Computing the solution of an ODE using power series
$begingroup$
I have a system of ODEs defined on $mathbb{R}timesmathbb{S}$,
$$begin{aligned}dot{x}={ }&y\dot{y}={ }&-0.2y+frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)},end{aligned}$$
This system has a saddle at $(0,0)$ and I want to compute the unstable manifold which is supposed to be pretty close to a homoclinic trajectory. So, I eliminate time, and solve the resulting DE
begin{equation}frac{dy}{dx}=-0.2+frac{1}{y}frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}tag{*}end{equation}
using power series $y(x)=a_1 x+a_2x^2+dots+a_kx^k +mathcal{O}(x^{k+1})$ for different $k$. That is, I plug my $y(x)$ into (*) and compute the coefficents $a_i$. I do so using symbolic mathematics, so there are shouldn't be any stupid mistakes (hopefully).
My problem: For all $k>3$, the solutions behave pretty well halfway to the point when it comes closer to the equilibrium. But then it diverge wildly from the expected trajectory. When I add more terms to the expansion (i.e., increase $k$), the divergence becomes even bigger. See the picture below.
Using numerical simulation I got a solution that leaves $(0,0)$ and makes an arc toward $(2pi,0)$. In this sense, the best result seems to be given by either $k=3$ or $k=7$, but they are still very far from what I expect.
Following @LutzL's suggestion I tried to determine the radius of convergence of the power series expansion of the fraction in the second DE. The denominator does not have any real zeroes, so I computed a complex zero, but it does not restrict the radius of convergence: $x_0approx -227.7655 + 3.15i$. However, I'm not sure if this pole is unique.
ordinary-differential-equations dynamical-systems approximation-theory
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
$endgroup$
|
show 9 more comments
$begingroup$
I have a system of ODEs defined on $mathbb{R}timesmathbb{S}$,
$$begin{aligned}dot{x}={ }&y\dot{y}={ }&-0.2y+frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)},end{aligned}$$
This system has a saddle at $(0,0)$ and I want to compute the unstable manifold which is supposed to be pretty close to a homoclinic trajectory. So, I eliminate time, and solve the resulting DE
begin{equation}frac{dy}{dx}=-0.2+frac{1}{y}frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}tag{*}end{equation}
using power series $y(x)=a_1 x+a_2x^2+dots+a_kx^k +mathcal{O}(x^{k+1})$ for different $k$. That is, I plug my $y(x)$ into (*) and compute the coefficents $a_i$. I do so using symbolic mathematics, so there are shouldn't be any stupid mistakes (hopefully).
My problem: For all $k>3$, the solutions behave pretty well halfway to the point when it comes closer to the equilibrium. But then it diverge wildly from the expected trajectory. When I add more terms to the expansion (i.e., increase $k$), the divergence becomes even bigger. See the picture below.
Using numerical simulation I got a solution that leaves $(0,0)$ and makes an arc toward $(2pi,0)$. In this sense, the best result seems to be given by either $k=3$ or $k=7$, but they are still very far from what I expect.
Following @LutzL's suggestion I tried to determine the radius of convergence of the power series expansion of the fraction in the second DE. The denominator does not have any real zeroes, so I computed a complex zero, but it does not restrict the radius of convergence: $x_0approx -227.7655 + 3.15i$. However, I'm not sure if this pole is unique.
ordinary-differential-equations dynamical-systems approximation-theory
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
$endgroup$
1
$begingroup$
What is the $f$ you used? It looks like the power series has its radius of convergence at around $1.5$. What radius of convergence does the power series for $f$ have? You could try to compute Padé approximants, they might converge on a larger interval.
$endgroup$
– LutzL
Dec 18 '18 at 12:17
1
$begingroup$
You only need to check the denominator for roots, also complex roots. These are poles of the function and thus obviously limit the radius of convergence. By Cauchy-Hadamard, this bound is exact.
$endgroup$
– LutzL
Dec 18 '18 at 13:08
1
$begingroup$
This function now has a pole at radius about $1.8$ close to $z = 0.26 + 1.8,i$. This is still inside the distance $pi$ to the next stationary point to derail the power series of the solution at half the desired distance, as you observed.
