Transforming Bessel's equation with change of variables
$begingroup$
Let bessel's equation be written as
$ z^2$$frac{d^2w}{dz^2}$ + $z$$frac{dw}{dz}$ + $(z^2 -p^2)w$ = $0$
and show that the change of variables defined by $z = ax^b$ and $w = yx^c$ (where a,b and c are constants) transforms it into
$ x^2$$frac{d^2y}{dx^2}$ + $(2c+1)xfrac {dy}{dx}$ + $[a^2b^2(x)^{2b} + (c^2 - p^2b^2)]y$ = $0$
Write the general solution of this equation in terms of Bessels function.
I also don't understand why we use this transformation. Any help would be apreciated.
ordinary-differential-equations bessel-functions
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
$endgroup$
add a comment |
$begingroup$
Let bessel's equation be written as
$ z^2$$frac{d^2w}{dz^2}$ + $z$$frac{dw}{dz}$ + $(z^2 -p^2)w$ = $0$
and show that the change of variables defined by $z = ax^b$ and $w = yx^c$ (where a,b and c are constants) transforms it into
$ x^2$$frac{d^2y}{dx^2}$ + $(2c+1)xfrac {dy}{dx}$ + $[a^2b^2(x)^{2b} + (c^2 - p^2b^2)]y$ = $0$
Write the general solution of this equation in terms of Bessels function.
I also don't understand why we use this transformation. Any help would be apreciated.
ordinary-differential-equations bessel-functions
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
$endgroup$
add a comment |
$begingroup$
Let bessel's equation be written as
$ z^2$$frac{d^2w}{dz^2}$ + $z$$frac{dw}{dz}$ + $(z^2 -p^2)w$ = $0$
and show that the change of variables defined by $z = ax^b$ and $w = yx^c$ (where a,b and c are constants) transforms it into
$ x^2$$frac{d^2y}{dx^2}$ + $(2c+1)xfrac {dy}{dx}$ + $[a^2b^2(x)^{2b} + (c^2 - p^2b^2)]y$ = $0$
Write the general solution of this equation in terms of Bessels function.
I also don't understand why we use this transformation. Any help would be apreciated.
ordinary-differential-equations bessel-functions
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
$endgroup$
Let bessel's equation be written as
$ z^2$$frac{d^2w}{dz^2}$ + $z$$frac{dw}{dz}$ + $(z^2 -p^2)w$ = $0$
and show that the change of variables defined by $z = ax^b$ and $w = yx^c$ (where a,b and c are constants) transforms it into
$ x^2$$frac{d^2y}{dx^2}$ + $(2c+1)xfrac {dy}{dx}$ + $[a^2b^2(x)^{2b} + (c^2 - p^2b^2)]y$ = $0$
Write the general solution of this equation in terms of Bessels function.
I also don't understand why we use this transformation. Any help would be apreciated.
ordinary-differential-equations bessel-functions
ordinary-differential-equations bessel-functions
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
asked Oct 17 '16 at 14:37
user34304user34304
1,27911230
1,27911230
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is because you are invited to chose $b$ and $c$ such that the constant $c^2 - p^2b^2$ is equal to $0$. In this way you obtain a simpler differential equation...
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
$endgroup$
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
add a comment |
$begingroup$
It is because the solution of the FIRST equation can be written as c1*F(z)+c2*G(z), where c1, c2 are computed based on the boundary conditions and F, G are Bessel functions.
Therefore, any equation that follows the SECOND equation pattern can be tranformed into an equivalent first equation, and thus be easily solved.
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is because you are invited to chose $b$ and $c$ such that the constant $c^2 - p^2b^2$ is equal to $0$. In this way you obtain a simpler differential equation...
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
$endgroup$
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
add a comment |
$begingroup$
It is because you are invited to chose $b$ and $c$ such that the constant $c^2 - p^2b^2$ is equal to $0$. In this way you obtain a simpler differential equation...
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
$endgroup$
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
add a comment |
$begingroup$
It is because you are invited to chose $b$ and $c$ such that the constant $c^2 - p^2b^2$ is equal to $0$. In this way you obtain a simpler differential equation...
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
$endgroup$
It is because you are invited to chose $b$ and $c$ such that the constant $c^2 - p^2b^2$ is equal to $0$. In this way you obtain a simpler differential equation...
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
answered Oct 17 '16 at 14:54
Jean MarieJean Marie
29.3k42050
29.3k42050
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
add a comment |
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
$begingroup$
But how would you accomplish that?
$endgroup$
– user34304
Oct 17 '16 at 15:21
add a comment |
$begingroup$
It is because the solution of the FIRST equation can be written as c1*F(z)+c2*G(z), where c1, c2 are computed based on the boundary conditions and F, G are Bessel functions.
Therefore, any equation that follows the SECOND equation pattern can be tranformed into an equivalent first equation, and thus be easily solved.
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
$endgroup$
add a comment |
$begingroup$
It is because the solution of the FIRST equation can be written as c1*F(z)+c2*G(z), where c1, c2 are computed based on the boundary conditions and F, G are Bessel functions.
Therefore, any equation that follows the SECOND equation pattern can be tranformed into an equivalent first equation, and thus be easily solved.
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
$endgroup$
add a comment |
$begingroup$
It is because the solution of the FIRST equation can be written as c1*F(z)+c2*G(z), where c1, c2 are computed based on the boundary conditions and F, G are Bessel functions.
Therefore, any equation that follows the SECOND equation pattern can be tranformed into an equivalent first equation, and thus be easily solved.
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
$endgroup$
It is because the solution of the FIRST equation can be written as c1*F(z)+c2*G(z), where c1, c2 are computed based on the boundary conditions and F, G are Bessel functions.
Therefore, any equation that follows the SECOND equation pattern can be tranformed into an equivalent first equation, and thus be easily solved.
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
answered Nov 9 '17 at 3:57
Michael NikolaouMichael Nikolaou
112
112
add a comment |
add a comment |
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