Euler equations with Y(1) = 0, Y(2) = 0

  • Thread starter gerald87
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  • #1
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Is it possible for an Euler equation to satisfy the boundary conditions Y(1)=0, Y(2)=0?

I have considered the three possibilities, distinct real roots, repeated roots and conjugate complex roots and cannot find any solutions.

Are there any other possibilities to consider?

Thanks
 

Answers and Replies

  • #2
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Why don't you write the equation here?
 
  • #3
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Hi Eli,

Sorry, I should have been more clear in my question.

I am not talking about a specific Euler equation, but the general Euler equation,

X^2Y'' + AXY' +BY = 0.

I understand that there are three possible solutions for Y;
Y=c1x^m1 + c2x^m2 for distinct roots
Y=[c1 + c2ln(x)]x^m for repeated roots
or Y=[c1cos(bln(x))+c2sin(bln(x))]x^a for complex conjugate roots

When I put the boundary conditions Y(1) = 0, Y(2) = 0 into each of the possible solutions of Y, the only solution I can get is Y=0.

Are there any other possible solutions to an euler-cauchy problem than the three I mentioned above that should be considered?

Thanks
 
  • #4
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Hi,

I have a question relating to the Euler equation,

X^2Y'' + AXY' +BY = 0.

I understand that there are three possible solutions for Y;
Y=c1x^m1 + c2x^m2 for distinct roots
Y=[c1 + c2ln(x)]x^m for repeated roots
or Y=[c1cos(bln(x))+c2sin(bln(x))]x^a for complex conjugate roots

I am to determine whether or not there are solutions (real or imaginary), other than Y=0 to an euler Equation which satisfies the boundary conditions Y(1) = 0, Y(2) = 0.

When I put the boundary conditions Y(1) = 0, Y(2) = 0 into each of the possible solutions of Y, the only solution I can get is Y=0.

Are there any other possible solutions to an euler-cauchy problem than the three I mentioned above that should be considered?

I have noticed in every boundary value euler cauchy problem has the boundary conditions with Y and the derivative of Y, rather than Y in both. Could this mean there are no solutions?

I hope this makes sense, I am a little confused by this problem.

Thanks
 
  • #5
LCKurtz
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Before you evaluate the boundary conditions, you have two linearly independent solutions in each case, which is all you can expect. If you show in each case that your conditions give only the trivial solution you are done.

What you have here is a boundary value problem (boundary conditions at two different points) as opposed to an initial value problem where the two conditions are at the same point. In a homogeneous boundary value problem like you have here, Y identically zero always works. And if you would have found a nontrivial solution satisfying the boundary conditions, then any constant times it would have also worked.
 
  • #6
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Thanks for the help.

What would happen if A and B in the initial equation were not real numbers?

Would it still be considered an euler equation? or would it have to be solved in a different way?

Thanks again.
 
  • #7
vela
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Can't you get a non-trivial solution in the last case if b turned out to be a multiple of [itex]\pi/\log 2[/itex]? I'm assuming in this problem you have the freedom to adjust A and B to get the roots you want.
 
  • #8
HallsofIvy
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No. In order to have pure trig function (of ln(x)) solution, without a "x-B" term, you would have to have B= 0. But in that case, the characteristic equation is r2+ (A- 1)r= 0 which has roots r= 0 and r= 1- A, still not giving a periodic solution.
 
  • #9
vela
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I didn't have a pure trig function solution in mind. I was thinking of something like [itex]y=c |x|^{\alpha} \sin(\pi \log |x|/\log 2)[/itex]. It would vanish at both x=1 and x=2.
 
  • #10
HallsofIvy
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I didn't have a pure trig function solution in mind. I was thinking of something like [itex]y=c |x|^{\alpha} \sin(\pi \log |x|/\log 2)[/itex]. It would vanish at both x=1 and x=2.
Ah, thanks.
 
  • #11
HallsofIvy
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This same question was asked in "Differential Equations" so I merged the two threads.
gerald87, please do NOT post the same question repeatedly.
 
  • #12
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thankyou so much for the help guys.

I think I have got it under control now.

And sorry about the double posts.
 

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