Mixing
Series solution of the second order ODE around a regular singular point
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Here is the ODE I want to integrate,
$$R''(y)-frac{2}{k-y}R'(y)-frac{l(l+1)}{(k-y)^{2}}R(y)=0$$
We see that it has a regular singular point at $y=k$ where $k<0$. Is there a way to obtain the solution of this ODE? What I am thinking is to obtain a series solution of the ODE of the form,
$$R(y)=c_{1}(y-k)^{alpha_{1}}(a_{1}+a_{2}(y-k)+a_{3}(y-k)^{2}+...)+c_{2}(y-k)^{alpha_{2}}(b_{1}+b_{2}(y-k)+b_{3}(y-k)^{2}+...)$$
and then find the indicial exponents $alpha_{1}$ and $alpha_{2}$ as well as the constants $a_{1},a_{2},...$ and $b_{1}, b_{2},...$. The crucial part is the computation of the constants $c_{1}$ and $c_{2}$ which I think can be obtained numerically by sampling points in the neighborhood of $y=k$ and then linearly fit with $R$ to obtain $c_{1}$ and $c_{2}$. But I am not sure of that. Is there a better efficient computational method of integrating the ODE?
integration ordinary-differential-equations power-series singularity frobenius-method
asked Feb 2 at 8:28
user583893user583893
336
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add a comment |
$begingroup$
Here is the ODE I want to integrate,
$$R''(y)-frac{2}{k-y}R'(y)-frac{l(l+1)}{(k-y)^{2}}R(y)=0$$
We see that it has a regular singular point at $y=k$ where $k<0$. Is there a way to obtain the solution of this ODE? What I am thinking is to obtain a series solution of the ODE of the form,
$$R(y)=c_{1}(y-k)^{alpha_{1}}(a_{1}+a_{2}(y-k)+a_{3}(y-k)^{2}+...)+c_{2}(y-k)^{alpha_{2}}(b_{1}+b_{2}(y-k)+b_{3}(y-k)^{2}+...)$$
and then find the indicial exponents $alpha_{1}$ and $alpha_{2}$ as well as the constants $a_{1},a_{2},...$ and $b_{1}, b_{2},...$. The crucial part is the computation of the constants $c_{1}$ and $c_{2}$ which I think can be obtained numerically by sampling points in the neighborhood of $y=k$ and then linearly fit with $R$ to obtain $c_{1}$ and $c_{2}$. But I am not sure of that. Is there a better efficient computational method of integrating the ODE?
integration ordinary-differential-equations power-series singularity frobenius-method
asked Feb 2 at 8:28
user583893user583893
336
$endgroup$
$begingroup$
This is a Cauchy-Euler equation. Solution has the form $R = (y-k)^r$
$endgroup$
– Dylan
Feb 2 at 9:04
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Hi @Dylan, how can one obtain the exponent $r$?
$endgroup$
– user583893
Feb 2 at 9:14
add a comment |
$begingroup$
Here is the ODE I want to integrate,
$$R''(y)-frac{2}{k-y}R'(y)-frac{l(l+1)}{(k-y)^{2}}R(y)=0$$
We see that it has a regular singular point at $y=k$ where $k<0$. Is there a way to obtain the solution of this ODE? What I am thinking is to obtain a series solution of the ODE of the form,
$$R(y)=c_{1}(y-k)^{alpha_{1}}(a_{1}+a_{2}(y-k)+a_{3}(y-k)^{2}+...)+c_{2}(y-k)^{alpha_{2}}(b_{1}+b_{2}(y-k)+b_{3}(y-k)^{2}+...)$$
and then find the indicial exponents $alpha_{1}$ and $alpha_{2}$ as well as the constants $a_{1},a_{2},...$ and $b_{1}, b_{2},...$. The crucial part is the computation of the constants $c_{1}$ and $c_{2}$ which I think can be obtained numerically by sampling points in the neighborhood of $y=k$ and then linearly fit with $R$ to obtain $c_{1}$ and $c_{2}$. But I am not sure of that. Is there a better efficient computational method of integrating the ODE?
integration ordinary-differential-equations power-series singularity frobenius-method
asked Feb 2 at 8:28
user583893user583893
336
$endgroup$
Here is the ODE I want to integrate,
$$R''(y)-frac{2}{k-y}R'(y)-frac{l(l+1)}{(k-y)^{2}}R(y)=0$$
We see that it has a regular singular point at $y=k$ where $k<0$. Is there a way to obtain the solution of this ODE? What I am thinking is to obtain a series solution of the ODE of the form,
$$R(y)=c_{1}(y-k)^{alpha_{1}}(a_{1}+a_{2}(y-k)+a_{3}(y-k)^{2}+...)+c_{2}(y-k)^{alpha_{2}}(b_{1}+b_{2}(y-k)+b_{3}(y-k)^{2}+...)$$
and then find the indicial exponents $alpha_{1}$ and $alpha_{2}$ as well as the constants $a_{1},a_{2},...$ and $b_{1}, b_{2},...$. The crucial part is the computation of the constants $c_{1}$ and $c_{2}$ which I think can be obtained numerically by sampling points in the neighborhood of $y=k$ and then linearly fit with $R$ to obtain $c_{1}$ and $c_{2}$. But I am not sure of that. Is there a better efficient computational method of integrating the ODE?
