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I can't get my head around the following term:

$(\frac{\partial}{\partial t} + v_o\nabla)^2 p(t,x)$

I know what $\frac{\partial}{\partial t}p(t,x)$ is supposed to mean and I assume that $v_o\nabla p(t,x)$ is simply a product of $v_o$ with $p(t,x)$ derivated in space. But how can I solve the bracket?

ruhig brauner
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  • Without an equality, it's not really possible to solve anything, so I don't know what you mean by that. Everything in the parenthesis are just operators acting on $p(x,t)$. You still need to square the operators (using FOIL or pascals triangle). $\nabla^2$ represents the spread (divergence) of the derivative of $p(x,t)$. The sole time derivative term is to the second power, not first, and more or less signifies the acceleration of $p(x,t)$. – infinitylord Dec 01 '16 at 18:02
  • But can I simplify the term, remove the brackets? Initially, I would just solve it with good old $(a + b)^2 = a^2 + ab + ba + b^2$ but I'm really not used to this kind of maths so I think my assumption is quite off. I don't really want to post the full issue because I feel i can solve the rest on my own. – ruhig brauner Dec 01 '16 at 18:05
  • In short, yes, you can expand out $(a+b)^2$ with operators just as you did. Operators can be a little tricky in the sense that they don't always commute. Clairaut's theorem guarantees that if the function $p(x,t)$ is continuously differentiable, the partial derivatives will commute. Assuming $v_0$ is a constant, you're all set. – infinitylord Dec 01 '16 at 18:09

1 Answers1

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First of all you have to consider that what you have inside the bracket is an operator.

This means that you cannot compute the normal square of the two objects like $a^2 + b^2 + 2ab$, but you will have to write $ab$ and $ba$ differently, because they will be two different objects.

To see what I mean, let's expand the square:

$$\left(\frac{\partial}{\partial t} + v_0\nabla\right)^2 = \frac{\partial^2}{\partial t^2} + v_0^2 \nabla^2 + v_0\frac{\partial}{\partial t}\nabla + v_0 \nabla\frac{\partial}{\partial t}$$

In which my only assumption has been that $v_0$ is a scalar, that is, a number.

So, in another way, you can collect it and get

$$\frac{\partial^2}{\partial t^2} + v_0^2 \nabla^2 + v_0\left(\frac{\partial}{\partial t}\nabla + \nabla\frac{\partial}{\partial t}\right)$$

Hence, when acting of the function $p(t,x)$ you will get

$$\frac{\partial^2 p(x, t)}{\partial t^2} + v_0^2 \nabla^2p(t, x) + v_0\left(\frac{\partial}{\partial t}\nabla + \nabla\frac{\partial}{\partial t}\right)p(t, x)$$

And the last fracked will obviously mean

$$v_0\left(\frac{\partial}{\partial t}\nabla p(t, x) + \nabla\frac{\partial}{\partial t}p(t, x)\right)$$

Then it clearly depends upon the function $p$ you have. Under certain conditions, the two operators will commute that is

$$\frac{\partial}{\partial t}\nabla = \nabla \frac{\partial}{\partial t}$$

And in this case you will be the normal double product, but it's not always guaranteed.

Enrico M.
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