BVP vs IVP when solving for eigenvalues

• I
• Mr Davis 97
In summary: This is not always the case, but it does demonstrate the importance of boundary conditions in determining the eigenvalues. In summary, the given problem is an eigenvalue problem with boundary conditions, where the value of lambda can be found by testing the different cases of lambda and solving for what lambda must be to have non-trivial solutions. This is necessary because initial value problems for this type of equation do not have non-trivial solutions, emphasizing the importance of boundary conditions in determining eigenvalues.
Mr Davis 97
I am being asked to find ##\lambda## such that ##y'' + \lambda y = 0; ~y(0) =0; ~y'( \pi ) = 0##. This is an eigenvalue problem where we are given boundary conditions. ##\lambda## can be found such that we don't have a trivial solution if we test the different cases when ##\lambda < 0, \lambda = 0##, or ##\lambda > 0##, and solve for what ##\lambda## must be such that we have non-trivial solutions.

I'm just curious as to why these problems are given with boundary conditions instead of initial conditions. Is there some theory behind why we need boundary conditions in order to determine the eigenvalues? Would being given initial conditions allow us to do the same thing?

Looks to me like a second-order linear differential equation. It also looks to me like an initial value problem, not a boundary value problem. Boundary value would be if you had ##y (x_1) = y_1,~ y(x_2) = y_2##. That is just one of the methods of solving for it.

TJGilb said:
It also looks to me like an initial value problem, not a boundary value problem.
Are you sure, TJGlib? Boundary value problems are specified at the boundaries; the 2nd order ODE as specified by Mr Davis 97 is specified ##0## and at ##\pi## and is then by definition a BVP, hence NOT an IVP. The derivatives of the boundary conditions do not dictate whether a problem is a BVP or IVP; but rather where the boundary conditions are specified.

boundary value problems (bvp) tend to be more versatile than initial value problems (ivp)
for example often for a situation an ivp has a unique solution
while a bvp may have a unique solution, infinite solutions, or no solution

often the bvp has arisen naturally
sometimes we convert the bvp to an ivp by finding one with the same solution
ie shooting method

often it is not possible or desirable to do this
in fact it is often helpful to convert an ivp to a bvp
for example if the ivp is sensitive a bvp may be less so

many methods involve breaking a function or region into pieces solving the parts and reassembling them
this can be done for both ivp and bvp but often bvp are more convenient
scale is often a problem in ivp such is in boundary layers and other situations when different effects dominate different regions

Mr Davis 97 said:
I am being asked to find ##\lambda## such that ##y'' + \lambda y = 0; ~y(0) =0; ~y'( \pi ) = 0##. This is an eigenvalue problem where we are given boundary conditions. ##\lambda## can be found such that we don't have a trivial solution if we test the different cases when ##\lambda < 0, \lambda = 0##, or ##\lambda > 0##, and solve for what ##\lambda## must be such that we have non-trivial solutions.

I'm just curious as to why these problems are given with boundary conditions instead of initial conditions. Is there some theory behind why we need boundary conditions in order to determine the eigenvalues? Would being given initial conditions allow us to do the same thing?
It's instructive to see what happen when we to solving the similar initial value problem: Find $\lambda$ such that $y'' +\lambda y =0$ for $y(0)=0$ and $y'(0)=0$.

For $\lambda = 0$ the general solution is $y= ax +b$ but when we apply the boundary conditions we find that $a=0, b=0$.

For $\lambda > 0$ the general solution is $y= a sin \sqrt{\lambda} x +b cos \sqrt{\lambda} x$, but when we apply the boundary conditions we again find that $a=0, b=0$.

For $\lambda < 0$ the general solution is $y= a exp (\sqrt{-\lambda} x) +b exp (-\sqrt{-\lambda} x)$, but when we apply the boundary conditions we again find that $a=0, b=0$.

You can see that in this case the initial value eigenvalue problem has zero nontrivial solutions.

1. What is the difference between BVP and IVP when solving for eigenvalues?

BVP (Boundary Value Problem) and IVP (Initial Value Problem) are two types of problems that can arise when solving for eigenvalues. BVP involves finding the values of the eigenvalues at specific boundary conditions, while IVP involves finding the values of the eigenvalues at a specific initial condition.

2. Which type of problem is more commonly used in solving for eigenvalues?

BVP is more commonly used in solving for eigenvalues because it allows for more flexibility in choosing the boundary conditions, and thus can provide more accurate and precise solutions.

3. How are the solutions for BVP and IVP different?

The solutions for BVP and IVP can differ in terms of the number of eigenvalues that are found and the accuracy of those values. BVP typically provides a more accurate and complete set of eigenvalues, while IVP may only provide a few eigenvalues depending on the initial condition chosen.

4. What are the advantages of using BVP over IVP when solving for eigenvalues?

BVP allows for more control over the boundary conditions, which can result in more accurate and precise solutions for the eigenvalues. It also provides a larger set of eigenvalues, which can be useful in certain applications.

5. In what situations would using IVP be more appropriate than using BVP when solving for eigenvalues?

IVP may be more appropriate in situations where the boundary conditions are unknown or not well-defined. In these cases, using IVP can help determine the eigenvalues based on a specific initial condition. IVP may also be more efficient to use when dealing with simpler systems, as it involves fewer computations compared to BVP.

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