Gauge transformation of differential equations I
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This is a follow-up question to Gauge transformation of differential equations. .
Let $y(x)$ be a solution to the following ODE:
begin{eqnarray}
y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
end{eqnarray}
Now define:
begin{equation}
g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
end{equation}
where
begin{equation}
r^{'}(x) + 1 - a_1(x) r(x)=0
end{equation}
Then:
begin{eqnarray}
&&g^{''}(x) + \
&&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
end{eqnarray}
In[7]:=
Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
x0 =.;
r[x_] = Exp[Integrate[a1[x], x]] C[1] -
Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
Simplify[r'[x] + 1 - a1[x] r[x]]
g[x_] = (y[x] + r[x] y'[x])/(
r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
Collect[(g''[x] +
1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
3 Derivative[1][a0][x]^2)/a0[x]^2 + (
2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
x] :> -a1[x] y'[x] - a0[x] y[x],
Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
a0[x] y'[x]}, {y[x], y'[x]}, Simplify]
Out[9]= 0
Out[11]= 0
Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
begin{eqnarray}
a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
a_1(x)&=& frac{2}{x}\
Longrightarrow\
r(x)&=& frac{x^2}{x_0} +x
end{eqnarray}
then define:
begin{eqnarray}
{mathfrak P}_0&:=&x_0^2 x_2^2\
{mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
{mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
{mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
{mathfrak P}_4&=&-8
C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
{mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
{mathfrak P}_6&=&-8 C^2
end{eqnarray}
then we have:
begin{equation}
g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
end{equation}
Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
begin{eqnarray}
g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
end{eqnarray}
Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.
differential-equations special-functions
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up vote
1
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This is a follow-up question to Gauge transformation of differential equations. .
Let $y(x)$ be a solution to the following ODE:
begin{eqnarray}
y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
end{eqnarray}
Now define:
begin{equation}
g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
end{equation}
where
begin{equation}
r^{'}(x) + 1 - a_1(x) r(x)=0
end{equation}
Then:
begin{eqnarray}
&&g^{''}(x) + \
&&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
end{eqnarray}
In[7]:=
Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
x0 =.;
r[x_] = Exp[Integrate[a1[x], x]] C[1] -
Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
Simplify[r'[x] + 1 - a1[x] r[x]]
g[x_] = (y[x] + r[x] y'[x])/(
r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
Collect[(g''[x] +
1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
3 Derivative[1][a0][x]^2)/a0[x]^2 + (
2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
x] :> -a1[x] y'[x] - a0[x] y[x],
Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
a0[x] y'[x]}, {y[x], y'[x]}, Simplify]
Out[9]= 0
Out[11]= 0
Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
begin{eqnarray}
a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
a_1(x)&=& frac{2}{x}\
Longrightarrow\
r(x)&=& frac{x^2}{x_0} +x
end{eqnarray}
then define:
begin{eqnarray}
{mathfrak P}_0&:=&x_0^2 x_2^2\
{mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
{mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
{mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
{mathfrak P}_4&=&-8
C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
{mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
{mathfrak P}_6&=&-8 C^2
end{eqnarray}
then we have:
begin{equation}
g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
end{equation}
Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
begin{eqnarray}
g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
end{eqnarray}
Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.
differential-equations special-functions
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This is a follow-up question to Gauge transformation of differential equations. .
Let $y(x)$ be a solution to the following ODE:
begin{eqnarray}
y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
end{eqnarray}
Now define:
begin{equation}
g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
end{equation}
where
begin{equation}
r^{'}(x) + 1 - a_1(x) r(x)=0
end{equation}
Then:
begin{eqnarray}
&&g^{''}(x) + \
&&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
end{eqnarray}
In[7]:=
Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
x0 =.;
r[x_] = Exp[Integrate[a1[x], x]] C[1] -
Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
Simplify[r'[x] + 1 - a1[x] r[x]]
g[x_] = (y[x] + r[x] y'[x])/(
r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
Collect[(g''[x] +
1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
3 Derivative[1][a0][x]^2)/a0[x]^2 + (
2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
x] :> -a1[x] y'[x] - a0[x] y[x],
Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
a0[x] y'[x]}, {y[x], y'[x]}, Simplify]
Out[9]= 0
Out[11]= 0
Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
begin{eqnarray}
a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
a_1(x)&=& frac{2}{x}\
Longrightarrow\
r(x)&=& frac{x^2}{x_0} +x
end{eqnarray}
then define:
begin{eqnarray}
{mathfrak P}_0&:=&x_0^2 x_2^2\
{mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
{mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
{mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
{mathfrak P}_4&=&-8
C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
{mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
{mathfrak P}_6&=&-8 C^2
end{eqnarray}
then we have:
begin{equation}
g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
end{equation}
Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
begin{eqnarray}
g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
end{eqnarray}
Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.
differential-equations special-functions
This is a follow-up question to Gauge transformation of differential equations. .