integration ordinary-differential-equations power-series singularity frobenius-method
integration ordinary-differential-equations power-series singularity frobenius-method
asked Feb 2 at 8:28
user583893user583893
336
asked Feb 2 at 8:28
user583893user583893
336
asked Feb 2 at 8:28
user583893user583893
336
asked Feb 2 at 8:28
user583893user583893
336
asked Feb 2 at 8:28
user583893user583893
336
336
$begingroup$
This is a Cauchy-Euler equation. Solution has the form $R = (y-k)^r$
$endgroup$
– Dylan
Feb 2 at 9:04
$begingroup$
Hi @Dylan, how can one obtain the exponent $r$?
$endgroup$
– user583893
Feb 2 at 9:14
add a comment |
$begingroup$
This is a Cauchy-Euler equation. Solution has the form $R = (y-k)^r$
$endgroup$
– Dylan
Feb 2 at 9:04
$begingroup$
Hi @Dylan, how can one obtain the exponent $r$?
$endgroup$
– user583893
Feb 2 at 9:14
$begingroup$
This is a Cauchy-Euler equation. Solution has the form $R = (y-k)^r$
$endgroup$
– Dylan
Feb 2 at 9:04
$begingroup$
This is a Cauchy-Euler equation. Solution has the form $R = (y-k)^r$
$endgroup$
– Dylan
Feb 2 at 9:04
$begingroup$
Hi @Dylan, how can one obtain the exponent $r$?
$endgroup$
– user583893
Feb 2 at 9:14
$begingroup$
Hi @Dylan, how can one obtain the exponent $r$?
$endgroup$
– user583893
Feb 2 at 9:14
add a comment |
1 Answer
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$begingroup$
Plugging the standard ansatz $R(y) = (y-k)^r$ into the ODE, we get the characteristic polynomial
$$ r(r-1) + 2r - l(l+1) = 0 $$
$$ implies (r-l)(r+l+1) = 0 $$
$$ implies r = l, -1-l $$
So the general solution is
$$ R(y) = c_1(y-k)^l + c_2(y-k)^{-1-l} $$
The constants are obtained from the initial conditions, which you do not have.
answered Feb 2 at 10:01
DylanDylan
14.3k31127
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1 Answer
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$begingroup$
Plugging the standard ansatz $R(y) = (y-k)^r$ into the ODE, we get the characteristic polynomial
$$ r(r-1) + 2r - l(l+1) = 0 $$
$$ implies (r-l)(r+l+1) = 0 $$
$$ implies r = l, -1-l $$
So the general solution is
$$ R(y) = c_1(y-k)^l + c_2(y-k)^{-1-l} $$
The constants are obtained from the initial conditions, which you do not have.
answered Feb 2 at 10:01
DylanDylan
14.3k31127
$endgroup$
add a comment |
$begingroup$
Plugging the standard ansatz $R(y) = (y-k)^r$ into the ODE, we get the characteristic polynomial
$$ r(r-1) + 2r - l(l+1) = 0 $$
$$ implies (r-l)(r+l+1) = 0 $$
$$ implies r = l, -1-l $$
So the general solution is
$$ R(y) = c_1(y-k)^l + c_2(y-k)^{-1-l} $$
The constants are obtained from the initial conditions, which you do not have.
answered Feb 2 at 10:01
DylanDylan
14.3k31127
$endgroup$
add a comment |
$begingroup$
Plugging the standard ansatz $R(y) = (y-k)^r$ into the ODE, we get the characteristic polynomial
$$ r(r-1) + 2r - l(l+1) = 0 $$
$$ implies (r-l)(r+l+1) = 0 $$
$$ implies r = l, -1-l $$
So the general solution is
$$ R(y) = c_1(y-k)^l + c_2(y-k)^{-1-l} $$
The constants are obtained from the initial conditions, which you do not have.
answered Feb 2 at 10:01
DylanDylan
14.3k31127
$endgroup$
Plugging the standard ansatz $R(y) = (y-k)^r$ into the ODE, we get the characteristic polynomial
$$ r(r-1) + 2r - l(l+1) = 0 $$
$$ implies (r-l)(r+l+1) = 0 $$
$$ implies r = l, -1-l $$
So the general solution is
$$ R(y) = c_1(y-k)^l + c_2(y-k)^{-1-l} $$
The constants are obtained from the initial conditions, which you do not have.
answered Feb 2 at 10:01
DylanDylan
14.3k31127
answered Feb 2 at 10:01
DylanDylan
14.3k31127
answered Feb 2 at 10:01
DylanDylan
14.3k31127
answered Feb 2 at 10:01
DylanDylan
14.3k31127
14.3k31127
add a comment |
add a comment |
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$begingroup$
This is a Cauchy-Euler equation. Solution has the form $R = (y-k)^r$
$endgroup$
– Dylan
Feb 2 at 9:04
$begingroup$
Hi @Dylan, how can one obtain the exponent $r$?
$endgroup$
– user583893
Feb 2 at 9:14