Why a solution for $c^2 Delta u = u_{tt}$ must have eigenfunctions as its series terms expansion?
I'm reading this which explains how to arrive at a solution for $u$ as a series expansion involving $J_0, J_1,cdots$ which are Bessel Functions.
It concludes at page 5 saying that $u$ is the following:
and that the terms $J_0$ and $J_i$ for $i=1,cdots$ are eigenfunctions of the Laplacian Operator. But why?
It isn't clear to me why the process of separating variables and arriving at a soluton will give me an expansion in terms of eigenfunctions. Maybe it's a theorem? Are the eigenvalues the terms multiplying $J_0, J_1,cdots$?
UPDATE:
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
real-analysis sequences-and-series pde bessel-functions wave-equation
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
add a comment |
I'm reading this which explains how to arrive at a solution for $u$ as a series expansion involving $J_0, J_1,cdots$ which are Bessel Functions.
It concludes at page 5 saying that $u$ is the following:
and that the terms $J_0$ and $J_i$ for $i=1,cdots$ are eigenfunctions of the Laplacian Operator. But why?
It isn't clear to me why the process of separating variables and arriving at a soluton will give me an expansion in terms of eigenfunctions. Maybe it's a theorem? Are the eigenvalues the terms multiplying $J_0, J_1,cdots$?
UPDATE:
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
real-analysis sequences-and-series pde bessel-functions wave-equation
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
you can expand the initial condition into the eigenfunction expansion. You can make the above informed guess for one solution, and then differentiate to verify it is indeed a solution. Then you can use a uniqueness theorem(conservation of energy) to show that its the only solution in some class of solutions
– Calvin Khor
Nov 26 '18 at 19:05
@CalvinKhor what do you mean by 'expand the initial condition into the eigenfunciton expansion'?
– Lucas Zanella
Nov 26 '18 at 19:13
@CalvinKhor I don't know about eigenfunction expansions, I want to understand why the above gives an eigenfunction expansion
– Lucas Zanella
Nov 26 '18 at 19:14
its written as a linear combination of eigenfunctions?
– Calvin Khor
Nov 26 '18 at 19:42
The separation parameters become eigenvalues, and the separated solutions are eigenfunctions of the separated ODES.
– DisintegratingByParts
Nov 27 '18 at 4:29
add a comment |
I'm reading this which explains how to arrive at a solution for $u$ as a series expansion involving $J_0, J_1,cdots$ which are Bessel Functions.
It concludes at page 5 saying that $u$ is the following:
and that the terms $J_0$ and $J_i$ for $i=1,cdots$ are eigenfunctions of the Laplacian Operator. But why?
It isn't clear to me why the process of separating variables and arriving at a soluton will give me an expansion in terms of eigenfunctions. Maybe it's a theorem? Are the eigenvalues the terms multiplying $J_0, J_1,cdots$?
UPDATE:
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
real-analysis sequences-and-series pde bessel-functions wave-equation
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
I'm reading this which explains how to arrive at a solution for $u$ as a series expansion involving $J_0, J_1,cdots$ which are Bessel Functions.
It concludes at page 5 saying that $u$ is the following:
and that the terms $J_0$ and $J_i$ for $i=1,cdots$ are eigenfunctions of the Laplacian Operator. But why?
It isn't clear to me why the process of separating variables and arriving at a soluton will give me an expansion in terms of eigenfunctions. Maybe it's a theorem? Are the eigenvalues the terms multiplying $J_0, J_1,cdots$?
UPDATE:
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
real-analysis sequences-and-series pde bessel-functions wave-equation
real-analysis sequences-and-series pde bessel-functions wave-equation
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
edited Nov 28 '18 at 14:25
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
asked Nov 26 '18 at 18:40
Lucas Zanella
92911330
92911330
you can expand the initial condition into the eigenfunction expansion. You can make the above informed guess for one solution, and then differentiate to verify it is indeed a solution. Then you can use a uniqueness theorem(conservation of energy) to show that its the only solution in some class of solutions
– Calvin Khor
Nov 26 '18 at 19:05
@CalvinKhor what do you mean by 'expand the initial condition into the eigenfunciton expansion'?
– Lucas Zanella
Nov 26 '18 at 19:13
@CalvinKhor I don't know about eigenfunction expansions, I want to understand why the above gives an eigenfunction expansion
– Lucas Zanella
Nov 26 '18 at 19:14
its written as a linear combination of eigenfunctions?
– Calvin Khor
Nov 26 '18 at 19:42
The separation parameters become eigenvalues, and the separated solutions are eigenfunctions of the separated ODES.
