# Fourier Transform and Partial Differential Equations

• sa1988
In summary, the conversation discusses transforming a function and using it to solve a partial differential equation. The final equation found is an ordinary differential equation, not a partial differential equation, as it only involves one variable. The confusion about the type of equation is clarified, and the conversation ends with the understanding that details are important in learning.
sa1988

## The Attempt at a Solution

First write ##\phi(x,t)## as its transform

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx} \widetilde{\phi}(p,t) \, \mathrm{d}p##

which I then plug into the PDE in the question to get:

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4} \, \mathrm{d}p = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}(p^4-p^2) \widetilde{\phi}(p,t) \, \mathrm{d}p##

Which neatens to:

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}\Big[\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4}-(p^4-p^2) \widetilde{\phi}(p,t)\Big] \, \mathrm{d}p = 0##

which can only hold if the part in the square bracket obeys

##\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4}=(p^4-p^2) \widetilde{\phi}(p,t)##

So this is where I am now. I've found a differential equation obeyed by ##\widetilde{\phi}(p,t)##, however it's a PDE, not an ODE. So I'm either mistaken with my understanding of what the precise definition of an ODE is, or I've gone wrong with my logic in the working out.

To be honest the wording of the question itself seem a touch ambiguous. Is it even possible for a multi-variable function to be part of an ODE?

Any help would be appreciated, thanks.

sa1988 said:

## The Attempt at a Solution

First write ##\phi(x,t)## as its transform

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx} \widetilde{\phi}(p,t) \, \mathrm{d}p##

which I then plug into the PDE in the question to get:

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4} \, \mathrm{d}p = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}(p^4-p^2) \widetilde{\phi}(p,t) \, \mathrm{d}p##

Which neatens to:

##\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \! e^{ipx}\Big[\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4}-(p^4-p^2) \widetilde{\phi}(p,t)\Big] \, \mathrm{d}p = 0##

which can only hold if the part in the square bracket obeys

##\frac{\partial^4\widetilde{\phi}(p,t)}{\partial t^4}=(p^4-p^2) \widetilde{\phi}(p,t)##

So this is where I am now. I've found a differential equation obeyed by ##\widetilde{\phi}(p,t)##, however it's a PDE, not an ODE. So I'm either mistaken with my understanding of what the precise definition of an ODE is, or I've gone wrong with my logic in the working out.

To be honest the wording of the question itself seem a touch ambiguous. Is it even possible for a multi-variable function to be part of an ODE?

Any help would be appreciated, thanks.

Your final equation is an ordinary differential equation, NOT a partial differential equation. You can tell that from the fact that the only derivative involved is taken with respect to ##t##, so you could (and should) write ##d/dt## instead of ##\partial/\partial t##. (You would have had a partial differential equation if you had ##\partial / \partial p## involved as well, but you do not.)

sa1988
Ray Vickson said:
Your final equation is an ordinary differential equation, NOT a partial differential equation. You can tell that from the fact that the only derivative involved is taken with respect to ##t##, so you could (and should) write ##d/dt## instead of ##\partial/\partial t##. (You would have had a partial differential equation if you had ##\partial / \partial p## involved as well, but you do not.)

Aaahh ok, so it only becomes a PDE when another partial differential is actually involved.

I thought it was PDE because it only dealt with one variable on a multi-variable function.

Thanks for clearing that up.

EDIT:

Having said that... the stock definition on wikipedia seems to say that my solution is a PDE..?

In mathematics, a partial differential equation (PDE) is a differential equation that contains unknown multivariable functions and their partial derivatives.

It goes on to say,
A relatively simple PDE is

Is this not pretty much the same as my answer above?

sa1988 said:
Aaahh ok, so it only becomes a PDE when another partial differential is actually involved.

I thought it was PDE because it only dealt with one variable on a multi-variable function.

Thanks for clearing that up.

EDIT:

Having said that... the stock definition on wikipedia seems to say that my solution is a PDE..?
It goes on to say,Is this not pretty much the same as my answer above?

Yes, but you are getting hung up on things that really do not matter! Whether you write ##\partial u(x,y) /\partial x = 0## or ##du(x,y)/dx = 0## depends on whether or not you regard ##y## as a variable (having the same type of status as ##x##) or as just a parameter that appears in the formula. For example, if I write ##f(x) = a x^2 + bx + c##, I would normally regard ##a,b,c## as some type of input parameters, and ##x## as the only variable involved. So, I would write ##df/dx## for the derivative, because the idea is that ##y = ax^2+ bx + c## is a curve in ##(x,y)##-space, and for fixed ##a,b,c## corresponds to a single fixed curve, about which I might ask several questions about aspects involving different values of ##x## (such its ##x##-intercepts, its maximum or minimum points and the like). For each input curve the parameters ##a,b,c## remain the same throughout the calculations.

sa1988
Ray Vickson said:
Yes, but you are getting hung up on things that really do not matter!

One does not learn without getting hung up on the details every now and again!

Thanks for clearing it up

## 1. What is a Fourier Transform?

A Fourier Transform is a mathematical operation that decomposes a function into its constituent frequencies. In other words, it converts a function from its original domain (such as time or position) to its frequency domain.

## 2. How is a Fourier Transform used in PDEs?

A Fourier Transform is commonly used in solving partial differential equations (PDEs). It allows us to convert the PDE into an algebraic equation in the frequency domain, which can then be solved more easily.

## 3. What is the relationship between the Fourier Transform and the Heat Equation?

The Heat Equation is a PDE that describes the diffusion of heat over time. The Fourier Transform can be used to solve the Heat Equation by converting it into an algebraic equation in the frequency domain. This allows us to find the solution for any point in time.

## 4. Can the Fourier Transform be used for non-linear PDEs?

No, the Fourier Transform can only be used for linear PDEs. This is because the Fourier Transform assumes that the underlying function is a linear combination of sinusoidal waves, which is not true for non-linear PDEs.

## 5. Are there any limitations to using the Fourier Transform with PDEs?

Yes, the Fourier Transform can only be used for PDEs with certain boundary conditions. If the PDE has non-zero boundary conditions, then the Fourier Transform will not work. Additionally, the Fourier Transform may not always provide a closed-form solution and numerical methods may need to be used to approximate the solution.

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