Mixing
Finding general solution to DE subject to initial condition
$begingroup$
How do we solve the following Differential Equation?
$$2 x''' + xx'' =0$$
Subject to conditions:
$$ x(0)=0$$
$$ x'(0)=0$$
$$ x'(infty)=1$$
Is there any numerical method to solve it or some general method??
ordinary-differential-equations numerical-methods numerical-calculus
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
$endgroup$
add a comment |
$begingroup$
How do we solve the following Differential Equation?
$$2 x''' + xx'' =0$$
Subject to conditions:
$$ x(0)=0$$
$$ x'(0)=0$$
$$ x'(infty)=1$$
Is there any numerical method to solve it or some general method??
ordinary-differential-equations numerical-methods numerical-calculus
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
$endgroup$
$begingroup$
My mistake while typing. It is just single derivative. Thankyou for pointing mistake
$endgroup$
– Abhi7731756
Dec 28 '18 at 15:38
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Do you have any reason to think this can be solved analytically? One certainly would not expect the Fourier transform or the Laplace transform to help, since the equation's not linear. In general non-linear differential equations are hard...
$endgroup$
– David C. Ullrich
Dec 28 '18 at 16:31
add a comment |
$begingroup$
How do we solve the following Differential Equation?
$$2 x''' + xx'' =0$$
Subject to conditions:
$$ x(0)=0$$
$$ x'(0)=0$$
$$ x'(infty)=1$$
Is there any numerical method to solve it or some general method??
ordinary-differential-equations numerical-methods numerical-calculus
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
$endgroup$
How do we solve the following Differential Equation?
$$2 x''' + xx'' =0$$
Subject to conditions:
$$ x(0)=0$$
$$ x'(0)=0$$
$$ x'(infty)=1$$
Is there any numerical method to solve it or some general method??
ordinary-differential-equations numerical-methods numerical-calculus
ordinary-differential-equations numerical-methods numerical-calculus
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
edited Dec 29 '18 at 6:11
Abhi7731756
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
asked Dec 28 '18 at 15:28
Abhi7731756Abhi7731756
114
114
$begingroup$
My mistake while typing. It is just single derivative. Thankyou for pointing mistake
$endgroup$
– Abhi7731756
Dec 28 '18 at 15:38
$begingroup$
Do you have any reason to think this can be solved analytically? One certainly would not expect the Fourier transform or the Laplace transform to help, since the equation's not linear. In general non-linear differential equations are hard...
$endgroup$
– David C. Ullrich
Dec 28 '18 at 16:31
add a comment |
$begingroup$
My mistake while typing. It is just single derivative. Thankyou for pointing mistake
$endgroup$
– Abhi7731756
Dec 28 '18 at 15:38
$begingroup$
Do you have any reason to think this can be solved analytically? One certainly would not expect the Fourier transform or the Laplace transform to help, since the equation's not linear. In general non-linear differential equations are hard...
$endgroup$
– David C. Ullrich
Dec 28 '18 at 16:31
$begingroup$
My mistake while typing. It is just single derivative. Thankyou for pointing mistake
$endgroup$
– Abhi7731756
Dec 28 '18 at 15:38
$begingroup$
My mistake while typing. It is just single derivative. Thankyou for pointing mistake
$endgroup$
– Abhi7731756
Dec 28 '18 at 15:38
$begingroup$
Do you have any reason to think this can be solved analytically? One certainly would not expect the Fourier transform or the Laplace transform to help, since the equation's not linear. In general non-linear differential equations are hard...
$endgroup$
– David C. Ullrich
Dec 28 '18 at 16:31
$begingroup$
Do you have any reason to think this can be solved analytically? One certainly would not expect the Fourier transform or the Laplace transform to help, since the equation's not linear. In general non-linear differential equations are hard...
$endgroup$
– David C. Ullrich
Dec 28 '18 at 16:31
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Too long for a comment.
In his answer, JJacquelin ended with
$$t(x)=int_0^xfrac{dxi}{sqrt{text{erf}(xi/2)}}$$ which does not have explicit solution.
However, we can make quite good approximation building the Padé approximant at $xi=0$
$$frac{1}{sqrt{text{erf}(xi/2)}}simeq sqrt[4]pi , frac{1+frac{363 }{3680}xi ^2+frac{491 }{264960}xi ^4-frac{23789 }{445132800}xi^6 }{sqrt xi left(1+frac{629 }{11040}xi ^2 right) }$$ which can be integrated to lead to the nasty (but practicable)
$$t(x)=-frac{23789 sqrt[4]{pi } x^{9/2}}{114125760}+frac{1085389 sqrt[4]{pi }
x^{5/2}}{55389740}+frac{3036269046 sqrt[4]{pi }
sqrt{x}}{1742007323}-frac{111936400 2^{3/4} sqrt[4]{frac{345 pi }{629}}
log left(sqrt{434010} x-4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}+frac{111936400 2^{3/4} sqrt[4]{frac{345
pi }{629}} log left(sqrt{434010} x+4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}-frac{223872800 2^{3/4} sqrt[4]{frac{345
pi }{629}} tan ^{-1}left(1-frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}right)}{1742007323}+frac{223872800 2^{3/4}
sqrt[4]{frac{345 pi }{629}} tan ^{-1}left(frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}+1right)}{1742007323}$$
which is not so bad as shown below
$$left(
begin{array}{ccc}
x & text{approximation} & text{exact} \
0.00 & 0.00000 & 0.00000 \
0.25 & 1.33203 & 1.33203 \
0.50 & 1.88671 & 1.88671 \
0.75 & 2.31671 & 2.31671 \
1.00 & 2.68470 & 2.68470 \
1.25 & 3.01528 & 3.01528 \
1.50 & 3.32122 & 3.32122 \
1.75 & 3.61020 & 3.61020 \
2.00 & 3.88724 & 3.88725 \
2.25 & 4.15580 & 4.15584 \
2.50 & 4.41833 & 4.41843 \
2.75 & 4.67656 & 4.67681 \
3.00 & 4.93172 & 4.93227 \
3.25 & 5.18460 & 5.18576 \
3.50 & 5.43566 & 5.43794 \
3.75 & 5.68507 & 5.68926 \
4.00 & 5.93273 & 5.94004 \
4.25 & 6.17831 & 6.19048 \
4.50 & 6.42126 & 6.44074 \
4.75 & 6.66082 & 6.69087 \
5.00 & 6.89601 & 6.94094
end{array}
right)$$
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
add a comment |
$begingroup$
The first step in a "general method" to solve this problem is to recognize that the given third-order ordinary differential equation (ODE) is autonomous.
