Volterra Eqn of 2nd Kind -> DEQ

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In summary, the conversation discusses converting a Volterra equation of the 2nd kind into a differential equation with appropriate boundary conditions. The attempt at a solution involves differentiating the equation and solving for the boundary conditions. There is also a mention of a sign error in the integral and the suggestion to differentiate again to obtain a differential equation.
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Elvex
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Volterra Eqn of 2nd Kind --> DEQ

Homework Statement


I need to convert y(x) = 1 - x + int[dt(x-t)y(t)] from 0 to x to a differential equation with the appropriate boundary conditions.





The Attempt at a Solution



OK I just had a problem converting a DEQ into an integral equation so I know the form it will take, I know it will be a 2nd order homogeneous equation... something like d^2y/dx^2 + y(x) = 0 with y(a) = b , dy/dx|c = d for some constants. I know it won't be a periodic boundary condition because that turns into a Fredholm Equation of the 2nd Kind.

So I applied d/dx to both sides getting

dy/dx = -1 + int[dty(t)] from 0 to x

I think this is valid since the integrand is with respect to t, I just applied the differentiation within the integral.

I'm not sure what to do from here though.
 
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  • #2
Ok so I missed the obvious step of solving for the boundary conditions by simply plugging in x= 0 into my y(x) and dy/dx equations

so y(0) = 1 and dy(0)/dx = -1

This is what I got from my sketchy inspection method prior to this. I know the answer now, but I still don't know how to show it will become a 2nd order homogeneous equation.
 
  • #3
First off there are two terms in the derivative of the integral since there are two sources of x dependence in the integral (one in the function and one in the limit). The one you have ignored does indeed vanish - is that why you ignored it? Once you get to this:

dy/dx = -1 + int[dty(t)] from 0 to x

you have a sign error in the integral. After fixing it you will have a correct equation, but it's still an integral equation. Suggests that you might want to differentiate again, yes?
 

FAQ: Volterra Eqn of 2nd Kind -> DEQ

What is the Volterra Equation of the 2nd kind?

The Volterra Equation of the 2nd kind is a type of integral equation that relates a function to its integral. It is written in the form of y(t) = f(t) + ∫atk(t,s)y(s)ds, where f(t) is the given function and k(t,s) is the kernel of the integral.

How is the Volterra Equation of the 2nd kind different from the 1st kind?

The main difference between the Volterra Equation of the 2nd kind and the 1st kind is the presence of the function f(t) in the equation. In the 1st kind, the function and the integral are both unknowns, while in the 2nd kind, the function is given and the integral is the unknown.

What is the significance of the Volterra Equation of the 2nd kind in mathematics?

The Volterra Equation of the 2nd kind has many applications in mathematical analysis, such as in the study of integral equations, differential equations, and functional analysis. It also has practical applications in fields such as physics, engineering, and economics.

How can the Volterra Equation of the 2nd kind be solved?

There are various methods for solving the Volterra Equation of the 2nd kind, such as the Duhamel principle, the method of successive approximations, and the resolvent kernel method. The choice of method depends on the specific form of the equation and the properties of the kernel function.

What are the limitations of the Volterra Equation of the 2nd kind?

One limitation of the Volterra Equation of the 2nd kind is that it can only be applied to linear problems. Additionally, the integral in the equation must be well-defined and the kernel function must satisfy certain conditions for the equation to have a unique solution. In some cases, it may also be difficult to find an analytic solution, requiring the use of numerical methods.

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