$endgroup$
– LutzL
Dec 19 '18 at 15:06
1
$begingroup$
I used a slightly different function to avoid irrational numbers like $sin(2),cos(2)$ and determined roots of the power series. So for the real denominator the root location will be close, as the function value you got also strongly indicates.
$endgroup$
– LutzL
Dec 19 '18 at 15:14
1
$begingroup$
The numbers for the actual denominator can be found in a few Newton iterations as radius $1.7911947013226999$ for $z=0.23817399466607489+1.7752891613231125,i$.
$endgroup$
– LutzL
Dec 19 '18 at 15:32
|
show 9 more comments
$begingroup$
I have a system of ODEs defined on $mathbb{R}timesmathbb{S}$,
$$begin{aligned}dot{x}={ }&y\dot{y}={ }&-0.2y+frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)},end{aligned}$$
This system has a saddle at $(0,0)$ and I want to compute the unstable manifold which is supposed to be pretty close to a homoclinic trajectory. So, I eliminate time, and solve the resulting DE
begin{equation}frac{dy}{dx}=-0.2+frac{1}{y}frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}tag{*}end{equation}
using power series $y(x)=a_1 x+a_2x^2+dots+a_kx^k +mathcal{O}(x^{k+1})$ for different $k$. That is, I plug my $y(x)$ into (*) and compute the coefficents $a_i$. I do so using symbolic mathematics, so there are shouldn't be any stupid mistakes (hopefully).
My problem: For all $k>3$, the solutions behave pretty well halfway to the point when it comes closer to the equilibrium. But then it diverge wildly from the expected trajectory. When I add more terms to the expansion (i.e., increase $k$), the divergence becomes even bigger. See the picture below.
Using numerical simulation I got a solution that leaves $(0,0)$ and makes an arc toward $(2pi,0)$. In this sense, the best result seems to be given by either $k=3$ or $k=7$, but they are still very far from what I expect.
Following @LutzL's suggestion I tried to determine the radius of convergence of the power series expansion of the fraction in the second DE. The denominator does not have any real zeroes, so I computed a complex zero, but it does not restrict the radius of convergence: $x_0approx -227.7655 + 3.15i$. However, I'm not sure if this pole is unique.
ordinary-differential-equations dynamical-systems approximation-theory
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
$endgroup$
I have a system of ODEs defined on $mathbb{R}timesmathbb{S}$,
$$begin{aligned}dot{x}={ }&y\dot{y}={ }&-0.2y+frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)},end{aligned}$$
This system has a saddle at $(0,0)$ and I want to compute the unstable manifold which is supposed to be pretty close to a homoclinic trajectory. So, I eliminate time, and solve the resulting DE
begin{equation}frac{dy}{dx}=-0.2+frac{1}{y}frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}tag{*}end{equation}
using power series $y(x)=a_1 x+a_2x^2+dots+a_kx^k +mathcal{O}(x^{k+1})$ for different $k$. That is, I plug my $y(x)$ into (*) and compute the coefficents $a_i$. I do so using symbolic mathematics, so there are shouldn't be any stupid mistakes (hopefully).
My problem: For all $k>3$, the solutions behave pretty well halfway to the point when it comes closer to the equilibrium. But then it diverge wildly from the expected trajectory. When I add more terms to the expansion (i.e., increase $k$), the divergence becomes even bigger. See the picture below.
Using numerical simulation I got a solution that leaves $(0,0)$ and makes an arc toward $(2pi,0)$. In this sense, the best result seems to be given by either $k=3$ or $k=7$, but they are still very far from what I expect.
Following @LutzL's suggestion I tried to determine the radius of convergence of the power series expansion of the fraction in the second DE. The denominator does not have any real zeroes, so I computed a complex zero, but it does not restrict the radius of convergence: $x_0approx -227.7655 + 3.15i$. However, I'm not sure if this pole is unique.
ordinary-differential-equations dynamical-systems approximation-theory
ordinary-differential-equations dynamical-systems approximation-theory
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
edited Dec 19 '18 at 15:05
Dmitry
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
asked Dec 18 '18 at 12:07
DmitryDmitry
716618
716618
1
$begingroup$
What is the $f$ you used? It looks like the power series has its radius of convergence at around $1.5$. What radius of convergence does the power series for $f$ have? You could try to compute Padé approximants, they might converge on a larger interval.