We'll use subscripts for derivatives in the following, because we are going to use derivatives with respect to the independent variable $t$ but also with respect to the dependent variable $x$.
Starting with $2 x_{ttt} + x x_{tt} = 0$ we write $x_t = w(x)$ with some unknown function $w$, to obtain a boundary-value problem (BVP) with a second-order ODE for $w$:
begin{equation}
w w_{xx} + w_x left( w_x + frac{x}{2} right) = 0, quad w(0) = 0, quad w(infty) = 1
end{equation}
(note that the derivatives of $w$ are taken with respect to $x$!).
Once the function $w$ is found we may solve the following initial-value problem (IVP) with a separable first-order ODE for $x$:
begin{equation}
x_t = w(x), quad x(0) = 0.
end{equation}
Thus we have split the initial-boundary-value problem (IBVP) with a third-order autonomous ODE into a BVP with a second-order ODE and an IVP with a separable first-order ODE.
Going for the inverse function $t(x)$ as suggested by JJacquelin we may write
begin{equation}
t(x) = int limits_0^x frac{1}{w(xi)} , mathrm{d}xi,
end{equation}
with the function $w$ from above. Thus it remains to solve the BVP for $w$ and then to compute the integral.
Both tasks may be difficult, and I believe that there is a mistake in the solution provided by JJacquelin (in the calculation of the third derivative of the inverse function). I would normally write a comment on this but I am not yet allowed.
answered Jan 2 at 9:54
ChristophChristoph
1916
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$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
add a comment |
$begingroup$
Numerically there does not seem to be any problem, $T=20$ as approximation of $T=infty$ appears large enough, the solution can be computed via boundary value solver, here python's scipy.integrate.solve_bvp:
T = 20
def x_ode(t,x): return [x[1], x[2], -0.5*x[0]*x[2]]
def x_bc(x0, xT): return [x0[0], x0[1], xT[1]-1]
s = np.linspace(0,1,11);
t_init = T*s
x_init = [ T*s , 1+0*s, 0+0*s ]
res = solve_bvp(x_ode, x_bc, t_init, x_init, tol=1e-5, max_nodes=80000)
print res.message
with the result
The algorithm converged to the desired accuracy.
The plot of function and derivative
is then produced via
if res.success:
plt.figure(figsize=(9,6))
plt.subplot(211); plt.plot(res.x, res.y[0], '-'); plt.grid();
plt.subplot(212); plt.plot(res.x, res.y[1], '-'); plt.grid();
plt.show()
second approach
To transform the problem to a finite interval the boundary conditions suggest to use $s=x'$, $sin[0,1]$, as new independent parameter, thus using the inverse function $u$ to $x'$, $t=u(x')$, $x=v_0(x')$, $v_1(s)=x'(u(s))=s$, $v_2(s)=x''(u(s))$ which leads to the derivatives
begin{align}
1=frac{dt}{dt}&=u'(x'(t))x''(t)\
frac{dt}{ds}&=u'(s)=frac1{v_2(s)}\
frac{dv_0(s)}{ds}&=x'(u(s))u'(s)=frac{s}{v_2(s)}\
frac{dv_2(s)}{ds}&=x'''(u(s))u'(s)=-frac{v_0(s)}{2}
end{align}
To get $v_0(s)toinfty$ for $sto 1$ we need $v_2(1)=0$ along with $u(0)=0$ and $v_0(0)=0$. To desingularize $1/v_2$ the approach using $v_2/(epsilon^2+v_2^2)$ works best.
eps = 1e-8
def uv_ode(s,y):
u, v0, v2 = y;
v2inv = v2/(eps**2+v2**2);
return [v2inv, s*v2inv, -0.5*v0]
def uv_bc(u0, u1):
return [ u0[0], u0[1], u1[2]]
s = np.linspace(0,1,11);
y_init = [ 16*s**2 , 14*s**2, (1-s)**2/4 ]
res = solve_bvp(uv_ode, uv_bc, s, y_init, tol=1e-5, max_nodes=80000)
print res.message
if True or res.success:
plt.figure(figsize=(9,9))
plt.subplot(311); plt.plot(res.x, res.y[0], '-o',ms=1); plt.ylabel('$u=t$'); plt.grid();
plt.subplot(312); plt.plot(res.x, res.y[1], '-o',ms=1); plt.ylabel('$v_0=x$'); plt.grid();
plt.subplot(313); plt.plot(res.x, res.y[2], '-o',ms=1); plt.ylabel('$v_2=x''$'); plt.grid(); plt.xlabel('$s=x'$');
plt.show()
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
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add a comment |
$begingroup$
Follows a MATHEMATICA script showing a numerical solution. Here $t = 200 approxinfty$
tmax = 200;
solx = NDSolve[{x1'[t] == x2[t],
x2'[t] == x3[t],
2 x3'[t] + x1[t] x3[t] == 0, x1[0] == 0, x2[0] == 0, x2[200] == 1},
{x1, x2, x3}, {t, 0, tmax}][[1]]
Plot[Evaluate[{x1[t], x2[t], x3[t]} /. solx], {t, 0, 20},
PlotRange -> {0, 2},
PlotStyle -> {{Black, Thick}, {Blue, Thick}, {Red, Thick}},
PlotLegends -> {Subscript[x, 1][t], Subscript[x, 2][t],
Subscript[x, 3][t]}]
$$
x_1(t) to mbox{black}\
x_2(t) to mbox{blue}\
x_3(t) to mbox{red}
$$
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Too long for a comment.