– DisintegratingByParts
Nov 27 '18 at 4:29
add a comment |
you can expand the initial condition into the eigenfunction expansion. You can make the above informed guess for one solution, and then differentiate to verify it is indeed a solution. Then you can use a uniqueness theorem(conservation of energy) to show that its the only solution in some class of solutions
– Calvin Khor
Nov 26 '18 at 19:05
@CalvinKhor what do you mean by 'expand the initial condition into the eigenfunciton expansion'?
– Lucas Zanella
Nov 26 '18 at 19:13
@CalvinKhor I don't know about eigenfunction expansions, I want to understand why the above gives an eigenfunction expansion
– Lucas Zanella
Nov 26 '18 at 19:14
its written as a linear combination of eigenfunctions?
– Calvin Khor
Nov 26 '18 at 19:42
The separation parameters become eigenvalues, and the separated solutions are eigenfunctions of the separated ODES.
– DisintegratingByParts
Nov 27 '18 at 4:29
you can expand the initial condition into the eigenfunction expansion. You can make the above informed guess for one solution, and then differentiate to verify it is indeed a solution. Then you can use a uniqueness theorem(conservation of energy) to show that its the only solution in some class of solutions
– Calvin Khor
Nov 26 '18 at 19:05
you can expand the initial condition into the eigenfunction expansion. You can make the above informed guess for one solution, and then differentiate to verify it is indeed a solution. Then you can use a uniqueness theorem(conservation of energy) to show that its the only solution in some class of solutions
– Calvin Khor
Nov 26 '18 at 19:05
@CalvinKhor what do you mean by 'expand the initial condition into the eigenfunciton expansion'?
– Lucas Zanella
Nov 26 '18 at 19:13
@CalvinKhor what do you mean by 'expand the initial condition into the eigenfunciton expansion'?
– Lucas Zanella
Nov 26 '18 at 19:13
@CalvinKhor I don't know about eigenfunction expansions, I want to understand why the above gives an eigenfunction expansion
– Lucas Zanella
Nov 26 '18 at 19:14
@CalvinKhor I don't know about eigenfunction expansions, I want to understand why the above gives an eigenfunction expansion
– Lucas Zanella
Nov 26 '18 at 19:14
its written as a linear combination of eigenfunctions?
– Calvin Khor
Nov 26 '18 at 19:42
its written as a linear combination of eigenfunctions?
– Calvin Khor
Nov 26 '18 at 19:42
The separation parameters become eigenvalues, and the separated solutions are eigenfunctions of the separated ODES.
– DisintegratingByParts
Nov 27 '18 at 4:29
The separation parameters become eigenvalues, and the separated solutions are eigenfunctions of the separated ODES.
– DisintegratingByParts
Nov 27 '18 at 4:29
add a comment |
1 Answer
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It arises from the Neumann spectral theorem which guarantees that Laplacian operator has basis of eigenfunctions and that basis is full and orthogonal. So every function in Hilbert space can be represented as a linear sum of eigenfunctions and that representation is unique. And since the time operator is linear too, all eigenfunctions evolve independently.
So the logic is the following.
- At each point $t$ the solution $tilde u$ can be uniquely represented as a sum of series of eigenfunctions $f_i(mathbf{r})$ with coefficients $a_i$.
- Thus, coefficients are some functions of $t$: $a_i = a_i(t)$ and the solution can be uniquely represented as a sum
$$
tilde u(mathbf r, t) = sum_i a_i(t)f_i(mathbf r)
$$
- Since both Laplacian and second time derivative are linear, then all terms $a_i(t)f_i(mathbf r)$ must satisfy original equation.
The eigenvalues are those $lambda_{n,m}$.
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
|
show 1 more comment
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1 Answer
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It arises from the Neumann spectral theorem which guarantees that Laplacian operator has basis of eigenfunctions and that basis is full and orthogonal. So every function in Hilbert space can be represented as a linear sum of eigenfunctions and that representation is unique. And since the time operator is linear too, all eigenfunctions evolve independently.
So the logic is the following.
- At each point $t$ the solution $tilde u$ can be uniquely represented as a sum of series of eigenfunctions $f_i(mathbf{r})$ with coefficients $a_i$.
- Thus, coefficients are some functions of $t$: $a_i = a_i(t)$ and the solution can be uniquely represented as a sum
$$
tilde u(mathbf r, t) = sum_i a_i(t)f_i(mathbf r)
$$
- Since both Laplacian and second time derivative are linear, then all terms $a_i(t)f_i(mathbf r)$ must satisfy original equation.
The eigenvalues are those $lambda_{n,m}$.
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
|
show 1 more comment
It arises from the Neumann spectral theorem which guarantees that Laplacian operator has basis of eigenfunctions and that basis is full and orthogonal. So every function in Hilbert space can be represented as a linear sum of eigenfunctions and that representation is unique. And since the time operator is linear too, all eigenfunctions evolve independently.
