Self-adjoint boundary value (Sturm-Liouville)

In summary, the boundary conditions are self-adjoint if r(f'\bar g-f\bar g ') = 0 for all f, g satisfying the boundary conditions.
  • #1
Incand
334
47

Homework Statement


Under what condition on the constant ##c## and ##c'## are the boundary conditions ##f(b) = cf(a)## and ##f'(b)=c'f'(a)## self-adjoint for the operator ##L(f) = (rf')'+pf## on ##[a,b]##? (Assume that ##r,p## are real.)

Homework Equations


The boundary conditions are self-adjoint (relative L) if
##\left[r(f'\bar g-f\bar g ')\right]_a^b=0##
for all ##f,g## satisfying the boundary conditions.
(Assuming that ##L## is formally self-adjoint as it is in this case.)

The Attempt at a Solution


##\left[r(f'\bar g-f\bar g ')\right]_a^b = r(b)\left( f'(b)\bar g (b) - f(b)\bar g '(b)\right) -r(a)\left( f'(a)\bar g(a)-f(a)\bar g '(a)\right) = f'(a)\bar g (a)\left(c'\bar c r(b)-r(a)\right) + f(a)\bar g(a)\left(-r(b)c\bar c'+r(a)\right)##

So a solution would be ##c'\bar c = c\bar c' = \frac{r(a)}{r(b)}##.

The answer to the question says only ##c\bar c' = \frac{r(a)}{r(b)}## but I got that ##c'\bar c## also has to have the same value. Am I correct here or did i make a mistake?
 
Physics news on Phys.org
  • #2
Incand said:

Homework Statement


Under what condition on the constant ##c## and ##c'## are the boundary conditions ##f(b) = cf(a)## and ##f'(b)=c'f'(a)## self-adjoint for the operator ##L(f) = (rf')'+pf## on ##[a,b]##? (Assume that ##r,p## are real.)

Homework Equations


The boundary conditions are self-adjoint (relative L) if
##\left[r(f'\bar g-f\bar g ')\right]_a^b=0##
for all ##f,g## satisfying the boundary conditions.
(Assuming that ##L## is formally self-adjoint as it is in this case.)

The Attempt at a Solution


##\left[r(f'\bar g-f\bar g ')\right]_a^b = r(b)\left( f'(b)\bar g (b) - f(b)\bar g '(b)\right) -r(a)\left( f'(a)\bar g(a)-f(a)\bar g '(a)\right) = f'(a)\bar g (a)\left(c'\bar c r(b)-r(a)\right) + f(a)\bar g(a)\left(-r(b)c\bar c'+r(a)\right)##

So a solution would be ##c'\bar c = c\bar c' = \frac{r(a)}{r(b)}##.

The answer to the question says only ##c\bar c' = \frac{r(a)}{r(b)}## but I got that ##c'\bar c## also has to have the same value. Am I correct here or did i make a mistake?

[itex]c'\bar c[/itex] is the complex conjugate of [itex]\bar c' c[/itex]. The two are equal as [itex]\frac{r(a)}{r(b)}[/itex] is real and therefore equal to its complex conjugate.
 
  • Like
Likes Incand
  • #3
pasmith said:
[itex]c'\bar c[/itex] is the complex conjugate of [itex]\bar c' c[/itex]. The two are equal as [itex]\frac{r(a)}{r(b)}[/itex] is real and therefore equal to its complex conjugate.
Thanks!
 

Related to Self-adjoint boundary value (Sturm-Liouville)

1. What is a self-adjoint boundary value problem?

A self-adjoint boundary value problem is a type of differential equation that involves solving for a function and its derivatives such that the function satisfies certain boundary conditions. The term "self-adjoint" refers to the fact that the differential operator in the equation is equal to its own adjoint, making the problem easier to solve mathematically.

2. What is the significance of self-adjoint boundary value problems in physics and engineering?

Self-adjoint boundary value problems arise frequently in physics and engineering because they can be used to model many physical phenomena, such as heat transfer, fluid flow, and quantum mechanics. They also have special properties that make them easier to solve and lead to more accurate predictions.

3. How are Sturm-Liouville problems related to self-adjoint boundary value problems?

Sturm-Liouville problems are a specific type of self-adjoint boundary value problem that involves a second-order linear differential equation with a variable coefficient. They are named after mathematicians Jacques Charles François Sturm and Joseph Liouville, who studied these types of problems in the 19th century.

4. Can self-adjoint boundary value problems have multiple solutions?

In general, self-adjoint boundary value problems have a unique solution. However, there are cases where they can have multiple solutions, such as when the boundary conditions are not well-defined or when there are multiple eigenvalues (values that satisfy the boundary conditions) for the differential equation.

5. How are self-adjoint boundary value problems solved?

Self-adjoint boundary value problems can be solved using various analytical and numerical methods, depending on the complexity of the problem. Some common techniques include separation of variables, eigenfunction expansions, and the method of Green's functions. Advanced computational methods, such as finite element analysis, can also be used to solve these types of problems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
856
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
503
  • Calculus and Beyond Homework Help
Replies
16
Views
775
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
797
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Special and General Relativity
Replies
2
Views
1K
Back
Top