In his answer, JJacquelin ended with
$$t(x)=int_0^xfrac{dxi}{sqrt{text{erf}(xi/2)}}$$ which does not have explicit solution.
However, we can make quite good approximation building the Padé approximant at $xi=0$
$$frac{1}{sqrt{text{erf}(xi/2)}}simeq sqrt[4]pi , frac{1+frac{363 }{3680}xi ^2+frac{491 }{264960}xi ^4-frac{23789 }{445132800}xi^6 }{sqrt xi left(1+frac{629 }{11040}xi ^2 right) }$$ which can be integrated to lead to the nasty (but practicable)
$$t(x)=-frac{23789 sqrt[4]{pi } x^{9/2}}{114125760}+frac{1085389 sqrt[4]{pi }
x^{5/2}}{55389740}+frac{3036269046 sqrt[4]{pi }
sqrt{x}}{1742007323}-frac{111936400 2^{3/4} sqrt[4]{frac{345 pi }{629}}
log left(sqrt{434010} x-4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}+frac{111936400 2^{3/4} sqrt[4]{frac{345
pi }{629}} log left(sqrt{434010} x+4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}-frac{223872800 2^{3/4} sqrt[4]{frac{345
pi }{629}} tan ^{-1}left(1-frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}right)}{1742007323}+frac{223872800 2^{3/4}
sqrt[4]{frac{345 pi }{629}} tan ^{-1}left(frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}+1right)}{1742007323}$$
which is not so bad as shown below
$$left(
begin{array}{ccc}
x & text{approximation} & text{exact} \
0.00 & 0.00000 & 0.00000 \
0.25 & 1.33203 & 1.33203 \
0.50 & 1.88671 & 1.88671 \
0.75 & 2.31671 & 2.31671 \
1.00 & 2.68470 & 2.68470 \
1.25 & 3.01528 & 3.01528 \
1.50 & 3.32122 & 3.32122 \
1.75 & 3.61020 & 3.61020 \
2.00 & 3.88724 & 3.88725 \
2.25 & 4.15580 & 4.15584 \
2.50 & 4.41833 & 4.41843 \
2.75 & 4.67656 & 4.67681 \
3.00 & 4.93172 & 4.93227 \
3.25 & 5.18460 & 5.18576 \
3.50 & 5.43566 & 5.43794 \
3.75 & 5.68507 & 5.68926 \
4.00 & 5.93273 & 5.94004 \
4.25 & 6.17831 & 6.19048 \
4.50 & 6.42126 & 6.44074 \
4.75 & 6.66082 & 6.69087 \
5.00 & 6.89601 & 6.94094
end{array}
right)$$
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
add a comment |
$begingroup$
Too long for a comment.
In his answer, JJacquelin ended with
$$t(x)=int_0^xfrac{dxi}{sqrt{text{erf}(xi/2)}}$$ which does not have explicit solution.
However, we can make quite good approximation building the Padé approximant at $xi=0$
$$frac{1}{sqrt{text{erf}(xi/2)}}simeq sqrt[4]pi , frac{1+frac{363 }{3680}xi ^2+frac{491 }{264960}xi ^4-frac{23789 }{445132800}xi^6 }{sqrt xi left(1+frac{629 }{11040}xi ^2 right) }$$ which can be integrated to lead to the nasty (but practicable)
$$t(x)=-frac{23789 sqrt[4]{pi } x^{9/2}}{114125760}+frac{1085389 sqrt[4]{pi }
x^{5/2}}{55389740}+frac{3036269046 sqrt[4]{pi }
sqrt{x}}{1742007323}-frac{111936400 2^{3/4} sqrt[4]{frac{345 pi }{629}}
log left(sqrt{434010} x-4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}+frac{111936400 2^{3/4} sqrt[4]{frac{345
pi }{629}} log left(sqrt{434010} x+4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}-frac{223872800 2^{3/4} sqrt[4]{frac{345
pi }{629}} tan ^{-1}left(1-frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}right)}{1742007323}+frac{223872800 2^{3/4}
sqrt[4]{frac{345 pi }{629}} tan ^{-1}left(frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}+1right)}{1742007323}$$
which is not so bad as shown below
$$left(
begin{array}{ccc}
x & text{approximation} & text{exact} \
0.00 & 0.00000 & 0.00000 \
0.25 & 1.33203 & 1.33203 \
0.50 & 1.88671 & 1.88671 \
0.75 & 2.31671 & 2.31671 \
1.00 & 2.68470 & 2.68470 \
1.25 & 3.01528 & 3.01528 \
1.50 & 3.32122 & 3.32122 \
1.75 & 3.61020 & 3.61020 \
2.00 & 3.88724 & 3.88725 \
2.25 & 4.15580 & 4.15584 \
2.50 & 4.41833 & 4.41843 \
2.75 & 4.67656 & 4.67681 \
3.00 & 4.93172 & 4.93227 \
3.25 & 5.18460 & 5.18576 \
3.50 & 5.43566 & 5.43794 \
3.75 & 5.68507 & 5.68926 \
4.00 & 5.93273 & 5.94004 \
4.25 & 6.17831 & 6.19048 \
4.50 & 6.42126 & 6.44074 \
4.75 & 6.66082 & 6.69087 \
5.00 & 6.89601 & 6.94094
end{array}
right)$$
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
add a comment |
$begingroup$
Too long for a comment.
In his answer, JJacquelin ended with
$$t(x)=int_0^xfrac{dxi}{sqrt{text{erf}(xi/2)}}$$ which does not have explicit solution.
