What Is the Adjoint of the Curl Operator in Differential Equations?

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In summary, the adjoint operator L* of the first order differential operator L = curl is found to be self-adjoint for the given inner product, meaning L* = L. The adjoint is found by integrating by parts for each component of the operator and then re-building it into a curl.
  • #1
shreddinglicks
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1. The problem statement, all variables and given/k
Find the adjoint operator L* to the first order differential operator L = curl defined to domain omega. The full problem is attached.

Homework Equations

The Attempt at a Solution


I've checked online. I am getting two different answers. Is the solution the conjugate transpose or the adjugate matrix?
 

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  • #2
Adjoint is defined relative to inner product. What is your inner product? I.e. your operator operates on some functions. What is the inner product of these functions? Also, what is domain ##\Omega##?
 
  • #3
The inner product would be:

∫u(x)*v(x) dx from -∞<Ω<∞
 
  • #4
In your scan I can see things being differentiated with respect to ##x, y, z## so how come your inner product is for functions of just one variable?

My strong suspicion is that ##\Omega## is the solid angle domain, which would make it two-dimensional. The reason I think so is because ##\mathbf{\nabla}\times\vec{r}=\vec{0}##, so the operator commutes with the 'position' operator, i.e. your ##L## operator acts on angles only.
 
  • #5
Cryo said:
In your scan I can see things being differentiated with respect to ##x, y, z## so how come your inner product is for functions of just one variable?

My strong suspicion is that ##\Omega## is the solid angle domain, which would make it two-dimensional. The reason I think so is because ##\mathbf{\nabla}\times\vec{r}=\vec{0}##, so the operator commutes with the 'position' operator, i.e. your ##L## operator acts on angles only.
I'm not sure I understand. I see functions that have x as an independent variable, such as u(x) and v(x).
 
  • #6
Your operator is curl ##\mathbf{L}=\mathbf{\nabla}\times.##. It acts on vector-valued functions. In your scan it is also shown that operator acts on functions of ##x,y, z##. So I would expect your functions, on which ##\mathbf{L}## acts, to be ##\mathbf{u}\left(\mathbf{r}\right)=\mathbf{u}\left(x, y, z\right)##, i.e. functions that take a position vector (##\mathbf{r}##) in, and give your some other vector (##\left(u_x, u_y, u_z\right)^T##) out.

Suitable candidate for the inner product is then:

##\langle \mathbf{u} | \mathbf{v}\rangle = \int d^3 r \, \mathbf{u}\left(\mathbf{r}\right)^{\dagger}.\mathbf{v}\left(\mathbf{r}\right)##,

but this is not the only candidate. Also, I am slightly worried by the fact that your domain is ##\Omega##. Often ##\Omega## means solid angle, which would suggest

##\langle \mathbf{u} | \mathbf{v}\rangle = \oint_\Omega d^2 \mathbf{\hat{r}} \, \mathbf{u}\left(\mathbf{r}\right)^{\dagger}.\mathbf{v}\left(\mathbf{r}\right)##

So how about defining the problem in a way that one does not have to guess?
 
  • #7
Cryo said:
Your operator is curl ##\mathbf{L}=\mathbf{\nabla}\times.##. It acts on vector-valued functions. In your scan it is also shown that operator acts on functions of ##x,y, z##. So I would expect your functions, on which ##\mathbf{L}## acts, to be ##\mathbf{u}\left(\mathbf{r}\right)=\mathbf{u}\left(x, y, z\right)##, i.e. functions that take a position vector (##\mathbf{r}##) in, and give your some other vector (##\left(u_x, u_y, u_z\right)^T##) out.

Suitable candidate for the inner product is then:

##\langle \mathbf{u} | \mathbf{v}\rangle = \int d^3 r \, \mathbf{u}\left(\mathbf{r}\right)^{\dagger}.\mathbf{v}\left(\mathbf{r}\right)##,

but this is not the only candidate. Also, I am slightly worried by the fact that your domain is ##\Omega##. Often ##\Omega## means solid angle, which would suggest

##\langle \mathbf{u} | \mathbf{v}\rangle = \oint_\Omega d^2 \mathbf{\hat{r}} \, \mathbf{u}\left(\mathbf{r}\right)^{\dagger}.\mathbf{v}\left(\mathbf{r}\right)##

So how about defining the problem in a way that one does not have to guess?

I wish I could define it better for my understanding and yours. This is a homework problem from my online course. My teacher is unresponsive and seems to not care. I have gotten as far as I have by reading constantly. I can't find any info regarding this problem in my books.

