Two-Point Boundary Value Problem

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Homework Help Overview

The discussion revolves around a two-point boundary value problem described by the differential equation y'' + ßy = 0, with boundary conditions y'(0)=0 and y'(L)=0. The participants are exploring the case when ß=0, questioning the implications of the boundary conditions on the existence of non-trivial solutions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the equation y'' = 0 and the resulting expressions for y' and y. There is a focus on whether the boundary conditions lead to only trivial solutions or if non-trivial solutions are possible.

Discussion Status

The discussion is ongoing, with participants sharing their reasoning and questioning the nature of the solutions under the given boundary conditions. Some guidance has been offered regarding the integration process, but no consensus has been reached on the existence of non-trivial solutions.

Contextual Notes

Participants note that the boundary conditions are specified in terms of the derivative y', which may influence the interpretation of the solutions. There is also mention of a reference to an answer in a textbook that suggests a non-trivial solution exists for ß=0, prompting further inquiry.

Jamin2112
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Homework Statement



y'' + ßy = 0, y'(0)=0, y'(L)=0

Homework Equations



Meh

The Attempt at a Solution



I so already did the ß>1 and ß<1; I'm stuck on the ß=0. It seems easy enough. y'' = 0 -----> y' = A -----> 0=A, 0=A (from the two initial conditions) ------> No non-trivial solution.

But...

The answer in the back of the book says ß0=0, y0(x)=1; ...


??
 
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Jamin2112 said:

Homework Statement



y'' + ßy = 0, y'(0)=0, y'(L)=0

Homework Equations



Meh

The Attempt at a Solution



I so already did the ß>1 and ß<1; I'm stuck on the ß=0. It seems easy enough. y'' = 0 -----> y' = A -----> 0=A, 0=A (from the two initial conditions) ------> No non-trivial solution.

Starting with y'' = 0 you need to integrate twice to get y, getting two constants.
 
LCKurtz said:
Starting with y'' = 0 you need to integrate twice to get y, getting two constants.

The boundaries it gives me are in terms of y' :wink:


Of course I know
y'' = 0 -----> y' = B ----> y = Ax + b
 
Jamin2112 said:
The boundaries it gives me are in terms of y' :wink:


Of course I know
y'' = 0 -----> y' = B ----> y = Ax + b

But the point is: does your boundary condition force only the trivial solution or can you get a non-trivial solution in this case?
 

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