However, we can make quite good approximation building the Padé approximant at $xi=0$
$$frac{1}{sqrt{text{erf}(xi/2)}}simeq sqrt[4]pi , frac{1+frac{363 }{3680}xi ^2+frac{491 }{264960}xi ^4-frac{23789 }{445132800}xi^6 }{sqrt xi left(1+frac{629 }{11040}xi ^2 right) }$$ which can be integrated to lead to the nasty (but practicable)
$$t(x)=-frac{23789 sqrt[4]{pi } x^{9/2}}{114125760}+frac{1085389 sqrt[4]{pi }
x^{5/2}}{55389740}+frac{3036269046 sqrt[4]{pi }
sqrt{x}}{1742007323}-frac{111936400 2^{3/4} sqrt[4]{frac{345 pi }{629}}
log left(sqrt{434010} x-4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}+frac{111936400 2^{3/4} sqrt[4]{frac{345
pi }{629}} log left(sqrt{434010} x+4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}-frac{223872800 2^{3/4} sqrt[4]{frac{345
pi }{629}} tan ^{-1}left(1-frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}right)}{1742007323}+frac{223872800 2^{3/4}
sqrt[4]{frac{345 pi }{629}} tan ^{-1}left(frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}+1right)}{1742007323}$$
which is not so bad as shown below
$$left(
begin{array}{ccc}
x & text{approximation} & text{exact} \
0.00 & 0.00000 & 0.00000 \
0.25 & 1.33203 & 1.33203 \
0.50 & 1.88671 & 1.88671 \
0.75 & 2.31671 & 2.31671 \
1.00 & 2.68470 & 2.68470 \
1.25 & 3.01528 & 3.01528 \
1.50 & 3.32122 & 3.32122 \
1.75 & 3.61020 & 3.61020 \
2.00 & 3.88724 & 3.88725 \
2.25 & 4.15580 & 4.15584 \
2.50 & 4.41833 & 4.41843 \
2.75 & 4.67656 & 4.67681 \
3.00 & 4.93172 & 4.93227 \
3.25 & 5.18460 & 5.18576 \
3.50 & 5.43566 & 5.43794 \
3.75 & 5.68507 & 5.68926 \
4.00 & 5.93273 & 5.94004 \
4.25 & 6.17831 & 6.19048 \
4.50 & 6.42126 & 6.44074 \
4.75 & 6.66082 & 6.69087 \
5.00 & 6.89601 & 6.94094
end{array}
right)$$
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
Too long for a comment.
In his answer, JJacquelin ended with
$$t(x)=int_0^xfrac{dxi}{sqrt{text{erf}(xi/2)}}$$ which does not have explicit solution.
However, we can make quite good approximation building the Padé approximant at $xi=0$
$$frac{1}{sqrt{text{erf}(xi/2)}}simeq sqrt[4]pi , frac{1+frac{363 }{3680}xi ^2+frac{491 }{264960}xi ^4-frac{23789 }{445132800}xi^6 }{sqrt xi left(1+frac{629 }{11040}xi ^2 right) }$$ which can be integrated to lead to the nasty (but practicable)
$$t(x)=-frac{23789 sqrt[4]{pi } x^{9/2}}{114125760}+frac{1085389 sqrt[4]{pi }
x^{5/2}}{55389740}+frac{3036269046 sqrt[4]{pi }
sqrt{x}}{1742007323}-frac{111936400 2^{3/4} sqrt[4]{frac{345 pi }{629}}
log left(sqrt{434010} x-4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}+frac{111936400 2^{3/4} sqrt[4]{frac{345
pi }{629}} log left(sqrt{434010} x+4 345^{3/4} sqrt[4]{1258}
sqrt{x}+2760right)}{1742007323}-frac{223872800 2^{3/4} sqrt[4]{frac{345
pi }{629}} tan ^{-1}left(1-frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}right)}{1742007323}+frac{223872800 2^{3/4}
sqrt[4]{frac{345 pi }{629}} tan ^{-1}left(frac{sqrt[4]{frac{629}{345}}
sqrt{x}}{2^{3/4}}+1right)}{1742007323}$$
which is not so bad as shown below
$$left(
begin{array}{ccc}
x & text{approximation} & text{exact} \
0.00 & 0.00000 & 0.00000 \
0.25 & 1.33203 & 1.33203 \
0.50 & 1.88671 & 1.88671 \
0.75 & 2.31671 & 2.31671 \
1.00 & 2.68470 & 2.68470 \
1.25 & 3.01528 & 3.01528 \
1.50 & 3.32122 & 3.32122 \
1.75 & 3.61020 & 3.61020 \
2.00 & 3.88724 & 3.88725 \
2.25 & 4.15580 & 4.15584 \
2.50 & 4.41833 & 4.41843 \
2.75 & 4.67656 & 4.67681 \
3.00 & 4.93172 & 4.93227 \
3.25 & 5.18460 & 5.18576 \
3.50 & 5.43566 & 5.43794 \
3.75 & 5.68507 & 5.68926 \
4.00 & 5.93273 & 5.94004 \
4.25 & 6.17831 & 6.19048 \
4.50 & 6.42126 & 6.44074 \
4.75 & 6.66082 & 6.69087 \
5.00 & 6.89601 & 6.94094
end{array}
right)$$
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
answered Dec 30 '18 at 16:01
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
add a comment |
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
$begingroup$
Hi Claude ! As observed by Christoph, there was a mistake in one of my derivations. So, the magnificent integral is not a correct solution. Sorry to have draw you to waste time. Still, happy new year !
$endgroup$
– JJacquelin
Jan 2 at 12:09
add a comment |
$begingroup$
The first step in a "general method" to solve this problem is to recognize that the given third-order ordinary differential equation (ODE) is autonomous.
We'll use subscripts for derivatives in the following, because we are going to use derivatives with respect to the independent variable $t$ but also with respect to the dependent variable $x$.
