this: $L\alpha=\frac{\partial^2}{\partial x^2}\alpha + \frac{\partial^2}{\partial y^2}\alpha$. Is that how we talk about operators and that we can get rid of the variable that's being acted on and just talk about the operation that's being performed?

Let's say I have an equation that has $p\alpha^2$. If I wanted to write the operator for this, how would I do it?

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For the first part of your question: indeed, if you define $L$ like that, the expression $L \alpha = \frac{\partial^2}{\partial x^2} \alpha + \frac{\partial^2}{\partial y^2} \alpha$ is correct. What is not so right is saying that "$L$ maps $\alpha$ to itself - it doesn't, in the same sense the derivative of a function is a new function, usually different from the original function.

Which brings me to the next point: I'm not too sure of how you interpret the term "variable" here, but in this case $\alpha$ must be a function, and not a regular variable. The idea behind working with operators like this is not so much talking about the operation being performed, but rather talking about properties the operator as such might have. It is, in a way, a generalization of finite-dimensional linear algebra. There you have a linear equation, say $Tx = y$, and you are concerned with what the eigenvalues of $T$ look like and how they determine the kind of possible solutions to that equation. It's similar here, you can still find the eigenvalues of the Laplacian. But the function spaces we deal with here are infinite-dimensional.

For the second part of your question: what is $p$? Is $\alpha$ still understood to be a function, as in the previous part?

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Yes, $\alpha(x,y)$ and $p$ is a constant. I apologize for the ambiguity and appreciate your response :). – user6390 Feb 1 '11 at 16:59
Ah, in this case the specified operation is not linear, so you can't factor the operator so "nicely" out of it, like you did with the Laplacian. It is actually defined by that functional equation. Glad to be of assistance. :) – Jose L. Lykón Feb 1 '11 at 17:21

You can think of an operator as being a map from some space of functions to itself or to another space of functions. For instance, if you are considering only smooth real-valued functions on ${\mathbb{R}}^2$, you might look at operators $O: C^\infty(\mathbb{R}^2)\rightarrow C^\infty(\mathbb{R}^2)$. If the domain and range are both vector spaces (over ${\mathbb{R}}$, say), then the linear operators are just those maps that satisfy $$ a(O\cdot f) + b(O\cdot g) = O\cdot(af + bg) $$ for all $a,b\in\mathbb{R}$ and functions $f$ and $g$. It is straightforward to verify that differential expressions like $\partial_x^{2} + \partial_y^{2}$ are linear operators on the appropriate function spaces. Nonlinear operators can also be defined, like the one you mentioned, by specifying their effect on arbitrary functions: e.g., $(O \cdot f)(x) = p f(x)^2$.

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