I am very familiar with problems of the form
y'(x) = f(r; x, y), y(a)=y0, y(b)=y1
where r is a parameter to be estimated using the boundary values.
But apart from this situation, has anyone found other reasons why we should a boundary value problem of the form
y'(x) = f(x, y), y(a)=y0, y(b)=y1
To me, I think a BVP should be of second-order or higher. So I think the simplest BVP should be of the form
y''(x)=f(x,y,y'), y(a)=y0, y(b)=y1
Hi.,
A boundary value problem has conditions specified at the extremes ("boundaries") of the independent variable in the equation whereas an initial value problem has all of the conditions specified at the same value of the independent variable (and that value is at the lower boundary of the domain, thus the term "initial"
For More info:
https://users.math.msu.edu/users/gnagy/teaching/ode.pdf
Hello,
Not sure if this is helpful, but ... the main instance I can think of is in the context of shooting methods, where a BVP featuring one independent variable is re-written as a first-order IVP, and the task transforms into iterating (e.g., via bisection etc.) on the initial value of high-order derivatives (i.e., y^(k)(a)=?) such as the retrieve the intended boundary value, y(b)=y1. In terms of the "why", it may be that your special choice of ODE solver + boundary iteration strategy yields an algorithm that is more efficient overall than a straightforward relaxation approach to the original BVP.
C K Gomathy and Ayaboe Edoh Thanks for the responses.
But, please the argument about the bvp is whether there can be a first order BVP or not.
Ayaboe Edoh mentioned that it can be rewritten as a system of first order ODE, I agree but that also means that the bvp must have been of order higher than 1.
I will appreciate more contributions. Thanks
- Badges
- Science topic