Mixing
solve a second order ODE
Can we solve analytically (find a closed form solution) the second order ODE
$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.
differential-equations
asked Nov 22 '18 at 3:00
MedoMedo
613213
|
show 3 more comments
Can we solve analytically (find a closed form solution) the second order ODE
$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.
differential-equations
asked Nov 22 '18 at 3:00
MedoMedo
613213
Fourier transform?
– Don Thousand
Nov 22 '18 at 3:19
Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 '18 at 3:23
@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 '18 at 3:28
Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 '18 at 3:29
@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 '18 at 3:32
|
show 3 more comments
Can we solve analytically (find a closed form solution) the second order ODE
$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.
differential-equations
asked Nov 22 '18 at 3:00
MedoMedo
613213
Can we solve analytically (find a closed form solution) the second order ODE
$$x^{alpha}y^{primeprime}=y,quad x>0$$
where $alphain,]0,1[$.
Consider the conditions
$$y(0)=1,quad y^{prime}(0)=1$$
This equation appears in a fractional model that describes viscoelasticity properties in certain materials.
differential-equations
differential-equations
asked Nov 22 '18 at 3:00
MedoMedo
613213
asked Nov 22 '18 at 3:00
MedoMedo
613213
edited Nov 22 '18 at 3:31
Medo
asked Nov 22 '18 at 3:00
MedoMedo
613213
asked Nov 22 '18 at 3:00
MedoMedo
613213
asked Nov 22 '18 at 3:00
MedoMedo
613213
613213
Fourier transform?
– Don Thousand
Nov 22 '18 at 3:19
Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 '18 at 3:23
@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 '18 at 3:28
Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 '18 at 3:29
@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 '18 at 3:32
|
show 3 more comments
Fourier transform?
– Don Thousand
Nov 22 '18 at 3:19
Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 '18 at 3:23
@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 '18 at 3:28
Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 '18 at 3:29
@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 '18 at 3:32
Fourier transform?
– Don Thousand
Nov 22 '18 at 3:19
Fourier transform?
– Don Thousand
Nov 22 '18 at 3:19
Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 '18 at 3:23
Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 '18 at 3:23
@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 '18 at 3:28
@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 '18 at 3:28
Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 '18 at 3:29
Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 '18 at 3:29
@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 '18 at 3:32
@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 '18 at 3:32
|
show 3 more comments
2 Answers
2
active
oldest
votes
Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$
answered Nov 22 '18 at 3:58
MedoMedo
613213
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
add a comment |
Too long for a comment.
As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.
Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
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
Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$
answered Nov 22 '18 at 3:58
MedoMedo
613213
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
add a comment |
Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$
answered Nov 22 '18 at 3:58
MedoMedo
613213
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
add a comment |
Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$
answered Nov 22 '18 at 3:58
MedoMedo
613213
Mathematica gives the answer in terms of the modified Bessel function of the first kind $I_{n}$
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
Actually this is the solution of $$x^{1-alpha}y^{primeprime}=y$$
answered Nov 22 '18 at 3:58
MedoMedo
613213
answered Nov 22 '18 at 3:58
MedoMedo
613213
answered Nov 22 '18 at 3:58
MedoMedo
613213
answered Nov 22 '18 at 3:58
MedoMedo
613213
613213
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
add a comment |
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
This does not seem to take into account the boundary conditions.
– Claude Leibovici
Nov 22 '18 at 4:01
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
Thanks a lot for your answer. Can you get the solution for a general $alpha$ ?
– Medo
Nov 22 '18 at 7:34
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
What I would like is to see how you could simplify the expression you give taking into account $x >0$. This would make the argument of Bessel functions to be $frac{2x^{frac{a+1}2}}{a+1}$ and a leading factor $sqrt x$. Similarly, make a full simplfication of the front coefficients. Finally, apply the boundary conditions; this should again give something much better. Please, add the results of all of that to your answer.
– Claude Leibovici
Nov 22 '18 at 7:44
add a comment |
Too long for a comment.
As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.
Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
Too long for a comment.
As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.
Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
Too long for a comment.
As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.
Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
Too long for a comment.
As you wrote in you answer, it seems that you obtained solutions in terms of Bessel $I$ functions (the expressions you gave could be simplified quite a lot). As I wrote in comments, I suppose that they would become even simpler when using the given boundary conditions.