Starting with $2 x_{ttt} + x x_{tt} = 0$ we write $x_t = w(x)$ with some unknown function $w$, to obtain a boundary-value problem (BVP) with a second-order ODE for $w$:
begin{equation}
w w_{xx} + w_x left( w_x + frac{x}{2} right) = 0, quad w(0) = 0, quad w(infty) = 1
end{equation}
(note that the derivatives of $w$ are taken with respect to $x$!).
Once the function $w$ is found we may solve the following initial-value problem (IVP) with a separable first-order ODE for $x$:
begin{equation}
x_t = w(x), quad x(0) = 0.
end{equation}
Thus we have split the initial-boundary-value problem (IBVP) with a third-order autonomous ODE into a BVP with a second-order ODE and an IVP with a separable first-order ODE.
Going for the inverse function $t(x)$ as suggested by JJacquelin we may write
begin{equation}
t(x) = int limits_0^x frac{1}{w(xi)} , mathrm{d}xi,
end{equation}
with the function $w$ from above. Thus it remains to solve the BVP for $w$ and then to compute the integral.
Both tasks may be difficult, and I believe that there is a mistake in the solution provided by JJacquelin (in the calculation of the third derivative of the inverse function). I would normally write a comment on this but I am not yet allowed.
answered Jan 2 at 9:54
ChristophChristoph
1916
$endgroup$
$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
add a comment |
$begingroup$
The first step in a "general method" to solve this problem is to recognize that the given third-order ordinary differential equation (ODE) is autonomous.
We'll use subscripts for derivatives in the following, because we are going to use derivatives with respect to the independent variable $t$ but also with respect to the dependent variable $x$.
Starting with $2 x_{ttt} + x x_{tt} = 0$ we write $x_t = w(x)$ with some unknown function $w$, to obtain a boundary-value problem (BVP) with a second-order ODE for $w$:
begin{equation}
w w_{xx} + w_x left( w_x + frac{x}{2} right) = 0, quad w(0) = 0, quad w(infty) = 1
end{equation}
(note that the derivatives of $w$ are taken with respect to $x$!).
Once the function $w$ is found we may solve the following initial-value problem (IVP) with a separable first-order ODE for $x$:
begin{equation}
x_t = w(x), quad x(0) = 0.
end{equation}
Thus we have split the initial-boundary-value problem (IBVP) with a third-order autonomous ODE into a BVP with a second-order ODE and an IVP with a separable first-order ODE.
Going for the inverse function $t(x)$ as suggested by JJacquelin we may write
begin{equation}
t(x) = int limits_0^x frac{1}{w(xi)} , mathrm{d}xi,
end{equation}
with the function $w$ from above. Thus it remains to solve the BVP for $w$ and then to compute the integral.
Both tasks may be difficult, and I believe that there is a mistake in the solution provided by JJacquelin (in the calculation of the third derivative of the inverse function). I would normally write a comment on this but I am not yet allowed.
answered Jan 2 at 9:54
ChristophChristoph
1916
$endgroup$
$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
add a comment |
$begingroup$
The first step in a "general method" to solve this problem is to recognize that the given third-order ordinary differential equation (ODE) is autonomous.
We'll use subscripts for derivatives in the following, because we are going to use derivatives with respect to the independent variable $t$ but also with respect to the dependent variable $x$.
Starting with $2 x_{ttt} + x x_{tt} = 0$ we write $x_t = w(x)$ with some unknown function $w$, to obtain a boundary-value problem (BVP) with a second-order ODE for $w$:
begin{equation}
w w_{xx} + w_x left( w_x + frac{x}{2} right) = 0, quad w(0) = 0, quad w(infty) = 1
end{equation}
(note that the derivatives of $w$ are taken with respect to $x$!).
Once the function $w$ is found we may solve the following initial-value problem (IVP) with a separable first-order ODE for $x$:
begin{equation}
x_t = w(x), quad x(0) = 0.
end{equation}
Thus we have split the initial-boundary-value problem (IBVP) with a third-order autonomous ODE into a BVP with a second-order ODE and an IVP with a separable first-order ODE.
Going for the inverse function $t(x)$ as suggested by JJacquelin we may write
begin{equation}
t(x) = int limits_0^x frac{1}{w(xi)} , mathrm{d}xi,
end{equation}
with the function $w$ from above. Thus it remains to solve the BVP for $w$ and then to compute the integral.
Both tasks may be difficult, and I believe that there is a mistake in the solution provided by JJacquelin (in the calculation of the third derivative of the inverse function). I would normally write a comment on this but I am not yet allowed.
answered Jan 2 at 9:54
ChristophChristoph
1916
$endgroup$
The first step in a "general method" to solve this problem is to recognize that the given third-order ordinary differential equation (ODE) is autonomous.
We'll use subscripts for derivatives in the following, because we are going to use derivatives with respect to the independent variable $t$ but also with respect to the dependent variable $x$.
Starting with $2 x_{ttt} + x x_{tt} = 0$ we write $x_t = w(x)$ with some unknown function $w$, to obtain a boundary-value problem (BVP) with a second-order ODE for $w$:
begin{equation}
w w_{xx} + w_x left( w_x + frac{x}{2} right) = 0, quad w(0) = 0, quad w(infty) = 1
end{equation}
(note that the derivatives of $w$ are taken with respect to $x$!).
Once the function $w$ is found we may solve the following initial-value problem (IVP) with a separable first-order ODE for $x$:
begin{equation}
x_t = w(x), quad x(0) = 0.
end{equation}
Thus we have split the initial-boundary-value problem (IBVP) with a third-order autonomous ODE into a BVP with a second-order ODE and an IVP with a separable first-order ODE.
Going for the inverse function $t(x)$ as suggested by JJacquelin we may write
begin{equation}
t(x) = int limits_0^x frac{1}{w(xi)} , mathrm{d}xi,
end{equation}
with the function $w$ from above. Thus it remains to solve the BVP for $w$ and then to compute the integral.