Let $y(x)$ be a solution to the following ODE:
begin{eqnarray}
y^{''}(x) + a_1(x) y^{'}(x)+a_0(x) y(x)=0
end{eqnarray}
Now define:
begin{equation}
g(x):= frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)} exp(-1/2 int a_1(x) dx)}
end{equation}
where
begin{equation}
r^{'}(x) + 1 - a_1(x) r(x)=0
end{equation}
Then:
begin{eqnarray}
&&g^{''}(x) + \
&&!!!!!!!!!!!!!!!!!! frac{1}{4} left(frac{2 a_0''(x)}{a_0(x)}+frac{a_0'(x) left(frac{4}{r(x)}-2 a_1(x)right)}{a_0(x)}-frac{3 a_0'(x)^2}{a_0(x)^2}+4 a_0(x)+2
a_1'(x)+frac{8 a_1(x)}{r(x)}-a_1(x)^2-frac{8}{r(x)^2}right)g(x)=0
end{eqnarray}
In[7]:=
Clear[a0]; Clear[a1]; Clear[y]; Clear[r]; Clear[g]; Clear[m]; x =.;
x0 =.;
r[x_] = Exp[Integrate[a1[x], x]] C[1] -
Exp[Integrate[a1[x], x]] Integrate[ Exp[-Integrate[a1[x], x]], x];
Simplify[r'[x] + 1 - a1[x] r[x]]
g[x_] = (y[x] + r[x] y'[x])/(
r[x] Sqrt[a0[x]] Exp[-1/2 Integrate[a1[x], x]]);
Collect[(g''[x] +
1/4 (4 a0[x] + Derivative[1][a0][x]/a0[x] (4/r[x] - 2 a1[x]) - (
3 Derivative[1][a0][x]^2)/a0[x]^2 + (
2 (a0^[Prime][Prime])[x])/a0[x] - a1[x]^2 + (8 a1[x])/r[x] +
2 Derivative[1][a1][x] - 8/r[x]^2) g[x]) //. {Derivative[2][y][
x] :> -a1[x] y'[x] - a0[x] y[x],
Derivative[3][y][x] :> -a1'[x] y'[x] - a1[x] y''[x] - a0'[x] y[x] -
a0[x] y'[x]}, {y[x], y'[x]}, Simplify]
Out[9]= 0
Out[11]= 0
Note that the result above can be used to generate ODEs whose solutions are known. For example let us take $j=1$ and $B=C x_1$, $A=C x_1/x_2$ and :
begin{eqnarray}
a_0(x)&=& (B C - A D)^2 frac{x^{j-1}}{4(B+A x)^2 (B-D+(A-C) x)^2(D+C x)^2}\
a_1(x)&=& frac{2}{x}\
Longrightarrow\
r(x)&=& frac{x^2}{x_0} +x
end{eqnarray}
then define:
begin{eqnarray}
{mathfrak P}_0&:=&x_0^2 x_2^2\
{mathfrak P}_1&:=&2 x_0 x_2 left(x_2-4 C^2 x_1 (x_0 (x_1+x_2)-x_1 x_2)right)\
{mathfrak P}_2&:=&x_2^2-8 C^2 x_0 left(x_0
left(x_1^2+5 x_1 x_2+x_2^2right)-x_1 x_2 (x_1+x_2)right)\
{mathfrak P}_3&:=&-16 C^2 x_0 (2 x_0 (x_1+x_2)+x_1 x_2)\
{mathfrak P}_4&=&-8
C^2 left(3 x_0^2+3 x_0 (x_1+x_2)+x_1 x_2right)\
{mathfrak P}_5&=&-8 C^2 (3 x_0+x_1+x_2)\
{mathfrak P}_6&=&-8 C^2
end{eqnarray}
then we have:
begin{equation}
g(x):= xcdot frac{y(x)+ r(x) y^{'}(x)}{r(x) sqrt{a_0(x)}}
end{equation}
Since from my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. we know that $y(x)$ is expressed through hypergeometric functions we automaticaly know the solution to the following rather complicated ODE:
begin{eqnarray}
g^{''}(x) + left( frac{sum_{j=0}^6 {mathfrak P}_j x^j}{4 C^2 x^2 (x+x_0)^2 (x+x_1)^2 (x+x_2)^2}right) g(x)=0
end{eqnarray}
Again my question in here would be find other cases where we can find close form solutions to ODEs which are too complicated to be handled using other methods.
differential-equations special-functions
differential-equations special-functions
asked Nov 16 at 19:11
Przemo
4,1171928
4,1171928
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