I don't think omega is solid angle in this case.

What is d in your inner product?
 
  • #8
Lets assume the first guess is correct. The inner product is ##\langle \mathbf{u} | \mathbf{v}\rangle = \int d^3 r \, \mathbf{u}\left(\mathbf{r}\right)^{\dagger}.\mathbf{v}\left(\mathbf{r}\right)##

The adjoint of ##\mathbf{L}##, which we shall denote ##\mathbf{L}^\dagger##, is defined by:

##\langle \mathbf{u} | \mathbf{L}\mathbf{v}\rangle=\langle\mathbf{L}^\dagger \mathbf{u} | \mathbf{v}\rangle##.

Integrating by parts (assuming ##\mathbf{u}\, \mathbf{v}## vanish at infinity):

##\langle \mathbf{u} | \mathbf{L}\mathbf{v}\rangle =\int d^3 r \, \mathbf{u}_\alpha^{\dagger}.\epsilon_{\alpha\beta\gamma}\partial_\beta\mathbf{v}_\gamma=-\epsilon_{\alpha\beta\gamma}\int d^3 r \, \partial_\beta \left(\mathbf{u}_\alpha^{\dagger}\right).\mathbf{v}_\gamma=\int d^3 r \, \epsilon_{\gamma\beta\alpha}\partial_\beta \left(\mathbf{u}_\alpha^{\dagger}\right).\mathbf{v}_\gamma=\int d^3 r \, \left(\mathbf{\nabla}\times\mathbf{u}^{\dagger}\right).\mathbf{v}=\langle \mathbf{L} \mathbf{u} | \mathbf{v}\rangle##

i.e. ##\mathbf{L}## is self-adjoint (for this inner product), so ##\mathbf{L}^\dagger=\mathbf{L}##. Above ##\epsilon_{\gamma\beta\alpha}## is the Levi-Civita (relative) tensor, and index repitition implies summation. If you are not confident with this notation, try doing integration by parts for all components of your operator independently, i.e. start with

##\langle \mathbf{u}_x | L_x\mathbf{v}\rangle + \langle \mathbf{u}_y | L_y\mathbf{v}\rangle + \langle \mathbf{u}_z | L_z\mathbf{v}\rangle =\int d^3 r \, u_x^{\dagger}.\left(\partial_y v_z - \partial_z v_y\right) + \dots = -\int d^3 r \, \partial_y u_x^{\dagger}.v_z + \int d^3 r \, \partial_z u_x^{\dagger}.v_y + \dots##
 
  • #9
So I am integrating the transpose of each partial with each component of the curl?
 
  • #10
Lets look at the first term

##\int d^3 r u_x^\dagger \partial_y v_z = \int dx \int dz \int^\infty_{-\infty} dy u_x^\dagger \partial_y v_z = \int dx \int dz \int^\infty_{-\infty} dy \partial_y \left(u_x^\dagger v_z\right)-\int dx \int dz \int^\infty_{-\infty} dy \partial_y\left(u_x^\dagger\right) v_z= \int dx \int dz \left[u_x^\dagger v_z\right]^\infty_{-\infty}-\int dx \int dz \int^\infty_{-\infty} dy \partial_y\left(u_x^\dagger\right) v_z=0-\int d^3 r\, \partial_y\left(u_x^\dagger\right) v_z##

Do it for the other 5 terms of the full expression, and then re-build it into a curl and a dot-product
 
  • #11
I want to make sure I understand the notation.

ux is the adjoint?

what is d3?
 
  • #12
##d^3\vec r## is the volume element. In Cartesian coordinates, it's shorthand for ##dx\,dy\,dz##.

If ##\vec u## and ##\vec v## are real, and you represent them as two 3x1 matrices, then you can write the dot product in terms of matrix multiplication:
$$\vec u \cdot \vec v = \vec u^T \vec v.$$ If ##\vec u## and ##\vec v## are complex, then you need to also take the complex conjugate of ##\vec u##, so you'd have
$$\vec u \cdot \vec v = \vec u^\dagger \vec v,$$ where ##\dagger## means conjugate-transpose.

##u_x^\dagger## is the ##x## component of the adjoint, so it's really ##(\vec u^\dagger)_x = u_x^*##
 
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  • #13
I have to take the dot product of the conjugate transpose of each component of u and form a dot product u⋅v

Then take an inner product ∫dx∫dy∫dz(u⋅v)

I do this with each pair of u⋅v for components x,y,z?
 
