Setting Up Inverse Problems

[member 428835]

Hi PF!

I'm reading an article and there is a differential equation cast as an operator equation: $$f_n-d_x^2 f_n = \lambda f$$ where $f_n = \partial_n f$, which is derivative of $f$ normal to a given parameterized curve. The author casts the ODE as $$B[f_n] = \lambda A[f_n]:\\ B[f_n] \equiv f_n-d_x^2 f_n,\\A[f_n] \equiv f.$$

So if we simply take $A^{-1}$ and $B^{-1}$ to both sides of the operator equation we have $$A^{-1}[f_n] = \lambda B^{-1}[f_n]$$

However, if we rewrite the operator equation as $$B[f_n] = \lambda f \implies\\ f_n = \lambda B^{-1}f.$$

Recall $A[f_n] \equiv f \implies A^{-1}[f] \equiv f_n$. Then the rewritten operator equation is $$A^{-1}[f] = \lambda B^{-1}[f].$$

Notice one inverse equation operates on $f$ and the other on $f_n$. How can I tell which is correct? (I know it is $f$, just not sure where the logic went wrong).

PS sorry, I can't figure out how to label equations here.

 

[andrewkirk]

There are two hidden assumptions embedded in the above steps:

1. The operators A and B both have inverses.
2. The inverse operators $A^{-1}$ and $B^{-1}$ commute with one another.

The second one is implicitly assumed when you say:

So if we simply take $A^{-1}$ and $B^{-1}$ to both sides of the operator equation we have $$A^{-1}[f_n] = \lambda B^{-1}[f_n]$$

If we apply $A^{-1}B^{-1}$ to both sides of the original equation we get:

$$A^{-1}B^{-1}B[f_n] = \lambda A^{-1}B^{-1} A[f_n]$$
which is
$$A^{-1}[f_n] = \lambda A^{-1}B^{-1}A[f_n]$$
but we can't eliminate $A^{-1}$ and $A$ from the RHS unless $A^{-1}$ commutes with $B^{-1}$.

We run into the same difficulty if we apply $B^{-1}A^{-1}$ to both sides.

Also, to number an equation set it between the delimiters \begin{equation} and \end{equation}. It will then be autonumbered. To number every line in a sequence of equations, use \begin{align} and \end{align}. I've used italics here so that the codes look like codes rather than functioning as codes. Don't use italics when you want to use these codes to make numbered equations.

 

[member 428835]

Thanks for the response! I had two follow-up questions: first, what determines "commutes"? I thought if the operators were linear they commuted; is this not the case?

Secondly, when I used the beginalign and endalign preferences and click PREVIEW it updated the equations on every preview click, so very quickly my equation count was in the 20s. Is this just a bug?

 

[andrewkirk]

Thanks for the response! I had two follow-up questions: first, what determines "commutes"? I thought if the operators were linear they commuted; is this not the case?

That's right. Matrices are linear operators when used to pre-multiply column vectors, but matrix multiplication is in general non-commutative.

A nice geometric example of commutativity failing is for the 2D Euclidean vector space, where linear operator A rotates a point by 90 degrees anti-clockwise around the origin, and linear operator B reflects a point in the y- axis. Then AB takes the point (1,1) to (-1,-1), via (-1,1), but BA takes (1,1) to itself, via (-1,1).
BA is in fact a reflection in the line y=x, while AB is a reflection in the line y=-x.

Secondly, when I used the beginalign and endalign preferences and click PREVIEW it updated the equations on every preview click, so very quickly my equation count was in the 20s. Is this just a bug?

That sounds like a bug in the MathJax engine that implements latex on the web. One solution might be to, once you're happy with the preview, copy the edit version of your draft post to an editor program (eg Notepad on Windows), delete the draft on the PF web page, then start a new post and copy the text back from the editor to the new draft post. That might reset the equation counter. If that fails, you might have to close the PF page on your browser before remaking the post.

 

[member 428835]

Hey, what a good example; thanks! Okay, so I think I have one last question on this matter for you. Let's say I'm solving the inverse problem $$A^{-1}[f] = \lambda B^{-1}[f].$$
We know $$B^{-1}[f] = \int_0^1 G f$$ where I assume $f=f(x):x\in[0,1]$. $G$ is constructed via $B[G] = \delta$, where $\delta$ is the Dirac delta function, and thus $G$ is the Green's function of $B$. When constructing $G$ we need boundary conditions. How would these two situations be different when constructing $G$, say $f_n = 0$ at $x=0,1$ versus $f = 0$ at $x=0,1$? Would the $f_n$ boundary conditions imply $G = 0$ at $x=0,1$ since $B$ itself operates on $f_n$? Then how would we deal with separate problem where boundary conditions are $f = 0$ at $x=0,1$?