How Do Boundary Conditions Affect Complex Eigenvalues in Differential Operators?

In summary: You're welcome!In summary, the problem deals with finding the complex eigenvalues of the first derivative operator d/dx subject to the single boundary condition X(0) = X(1). The solution involves using the fact that ##e^{2 \pi i}=1## and applying Euler's formula. The function is put in the form X'(x) = λX in order to solve for the eigenvalues, which follows the same concept as eigenvalues in linear algebra.
  • #1
trap101
342
0
Find the complex eignevalues of the first derivative operator d/dx subject to the single boundary condition X(0) = X(1).

So this has to do with PDEs and separation of variables:

I get to the point of using the BC and I am left with an expression:

1 = eλ, this is where my issue falls. How do I convert it into a complex eignvalue and then also get that eigenfunction? I know how to do this when dealing with the Fourier series in terms of sines and cosine, but complex has me baffled and I know our professor is going to ask us about the complex form on our test in 2 days, he's been talking about it all semester.
 
Physics news on Phys.org
  • #2
trap101 said:
Find the complex eignevalues of the first derivative operator d/dx subject to the single boundary condition X(0) = X(1).

So this has to do with PDEs and separation of variables:

I get to the point of using the BC and I am left with an expression:

1 = eλ, this is where my issue falls. How do I convert it into a complex eignvalue and then also get that eigenfunction? I know how to do this when dealing with the Fourier series in terms of sines and cosine, but complex has me baffled and I know our professor is going to ask us about the complex form on our test in 2 days, he's been talking about it all semester.

Well, ##e^0=1## and ##e^{2 \pi i}=1##. Does that give you any ideas?
 
  • #3
Dick said:
Well, ##e^0=1## and ##e^{2 \pi i}=1##. Does that give you any ideas?

erck...kinda sorta: I suppose I could set λ = 2[itex]\pi[/itex]i and the only way that equals 0 is when it is 2n[itex]\pi[/itex]i, but that is me reaching, the issue I think falls with me not even knowing how to manipulate the complex representation too well.

How is ##e^{2 \pi i}=1##? Should I take it in terms of euler's formula to see the relation?
 
  • #4
trap101 said:
erck...kinda sorta: I suppose I could set λ = 2[itex]\pi[/itex]i and the only way that equals 0 is when it is 2n[itex]\pi[/itex]i, but that is me reaching, the issue I think falls with me not even knowing how to manipulate the complex representation too well.

How is ##e^{2 \pi i}=1##? Should I take it in terms of euler's formula to see the relation?

Yes, use Euler's formula. If ##\lambda=a+bi## with a and b real then ##e^{a+bi}=e^a e^{bi}=e^a (\cos(b)+i\sin(b))##. Set that equal to 1 and figure out what a and b can be.
 
  • #5
Dick said:
Yes, use Euler's formula. If ##\lambda=a+bi## with a and b real then ##e^{a+bi}=e^a e^{bi}=e^a (\cos(b)+i\sin(b))##. Set that equal to 1 and figure out what a and b can be.

Ok thanks. I actually had another quaetion with regards to this problem. In finding the solution I followed a procedure they did in the book:

X'(x) = λX and then solved for the solution, but my question is how or why did they put the function in this form first?
 
  • #6
trap101 said:
Ok thanks. I actually had another quaetion with regards to this problem. In finding the solution I followed a procedure they did in the book:

X'(x) = λX and then solved for the solution, but my question is how or why did they put the function in this form first?

If X(x) is an eigenfunction of an operator O with eigenvalue λ, then O(X(x))=λX(x). That's the definition of eigenfunction and eigenvalue. Now put O=d/dx.
 
  • #7
Dick said:
If X(x) is an eigenfunction of an operator O with eigenvalue λ, then O(X(x))=λX(x). That's the definition of eigenfunction and eigenvalue. Now put O=d/dx.


Ohhhh, now if I consolidate this with what I previoiusly studied in linear algebra:

that would mean AX = λX, but in this case we're dealing with functions so that A matrix is now an operator. Awesome. Aha moments... thanks.
 
  • #8
trap101 said:
Ohhhh, now if I consolidate this with what I previoiusly studied in linear algebra:

that would mean AX = λX, but in this case we're dealing with functions so that A matrix is now an operator. Awesome. Aha moments... thanks.

Yes, it's same concept as eigenvalues in linear algebra.
 

Related to How Do Boundary Conditions Affect Complex Eigenvalues in Differential Operators?

1. What are complex eigenvalues?

Complex eigenvalues are values that can be obtained when solving for the eigenvalues of a square matrix. These values can have a real and imaginary component, and are represented as a + bi, where a and b are real numbers and i is the imaginary unit.

2. Why do we need to find complex eigenvalues?

Complex eigenvalues are important in many areas of mathematics, physics, and engineering. They are used in solving systems of differential equations, understanding the behavior of dynamical systems, and in the analysis of quantum mechanics. They also provide insight into the geometric properties of a matrix and its transformations.

3. How do you find complex eigenvalues?

To find complex eigenvalues, you first need to compute the eigenvalues of the matrix. This can be done by finding the roots of the characteristic polynomial of the matrix. Once the eigenvalues are found, you can determine if they are complex by checking if the discriminant of the characteristic polynomial is negative. If it is negative, the eigenvalues will have imaginary components.

4. Are complex eigenvalues always present in a matrix?

No, complex eigenvalues are not always present in a matrix. The presence of complex eigenvalues depends on the properties of the matrix, such as its size, symmetry, and determinant. For example, a 2x2 matrix with real entries will always have real eigenvalues, while a 3x3 matrix can have both real and complex eigenvalues.

5. Can complex eigenvalues be found for non-square matrices?

No, complex eigenvalues can only be found for square matrices. This is because the eigenvalues of a matrix are defined as the values that satisfy the characteristic polynomial, which can only be computed for square matrices. Non-square matrices do not have eigenvalues, but they do have singular values which serve a similar purpose in matrix analysis.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
Replies
13
Views
1K
Replies
2
Views
755
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Quantum Physics
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
Replies
22
Views
572
Replies
16
Views
404
  • Calculus and Beyond Homework Help
Replies
15
Views
3K
Back
Top