Differential equation with singular boundary conditions

In summary, the conversation discusses solving a Sturm-Liouville problem with the given boundary conditions by substituting f=y/x. This results in a solution of u''+au=0 and y=xu=c1*x*cos(kx)+c2*x*sin(kx). However, since y'(0)=0 and y(1)=0, the only solution is y=0, making it a trivial problem. The conversation also mentions trying different methods such as using orthogonality of solutions and series expansions, but they were not successful.
  • #1
FunkyDwarf
489
0
Hey guys, just need some hints with this doosey

Homework Statement


We have
(x^2 y')' + ax^2y = 0 where a the eigenvalue (a sturm-lioville problem) (sp?)
with y'(0)=y(1) = 0 and we get the hint to substitute f = y/x.

The Attempt at a Solution



Ok so i get the general solution being a sum of cosines and sines and i fiddled around a LOT with limits and exponentials and small angle approximations etc to try and get around the fact that the function blows up at x = 0 and were supposed to fit a boundary condition there. In all my efforts, the ones that make SOME sense, i get the eigen values = 0 which kind of makes it a trivial problem. I also tried taking advantage of the orthogonality of solutions given we know the weighting function x^2 but that required me to know what at least one eigenvalue was, and when i tried zero i had trouble solving exactly for the second but graphically it appeared to be close to root 2 but not quite.

Im sure I am missing something stupid so if you could, gently, point that out that would be super =)

Cheers
-Z

EDIT: i also tried a series expansion with the first 3 terms for sine and cos but got lost =( i thought about trying a general series solution but wasnt sure
 
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  • #2
Well, using hint would be a good start! If you let u= y/x, so that y= xu, what does the equation become?
 
  • #3
With the substitution u=y/x equation becomes
u''+au=0
which has solution
u=c1*cos(kx)+c2*sin(kx) [k=sqrt(a)]
so that
y=xu=c1*x*cos(kx)+c2*x*sin(kx)
So doing y'(0)=c1=0 y(1)=c1*cos(k)+c2*sin(k)=0
i.e. y=0, onlythe trivial solution ?
 
  • #4
Thats what i got (sorry if it wasnt obvious, yeh i used the hint =) )
 

FAQ: Differential equation with singular boundary conditions

1. What are singular boundary conditions in a differential equation?

Singular boundary conditions in a differential equation occur when the boundary values are infinite or do not exist. This means that the solution to the differential equation may not be continuous at the boundary.

2. How do singular boundary conditions affect the solution of a differential equation?

Singular boundary conditions can make it more difficult to find a unique solution to a differential equation. They may also lead to more complex solutions, such as piecewise functions.

3. What are some examples of differential equations with singular boundary conditions?

One example is the Bessel's equation, which arises in problems involving circular or spherical symmetry. Another example is the Airy's equation, which describes the motion of a particle in a potential field.

4. How do you solve a differential equation with singular boundary conditions?

The method for solving a differential equation with singular boundary conditions depends on the specific equation and boundary conditions. In some cases, it may involve using special techniques such as the Frobenius method or the Laplace transform.

5. What are the applications of differential equations with singular boundary conditions?

Differential equations with singular boundary conditions have various applications in physics, engineering, and other fields. They are used to model systems that exhibit discontinuous behavior, such as in the study of phase transitions or in the analysis of electronic circuits.

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