We compute a system of PDE's for J[1](alpha x)
f:=BesselJ(1,alpha*x):
by first computing an ODE with respect to the variable x
using gfun[holexprtodiffeq] , and next considering symmetries of
f to derive a complete system.
ff:=holexprtodiffeq(f,y(x));
// 2 2\ / d \ 2 / d / d \\ 1
{ \-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||, D(y)(0) = - alpha
\ \ dx / \ dx \ dx // 2
\
}
/
diffeq:=op( select(has,ff,x) );
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||
\ dx / \ dx \ dx //
diffeqx:=subs(y(x)=h(x,alpha),diffeq);
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / h(x, alpha) + x |--- h(x, alpha)| + x |--- |--- h(x, alpha)||
\ dx / \ dx \ dx //
diffeqalpha:=subs({x=alpha,alpha=x,y(x)=h(x,alpha)},diffeq);
/ 2 2\ / d \
\-1 + alpha x / h(x, alpha) + alpha |------- h(x, alpha)|
\ dalpha /
2 / d / d \\
+ alpha |------- |------- h(x, alpha)||
\ dalpha \ dalpha //
sol:=pdsolve([diffeqx,diffeqalpha]);
{h(x, alpha) = _C1 BesselJ(1, alpha x) + _C2 BesselY(1, alpha x)}
diffeqxalpha:=x*D[1](h)(x,alpha)-alpha*D[2](h)(x,alpha);
x D[1](h)(x, alpha) - alpha D[2](h)(x, alpha)
solall:=pdsolve([diffeqx,diffeqalpha,diffeqxalpha]);
{h(x, alpha) = _C2 BesselJ(1, alpha x) + _C1 BesselY(1, alpha x)}
We compute a system of PDE's for J[1](alpha x)
f:=BesselJ(1,alpha*x):
by first computing an ODE with respect to the variable x
using gfun[holexprtodiffeq] , and next considering symmetries of
f to derive a complete system.
ff:=holexprtodiffeq(f,y(x));
// 2 2\ / d \ 2 / d / d \\ 1
{ \-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||, D(y)(0) = - alpha
\ \ dx / \ dx \ dx // 2
\
}
/
diffeq:=op( select(has,ff,x) );
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||
\ dx / \ dx \ dx //
diffeqx:=subs(y(x)=h(x,alpha),diffeq);
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / h(x, alpha) + x |--- h(x, alpha)| + x |--- |--- h(x, alpha)||
\ dx / \ dx \ dx //
diffeqalpha:=subs({x=alpha,alpha=x,y(x)=h(x,alpha)},diffeq);
/ 2 2\ / d \
\-1 + alpha x / h(x, alpha) + alpha |------- h(x, alpha)|
\ dalpha /
2 / d / d \\
+ alpha |------- |------- h(x, alpha)||
\ dalpha \ dalpha //
sol:=pdsolve([diffeqx,diffeqalpha]);
{h(x, alpha) = _C1 BesselJ(1, alpha x) + _C2 BesselY(1, alpha x)}
diffeqxalpha:=x*D[1](h)(x,alpha)-alpha*D[2](h)(x,alpha);
x D[1](h)(x, alpha) - alpha D[2](h)(x, alpha)
solall:=pdsolve([diffeqx,diffeqalpha,diffeqxalpha]);
{h(x, alpha) = _C2 BesselJ(1, alpha x) + _C1 BesselY(1, alpha x)}
I see what you say, although I don't know the Lie algebraic methods for pdes.
In recent case however the solution sets are the same.
f:=BesselJ(1,alpha*x):
ff:=holexprtodiffeq(f,y(x));
print({(-1+alpha^2*x^2)*y(x)+x*(diff(y(x), x))+x^2*(diff(y(x), x, x)), (D(y))(0) = (1/2)*alpha});
// 2 2\ / d \ 2 / d / d \\ 1 \
{ \-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||, D(y)(0) = - alpha }
\ \ dx / \ dx \ dx // 2 /
diffeq:=op( select(has,ff,x) );
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||
\ dx / \ dx \ dx //
diffeqx:=subs(y(x)=h(x,alpha),diffeq);
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / h(x, alpha) + x |--- h(x, alpha)| + x |--- |--- h(x, alpha)||
\ dx / \ dx \ dx //
diffeqalpha:=subs({x=alpha,alpha=x,y(x)=h(x,alpha)},diffeq);
/ 2 2\ / d \
\-1 + alpha x / h(x, alpha) + alpha |------- h(x, alpha)|
\ dalpha /
2 / d / d \\
+ alpha |------- |------- h(x, alpha)||
\ dalpha \ dalpha //
sol:=pdsolve([diffeqx,diffeqalpha]);
{h(x, alpha) = _C1 BesselJ(1, alpha x) + _C2 BesselY(1, alpha x)}
diffeqxalpha:=x*D[1](h)(x,alpha)-alpha*D[2](h)(x,alpha);
x D[1](h)(x, alpha) - alpha D[2](h)(x, alpha)
solall:=pdsolve([diffeqx,diffeqalpha,diffeqxalpha]);
{h(x, alpha) = _C2 BesselJ(1, alpha x) + _C1 BesselY(1, alpha x)}
I see what you say, although I don't know the Lie algebraic methods for pdes.
