Orthonormal basis for L2 (0,1) by using Laplacian's eigenfunctions.
A standard orthonormal basis for L2 (0,1) is given by the Fourier expansion, as described here, for example (Orthonormal Basis of $L^2$). On the other hand, it seems a standard result that the Laplacians eigenfunctions form an orthonormal basis, too.
Since I needed that basis for speeding-up a numerical simulation, I computed them directly: $phi_j (x) = sqrt{2} sin( j pi x )$ for positive natural $j$, eigenvalues $-(pi j)^2$.
On the other hand, the standard basis previously linked works very well in my script, while $phi$ causes some troubles. So, before digging into many lines of code, let's start from a simple point. I am honestly not to sure that $phi$ truly consituted a basis, maybe I just misunderstood this informal claim I remember from some past course. So I ask:
Does the family $phi$ constitutes a basis for L2(0,1)?
real-analysis numerical-methods approximation-theory laplacian
asked Dec 4 '18 at 15:36
user233650
48128
add a comment |
A standard orthonormal basis for L2 (0,1) is given by the Fourier expansion, as described here, for example (Orthonormal Basis of $L^2$). On the other hand, it seems a standard result that the Laplacians eigenfunctions form an orthonormal basis, too.
Since I needed that basis for speeding-up a numerical simulation, I computed them directly: $phi_j (x) = sqrt{2} sin( j pi x )$ for positive natural $j$, eigenvalues $-(pi j)^2$.
On the other hand, the standard basis previously linked works very well in my script, while $phi$ causes some troubles. So, before digging into many lines of code, let's start from a simple point. I am honestly not to sure that $phi$ truly consituted a basis, maybe I just misunderstood this informal claim I remember from some past course. So I ask:
Does the family $phi$ constitutes a basis for L2(0,1)?
real-analysis numerical-methods approximation-theory laplacian
asked Dec 4 '18 at 15:36
user233650
48128
add a comment |
A standard orthonormal basis for L2 (0,1) is given by the Fourier expansion, as described here, for example (Orthonormal Basis of $L^2$). On the other hand, it seems a standard result that the Laplacians eigenfunctions form an orthonormal basis, too.
Since I needed that basis for speeding-up a numerical simulation, I computed them directly: $phi_j (x) = sqrt{2} sin( j pi x )$ for positive natural $j$, eigenvalues $-(pi j)^2$.
On the other hand, the standard basis previously linked works very well in my script, while $phi$ causes some troubles. So, before digging into many lines of code, let's start from a simple point. I am honestly not to sure that $phi$ truly consituted a basis, maybe I just misunderstood this informal claim I remember from some past course. So I ask:
Does the family $phi$ constitutes a basis for L2(0,1)?
real-analysis numerical-methods approximation-theory laplacian
asked Dec 4 '18 at 15:36
user233650
48128
A standard orthonormal basis for L2 (0,1) is given by the Fourier expansion, as described here, for example (Orthonormal Basis of $L^2$). On the other hand, it seems a standard result that the Laplacians eigenfunctions form an orthonormal basis, too.
Since I needed that basis for speeding-up a numerical simulation, I computed them directly: $phi_j (x) = sqrt{2} sin( j pi x )$ for positive natural $j$, eigenvalues $-(pi j)^2$.
On the other hand, the standard basis previously linked works very well in my script, while $phi$ causes some troubles. So, before digging into many lines of code, let's start from a simple point. I am honestly not to sure that $phi$ truly consituted a basis, maybe I just misunderstood this informal claim I remember from some past course. So I ask:
Does the family $phi$ constitutes a basis for L2(0,1)?
real-analysis numerical-methods approximation-theory laplacian
real-analysis numerical-methods approximation-theory laplacian
asked Dec 4 '18 at 15:36
user233650
48128
asked Dec 4 '18 at 15:36
user233650
48128
asked Dec 4 '18 at 15:36
user233650
48128
asked Dec 4 '18 at 15:36
user233650
48128
asked Dec 4 '18 at 15:36
user233650
48128
48128
add a comment |
add a comment |
1 Answer
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Assuming you are restricting yourself to functions which vanish at the endpoints: Yes. Indeed, in this case, the Laplacian eigenfunctions coincide with the fourier basis (the cosines are excluded because they do not satisfy the boundary conditions). But the same fact holds in general.
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
1
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
add a comment |
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1 Answer
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1 Answer
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Assuming you are restricting yourself to functions which vanish at the endpoints: Yes. Indeed, in this case, the Laplacian eigenfunctions coincide with the fourier basis (the cosines are excluded because they do not satisfy the boundary conditions). But the same fact holds in general.
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
1
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
add a comment |
Assuming you are restricting yourself to functions which vanish at the endpoints: Yes. Indeed, in this case, the Laplacian eigenfunctions coincide with the fourier basis (the cosines are excluded because they do not satisfy the boundary conditions). But the same fact holds in general.
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
1
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
add a comment |
Assuming you are restricting yourself to functions which vanish at the endpoints: Yes. Indeed, in this case, the Laplacian eigenfunctions coincide with the fourier basis (the cosines are excluded because they do not satisfy the boundary conditions). But the same fact holds in general.
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
Assuming you are restricting yourself to functions which vanish at the endpoints: Yes. Indeed, in this case, the Laplacian eigenfunctions coincide with the fourier basis (the cosines are excluded because they do not satisfy the boundary conditions). But the same fact holds in general.
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
answered Dec 4 '18 at 16:03
Mike Hawk
1,500110
1,500110
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
1
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
add a comment |
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
1
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
Excellent answer, thank you so much. I forgot these boundary conditions. To drop them I can use the full Fourier basis, which actually is the Laplacian eigenfunction expansion. In other words, the two basis I am comparing, standard Fourier and Laplacian, are the same. Did I understand correctly?
– user233650
Dec 4 '18 at 16:44
1
1
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
@ user233650: Yes
– Mike Hawk
Dec 4 '18 at 17:29
add a comment |
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