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Pietro Majer
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This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$-\ddot u(x)+ q(x)u(x)=\lambda u(x)\qquad $ (SL)

$u(0)=0, u(\pi)=0.$$$ {(SL)} \qquad\begin{cases}-\ddot u(x)+ q(x)u(x)=\lambda u(x\\ u(0)=0\\ u(\pi)=0.\end{cases}$$

The fact that the n$n$-th eigenfunction has exactly n+1$n+1$ zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^2([0,\pi])$ zero on the boundary, the subset A$A$ of those all whose zeros are simple, is an open set with infinitely many connected components, and each component is characterized by the (finite) common number of zeros of functions in it.

  3. the n$n$-th (normalized) eigenfunction of (SL) depends continuously on the coefficient q(x)$q(x)$; therefore, for any n, the n-th eigenfunctions are in the same connected component of A$A$, irrespectively on q(x)$q(x)$. Thus they have the same number of zeros, which is the same as the n$n$-th eigenfunction corresponding to q= 1 $q= 1 $(constant), that is n+1.$n+1.$

This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$-\ddot u(x)+ q(x)u(x)=\lambda u(x)\qquad $ (SL)

$u(0)=0, u(\pi)=0.$

The fact that the n-th eigenfunction has exactly n+1 zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^2([0,\pi])$ zero on the boundary, the subset A of those all whose zeros are simple, is an open set with infinitely many connected components, and each component is characterized by the (finite) common number of zeros of functions in it.

  3. the n-th (normalized) eigenfunction of (SL) depends continuously on the coefficient q(x); therefore, for any n, the n-th eigenfunctions are in the same connected component of A, irrespectively on q(x). Thus they have the same number of zeros, which is the same as the n-th eigenfunction corresponding to q= 1 (constant), that is n+1.

This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$$ {(SL)} \qquad\begin{cases}-\ddot u(x)+ q(x)u(x)=\lambda u(x\\ u(0)=0\\ u(\pi)=0.\end{cases}$$

The fact that the $n$-th eigenfunction has exactly $n+1$ zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^2([0,\pi])$ zero on the boundary, the subset $A$ of those all whose zeros are simple, is an open set with infinitely many connected components, and each component is characterized by the (finite) common number of zeros of functions in it.

  3. the $n$-th (normalized) eigenfunction of (SL) depends continuously on the coefficient $q(x)$; therefore, for any n, the n-th eigenfunctions are in the same connected component of $A$, irrespectively on $q(x)$. Thus they have the same number of zeros, which is the same as the $n$-th eigenfunction corresponding to $q= 1 $(constant), that is $n+1.$

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Pietro Majer
  • 60.6k
  • 4
  • 122
  • 269

This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$-\ddot u(x)+ q(x)u(x)=\lambda u(x)\qquad $ (SL)

$u(0)=0, u(\pi)=0.$

The fact that the n-th eigenfunction has exactly n+1 zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^1([0,\pi])$$C^2([0,\pi])$ zero on the boundary, the subset A of those all whose zeros are simple zeros are, is an open set with infinitely many connected components, and each component isis characterized by the (finite) common number of zeros of its elements;functions in it.

  3. the n-th (normalized) eigenfunction of (SL) depends continuously on the coefficient q(x); therefore all, for any n, the n-th eigenfunctions are in the same connected component of themA, irrespectively on q(x). Thus they have the same number of zeros, which is the same as the n-th eigenfunction corresponding to q= 1 (constant), that is n+1.

This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$-\ddot u(x)+ q(x)u(x)=\lambda u(x)\qquad $ (SL)

$u(0)=0, u(\pi)=0.$

The fact that the n-th eigenfunction has exactly n+1 zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^1([0,\pi])$ zero on the boundary, those all whose zeros are simple zeros are an open set with infinitely many connected components, and each component is characterized by the common number of zeros of its elements;

  3. the n-th (normalized) eigenfunction of (SL) depends continuously on the coefficient q(x); therefore all of them have the same number of zeros, which is the same as the n-th eigenfunction corresponding to q= 1 (constant), that is n+1.

This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$-\ddot u(x)+ q(x)u(x)=\lambda u(x)\qquad $ (SL)

$u(0)=0, u(\pi)=0.$

The fact that the n-th eigenfunction has exactly n+1 zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^2([0,\pi])$ zero on the boundary, the subset A of those all whose zeros are simple, is an open set with infinitely many connected components, and each component is characterized by the (finite) common number of zeros of functions in it.

  3. the n-th (normalized) eigenfunction of (SL) depends continuously on the coefficient q(x); therefore, for any n, the n-th eigenfunctions are in the same connected component of A, irrespectively on q(x). Thus they have the same number of zeros, which is the same as the n-th eigenfunction corresponding to q= 1 (constant), that is n+1.

Source Link
Pietro Majer
  • 60.6k
  • 4
  • 122
  • 269

This is vintage ODE, but may be nice according to the level of the course. Consider a Sturm -Liouville eigenvalue problem, say, not to exceed in generality, $q\in C^0([0,\pi])$ and

$-\ddot u(x)+ q(x)u(x)=\lambda u(x)\qquad $ (SL)

$u(0)=0, u(\pi)=0.$

The fact that the n-th eigenfunction has exactly n+1 zeros in $[0, \pi]$ may be seen as a consequence that

  1. non trivial solution of a second order linear ODE, in particular the eigenfunctions of (SL), must have simple zeros (by the unicity of the solution of the Cauchy problem);

  2. in the space of all functions in $C^1([0,\pi])$ zero on the boundary, those all whose zeros are simple zeros are an open set with infinitely many connected components, and each component is characterized by the common number of zeros of its elements;

  3. the n-th (normalized) eigenfunction of (SL) depends continuously on the coefficient q(x); therefore all of them have the same number of zeros, which is the same as the n-th eigenfunction corresponding to q= 1 (constant), that is n+1.