# Green's functions: Logic behind this step

• WWCY
In summary, the steps are:-find the heat equation-apply the Green's function-derive the equation for ##\hat{G}##
WWCY

## Homework Statement

Hi all,

I came across these steps in my notes, relating to a step whereby,
$$\hat{G} (k, t - t') = \int_{-\infty}^{\infty} e^{-ik(x - x')}G(x-x' , t-t')dx$$
and performing the following operation on ##\hat{G}## gives the following expression,
$$[\frac{\partial}{\partial t} - D\frac{\partial ^2}{\partial x^2}] \hat{G} = \frac{\partial \hat{G}}{\partial t} + Dk^2 \hat{G}$$

These steps are in the context of a more complex problem of solving the Heat Equation using the Green's function.

## The Attempt at a Solution

I can partially understand the writing of ##\frac{\partial \hat{G}}{\partial t}## at the LHS as the same thing, ##\frac{\partial \hat{G}}{\partial t}## at the RHS; we do not know how ##G## (found in the first integral) depends on time exactly.

But the other terms that imply ##- D\frac{\partial ^2 \hat{G}}{\partial x^2} = Dk^2 \hat{G}## I can't follow. First of all, we don't know how ##G## varies with ##x##. Add to that that the fact that the integral isn't independent of ##x##, I don't see how we can apply differentiation under the integral sign to bring down the factor of ##(-ik)^2## down from ##e^{-ik(x-x')}##

Could someone explain the logic behind the steps? Help is greatly appreciated!

##\hat G## does not depend on ##x## (it is an internal integration variable). The equation you quoted is not true. What is true is that
$$\partial_t \hat G + k^2 D \hat G$$
is the Fourier transform of the left-hand side of the heat equation.

Orodruin said:
##\hat G## does not depend on ##x## (it is an internal integration variable). The equation you quoted is not true. What is true is that
$$\partial_t \hat G + k^2 D \hat G$$
is the Fourier transform of the left-hand side of the heat equation.
Hi @Orodruin , am I right to say that ##\hat{G}## has no explicit dependence of ##x## as it is the Fourier transform of ##G##?

Do you also mind illustrating how the factor of ##(-ik)^2## was brought down?

WWCY said:
Hi @Orodruin , am I right to say that ##\hat{G}## has no explicit dependence of ##x## as it is the Fourier transform of ##G##?

Yes.

Do you also mind illustrating how the factor of ##(-ik)^2## was brought down?
This is a general property of the Fourier transform. In general, for any function ##f(x)##, it holds that
$$F[f'] = \int f'(x) e^{-ikx} dx = - \int f(x) \frac{de^{-ikx}}{dx} dx = ik \int f(x) e^{-ikx} dx = ik F[f],$$
where ##F[f]## is the Fourier transform of ##f##.

WWCY
Orodruin said:
Yes.This is a general property of the Fourier transform. In general, for any function ##f(x)##, it holds that
$$F[f'] = \int f'(x) e^{-ikx} dx = - \int f(x) \frac{de^{-ikx}}{dx} dx = ik \int f(x) e^{-ikx} dx = ik F[f],$$
where ##F[f]## is the Fourier transform of ##f##.
Thank you!

## 1. What is the purpose of using Green's functions in scientific research?

Green's functions are mathematical tools used to solve differential equations in physics and engineering. They provide a systematic way of finding solutions to these equations, which can be used to model and predict real-world phenomena.

## 2. How do Green's functions work?

Green's functions work by decomposing a given differential equation into simpler, solvable components. These components are then combined using a convolution integral, resulting in a solution for the original equation.

## 3. What is the logic behind using Green's functions in solving differential equations?

The logic behind using Green's functions is based on the principle of superposition. This means that the solution to a differential equation can be obtained by summing the solutions to simpler, individual components of the equation. Green's functions provide a way to calculate these individual components, making it easier to find the overall solution.

## 4. Can Green's functions be used for any type of differential equation?

Green's functions are most commonly used for linear differential equations, as they rely on the principle of superposition. However, they can also be used for some non-linear equations, depending on the specific problem at hand.

## 5. Are there any limitations or drawbacks to using Green's functions?

One limitation of using Green's functions is that they may not always provide a closed-form solution, meaning they cannot be expressed in terms of elementary functions. Additionally, their use may be limited to specific boundary conditions and may not be applicable to all types of differential equations.

• Calculus and Beyond Homework Help
Replies
3
Views
559
• Calculus and Beyond Homework Help
Replies
6
Views
690
• Calculus and Beyond Homework Help
Replies
1
Views
571
• Calculus and Beyond Homework Help
Replies
9
Views
722
• Calculus and Beyond Homework Help
Replies
1
Views
423
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
818
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
10
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
683