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How to get nontrivial solution of differential equation (Euler buckling load)?

Ralph Schenn
Posted 6 months ago 
DSolve[{yy''''[x]+Mu^2*yy''[x]==0,
yy[0]==0,yy[l]==0,
yy''[0]==0, yy''[l]==0},
yy[x],x]

returned result is {{yy(x)->0}}

But i want a solution like 

yy[x] == C1*Sin[(C2*Pi)/l*x] (0 <= x <= l)

POSTED BY: Ralph Schenn
8 Replies
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Ralph Schenn
Ralph Schenn
Posted 5 months ago

Im afraid that i have to revisit this issue later when i have gained more experience with Mathematica. I have much more experience with AMS on TI-89 and compatibles in Basic and also with lower level programming for that System in C but at the moment this ist to hard for me. I was thinking it would be much simpler, probably just something like adding an assumption.

POSTED BY: Ralph Schenn
POSTED BY: Michael Rogers
Ralph Schenn
Ralph Schenn
Posted 6 months ago

Thanks for the replies. Will try to understand them when i have more time then at the moment
POSTED BY: Ralph Schenn

I guess I could add that DSolve[] treats μ as a given parameter, not as an unknown to be solved for. The only solution that is generically true — that is, true for almost all values of μ and l — is the trivial solution. You have to take special steps, such as you've shown, to solve the eigenvalue problem. Was that what you were trying to demonstrate?
POSTED BY: Michael Rogers

In the second order case DSolve provides conditions on the parameters:

DSolve[{yy''[x] + Mu^2*yy[x] == 0,
  yy[0] == 0, yy[l] == 0},
 yy[x], x]

In the fourth order case, we may impose the initial conditions first and the final conditions later:

sol0 = DSolve[{yy''''[x] + Mu^2*yy''[x] == 0,
   yy[0] == 0, yy''[0] == 0}, yy, x]
conditionsAtL = {yy[l] == 0, yy''[l] == 0, Mu > 0, l > 0} /. sol0[[1]]
paramConditions = conditionsAtL // Solve
finalSols = yy[x] /. sol0 /. paramConditions
POSTED BY: Gianluca Gorni
POSTED BY: Michael Rogers
Sangdon Lee
Sangdon Lee
Posted 6 months ago
POSTED BY: Sangdon Lee

This post is tagged as a question, and the title is a question; but the code seems to be a solution. Is it good enough, or are you looking for some kind of help?
POSTED BY: Michael Rogers
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