A Why Must the Constant in Hilbert Space Function B[f(x)] Be Defined as Shown?

[member 428835]

Hi PF!

Given a function $B$ defined as $$B[f(x)]\equiv f''(x) + f(x) + const.$$ Evidently in order for this function to be in the real Hilbert space $H$ we know $$const. = -\frac{1}{x_1-x_0}\int_{x_0}^{x_1} (f''(x) + f(x))\,dx.$$ Can someone please explain why?

I can elaborate further if more info is needed! PLEASE!

 

[member 587159]

What are $x_0, x_1$? On what Hilbert space are you working? (I.e. what set, what inner product?)

 

[member 428835]

What are $x_0, x_1$?

Sorry, $x\in[x_0,x_1]:x_0,x_1\in R$. So they are real numbers.

 

[member 587159]

Sorry, $x\in[x_0,x_1]:x_0,x_1\in R$. So they are real numbers.

And on what Hilbert space (= what is the set H) with what inner product are we working?

 

[member 428835]

And on what Hilbert space (= which set) with what inner product are we working?

$L_2(s)$ engendered with scalar product $(u,v) = \int_s uv$ for ease of notation, where I define $s = [x_0,x_1]$.

 

[member 587159]

Okay, you should have included this information from the beginning.

If $B$ is in $L_2$, the inner product $(B,1)$ must exist, right?

Write out this inner product. What can you conclude?

EDIT: this might be the wrong approach. Is the constant in line 1 the same as the constant in the second line?

I think the question is ill posed, if this is the exact wording.

 

[member 428835]

EDIT: this might be the wrong approach. Is the constant in line 1 the same as the constant in the second line?

I think the question is ill posed, if this is the exact wording.

The two constants are the same. I have summarized from the paper, but I don't think I've missed anything. Here's the link: https://link.springer.com/content/pdf/10.1007/BF01013829.pdf
and you'll see my question just after (2.12).

Edit: I'm waiting your response before following your initial instructions.

 

[member 428835]

At any rate, what we can conclude is that $(B,1)=0$.

 

[member 587159]

At any rate, what we can conclude is that $(B,1)=0$.

Well, if you say that $(B,1) = 0$ (I don't see why, but it is you who has to understand it ;)), it follows that

$0 = \int_c (f''(x) + f(x) + constant) dx = \int_c (f''(x) + f(x))dx + constant (x_1 - x_0)$

from which the desired equality follows.

 

[member 428835]

Well, if you say that $(B,1) = 0$ (I don't see why, but it is you who has to understand it ;)), it follows that

$0 = \int_c f''(x) + f(x) + constant dx = \int_c f''(x) + f(x)dx + constant (x_1 - x_0)$

from which the desired equality follows.

Sorry, I meant if the constant is equal to the average of $B[f(x)]$ along $s$. I don't understand why (before knowing that) the inner product should be zero.

The author uses the phrase "clearly" when defining the constant, so I'm assuming it's evident to most people?

 

[member 587159]

I don't know. I am no physicist and I also didn't read the paper. Sometimes people write clearly because they are too lazy to explain thinhs. But the result seems to follow if you manage to show that that inner product is 0 (the claim is equivalent with it), so you might want to investigate why that is true.

 

[member 428835]

Thanks for your help!

 

[George Jones]

At any rate, what we can conclude is that $(B,1)=0$.

I am not sure about your notation, as $B$ is an operator on the Hilbert space, with action defined by equation (2.8). Just before (2.8), the paper states "Let us consider in the space $H$ the seff-adjoint operator $B$ ...".

Given a function $B$

Again, $B$ is an operator on Hilbert space $H$, not a function in $H$ (actually, equivalence classes of functions, but I will stick to representatives).

defined as $$B[f(x)]\equiv f''(x) + f(x) + const.$$

I think (a simplified version of) equation (2.8) defines $B$ as
$$Bf = f - f' -\frac{1}{x_1 - x_0} \int^{x_1}_{x_0} \left( f\left(x\right) - f'\left(x\right) \right) dx$$

Now define the function $u$ by $u\left(x\right) = 1$ for every $x$. Then, $Bu = 0$, i.e., $B$ acting on the constant unity function gives the constant zero function. I think that this is what you mean by $\left( B , 1 \right) =0$.

$$const. = -\frac{1}{x_1-x_0}\int_{x_0}^{x_1} (f''(x) + f(x)) dx.$$ Can someone please explain why?

As @Math_QED has already written, this is just a property of a definite integral.

What follows is my guess at things, which could be really, really wrong.

I think that your questions is "Why is the operator $B$ defined such that it spits out zero when it operates on any constant function?"

The paper defines $f$ as "the magnitude of the deviation of the free surface $S'$ from the equilibrium surface $S$". Also, after (2.13), the paper characterizes $B$ as a potential energy operator.

To simplify things, consider a horizontal (guitar) string that has fixed endpoints. The string as a straight horizontal line is in the equilibrium position. The string in this configuration has no potential to do (vibrational) work. Now pull one point of the string up. In this configuration, the string has the potential to do work, i.e., if you let go, the string will vibrate.

Let $f$ represent, at any $x$ along the string, the vertical displacement from the original horizontal equilibrium. If $f$ is constant along the entire length of the string, we have just moved the entire string up our down by a constant amount, and the string still has the shape of a straight horizontal line, and this configuration still has no potential to vibrate.

Consequently, if $B$ is potential energy, and $f$ is displacement, we want $Bf=0$ when $f$ is a constant function.

 

[member 428835]

You reworded my question and I think gave a very good intuitive answer? Thanks!

 

[zinq]

B[f(x)] ≡ f′′(x) + f(x) + const.

This doesn't make sense as written. B is evaluated at the point f(x), so there seems no way that B could depend on f'(x) and f''(x) (unless these are functions of f(x), which is very rarely the case).