In recent case however the solution sets are the same.
f:=BesselJ(1,alpha*x):
ff:=holexprtodiffeq(f,y(x));
print({(-1+alpha^2*x^2)*y(x)+x*(diff(y(x), x))+x^2*(diff(y(x), x, x)), (D(y))(0) = (1/2)*alpha});
// 2 2\ / d \ 2 / d / d \\ 1 \
{ \-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||, D(y)(0) = - alpha }
\ \ dx / \ dx \ dx // 2 /
diffeq:=op( select(has,ff,x) );
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / y(x) + x |--- y(x)| + x |--- |--- y(x)||
\ dx / \ dx \ dx //
diffeqx:=subs(y(x)=h(x,alpha),diffeq);
/ 2 2\ / d \ 2 / d / d \\
\-1 + alpha x / h(x, alpha) + x |--- h(x, alpha)| + x |--- |--- h(x, alpha)||
\ dx / \ dx \ dx //
diffeqalpha:=subs({x=alpha,alpha=x,y(x)=h(x,alpha)},diffeq);
/ 2 2\ / d \
\-1 + alpha x / h(x, alpha) + alpha |------- h(x, alpha)|
\ dalpha /
2 / d / d \\
+ alpha |------- |------- h(x, alpha)||
\ dalpha \ dalpha //
sol:=pdsolve([diffeqx,diffeqalpha]);
{h(x, alpha) = _C1 BesselJ(1, alpha x) + _C2 BesselY(1, alpha x)}
diffeqxalpha:=x*D[1](h)(x,alpha)-alpha*D[2](h)(x,alpha);
x D[1](h)(x, alpha) - alpha D[2](h)(x, alpha)
solall:=pdsolve([diffeqx,diffeqalpha,diffeqxalpha]);
{h(x, alpha) = _C2 BesselJ(1, alpha x) + _C1 BesselY(1, alpha x)}
I see.
Thanks.
I began to rework the original worksheet.
The PDE of x is ok.
The PDE of BesselJ(1,a*x) is problematical.
I have a partial diff eq system
subs(y(x)=h(x,a),deq);
and
subs({x=a,a=x,y(x)=h(x,a)},deq);
It's ok.
But what is the meaning of the line
x(D_x h(x,a)) - a(D_a h(x,a)) ?
Why don't they take the pde system
sys2:={ x^2*(D[1,1](h)(x,a))+x*(D[1](h)(x,a))+(-1+a^2*x^2),
a^2*(D[2,2](h)(x,a))+a*(D[2](h)(x,a))+(-1+a^2*x^2)};
I see.
Thanks.
I began to rework the original worksheet.
The PDE of x is ok.
The PDE of BesselJ(1,a*x) is problematical.
I have a partial diff eq system
subs(y(x)=h(x,a),deq);
and
subs({x=a,a=x,y(x)=h(x,a)},deq);
It's ok.
But what is the meaning of the line
x(D_x h(x,a)) - a(D_a h(x,a)) ?
Why don't they take the pde system
sys2:={ x^2*(D[1,1](h)(x,a))+x*(D[1](h)(x,a))+(-1+a^2*x^2),
a^2*(D[2,2](h)(x,a))+a*(D[2](h)(x,a))+(-1+a^2*x^2)};
I also guessed that h(x,a)=x. :-)
sys1:={ x*(d/dx h(x,a)) -1, d/da h(x,a) }
Then d/da h(x,a)=0. Right.
But
x*(d/dx h(x,a)) -1 not=0.
What is zero, is
d/dx h(x,a) -1
So I think this worksheet is not only too outdated, but
there is a lot of error that probably can be corrected.
However, the idea is clear and when it needed I will be able
to apply.
I also guessed that h(x,a)=x. :-)
sys1:={ x*(d/dx h(x,a)) -1, d/da h(x,a) }
Then d/da h(x,a)=0. Right.
But
x*(d/dx h(x,a)) -1 not=0.
What is zero, is
d/dx h(x,a) -1
So I think this worksheet is not only too outdated, but
there is a lot of error that probably can be corrected.
However, the idea is clear and when it needed I will be able
to apply.
Many thanks.
Recently I'm working on Bessel functions and any
help is appreciated.
Which packages you recommend to use?
Many thanks.
Recently I'm working on Bessel functions and any
help is appreciated.
Which packages you recommend to use?
The origin of my question can be found here
http://algo.inria.fr/libraries/autocomb/FourBessel.mws
http://algo.inria.fr/libraries/autocomb/FourBessel.ps
http://algo.inria.fr/libraries/autocomb/FourBessel-html/FourBessel.html
They denote by kappa the integral in my question.
To tell the truth I don't understand the very first step, namely,
The identity function x trivially satisfies the following differential system
> sys1:={ x*(d/dx h(x,a)) -1, d/da h(x,a) }
where each entry expr in the set denotes the equation expr=0.
What does h(x,a) mean? x? Then the first equation is wrong.
The origin of my question can be found here
http://algo.inria.fr/libraries/autocomb/FourBessel.mws
http://algo.inria.fr/libraries/autocomb/FourBessel.ps
http://algo.inria.fr/libraries/autocomb/FourBessel-html/FourBessel.html
They denote by kappa the integral in my question.
To tell the truth I don't understand the very first step, namely,
The identity function x trivially satisfies the following differential system
> sys1:={ x*(d/dx h(x,a)) -1, d/da h(x,a) }
where each entry expr in the set denotes the equation expr=0.
What does h(x,a) mean? x? Then the first equation is wrong.
Dear JacquesC, Axel,
This solution is indeed very nice, unfortunately I don't know Mellin transformation, so I will
learn its basic ideas.
That paper is very interesting.
Hopefully in the near future in Maple xx it will be involved.
Many thanks.
Dear JacquesC, Axel,
This solution is indeed very nice, unfortunately I don't know Mellin transformation, so I will
learn its basic ideas.
That paper is very interesting.
Hopefully in the near future in Maple xx it will be involved.
Many thanks.
The Next is very efficient.
My proc is very fast.
Thanks again.
Sandor
