SVD of a differential operator

In summary, the conversation discusses finding the eigenfunction and singular value decomposition for a given operator in \mathbb{L}_2([0,1]). The steps for finding the eigenfunction and singular values are provided, along with a suggestion for a good introductory book on the topic.
  • #1
DivGradCurl
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I'm just checking my work; please correct me if I'm wrong. Also, if someone has a suggestion for an intro book that covers SVD of operators, please let me know! I'm not fully confident with my procedure for general problems yet, so the more info, the better. Many thanks.

Homework Statement



Given

[tex]\mathscr{D}\, f = \frac{df}{dx} + cf [/tex]

defined in [itex]\mathbb{L}_2([0,1])[/itex] with [itex]f(0)=f(1)=0[/itex], obtain the eigenfunction for the operator and its singular value decomposition.

Homework Equations



N/A

The Attempt at a Solution



1) Test this function:
[tex]f(x)=A\sin (Bx)[/tex]
If [itex]f(0)=A\sin (B0) = 0[/itex] OK
If [itex]f(1)=A\sin (B) = 0 \therefore B = m\pi[/itex]
Then [itex]f(x)=A\sin (m\pi \, x)[/itex] satisfies the boundary conditions and is an eigenfunction

2) Find [itex]A[/itex] by taking the inner product of the eigenfunction [itex]f(x)=A\sin (m\pi \, x)[/itex] with itself in [itex]\mathbb{L}_2([0,1])[/itex]:

[tex]\int _0 ^1 \left[ A \sin (m\pi \, x) \right] \left[ A \sin (m\pi \, x) \right] \, dx = 1[/tex]
[tex]\frac{A^2}{2(m\pi \, x)}\left[ m\pi \, x - \cos (m\pi \, x) \sin (m\pi \, x) \right] = 1[/tex]
[tex]\frac{A^2}{2} = 1 \therefore A = \pm \sqrt{2}[/tex]

Arbitrarily choose the positive root, so [itex]f(x)=\sqrt{2}\sin (m\pi \, x)[/itex].

3) Use the eigenvalue equation for the [itex]\mathscr{D} ^{\dagger} \mathscr{D}[/itex]:

[tex]\mathscr{D} ^{\dagger} \mathscr{D} \, \vec{u}_n = \mu_n \vec{u}_n[/tex]

That is, if

[tex]\mathscr{D}\, f = \frac{df}{dx} + cf = \sqrt{2} (m\pi) \cos (m\pi x) + c \sqrt{2} \sin (m\pi x) = f^{\prime} [/tex]

then because in this case [itex]\mathscr{D} ^{\dagger} = \mathscr{D}[/itex]

[tex] \mathscr{D}^{\dagger}\, \mathscr{D}\, f = \mathscr{D}\, f^{\prime} = \frac{df^{\prime}}{dx} + cf^{\prime} = 2\sqrt{2} c(m\pi) \cos (m\pi x) + (c^2 - m^2 \pi ^2) \sqrt{2} \sin (m\pi x) [/tex]

But if [itex]u_n = f(x) =\sqrt{2} \sin (m\pi x) [/itex] is one of the two singular vectors, then the cos(.) term must be set to zero (VALID STEP?), and we have [itex]\mu_n = c^2 - m^2 \pi ^2[/itex] as the singular value.

To find the other singular vector, use the relationship:

[tex]v_n = \frac{1}{\sqrt{\mu_n}} \mathscr{D} u_n[/tex]

so


[tex]v_n = \frac{1}{\sqrt{c^2 - m^2 \pi ^2}} \sqrt{2}\left[ (m\pi) \cos (m\pi x) + c \sin (m\pi x) \right][/tex]

and this completes the SVD.
 
Last edited:
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  • #2
Yes, your procedure is correct. As for a good introductory book on the SVD of operators, I would recommend the book "Linear Algebra and its Applications" by Gilbert Strang. It has a good introduction to the topic, with plenty of examples and exercises.
 

1. What is the SVD of a differential operator?

The Singular Value Decomposition (SVD) of a differential operator is a mathematical technique used to break down a linear differential operator into simpler, more manageable components. It involves representing the operator as a product of three matrices: the left singular matrix, the diagonal matrix of singular values, and the right singular matrix. This decomposition allows for a better understanding and analysis of the operator.

2. Why is the SVD of a differential operator useful?

The SVD of a differential operator is useful because it provides a way to solve differential equations that cannot be solved by traditional methods. It also allows for a better understanding of the behavior of the operator and its effects on a given system. Additionally, it can be used for data compression and noise reduction in applications such as image and signal processing.

3. How is the SVD of a differential operator calculated?

The SVD of a differential operator can be calculated using various numerical methods such as the power method, the inverse power method, and the QR algorithm. These methods involve iterating through the matrix representations of the operator to find its eigenvalues and eigenvectors, which are then used to construct the singular value decomposition.

4. What are the applications of the SVD of a differential operator?

The SVD of a differential operator has many applications in mathematics, engineering, and other scientific fields. Some examples include system identification, model reduction, control theory, and image and signal processing. It is also used in machine learning algorithms and data analysis techniques.

5. Can the SVD of a differential operator be extended to non-linear operators?

Yes, the SVD can be extended to non-linear operators through the use of the generalized SVD (GSVD) or the extended SVD (ESVD). These techniques involve representing the non-linear operator as a linear operator in a higher-dimensional space, allowing for the use of traditional SVD methods. However, the calculation of the GSVD or ESVD can be more complex and computationally expensive compared to the SVD of a linear operator.

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