Both tasks may be difficult, and I believe that there is a mistake in the solution provided by JJacquelin (in the calculation of the third derivative of the inverse function). I would normally write a comment on this but I am not yet allowed.
answered Jan 2 at 9:54
ChristophChristoph
1916
edited Jan 2 at 10:11
answered Jan 2 at 9:54
ChristophChristoph
1916
answered Jan 2 at 9:54
ChristophChristoph
1916
answered Jan 2 at 9:54
ChristophChristoph
1916
1916
$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
add a comment |
$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
$begingroup$
You are right, there is a mistake in the third derivative (I forgot a factor $1/t'$). As a consequence my solution is false. The corrected ODE with unknown $t(x)$ is no longer solvable. I delete my answer. In both methods (your and mine) the transformed ODE are not solvable for a closed form. The numerical method remains probably the only way.
$endgroup$
– JJacquelin
Jan 2 at 11:57
add a comment |
$begingroup$
Numerically there does not seem to be any problem, $T=20$ as approximation of $T=infty$ appears large enough, the solution can be computed via boundary value solver, here python's scipy.integrate.solve_bvp:
T = 20
def x_ode(t,x): return [x[1], x[2], -0.5*x[0]*x[2]]
def x_bc(x0, xT): return [x0[0], x0[1], xT[1]-1]
s = np.linspace(0,1,11);
t_init = T*s
x_init = [ T*s , 1+0*s, 0+0*s ]
res = solve_bvp(x_ode, x_bc, t_init, x_init, tol=1e-5, max_nodes=80000)
print res.message
with the result
The algorithm converged to the desired accuracy.
The plot of function and derivative
is then produced via
if res.success:
plt.figure(figsize=(9,6))
plt.subplot(211); plt.plot(res.x, res.y[0], '-'); plt.grid();
plt.subplot(212); plt.plot(res.x, res.y[1], '-'); plt.grid();
plt.show()
second approach
To transform the problem to a finite interval the boundary conditions suggest to use $s=x'$, $sin[0,1]$, as new independent parameter, thus using the inverse function $u$ to $x'$, $t=u(x')$, $x=v_0(x')$, $v_1(s)=x'(u(s))=s$, $v_2(s)=x''(u(s))$ which leads to the derivatives
begin{align}
1=frac{dt}{dt}&=u'(x'(t))x''(t)\
frac{dt}{ds}&=u'(s)=frac1{v_2(s)}\
frac{dv_0(s)}{ds}&=x'(u(s))u'(s)=frac{s}{v_2(s)}\
frac{dv_2(s)}{ds}&=x'''(u(s))u'(s)=-frac{v_0(s)}{2}
end{align}
To get $v_0(s)toinfty$ for $sto 1$ we need $v_2(1)=0$ along with $u(0)=0$ and $v_0(0)=0$. To desingularize $1/v_2$ the approach using $v_2/(epsilon^2+v_2^2)$ works best.
eps = 1e-8
def uv_ode(s,y):
u, v0, v2 = y;
v2inv = v2/(eps**2+v2**2);
return [v2inv, s*v2inv, -0.5*v0]
def uv_bc(u0, u1):
return [ u0[0], u0[1], u1[2]]
s = np.linspace(0,1,11);
y_init = [ 16*s**2 , 14*s**2, (1-s)**2/4 ]
res = solve_bvp(uv_ode, uv_bc, s, y_init, tol=1e-5, max_nodes=80000)
print res.message
if True or res.success:
plt.figure(figsize=(9,9))
plt.subplot(311); plt.plot(res.x, res.y[0], '-o',ms=1); plt.ylabel('$u=t$'); plt.grid();
plt.subplot(312); plt.plot(res.x, res.y[1], '-o',ms=1); plt.ylabel('$v_0=x$'); plt.grid();
plt.subplot(313); plt.plot(res.x, res.y[2], '-o',ms=1); plt.ylabel('$v_2=x''$'); plt.grid(); plt.xlabel('$s=x'$');
plt.show()
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
$endgroup$
add a comment |
$begingroup$
Numerically there does not seem to be any problem, $T=20$ as approximation of $T=infty$ appears large enough, the solution can be computed via boundary value solver, here python's scipy.integrate.solve_bvp:
T = 20
def x_ode(t,x): return [x[1], x[2], -0.5*x[0]*x[2]]
def x_bc(x0, xT): return [x0[0], x0[1], xT[1]-1]
s = np.linspace(0,1,11);
t_init = T*s
x_init = [ T*s , 1+0*s, 0+0*s ]
res = solve_bvp(x_ode, x_bc, t_init, x_init, tol=1e-5, max_nodes=80000)
print res.message
with the result
The algorithm converged to the desired accuracy.
The plot of function and derivative
is then produced via
if res.success:
plt.figure(figsize=(9,6))
plt.subplot(211); plt.plot(res.x, res.y[0], '-'); plt.grid();
plt.subplot(212); plt.plot(res.x, res.y[1], '-'); plt.grid();
plt.show()
second approach
To transform the problem to a finite interval the boundary conditions suggest to use $s=x'$, $sin[0,1]$, as new independent parameter, thus using the inverse function $u$ to $x'$, $t=u(x')$, $x=v_0(x')$, $v_1(s)=x'(u(s))=s$, $v_2(s)=x''(u(s))$ which leads to the derivatives
begin{align}
1=frac{dt}{dt}&=u'(x'(t))x''(t)\
frac{dt}{ds}&=u'(s)=frac1{v_2(s)}\
frac{dv_0(s)}{ds}&=x'(u(s))u'(s)=frac{s}{v_2(s)}\
frac{dv_2(s)}{ds}&=x'''(u(s))u'(s)=-frac{v_0(s)}{2}
end{align}
To get $v_0(s)toinfty$ for $sto 1$ we need $v_2(1)=0$ along with $u(0)=0$ and $v_0(0)=0$. To desingularize $1/v_2$ the approach using $v_2/(epsilon^2+v_2^2)$ works best.
