# Adjoint of a Differential Operator

• CoachZ
In summary, the conversation discusses finding D*, the adjoint operator, in a vector space V with dimension less than or equal to 3. The inner product is defined as the integral of two functions over the interval [0,1]. The goal is to find a closed form for D* that satisfies the equation <Du, v>= <u, D*v>. The basis for V is given as {1, x, (x^2)/2, (x^3)/3} and it is known that D* is defined as i times the derivative of g(x) if g(x) has degree greater than 0. However, a closed form for D* that works for all elements is not readily apparent. Suggestions are
CoachZ
If the inner product is defined on V with dimension less than or equal to 3 as $$\left\langle f,g \right\rangle = \int_{0}^{1}f(x)g(x)dx$$, I'm trying to find D* such that $$\left\langle Df,g \right\rangle$$ = $$\left\langle f,D*g \right\rangle$$, and I thought I had a closed form of D*. If {1, x, (x^2)/2, (x^3)/3}. Call each element e_i, i = 0,...,3 form a basis for V, then $$\left\langle Df,g \right\rangle$$ = $$\left\langle f,D*g \right\rangle$$ holds given D* is defined as i times the (derivative of g(x)) if g(x) has degree greater than 0, but I can't think of a closed form so that it will work for all elements. Does anyone have any suggestions?

Would $\delta(x-1)-\delta(x)-D$ work?

$$\int_0^1 Df(x) g(x)dx = \int_0^1 (df(x)/dx) g(x)dx = f(x) g(x) |^1_0 - \int^1_0 f(x) Dg(x) dx = \int^1_0 f(x) [(\delta(x-1)-\delta(x)) - D]g(x) dx$$

CoachZ said:
If the inner product is defined on V with dimension less than or equal to 3 as $$\left\langle f,g \right\rangle = \int_{0}^{1}f(x)g(x)dx$$, I'm trying to find D* such that $$\left\langle Df,g \right\rangle$$ = $$\left\langle f,D*g \right\rangle$$, and I thought I had a closed form of D*. If {1, x, (x^2)/2, (x^3)/3}. Call each element e_i, i = 0,...,3 form a basis for V, then $$\left\langle Df,g \right\rangle$$ = $$\left\langle f,D*g \right\rangle$$ holds given D* is defined as i times the (derivative of g(x)) if g(x) has degree greater than 0, but I can't think of a closed form so that it will work for all elements. Does anyone have any suggestions?
? Please be careful to tell us exactly what you are doing! You start by talking about a vector space "V with dimension less than or equal to 3" but then start integrating? A vector space, of any dimension, does not necessarily contain functions at all. "A vector space with dimension less than or equal to 3" is NOT the same as "polynomials of degree 3 or less". (In fact, the space of polynomials of degree 3 or less has dimension four, not three, as you make clear by giving a basis consisting of four functions.)

In any case, what D* is such that <Du, v>= <u, D*v> depends upon what D is! Do you mean that D is the derivative operator?

HallsofIvy said:
? Please be careful to tell us exactly what you are doing! You start by talking about a vector space "V with dimension less than or equal to 3" but then start integrating? A vector space, of any dimension, does not necessarily contain functions at all. "A vector space with dimension less than or equal to 3" is NOT the same as "polynomials of degree 3 or less". (In fact, the space of polynomials of degree 3 or less has dimension four, not three, as you make clear by giving a basis consisting of four functions.)

In any case, what D* is such that <Du, v>= <u, D*v> depends upon what D is! Do you mean that D is the derivative operator?

Exactly: D is the derivative operator and I'm trying to find D*, which is simply the adjoint operator. I apologize for writing it so messily. I'm a first year graduate student still working on my writing skills, so if you have any suggestions, either for finding D* or improving writing skills, it would be greatly appreciated.

Figured this out. Since every operator takes one some matrix representation, and we consider the orthonormal basis with respect to the differentiation operator, we can extend that basis to matrix form. But there's a theorem that states the matrix representation of D* is simply the transpose of the matrix represented by D. Therefore, I'm all set. Thanks for the suggestions!

## What is an adjoint of a differential operator?

The adjoint of a differential operator is a mathematical concept that is used to find the solution to a differential equation. It is a mathematical construct that is defined in such a way that it is easy to find the solution to a given differential equation. Essentially, it is a way to find the inverse of a differential operator.

## Why is the concept of adjoint important in differential equations?

The concept of adjoint is important in differential equations because it allows scientists to find the solution to a differential equation without having to solve the equation directly. This can save time and effort, especially when dealing with complex equations. It also provides a way to check the accuracy of the solution obtained through other methods.

## How is the adjoint of a differential operator calculated?

The adjoint of a differential operator is calculated using a process known as integration by parts. This involves breaking down the operator into simpler parts and applying the integration by parts formula. The result is a new operator that is the adjoint of the original operator.

## What are some real-world applications of the adjoint of a differential operator?

The adjoint of a differential operator has many real-world applications, including in physics, engineering, and finance. It is used to solve differential equations that describe the behavior of physical systems, such as fluid flow, heat transfer, and electrical circuits. It is also used in signal processing and image processing to enhance and analyze data.

## Is the adjoint of a differential operator unique?

Yes, the adjoint of a differential operator is unique. This means that there is only one operator that is the adjoint of a given differential operator. This uniqueness is important in ensuring that the solution obtained through the adjoint method is accurate and reliable.

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