Playing with the case where $alpha=frac mn$ ($m$ and $n$ being integers) and using the boundary conditions, it seems that they also can express in terms of hypergeometric functions where patterns seem to appear (I let you finding them).
$$left(
begin{array}{ccc}
m & n & y(x) \
1 & 2 & , _0F_1left(;frac{1}{3};frac{4
x^{3/2}}{9}right)+x , _0F_1left(;frac{5}{3};frac{4
x^{3/2}}{9}right)\
1 & 3 & , _0F_1left(;frac{2}{5};frac{9
x^{5/3}}{25}right)+x , _0F_1left(;frac{8}{5};frac{9
x^{5/3}}{25}right)\
1 & 4 & , _0F_1left(;frac{3}{7};frac{16
x^{7/4}}{49}right)+x , _0F_1left(;frac{11}{7};frac{16
x^{7/4}}{49}right)\
1 & 5 & , _0F_1left(;frac{4}{9};frac{25
x^{9/5}}{81}right)+x , _0F_1left(;frac{14}{9};frac{25
x^{9/5}}{81}right)\
1 & 6 & , _0F_1left(;frac{5}{11};frac{36
x^{11/6}}{121}right)+x , _0F_1left(;frac{17}{11};frac{36
x^{11/6}}{121}right)\
1 & 7 & , _0F_1left(;frac{6}{13};frac{49
x^{13/7}}{169}right)+x , _0F_1left(;frac{20}{13};frac{49
x^{13/7}}{169}right) \
2 & 3 & , _0F_1left(;frac{1}{4};frac{9
x^{4/3}}{16}right)+x , _0F_1left(;frac{7}{4};frac{9
x^{4/3}}{16}right)\
2 & 5 & , _0F_1left(;frac{3}{8};frac{25
x^{8/5}}{64}right)+x , _0F_1left(;frac{13}{8};frac{25
x^{8/5}}{64}right)\
2 & 7 & , _0F_1left(;frac{5}{12};frac{49
x^{12/7}}{144}right)+x , _0F_1left(;frac{19}{12};frac{49
x^{12/7}}{144}right)\
3 & 4 & , _0F_1left(;frac{1}{5};frac{16
x^{5/4}}{25}right)+x , _0F_1left(;frac{9}{5};frac{16
x^{5/4}}{25}right)\
3 & 5 & , _0F_1left(;frac{2}{7};frac{25
x^{7/5}}{49}right)+x , _0F_1left(;frac{12}{7};frac{25
x^{7/5}}{49}right)\
3 & 7 & , _0F_1left(;frac{4}{11};frac{49
x^{11/7}}{121}right)+x , _0F_1left(;frac{18}{11};frac{49
x^{11/7}}{121}right)\
3 & 8 & , _0F_1left(;frac{5}{13};frac{64
x^{13/8}}{169}right)+x , _0F_1left(;frac{21}{13};frac{64
x^{13/8}}{169}right)\
4 & 5 & , _0F_1left(;frac{1}{6};frac{25
x^{6/5}}{36}right)+x , _0F_1left(;frac{11}{6};frac{25
x^{6/5}}{36}right)\
4 & 7 & , _0F_1left(;frac{3}{10};frac{49
x^{10/7}}{100}right)+x , _0F_1left(;frac{17}{10};frac{49
x^{10/7}}{100}right)\
4 & 9 & , _0F_1left(;frac{5}{14};frac{81
x^{14/9}}{196}right)+x , _0F_1left(;frac{23}{14};frac{81
x^{14/9}}{196}right)
end{array}
right) $$
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 22 '18 at 5:11
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
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Fourier transform?
– Don Thousand
Nov 22 '18 at 3:19
Could you precise the boundary conditions ?
– Claude Leibovici
Nov 22 '18 at 3:23
@ Rushabh Mehta. Fourier transform is not easy due to the fractional exponent.
– Medo
Nov 22 '18 at 3:28
Claude Leibovici. I added the initial conditions, though not sure how that helps find a general solution.
– Medo
Nov 22 '18 at 3:29
@ Mattos. That at best would give a special function. Can we go around that?
– Medo
Nov 22 '18 at 3:32