# Deriving expressions for Fourier Transforms of Partial Derivatives

• N00813
In summary, using the formal limit definition of the derivative, we can derive expressions for the Fourier Transforms with respect to x of the partial derivatives \frac{\partial u}{\partial t} and \frac{\partial u}{\partial x} as \tilde{\frac{\partial u}{\partial t}}(k,t) = ik \tilde{u}(k,t).
N00813

## Homework Statement

Using the formal limit definition of the derivative, derive expressions for the Fourier Transforms with respect to x of the partial derivatives $\frac{\partial u}{\partial t}$ and $\frac {\partial u}{\partial x}$.

## Homework Equations

The Fourier Transform of a function $u(x,t)$ is:
$$\tilde{u}(k,t) = \int_{-\infty}^{\infty} u(x,t) e^{-ikx} dx$$

## The Attempt at a Solution

I attempted to write the FT of the derivative with respect to time as $\lim_{dt \to 0} \frac{u(x,t+dt) - u(x,t)}{dt}$ and then Fourier Transform it. Pulling the limit outside the integral, I got $FT of (\frac{\partial u}{\partial t}) = \tilde{\frac {\partial u}{\partial t}}(k,t).$Now, with the partial derivative of x, I'm not so sure. The partial derivative wrt x is $\lim_{dx \to 0} \frac{u(x + dx,t) - u(x,t)}{dx}$. When placed in the Fourier transform, I don't know what I can do. I don't think the dx at the end of the integral cancels out the dx in the definition of the derivative.

N00813 said:

## Homework Statement

Using the formal limit definition of the derivative, derive expressions for the Fourier Transforms with respect to x of the partial derivatives $\frac{\partial u}{\partial t}$ and $\frac {\partial u}{\partial x}$.

## Homework Equations

The Fourier Transform of a function $u(x,t)$ is:
$$\tilde{u}(k,t) = \int_{-\infty}^{\infty} u(x,t) e^{-ikx} dx$$

## The Attempt at a Solution

I attempted to write the FT of the derivative with respect to time as $\lim_{dt \to 0} \frac{u(x,t+dt) - u(x,t)}{dt}$ and then Fourier Transform it. Pulling the limit outside the integral, I got $FT of (\frac{\partial u}{\partial t}) = \tilde{\frac {\partial u}{\partial t}}(k,t).$

Now, with the partial derivative of x, I'm not so sure. The partial derivative wrt x is $\lim_{dx \to 0} \frac{u(x + dx,t) - u(x,t)}{dx}$. When placed in the Fourier transform, I don't know what I can do. I don't think the dx at the end of the integral cancels out the dx in the definition of the derivative.

You should avoid using "dx" and so forth as anything other than part of a derivative or integral (or as a differential form). Otherwise you run into this sort of confusion. When taking limits, use either $\delta x$ or another letter.

You want to calculate
$$\int_{-\infty}^\infty \lim_{h \to 0} \frac{u(x+h, t) - u(x,t)}h e^{-ikx}\,dx = \lim_{h \to 0} \frac1h \left( \int_{-\infty}^\infty u(x + h, t)e^{-ikx}\,dx - \int_{-\infty}^\infty u(x, t)e^{-ikx}\,dx \right)$$

1 person
pasmith said:
You should avoid using "dx" and so forth as anything other than part of a derivative or integral (or as a differential form). Otherwise you run into this sort of confusion. When taking limits, use either $\delta x$ or another letter.

You want to calculate
$$\int_{-\infty}^\infty \lim_{h \to 0} \frac{u(x+h, t) - u(x,t)}h e^{-ikx}\,dx = \lim_{h \to 0} \frac1h \left( \int_{-\infty}^\infty u(x + h, t)e^{-ikx}\,dx - \int_{-\infty}^\infty u(x, t)e^{-ikx}\,dx \right)$$

Thanks, I must have derped out last night.

I'm thinking about turning the 2nd term into $\lim_{h \to \infty} \frac{1}{h} \tilde{u}(k,t)$ and substituting the y = x + h into the first term, to give a final result of:
$$\lim_{h \to 0} \frac{1}{h} ((e^{ik(0+h)}-e^{ik0}) \tilde{u}) = ike^{0} \tilde{u}$$

## 1. What is the purpose of deriving expressions for Fourier Transforms of Partial Derivatives?

The purpose of deriving expressions for Fourier Transforms of Partial Derivatives is to understand the relationship between a function and its Fourier Transform when the function has multiple independent variables. This allows us to analyze and manipulate functions in the frequency domain, which has many practical applications in fields such as signal processing and image analysis.

## 2. How do you derive expressions for Fourier Transforms of Partial Derivatives?

The expressions for Fourier Transforms of Partial Derivatives can be derived using the properties of Fourier Transforms, such as linearity, time-shifting, and frequency-shifting. These properties allow us to break down the function into simpler components and then use the definitions of Fourier Transforms to find the transform of each component. The final expression is then obtained by combining these transforms using the appropriate property.

## 3. What are some common examples of functions that require deriving expressions for Fourier Transforms of Partial Derivatives?

Functions that are commonly encountered in physics and engineering, such as electromagnetic fields, heat conduction, and fluid flow, often involve multiple independent variables and thus require deriving expressions for Fourier Transforms of Partial Derivatives. Other examples include sound waves, image processing, and financial data analysis.

## 4. What are the benefits of using Fourier Transforms of Partial Derivatives?

Using Fourier Transforms of Partial Derivatives allows us to analyze functions in the frequency domain, which can reveal important information about the behavior of the function. It also simplifies the mathematical manipulation of functions with multiple independent variables, making it easier to solve complex problems and equations. Additionally, it has applications in signal filtering, data compression, and solving partial differential equations.

## 5. Are there any limitations to using Fourier Transforms of Partial Derivatives?

While Fourier Transforms of Partial Derivatives are a powerful tool for analyzing functions with multiple independent variables, they do have some limitations. They may not be applicable to functions with discontinuities or functions that are not integrable. Additionally, the inverse Fourier Transform of a partial derivative may not always exist, making it impossible to recover the original function in some cases. Care must also be taken when applying the properties of Fourier Transforms, as they may not hold for all functions.

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