$endgroup$
– LutzL
Dec 18 '18 at 12:17
1
$begingroup$
You only need to check the denominator for roots, also complex roots. These are poles of the function and thus obviously limit the radius of convergence. By Cauchy-Hadamard, this bound is exact.
$endgroup$
– LutzL
Dec 18 '18 at 13:08
1
$begingroup$
This function now has a pole at radius about $1.8$ close to $z = 0.26 + 1.8,i$. This is still inside the distance $pi$ to the next stationary point to derail the power series of the solution at half the desired distance, as you observed.
$endgroup$
– LutzL
Dec 19 '18 at 15:06
1
$begingroup$
I used a slightly different function to avoid irrational numbers like $sin(2),cos(2)$ and determined roots of the power series. So for the real denominator the root location will be close, as the function value you got also strongly indicates.
$endgroup$
– LutzL
Dec 19 '18 at 15:14
1
$begingroup$
The numbers for the actual denominator can be found in a few Newton iterations as radius $1.7911947013226999$ for $z=0.23817399466607489+1.7752891613231125,i$.
$endgroup$
– LutzL
Dec 19 '18 at 15:32
|
show 9 more comments
1
$begingroup$
What is the $f$ you used? It looks like the power series has its radius of convergence at around $1.5$. What radius of convergence does the power series for $f$ have? You could try to compute Padé approximants, they might converge on a larger interval.
$endgroup$
– LutzL
Dec 18 '18 at 12:17
1
$begingroup$
You only need to check the denominator for roots, also complex roots. These are poles of the function and thus obviously limit the radius of convergence. By Cauchy-Hadamard, this bound is exact.
$endgroup$
– LutzL
Dec 18 '18 at 13:08
1
$begingroup$
This function now has a pole at radius about $1.8$ close to $z = 0.26 + 1.8,i$. This is still inside the distance $pi$ to the next stationary point to derail the power series of the solution at half the desired distance, as you observed.
$endgroup$
– LutzL
Dec 19 '18 at 15:06
1
$begingroup$
I used a slightly different function to avoid irrational numbers like $sin(2),cos(2)$ and determined roots of the power series. So for the real denominator the root location will be close, as the function value you got also strongly indicates.
$endgroup$
– LutzL
Dec 19 '18 at 15:14
1
$begingroup$
The numbers for the actual denominator can be found in a few Newton iterations as radius $1.7911947013226999$ for $z=0.23817399466607489+1.7752891613231125,i$.
$endgroup$
– LutzL
Dec 19 '18 at 15:32
1
1
$begingroup$
What is the $f$ you used? It looks like the power series has its radius of convergence at around $1.5$. What radius of convergence does the power series for $f$ have? You could try to compute Padé approximants, they might converge on a larger interval.
$endgroup$
– LutzL
Dec 18 '18 at 12:17
$begingroup$
What is the $f$ you used? It looks like the power series has its radius of convergence at around $1.5$. What radius of convergence does the power series for $f$ have? You could try to compute Padé approximants, they might converge on a larger interval.
$endgroup$
– LutzL
Dec 18 '18 at 12:17
1
1
$begingroup$
You only need to check the denominator for roots, also complex roots. These are poles of the function and thus obviously limit the radius of convergence. By Cauchy-Hadamard, this bound is exact.
$endgroup$
– LutzL
Dec 18 '18 at 13:08
$begingroup$
You only need to check the denominator for roots, also complex roots. These are poles of the function and thus obviously limit the radius of convergence. By Cauchy-Hadamard, this bound is exact.
$endgroup$
– LutzL
Dec 18 '18 at 13:08
1
1
$begingroup$
This function now has a pole at radius about $1.8$ close to $z = 0.26 + 1.8,i$. This is still inside the distance $pi$ to the next stationary point to derail the power series of the solution at half the desired distance, as you observed.
$endgroup$
– LutzL
Dec 19 '18 at 15:06
$begingroup$
This function now has a pole at radius about $1.8$ close to $z = 0.26 + 1.8,i$. This is still inside the distance $pi$ to the next stationary point to derail the power series of the solution at half the desired distance, as you observed.