So the logic is the following.
- At each point $t$ the solution $tilde u$ can be uniquely represented as a sum of series of eigenfunctions $f_i(mathbf{r})$ with coefficients $a_i$.
- Thus, coefficients are some functions of $t$: $a_i = a_i(t)$ and the solution can be uniquely represented as a sum
$$
tilde u(mathbf r, t) = sum_i a_i(t)f_i(mathbf r)
$$
- Since both Laplacian and second time derivative are linear, then all terms $a_i(t)f_i(mathbf r)$ must satisfy original equation.
The eigenvalues are those $lambda_{n,m}$.
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
|
show 1 more comment
It arises from the Neumann spectral theorem which guarantees that Laplacian operator has basis of eigenfunctions and that basis is full and orthogonal. So every function in Hilbert space can be represented as a linear sum of eigenfunctions and that representation is unique. And since the time operator is linear too, all eigenfunctions evolve independently.
So the logic is the following.
- At each point $t$ the solution $tilde u$ can be uniquely represented as a sum of series of eigenfunctions $f_i(mathbf{r})$ with coefficients $a_i$.
- Thus, coefficients are some functions of $t$: $a_i = a_i(t)$ and the solution can be uniquely represented as a sum
$$
tilde u(mathbf r, t) = sum_i a_i(t)f_i(mathbf r)
$$
- Since both Laplacian and second time derivative are linear, then all terms $a_i(t)f_i(mathbf r)$ must satisfy original equation.
The eigenvalues are those $lambda_{n,m}$.
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
It arises from the Neumann spectral theorem which guarantees that Laplacian operator has basis of eigenfunctions and that basis is full and orthogonal. So every function in Hilbert space can be represented as a linear sum of eigenfunctions and that representation is unique. And since the time operator is linear too, all eigenfunctions evolve independently.
So the logic is the following.
- At each point $t$ the solution $tilde u$ can be uniquely represented as a sum of series of eigenfunctions $f_i(mathbf{r})$ with coefficients $a_i$.
- Thus, coefficients are some functions of $t$: $a_i = a_i(t)$ and the solution can be uniquely represented as a sum
$$
tilde u(mathbf r, t) = sum_i a_i(t)f_i(mathbf r)
$$
- Since both Laplacian and second time derivative are linear, then all terms $a_i(t)f_i(mathbf r)$ must satisfy original equation.
The eigenvalues are those $lambda_{n,m}$.
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
edited Nov 27 '18 at 11:32
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
answered Nov 26 '18 at 19:43
Vasily Mitch
1,33837
1,33837
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
|
show 1 more comment
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Can you explain it better why the equation is $u_{tt}-c^2nabla u = 0 $? If we were to find eigenvectors of the laplacian wouldn't I need to search for solutions of $cu-nabla u = 0 $?
– Lucas Zanella
Nov 28 '18 at 14:24
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
Equation $u_{tt}=c^2Delta u$ is what we a trying to solve. To find eigenfucntions of Laplacian operator we indeed need to solve $Delta u - lambda u = 0$
– Vasily Mitch
Nov 28 '18 at 14:57
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
But in the PDF he solves $u_{tt} = c^2 nabla u$ and says that the solution for that are the laplacian eigenvalues. Why exactly?
– Lucas Zanella
Nov 28 '18 at 14:59
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I explained it in the answer. What part of it is confusing?
– Vasily Mitch
Nov 28 '18 at 15:04
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
I think the 3rd step is where you explained, right? If so, what I understood is that each term of the series must satisfy $u_{tt}-c^2nabla u=0$ which is not $nabla u - cu = 0$
– Lucas Zanella
Nov 28 '18 at 15:12
|
show 1 more comment
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you can expand the initial condition into the eigenfunction expansion. You can make the above informed guess for one solution, and then differentiate to verify it is indeed a solution. Then you can use a uniqueness theorem(conservation of energy) to show that its the only solution in some class of solutions
– Calvin Khor
Nov 26 '18 at 19:05
@CalvinKhor what do you mean by 'expand the initial condition into the eigenfunciton expansion'?
– Lucas Zanella
Nov 26 '18 at 19:13
@CalvinKhor I don't know about eigenfunction expansions, I want to understand why the above gives an eigenfunction expansion
– Lucas Zanella
Nov 26 '18 at 19:14
its written as a linear combination of eigenfunctions?
– Calvin Khor
Nov 26 '18 at 19:42
The separation parameters become eigenvalues, and the separated solutions are eigenfunctions of the separated ODES.
– DisintegratingByParts
Nov 27 '18 at 4:29