Integral equations of convolution type

In summary, the conversation discusses two questions involving finding the transform and inverse transform of functions using convolution theorem and inversion formula. The first question involves finding the function f(x) given the integral equation and using differentiation and integration to derive the solution. The second question also uses the convolution theorem and contour integration to derive the transform of the function and it is shown that the inverse transform is equal to (e^-1)/(2pi)^1/2. The question concludes with the inquiry about the inverse of an operator Uφ for a given function φ ∈ L_2(−∞,∞).
  • #1
catcherintherye
48
0
i am asked to find f(x) s.t. exp(-xsqd/2) =1/2the integral (-inf to +inf) of exp[-|x-u|f(u)du. I have got as far as to show that the transform f(k)=(1+ksqd)exp[-ksqd/2) and my notes show that this implies the next line which is f(x)=exp[-xsqd/2] -[exp(-xsqd/2)]'' {'' denotes twice differentiation w.r.t x} and this in turn is equal to (2-xsqd)exp[-xsqd/2]. I am not sure how these last two lines where reached, was the inversion formula applied to f(k) and then some integration done? or was some other result used?





I have a simliar question where i must use the convolution theorem to show that the solution to the following integral equation

integral(-inf to + inf) of {f(u)/1+(x-u)^2}du=1/(xsqd+4) is


f(x)=1/2pi.1/(1+xsqd)


i have got as far as to show transform of 1/(xsqd+4)=(2pi)^1/2.transform of1/(1+xsqd).transf and i have used contour integration to show transform of 1/(4+xsqd)=(pi/4).exp[-2]. I have also shown that trans 1/(1+xsqd) =(pi/2)exp[-1]


so I have derived that transform f =(e^-1)/(2pi)^1/2



is this correct up to this point? and where do I go from here, I'm stuck at the same point as the first question...
 
Last edited:
Physics news on Phys.org
  • #2
what is the inverse of the operator

Uφ ≡ φ(x) − 2e^{-x}\int_{-\infty}^{x} e^{t}φ(t)dt

where φ ∈ L_2(−∞,∞)?
 

1. What are integral equations of convolution type?

Integral equations of convolution type are mathematical equations that involve the convolution operator, which is a mathematical operation that combines two functions to produce a third function.

2. What is the purpose of integral equations of convolution type?

The purpose of integral equations of convolution type is to solve problems in which the unknown function appears in an integral expression.

3. What types of problems can be solved using integral equations of convolution type?

Integral equations of convolution type can be used to solve problems in various fields, such as physics, engineering, and signal processing. These equations are particularly useful for solving differential equations with boundary conditions.

4. Are there different methods for solving integral equations of convolution type?

Yes, there are various methods for solving integral equations of convolution type, including the method of successive approximations, the method of eigenfunction expansions, and the method of moments.

5. Can integral equations of convolution type be solved analytically?

In some cases, integral equations of convolution type can be solved analytically using the aforementioned methods. However, in many cases, numerical methods such as the finite element method or the boundary element method are used to obtain approximate solutions.

Similar threads

  • Calculus
Replies
6
Views
1K
Replies
4
Views
739
Replies
8
Views
1K
Replies
3
Views
1K
Replies
31
Views
915
Replies
20
Views
2K
Replies
1
Views
2K
Replies
16
Views
2K
Back
Top