$endgroup$
– LutzL
Dec 19 '18 at 15:06
1
1
$begingroup$
I used a slightly different function to avoid irrational numbers like $sin(2),cos(2)$ and determined roots of the power series. So for the real denominator the root location will be close, as the function value you got also strongly indicates.
$endgroup$
– LutzL
Dec 19 '18 at 15:14
$begingroup$
I used a slightly different function to avoid irrational numbers like $sin(2),cos(2)$ and determined roots of the power series. So for the real denominator the root location will be close, as the function value you got also strongly indicates.
$endgroup$
– LutzL
Dec 19 '18 at 15:14
1
1
$begingroup$
The numbers for the actual denominator can be found in a few Newton iterations as radius $1.7911947013226999$ for $z=0.23817399466607489+1.7752891613231125,i$.
$endgroup$
– LutzL
Dec 19 '18 at 15:32
$begingroup$
The numbers for the actual denominator can be found in a few Newton iterations as radius $1.7911947013226999$ for $z=0.23817399466607489+1.7752891613231125,i$.
$endgroup$
– LutzL
Dec 19 '18 at 15:32
|
show 9 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The problem was originally formulated for the general system
begin{align}dot{x}={ }&y\dot{y}={ }&-ky+f(x),end{align}
with $2pi$-periodic $f$, $f(2kpi)=0$, $f'(0)>0$ so that the system has a saddle point at $(0,0)$. The solution curves as graphs of functions $y(x)$ also satisfy the DE
$$
frac{dy}{dx}=-k+frac{f(x)}{y(x)}
$$
The solutions with limit in the saddle point have initial condition $y(0)=0$, which is a singular point, and are determined by $y'(0)=r$ where $r$ is a solution of $r^2+kr=f'(0)$.
After multiplying with $y$ an alternative form of the equation is
$$
frac12frac{d(y^2)}{dx} = -ky+f(x),~~ y(x)=sqrt{2int_0^x[-ky(s)+f(s)]ds}
$$
which introduces a different kind of singularities (limiting the radius of converngence) at the roots of the integral on the right, which are branching points of the square root. At the same time they are also roots of $y$.
Due to the structure of that equation one has to expect trouble at the roots of $y$ and of course at the poles of $f$. Not all of them are seen in the numerical solution, but the power series is also valid as solution for the differential equation over the complex plane
The problem as specified now uses $k=0.2$ and
$$
f(x)=frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}.
$$
This leads to an initial slope around $r=14$.
To get rational coefficients to use in the CAS Magma which can deal in truncated power series with rational, but not with floating point coefficients, change $f$ to
$$
f(x)=(r^2-k^2/4) ,frac{frac{33}{20} , sin(x)}{frac{12}5-frac34cos(x)-frac15sin(x)}
$$
where $r=14$ and $k=frac15$. The modified coefficients are within $5%$ of the original. Compute the power series for $y$ up to $O(x^{60})$ and with that compare the partial series to the numerical solution. Plot these in groups, here of 12 graphs. Alongside compute and plot the roots of the power series with a circle of radius $1.9$.
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
$endgroup$
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
add a comment |
$begingroup$
As @Lutzl commented, consider thge denominator and expand $cos(x-2)$. Using whole numbers, it becomes propertional to
$$9 cos (2) cos (x)-sin (2) sin (x)+frac{117}{10}$$ Using the tangent half-angle substitution, this becomes
$$left(frac{117}{10}+9 cos (2)right)-2 sin (2),t+ left(frac{117}{10}-9 cos
(2)right),t^2$$ the roots of which being
$$t_pm=frac{10 sin (2)pm i sqrt{13689-100 sin ^2(2)-8100 cos ^2(2)}}{9 (13-10 cos
(2))}$$ So $x_pm=2 tan ^{-1}(t_pm)$ leads to the solutions $x_pm=(0.238174 pm 1.77529, i)$ as given by Lutzl..