eps = 1e-8
def uv_ode(s,y):
u, v0, v2 = y;
v2inv = v2/(eps**2+v2**2);
return [v2inv, s*v2inv, -0.5*v0]
def uv_bc(u0, u1):
return [ u0[0], u0[1], u1[2]]
s = np.linspace(0,1,11);
y_init = [ 16*s**2 , 14*s**2, (1-s)**2/4 ]
res = solve_bvp(uv_ode, uv_bc, s, y_init, tol=1e-5, max_nodes=80000)
print res.message
if True or res.success:
plt.figure(figsize=(9,9))
plt.subplot(311); plt.plot(res.x, res.y[0], '-o',ms=1); plt.ylabel('$u=t$'); plt.grid();
plt.subplot(312); plt.plot(res.x, res.y[1], '-o',ms=1); plt.ylabel('$v_0=x$'); plt.grid();
plt.subplot(313); plt.plot(res.x, res.y[2], '-o',ms=1); plt.ylabel('$v_2=x''$'); plt.grid(); plt.xlabel('$s=x'$');
plt.show()
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
$endgroup$
add a comment |
$begingroup$
Numerically there does not seem to be any problem, $T=20$ as approximation of $T=infty$ appears large enough, the solution can be computed via boundary value solver, here python's scipy.integrate.solve_bvp:
T = 20
def x_ode(t,x): return [x[1], x[2], -0.5*x[0]*x[2]]
def x_bc(x0, xT): return [x0[0], x0[1], xT[1]-1]
s = np.linspace(0,1,11);
t_init = T*s
x_init = [ T*s , 1+0*s, 0+0*s ]
res = solve_bvp(x_ode, x_bc, t_init, x_init, tol=1e-5, max_nodes=80000)
print res.message
with the result
The algorithm converged to the desired accuracy.
The plot of function and derivative
is then produced via
if res.success:
plt.figure(figsize=(9,6))
plt.subplot(211); plt.plot(res.x, res.y[0], '-'); plt.grid();
plt.subplot(212); plt.plot(res.x, res.y[1], '-'); plt.grid();
plt.show()
second approach
To transform the problem to a finite interval the boundary conditions suggest to use $s=x'$, $sin[0,1]$, as new independent parameter, thus using the inverse function $u$ to $x'$, $t=u(x')$, $x=v_0(x')$, $v_1(s)=x'(u(s))=s$, $v_2(s)=x''(u(s))$ which leads to the derivatives
begin{align}
1=frac{dt}{dt}&=u'(x'(t))x''(t)\
frac{dt}{ds}&=u'(s)=frac1{v_2(s)}\
frac{dv_0(s)}{ds}&=x'(u(s))u'(s)=frac{s}{v_2(s)}\
frac{dv_2(s)}{ds}&=x'''(u(s))u'(s)=-frac{v_0(s)}{2}
end{align}
To get $v_0(s)toinfty$ for $sto 1$ we need $v_2(1)=0$ along with $u(0)=0$ and $v_0(0)=0$. To desingularize $1/v_2$ the approach using $v_2/(epsilon^2+v_2^2)$ works best.
eps = 1e-8
def uv_ode(s,y):
u, v0, v2 = y;
v2inv = v2/(eps**2+v2**2);
return [v2inv, s*v2inv, -0.5*v0]
def uv_bc(u0, u1):
return [ u0[0], u0[1], u1[2]]
s = np.linspace(0,1,11);
y_init = [ 16*s**2 , 14*s**2, (1-s)**2/4 ]
res = solve_bvp(uv_ode, uv_bc, s, y_init, tol=1e-5, max_nodes=80000)
print res.message
if True or res.success:
plt.figure(figsize=(9,9))
plt.subplot(311); plt.plot(res.x, res.y[0], '-o',ms=1); plt.ylabel('$u=t$'); plt.grid();
plt.subplot(312); plt.plot(res.x, res.y[1], '-o',ms=1); plt.ylabel('$v_0=x$'); plt.grid();
plt.subplot(313); plt.plot(res.x, res.y[2], '-o',ms=1); plt.ylabel('$v_2=x''$'); plt.grid(); plt.xlabel('$s=x'$');
plt.show()
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
$endgroup$
Numerically there does not seem to be any problem, $T=20$ as approximation of $T=infty$ appears large enough, the solution can be computed via boundary value solver, here python's scipy.integrate.solve_bvp:
T = 20
def x_ode(t,x): return [x[1], x[2], -0.5*x[0]*x[2]]
def x_bc(x0, xT): return [x0[0], x0[1], xT[1]-1]
s = np.linspace(0,1,11);
t_init = T*s
x_init = [ T*s , 1+0*s, 0+0*s ]
res = solve_bvp(x_ode, x_bc, t_init, x_init, tol=1e-5, max_nodes=80000)
print res.message
with the result
The algorithm converged to the desired accuracy.
The plot of function and derivative
is then produced via
if res.success:
plt.figure(figsize=(9,6))
plt.subplot(211); plt.plot(res.x, res.y[0], '-'); plt.grid();
plt.subplot(212); plt.plot(res.x, res.y[1], '-'); plt.grid();
plt.show()
second approach
To transform the problem to a finite interval the boundary conditions suggest to use $s=x'$, $sin[0,1]$, as new independent parameter, thus using the inverse function $u$ to $x'$, $t=u(x')$, $x=v_0(x')$, $v_1(s)=x'(u(s))=s$, $v_2(s)=x''(u(s))$ which leads to the derivatives
begin{align}
1=frac{dt}{dt}&=u'(x'(t))x''(t)\
frac{dt}{ds}&=u'(s)=frac1{v_2(s)}\
frac{dv_0(s)}{ds}&=x'(u(s))u'(s)=frac{s}{v_2(s)}\
frac{dv_2(s)}{ds}&=x'''(u(s))u'(s)=-frac{v_0(s)}{2}
end{align}
To get $v_0(s)toinfty$ for $sto 1$ we need $v_2(1)=0$ along with $u(0)=0$ and $v_0(0)=0$. To desingularize $1/v_2$ the approach using $v_2/(epsilon^2+v_2^2)$ works best.
