General distributional solution of the Airy Equation
2
$begingroup$
How can I prove that the Airy equation
$$ frac{d^2u}{dx^2}-xu = 0 $$
has at least two linear independent solutions?
Once I've found it how can I prove the existence of two independent DISTRIBUTIONAL solutions
ordinary-differential-equations
asked Jan 6 at 18:41
MargeAMargeA
162
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migrated from physics.stackexchange.com Jan 6 at 19:37
This question came from our site for active researchers, academics and students of physics.
add a comment |
2
$begingroup$
How can I prove that the Airy equation
$$ frac{d^2u}{dx^2}-xu = 0 $$
has at least two linear independent solutions?
Once I've found it how can I prove the existence of two independent DISTRIBUTIONAL solutions
ordinary-differential-equations
asked Jan 6 at 18:41
MargeAMargeA
162
$endgroup$
migrated from physics.stackexchange.com Jan 6 at 19:37
This question came from our site for active researchers, academics and students of physics.
add a comment |
2
2
2
$begingroup$
How can I prove that the Airy equation
$$ frac{d^2u}{dx^2}-xu = 0 $$
has at least two linear independent solutions?
Once I've found it how can I prove the existence of two independent DISTRIBUTIONAL solutions
ordinary-differential-equations
asked Jan 6 at 18:41
MargeAMargeA
162
$endgroup$
How can I prove that the Airy equation
$$ frac{d^2u}{dx^2}-xu = 0 $$
has at least two linear independent solutions?
Once I've found it how can I prove the existence of two independent DISTRIBUTIONAL solutions
ordinary-differential-equations
ordinary-differential-equations
asked Jan 6 at 18:41
MargeAMargeA
162
asked Jan 6 at 18:41
MargeAMargeA
162
edited Jan 6 at 20:29
MargeA
asked Jan 6 at 18:41
MargeAMargeA
162
asked Jan 6 at 18:41
MargeAMargeA
162
asked Jan 6 at 18:41
MargeAMargeA
162
162
migrated from physics.stackexchange.com Jan 6 at 19:37
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Jan 6 at 19:37
This question came from our site for active researchers, academics and students of physics.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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$begingroup$
There is nothing special about the Airy equation. It is second-order linear equation. All such equations have two linearly independent solutions. Any book on differential equations will prove this.
answered Jan 6 at 19:00
mike stonemike stone
34317
$endgroup$
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
1
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
There is nothing special about the Airy equation. It is second-order linear equation. All such equations have two linearly independent solutions. Any book on differential equations will prove this.
answered Jan 6 at 19:00
mike stonemike stone
34317
$endgroup$
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
1
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
|
show 3 more comments
1
$begingroup$
There is nothing special about the Airy equation. It is second-order linear equation. All such equations have two linearly independent solutions. Any book on differential equations will prove this.
answered Jan 6 at 19:00
mike stonemike stone
34317
$endgroup$
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
1
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
|
show 3 more comments
1
1
1
$begingroup$
There is nothing special about the Airy equation. It is second-order linear equation. All such equations have two linearly independent solutions. Any book on differential equations will prove this.
answered Jan 6 at 19:00
mike stonemike stone
34317
$endgroup$
There is nothing special about the Airy equation. It is second-order linear equation. All such equations have two linearly independent solutions. Any book on differential equations will prove this.
answered Jan 6 at 19:00
mike stonemike stone
34317
answered Jan 6 at 19:00
mike stonemike stone
34317
answered Jan 6 at 19:00
mike stonemike stone
34317
answered Jan 6 at 19:00
mike stonemike stone
34317
34317
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
1
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
|
show 3 more comments
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
1
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
Even if it is a distribution?
$endgroup$
– MargeA
Jan 6 at 19:02
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
The two solutions are nice smooth -- even analytic --- functions. They can be regarded as distributions of course, but I feel that I am not understanding what it is that you are worried about . Why do distributionbs come into this? Are you perhaps thinking of them as some non-normalizable "eigenfunctions" of Schroedinger equation? Such generalized eigenfunctions can be thought of as distributions in a rigged Hilbert space.
$endgroup$
– mike stone
Jan 6 at 19:05
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
I asked about distributional solutions of the Airy equation. I found the solution thanks to the fourier transform, but I am not sure how to demonstrate that there are at least two independent solutions
$endgroup$
– MargeA
Jan 6 at 19:07
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
$begingroup$
@MargeA : Then please demonstrate this solution process (edit the question text) and highlight the point where you are uncertain about the existence of two basis solutions.
$endgroup$
– LutzL
Jan 6 at 19:42
1
1
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
$begingroup$
You get two linear independent solutions by replacing your integral's limits of $pm infty$ contours that run between any pair of the points $infty, e^{2pi i/3} infty, e^{-2pi i/3}infty$. The sum of all three contours is zero because the integrand is an entire function. I don't think that there are any singular distributional solutions in the weak sense of the equation.
$endgroup$
– mike stone
Jan 6 at 20:47
|
show 3 more comments
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