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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votes
$begingroup$
The problem was originally formulated for the general system
begin{align}dot{x}={ }&y\dot{y}={ }&-ky+f(x),end{align}
with $2pi$-periodic $f$, $f(2kpi)=0$, $f'(0)>0$ so that the system has a saddle point at $(0,0)$. The solution curves as graphs of functions $y(x)$ also satisfy the DE
$$
frac{dy}{dx}=-k+frac{f(x)}{y(x)}
$$
The solutions with limit in the saddle point have initial condition $y(0)=0$, which is a singular point, and are determined by $y'(0)=r$ where $r$ is a solution of $r^2+kr=f'(0)$.
After multiplying with $y$ an alternative form of the equation is
$$
frac12frac{d(y^2)}{dx} = -ky+f(x),~~ y(x)=sqrt{2int_0^x[-ky(s)+f(s)]ds}
$$
which introduces a different kind of singularities (limiting the radius of converngence) at the roots of the integral on the right, which are branching points of the square root. At the same time they are also roots of $y$.
Due to the structure of that equation one has to expect trouble at the roots of $y$ and of course at the poles of $f$. Not all of them are seen in the numerical solution, but the power series is also valid as solution for the differential equation over the complex plane
The problem as specified now uses $k=0.2$ and
$$
f(x)=frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}.
$$
This leads to an initial slope around $r=14$.
To get rational coefficients to use in the CAS Magma which can deal in truncated power series with rational, but not with floating point coefficients, change $f$ to
$$
f(x)=(r^2-k^2/4) ,frac{frac{33}{20} , sin(x)}{frac{12}5-frac34cos(x)-frac15sin(x)}
$$
where $r=14$ and $k=frac15$. The modified coefficients are within $5%$ of the original. Compute the power series for $y$ up to $O(x^{60})$ and with that compare the partial series to the numerical solution. Plot these in groups, here of 12 graphs. Alongside compute and plot the roots of the power series with a circle of radius $1.9$.
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
$endgroup$
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
add a comment |
$begingroup$
The problem was originally formulated for the general system
begin{align}dot{x}={ }&y\dot{y}={ }&-ky+f(x),end{align}
with $2pi$-periodic $f$, $f(2kpi)=0$, $f'(0)>0$ so that the system has a saddle point at $(0,0)$. The solution curves as graphs of functions $y(x)$ also satisfy the DE
$$
frac{dy}{dx}=-k+frac{f(x)}{y(x)}
$$
The solutions with limit in the saddle point have initial condition $y(0)=0$, which is a singular point, and are determined by $y'(0)=r$ where $r$ is a solution of $r^2+kr=f'(0)$.
After multiplying with $y$ an alternative form of the equation is
$$
frac12frac{d(y^2)}{dx} = -ky+f(x),~~ y(x)=sqrt{2int_0^x[-ky(s)+f(s)]ds}
$$
which introduces a different kind of singularities (limiting the radius of converngence) at the roots of the integral on the right, which are branching points of the square root. At the same time they are also roots of $y$.
Due to the structure of that equation one has to expect trouble at the roots of $y$ and of course at the poles of $f$. Not all of them are seen in the numerical solution, but the power series is also valid as solution for the differential equation over the complex plane
The problem as specified now uses $k=0.2$ and
$$
f(x)=frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}.
$$
This leads to an initial slope around $r=14$.
To get rational coefficients to use in the CAS Magma which can deal in truncated power series with rational, but not with floating point coefficients, change $f$ to
$$
f(x)=(r^2-k^2/4) ,frac{frac{33}{20} , sin(x)}{frac{12}5-frac34cos(x)-frac15sin(x)}
$$
where $r=14$ and $k=frac15$. The modified coefficients are within $5%$ of the original. Compute the power series for $y$ up to $O(x^{60})$ and with that compare the partial series to the numerical solution. Plot these in groups, here of 12 graphs. Alongside compute and plot the roots of the power series with a circle of radius $1.9$.
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
$endgroup$
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
add a comment |
$begingroup$
The problem was originally formulated for the general system
begin{align}dot{x}={ }&y\dot{y}={ }&-ky+f(x),end{align}
with $2pi$-periodic $f$, $f(2kpi)=0$, $f'(0)>0$ so that the system has a saddle point at $(0,0)$. The solution curves as graphs of functions $y(x)$ also satisfy the DE
$$
frac{dy}{dx}=-k+frac{f(x)}{y(x)}
$$
The solutions with limit in the saddle point have initial condition $y(0)=0$, which is a singular point, and are determined by $y'(0)=r$ where $r$ is a solution of $r^2+kr=f'(0)$.