eps = 1e-8
def uv_ode(s,y):
u, v0, v2 = y;
v2inv = v2/(eps**2+v2**2);
return [v2inv, s*v2inv, -0.5*v0]
def uv_bc(u0, u1):
return [ u0[0], u0[1], u1[2]]
s = np.linspace(0,1,11);
y_init = [ 16*s**2 , 14*s**2, (1-s)**2/4 ]
res = solve_bvp(uv_ode, uv_bc, s, y_init, tol=1e-5, max_nodes=80000)
print res.message
if True or res.success:
plt.figure(figsize=(9,9))
plt.subplot(311); plt.plot(res.x, res.y[0], '-o',ms=1); plt.ylabel('$u=t$'); plt.grid();
plt.subplot(312); plt.plot(res.x, res.y[1], '-o',ms=1); plt.ylabel('$v_0=x$'); plt.grid();
plt.subplot(313); plt.plot(res.x, res.y[2], '-o',ms=1); plt.ylabel('$v_2=x''$'); plt.grid(); plt.xlabel('$s=x'$');
plt.show()
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
edited Jan 2 at 10:29
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
answered Dec 28 '18 at 16:21
LutzLLutzL
56.9k42054
56.9k42054
add a comment |
add a comment |
$begingroup$
Follows a MATHEMATICA script showing a numerical solution. Here $t = 200 approxinfty$
tmax = 200;
solx = NDSolve[{x1'[t] == x2[t],
x2'[t] == x3[t],
2 x3'[t] + x1[t] x3[t] == 0, x1[0] == 0, x2[0] == 0, x2[200] == 1},
{x1, x2, x3}, {t, 0, tmax}][[1]]
Plot[Evaluate[{x1[t], x2[t], x3[t]} /. solx], {t, 0, 20},
PlotRange -> {0, 2},
PlotStyle -> {{Black, Thick}, {Blue, Thick}, {Red, Thick}},
PlotLegends -> {Subscript[x, 1][t], Subscript[x, 2][t],
Subscript[x, 3][t]}]
$$
x_1(t) to mbox{black}\
x_2(t) to mbox{blue}\
x_3(t) to mbox{red}
$$
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
$endgroup$
add a comment |
$begingroup$
Follows a MATHEMATICA script showing a numerical solution. Here $t = 200 approxinfty$
tmax = 200;
solx = NDSolve[{x1'[t] == x2[t],
x2'[t] == x3[t],
2 x3'[t] + x1[t] x3[t] == 0, x1[0] == 0, x2[0] == 0, x2[200] == 1},
{x1, x2, x3}, {t, 0, tmax}][[1]]
Plot[Evaluate[{x1[t], x2[t], x3[t]} /. solx], {t, 0, 20},
PlotRange -> {0, 2},
PlotStyle -> {{Black, Thick}, {Blue, Thick}, {Red, Thick}},
PlotLegends -> {Subscript[x, 1][t], Subscript[x, 2][t],
Subscript[x, 3][t]}]
$$
x_1(t) to mbox{black}\
x_2(t) to mbox{blue}\
x_3(t) to mbox{red}
$$
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
$endgroup$
add a comment |
$begingroup$
Follows a MATHEMATICA script showing a numerical solution. Here $t = 200 approxinfty$
tmax = 200;
solx = NDSolve[{x1'[t] == x2[t],
x2'[t] == x3[t],
2 x3'[t] + x1[t] x3[t] == 0, x1[0] == 0, x2[0] == 0, x2[200] == 1},
{x1, x2, x3}, {t, 0, tmax}][[1]]
Plot[Evaluate[{x1[t], x2[t], x3[t]} /. solx], {t, 0, 20},
PlotRange -> {0, 2},
PlotStyle -> {{Black, Thick}, {Blue, Thick}, {Red, Thick}},
PlotLegends -> {Subscript[x, 1][t], Subscript[x, 2][t],
Subscript[x, 3][t]}]
$$
x_1(t) to mbox{black}\
x_2(t) to mbox{blue}\
x_3(t) to mbox{red}
$$
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
$endgroup$
Follows a MATHEMATICA script showing a numerical solution. Here $t = 200 approxinfty$
tmax = 200;
solx = NDSolve[{x1'[t] == x2[t],
x2'[t] == x3[t],
2 x3'[t] + x1[t] x3[t] == 0, x1[0] == 0, x2[0] == 0, x2[200] == 1},
{x1, x2, x3}, {t, 0, tmax}][[1]]
Plot[Evaluate[{x1[t], x2[t], x3[t]} /. solx], {t, 0, 20},
PlotRange -> {0, 2},
PlotStyle -> {{Black, Thick}, {Blue, Thick}, {Red, Thick}},
PlotLegends -> {Subscript[x, 1][t], Subscript[x, 2][t],
Subscript[x, 3][t]}]
$$
x_1(t) to mbox{black}\
x_2(t) to mbox{blue}\
x_3(t) to mbox{red}
$$
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
answered Dec 29 '18 at 10:34
CesareoCesareo
8,4563516
8,4563516
add a comment |
add a comment |
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$begingroup$
My mistake while typing. It is just single derivative. Thankyou for pointing mistake
$endgroup$
– Abhi7731756
Dec 28 '18 at 15:38
$begingroup$
Do you have any reason to think this can be solved analytically? One certainly would not expect the Fourier transform or the Laplace transform to help, since the equation's not linear. In general non-linear differential equations are hard...
$endgroup$
– David C. Ullrich
Dec 28 '18 at 16:31