  • #14
Let ##\mathbf{u}=\mathbf{\hat{x}}\left(u^{(r)}_x+i u^{(i)}_x\right) + \mathbf{\hat{y}}\left(u^{(r)}_y+i u^{(i)}_y\right) + \mathbf{\hat{z}}\left(u^{(r)}_z+i u^{(i)}_z\right) ##

Where ##u^{(r)}_x, u^{(i)}_x, u^{(r)}_y, \dots## are real-valued functions, ##\mathbf{\hat{x}}, \mathbf{\hat{y}}, \mathbf{\hat{z}}## are Cartesian unit vectors, and ##i^2=-1##. Let ##\mathbf{v}## be defined in the same way

Then ##\mathbf{u}^\dagger \mathbf{v}=\left(u^{(r)}_x-i u^{(i)}_x\right)\cdot\left(v^{(r)}_x+i v^{(i)}_x\right)+\left(u^{(r)}_y-i u^{(i)}_y\right)\cdot\left(v^{(r)}_y+i v^{(i)}_y\right)+\left(u^{(r)}_z-i u^{(i)}_z\right)\cdot\left(v^{(r)}_z+i v^{(i)}_z\right)##

Substitute this into the integral, and proceed.
 
  • #15
OK, so I have:

2∫∫∫dxdydz (uxvx)+(uyvy)+(uzvz)
 
  • #16
I cannot guess what you started with. Please provide your starting conditions, intermediate steps and result if you want me to check this. Also, please do so in LaTeX form, as did I, it makes it much easier to read.
 
  • #17
I can't seem to upload the images by copy and paste.

When I get to the third line, I am unsure how to proceed. If I integrate with respect to x the first term becomes uv but what about the other two terms?
https://www.physicsforums.com/file:///C:/Users/SOLIDW~1/AppData/Local/Temp/msohtmlclip1/01/clip_image004.pnghttps://www.physicsforums.com/file:///C:/Users/SOLIDW~1/AppData/Local/Temp/msohtmlclip1/01/clip_image006.png
 

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  • #18
Why not use built-in LaTeX (https://www.physicsforums.com/help/latexhelp/)?

I don't understand. Why are you trying to integrate unknown functions ##u,v## (also you seem to have dropped my superscripts ##\dots^{(r,i)}## for some reason)? The aim, I thought, was to use integration by parts to find the adjoint of the curl-operator. You seem to be quite lost. Maybe it is better to drop the complicated 3d case, and go back to basics.

Consider one-dimensional functions, e.g. ##u=u\left(x\right)##, and operator ##p=\frac{d}{dx}##. Let the inner product of two functions be:

##\langle u | v \rangle=\int_{-\infty}^\infty dx\, u\left(x\right)^* v\left(x\right) ##, here ##\dots^*## denotes complex conjugation

Can you find the adjoint of ##p##? i.e. can you find ##p^{\dagger}## such that

##\langle p^\dagger u | v \rangle = \int dx\, p^\dagger u\left(x\right)^* v\left(x\right)=\langle u | p v \rangle = \int dx\, u\left(x\right)^* p v\left(x\right) = \int dx\, u\left(x\right)^* \frac{d v}{d x}\left(x\right)##

Assume all functions vanish as ##x\to\pm \infty## and use integration by parts (https://en.wikipedia.org/wiki/Integration_by_parts)
 

Related to What Is the Adjoint of the Curl Operator in Differential Equations?

What is the adjoint operator?

The adjoint operator is a mathematical concept used in linear algebra to determine the inverse of a linear transformation. It is denoted by a dagger symbol (†) and is defined as the transpose of the conjugate of the original linear transformation.

Why is finding the adjoint operator important?

Finding the adjoint operator is important because it allows us to find the inverse of a linear transformation, which is crucial in many applications of linear algebra. It also helps us solve systems of equations and understand the properties of the original linear transformation.

How do you find the adjoint operator?

The process of finding the adjoint operator involves taking the transpose of the matrix representation of the original linear transformation and then taking the conjugate of each entry. This can be done using the adjugate matrix, which is the transpose of the cofactor matrix.

What is the relationship between the adjoint operator and the inner product?

The adjoint operator is closely related to the inner product in linear algebra. In fact, the inner product of two vectors can be represented as the product of one vector and the adjoint of the other vector. This relationship is known as the adjoint property of inner products.

Can the adjoint operator be applied to non-linear transformations?

No, the adjoint operator is only defined for linear transformations. Non-linear transformations do not have a unique inverse, so the concept of an adjoint operator does not apply to them.

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