After multiplying with $y$ an alternative form of the equation is
$$
frac12frac{d(y^2)}{dx} = -ky+f(x),~~ y(x)=sqrt{2int_0^x[-ky(s)+f(s)]ds}
$$
which introduces a different kind of singularities (limiting the radius of converngence) at the roots of the integral on the right, which are branching points of the square root. At the same time they are also roots of $y$.
Due to the structure of that equation one has to expect trouble at the roots of $y$ and of course at the poles of $f$. Not all of them are seen in the numerical solution, but the power series is also valid as solution for the differential equation over the complex plane
The problem as specified now uses $k=0.2$ and
$$
f(x)=frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}.
$$
This leads to an initial slope around $r=14$.
To get rational coefficients to use in the CAS Magma which can deal in truncated power series with rational, but not with floating point coefficients, change $f$ to
$$
f(x)=(r^2-k^2/4) ,frac{frac{33}{20} , sin(x)}{frac{12}5-frac34cos(x)-frac15sin(x)}
$$
where $r=14$ and $k=frac15$. The modified coefficients are within $5%$ of the original. Compute the power series for $y$ up to $O(x^{60})$ and with that compare the partial series to the numerical solution. Plot these in groups, here of 12 graphs. Alongside compute and plot the roots of the power series with a circle of radius $1.9$.
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
$endgroup$
The problem was originally formulated for the general system
begin{align}dot{x}={ }&y\dot{y}={ }&-ky+f(x),end{align}
with $2pi$-periodic $f$, $f(2kpi)=0$, $f'(0)>0$ so that the system has a saddle point at $(0,0)$. The solution curves as graphs of functions $y(x)$ also satisfy the DE
$$
frac{dy}{dx}=-k+frac{f(x)}{y(x)}
$$
The solutions with limit in the saddle point have initial condition $y(0)=0$, which is a singular point, and are determined by $y'(0)=r$ where $r$ is a solution of $r^2+kr=f'(0)$.
After multiplying with $y$ an alternative form of the equation is
$$
frac12frac{d(y^2)}{dx} = -ky+f(x),~~ y(x)=sqrt{2int_0^x[-ky(s)+f(s)]ds}
$$
which introduces a different kind of singularities (limiting the radius of converngence) at the roots of the integral on the right, which are branching points of the square root. At the same time they are also roots of $y$.
Due to the structure of that equation one has to expect trouble at the roots of $y$ and of course at the poles of $f$. Not all of them are seen in the numerical solution, but the power series is also valid as solution for the differential equation over the complex plane
The problem as specified now uses $k=0.2$ and
$$
f(x)=frac{300cos(2)sin(x)}{1.8(1.3+cos(x-2))-2sin(x)sin(2)}.
$$
This leads to an initial slope around $r=14$.
To get rational coefficients to use in the CAS Magma which can deal in truncated power series with rational, but not with floating point coefficients, change $f$ to
$$
f(x)=(r^2-k^2/4) ,frac{frac{33}{20} , sin(x)}{frac{12}5-frac34cos(x)-frac15sin(x)}
$$
where $r=14$ and $k=frac15$. The modified coefficients are within $5%$ of the original. Compute the power series for $y$ up to $O(x^{60})$ and with that compare the partial series to the numerical solution. Plot these in groups, here of 12 graphs. Alongside compute and plot the roots of the power series with a circle of radius $1.9$.
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
answered Dec 22 '18 at 9:26
LutzLLutzL
59.6k42057
59.6k42057
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
add a comment |
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
$begingroup$
Very nice analysis, for sure. $to +1$
$endgroup$
– Claude Leibovici
Dec 22 '18 at 17:00
add a comment |
$begingroup$
As @Lutzl commented, consider thge denominator and expand $cos(x-2)$. Using whole numbers, it becomes propertional to
$$9 cos (2) cos (x)-sin (2) sin (x)+frac{117}{10}$$ Using the tangent half-angle substitution, this becomes
$$left(frac{117}{10}+9 cos (2)right)-2 sin (2),t+ left(frac{117}{10}-9 cos
(2)right),t^2$$ the roots of which being
$$t_pm=frac{10 sin (2)pm i sqrt{13689-100 sin ^2(2)-8100 cos ^2(2)}}{9 (13-10 cos
(2))}$$ So $x_pm=2 tan ^{-1}(t_pm)$ leads to the solutions $x_pm=(0.238174 pm 1.77529, i)$ as given by Lutzl..
