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### Boundary Value Problem

```Date: 11/02/2002 at 16:07:07
From: Alex Brigden
Subject: Second Order linear Homogeneous DE's Boundary Value Problem

Differential Equation:

d^2y/dx^2 + P/EI*y = Eulers buckling DE for a pin-ended strut

Where:
P = P critical load for stability
E = Young's Modulus
I = Moment of Inertia

General solution:

y = A sin (x sqrt(P/EI)) + B sin (x sqrt(P/EI))

My question is how to apply the boundary conditions to get the
solution.

x=0 y=0

This is easy, as sin 0=0 and cos 0=1; therefore B=0.

However, the second boundary condition is

x=L y=0

Most textbooks give the non-trivial result as

sin (L sqrt(P/EI))=0

How is this obtained? Do we simply carry the result from the first
boundary condition into the second and assume B=0? We therefore only
consider

A sin (x sqrt(P/EI))? and apply L to this?

Is this the general way to apply boundary conditions in a boundary
value problem?

```

```
Date: 11/03/2002 at 11:02:25
From: Doctor Douglas
Subject: Re: Second Order linear Homogeneous DE's Boundary Value
Problem

Hi Alex,

Thanks for writing!

Correct, although of course you meant B cos(...)  here.

>My question is how to apply the boundary conditions to get the
>solution x=0 y=0. This is easy as sin 0=0 and cos 0=1 therefore B=0

yes, absolutely.

>However the second boundary condition is x=L y=0
>Most textbooks give the non-trivial result as
>sin (L sqrt(P/EI))=0
>
>How is this obtained? Do we simply carry the result from the first
>boundary condition into the second and assume B=0?

This equation is obtained by writing down the equation that holds at
the other boundary (at x=L). For the moment let's assume that we don't
know anything about the x=0 condition. We could still write down the
following fact at x=L:

yL = A sin(L sqrt(P/EI)) + B cos(L sqrt(P/EI))

where yL is the value of y specified at the boundary (usually it is
zero, but in some problems it might not be). In this problem, we can
go further, because we already know from the other BC at x=0 that B=0:

yL = A sin(L sqrt(P/EI))

If B were not zero, we would have an additional term from the cosine

Let's make the further assumption that the boundary condition did in
fact specify that yL=0.  Then

0 = A sin(L sqrt(P/EI))
or
0 = sin(L sqrt(P/EI))

because we see that if A were in fact zero, then our original solution
would have A=B=0, which vanishes and is just a trivial solution and
not very useful.

This illustrates the general method. Eventually, as you become more
experienced with such problems that have BCs of the form y(x=0) =
y(x=L) = 0" (this is a quite common form), you'll be able to establish
the values of A and B quickly without going through all of the steps.

But you'll have to be careful. In the general case with BCs of
slightly different form (e.g. y(x=0) = 1 and y(x=L) = -3, you'll get a
system of two linear equations in two unknowns, and you'll have to
solve this system explicitly for A and B.

I hope this helps.

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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