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
As @Lutzl commented, consider thge denominator and expand $cos(x-2)$. Using whole numbers, it becomes propertional to
$$9 cos (2) cos (x)-sin (2) sin (x)+frac{117}{10}$$ Using the tangent half-angle substitution, this becomes
$$left(frac{117}{10}+9 cos (2)right)-2 sin (2),t+ left(frac{117}{10}-9 cos
(2)right),t^2$$ the roots of which being
$$t_pm=frac{10 sin (2)pm i sqrt{13689-100 sin ^2(2)-8100 cos ^2(2)}}{9 (13-10 cos
(2))}$$ So $x_pm=2 tan ^{-1}(t_pm)$ leads to the solutions $x_pm=(0.238174 pm 1.77529, i)$ as given by Lutzl..
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
As @Lutzl commented, consider thge denominator and expand $cos(x-2)$. Using whole numbers, it becomes propertional to
$$9 cos (2) cos (x)-sin (2) sin (x)+frac{117}{10}$$ Using the tangent half-angle substitution, this becomes
$$left(frac{117}{10}+9 cos (2)right)-2 sin (2),t+ left(frac{117}{10}-9 cos
(2)right),t^2$$ the roots of which being
$$t_pm=frac{10 sin (2)pm i sqrt{13689-100 sin ^2(2)-8100 cos ^2(2)}}{9 (13-10 cos
(2))}$$ So $x_pm=2 tan ^{-1}(t_pm)$ leads to the solutions $x_pm=(0.238174 pm 1.77529, i)$ as given by Lutzl..
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
As @Lutzl commented, consider thge denominator and expand $cos(x-2)$. Using whole numbers, it becomes propertional to
$$9 cos (2) cos (x)-sin (2) sin (x)+frac{117}{10}$$ Using the tangent half-angle substitution, this becomes
$$left(frac{117}{10}+9 cos (2)right)-2 sin (2),t+ left(frac{117}{10}-9 cos
(2)right),t^2$$ the roots of which being
$$t_pm=frac{10 sin (2)pm i sqrt{13689-100 sin ^2(2)-8100 cos ^2(2)}}{9 (13-10 cos
(2))}$$ So $x_pm=2 tan ^{-1}(t_pm)$ leads to the solutions $x_pm=(0.238174 pm 1.77529, i)$ as given by Lutzl..
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
answered Dec 22 '18 at 7:13
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
add a comment |
add a comment |
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$begingroup$
What is the $f$ you used? It looks like the power series has its radius of convergence at around $1.5$. What radius of convergence does the power series for $f$ have? You could try to compute Padé approximants, they might converge on a larger interval.
$endgroup$
– LutzL
Dec 18 '18 at 12:17
1
$begingroup$
You only need to check the denominator for roots, also complex roots. These are poles of the function and thus obviously limit the radius of convergence. By Cauchy-Hadamard, this bound is exact.
$endgroup$
– LutzL
Dec 18 '18 at 13:08
1
$begingroup$
This function now has a pole at radius about $1.8$ close to $z = 0.26 + 1.8,i$. This is still inside the distance $pi$ to the next stationary point to derail the power series of the solution at half the desired distance, as you observed.
$endgroup$
– LutzL
Dec 19 '18 at 15:06
1
$begingroup$
I used a slightly different function to avoid irrational numbers like $sin(2),cos(2)$ and determined roots of the power series. So for the real denominator the root location will be close, as the function value you got also strongly indicates.
$endgroup$
– LutzL
Dec 19 '18 at 15:14
1
$begingroup$
The numbers for the actual denominator can be found in a few Newton iterations as radius $1.7911947013226999$ for $z=0.23817399466607489+1.7752891613231125,i$.
$endgroup$
– LutzL
Dec 19 '18 at 15:32