How to solve this differential equation with singularity?
$begingroup$
I want to find the solution of the following differential equation.
$$
frac{mathrm{d}^{2}x}{mathrm{d}t^{2}} =
−frac{1}{ax + mathrm{d}x/mathrm{d}t}
$$
where a is a positive constant and x is a position of a free object that is not bounded or tied to a fixed object (e.g., a wall).
I think that the solution may behave like an oscillator, but it seem to be slightly different from that of the oscillator. However, due to the terms in the denominator, it has the singularity. How to solve this type of equation?
** As I studied differential equations for a while 20 years ago, I do not know how to solve this problem. Since the solution of the above differential equation is an important issue for me, I would appreciate it if you help.
ordinary-differential-equations wave-equation
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
asked Jan 6 at 12:32
SOQEHSOQEH
134
$endgroup$
add a comment |
$begingroup$
I want to find the solution of the following differential equation.
$$
frac{mathrm{d}^{2}x}{mathrm{d}t^{2}} =
−frac{1}{ax + mathrm{d}x/mathrm{d}t}
$$
where a is a positive constant and x is a position of a free object that is not bounded or tied to a fixed object (e.g., a wall).
I think that the solution may behave like an oscillator, but it seem to be slightly different from that of the oscillator. However, due to the terms in the denominator, it has the singularity. How to solve this type of equation?
** As I studied differential equations for a while 20 years ago, I do not know how to solve this problem. Since the solution of the above differential equation is an important issue for me, I would appreciate it if you help.
ordinary-differential-equations wave-equation
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
asked Jan 6 at 12:32
SOQEHSOQEH
134
$endgroup$
add a comment |
$begingroup$
I want to find the solution of the following differential equation.
$$
frac{mathrm{d}^{2}x}{mathrm{d}t^{2}} =
−frac{1}{ax + mathrm{d}x/mathrm{d}t}
$$
where a is a positive constant and x is a position of a free object that is not bounded or tied to a fixed object (e.g., a wall).
I think that the solution may behave like an oscillator, but it seem to be slightly different from that of the oscillator. However, due to the terms in the denominator, it has the singularity. How to solve this type of equation?
** As I studied differential equations for a while 20 years ago, I do not know how to solve this problem. Since the solution of the above differential equation is an important issue for me, I would appreciate it if you help.
ordinary-differential-equations wave-equation
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
asked Jan 6 at 12:32
SOQEHSOQEH
134
$endgroup$
I want to find the solution of the following differential equation.
$$
frac{mathrm{d}^{2}x}{mathrm{d}t^{2}} =
−frac{1}{ax + mathrm{d}x/mathrm{d}t}
$$
where a is a positive constant and x is a position of a free object that is not bounded or tied to a fixed object (e.g., a wall).
I think that the solution may behave like an oscillator, but it seem to be slightly different from that of the oscillator. However, due to the terms in the denominator, it has the singularity. How to solve this type of equation?
** As I studied differential equations for a while 20 years ago, I do not know how to solve this problem. Since the solution of the above differential equation is an important issue for me, I would appreciate it if you help.
ordinary-differential-equations wave-equation
ordinary-differential-equations wave-equation
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
asked Jan 6 at 12:32
SOQEHSOQEH
134
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
asked Jan 6 at 12:32
SOQEHSOQEH
134
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
edited Jan 6 at 20:16
Felix Marin
67.3k7107141
67.3k7107141
asked Jan 6 at 12:32
SOQEHSOQEH
134
asked Jan 6 at 12:32
SOQEHSOQEH
134
asked Jan 6 at 12:32
SOQEHSOQEH
134
134
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$frac{d^2x}{dt^2}=−frac{1}{a x + frac{dx}{dt}} tag 1$$
The usual change of function to reduce the order of this autonomous ODE is
$frac{dx}{dt}=y(x) quad;quad frac{d^2x}{dt^2}=frac{dy}{dx}frac{dx}{dt}=yfrac{dy}{dx}$
$$yfrac{dy}{dx}=−frac{1}{a x + y}$$
Consider the inverse function $x(y)$
$$frac{dx}{dy}=−(a x + y)y$$
This is a first order linear ODE with respect to the function $x(y)$. The solution is :
$$x(y)=-frac{y}{a}+sqrt{frac{pi}{2a^3}}e^{-ay^2/2}text{erfi}
left(sqrt{frac{a}{2}}:yright)+c_1 e^{-ay^2/2}$$
Unfortunately there is no closed form for the inverse function $y(x)$ in general (for arbitrary values of $a$ and $c_1$).
A fortiori no closed form for $t=intfrac{dx}{y(x)}$.
Of course, the solutions of equation $(1)$ exist but in general they cannot be expressed with a finite number of standard functions.
As a consequence, in practice solve the ODE $(1)$ with numerical calculus. It should be a waste of time to try to solve it analytically in general.
Nevertheless, it might be possible to solve the problem analytically is some particular cases, depending on the initial conditions of the ODE and/or with particular values of the parameter $a$.
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
$endgroup$
1
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
1
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
|
show 2 more comments
Your Answer
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1 Answer
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$begingroup$
$$frac{d^2x}{dt^2}=−frac{1}{a x + frac{dx}{dt}} tag 1$$
The usual change of function to reduce the order of this autonomous ODE is
$frac{dx}{dt}=y(x) quad;quad frac{d^2x}{dt^2}=frac{dy}{dx}frac{dx}{dt}=yfrac{dy}{dx}$
$$yfrac{dy}{dx}=−frac{1}{a x + y}$$
Consider the inverse function $x(y)$
$$frac{dx}{dy}=−(a x + y)y$$
This is a first order linear ODE with respect to the function $x(y)$. The solution is :
$$x(y)=-frac{y}{a}+sqrt{frac{pi}{2a^3}}e^{-ay^2/2}text{erfi}
left(sqrt{frac{a}{2}}:yright)+c_1 e^{-ay^2/2}$$
Unfortunately there is no closed form for the inverse function $y(x)$ in general (for arbitrary values of $a$ and $c_1$).
A fortiori no closed form for $t=intfrac{dx}{y(x)}$.
Of course, the solutions of equation $(1)$ exist but in general they cannot be expressed with a finite number of standard functions.
As a consequence, in practice solve the ODE $(1)$ with numerical calculus. It should be a waste of time to try to solve it analytically in general.
Nevertheless, it might be possible to solve the problem analytically is some particular cases, depending on the initial conditions of the ODE and/or with particular values of the parameter $a$.
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
$endgroup$
1
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
1
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
|
show 2 more comments
$begingroup$
$$frac{d^2x}{dt^2}=−frac{1}{a x + frac{dx}{dt}} tag 1$$
The usual change of function to reduce the order of this autonomous ODE is
$frac{dx}{dt}=y(x) quad;quad frac{d^2x}{dt^2}=frac{dy}{dx}frac{dx}{dt}=yfrac{dy}{dx}$
$$yfrac{dy}{dx}=−frac{1}{a x + y}$$
Consider the inverse function $x(y)$
$$frac{dx}{dy}=−(a x + y)y$$
This is a first order linear ODE with respect to the function $x(y)$. The solution is :
$$x(y)=-frac{y}{a}+sqrt{frac{pi}{2a^3}}e^{-ay^2/2}text{erfi}
left(sqrt{frac{a}{2}}:yright)+c_1 e^{-ay^2/2}$$
Unfortunately there is no closed form for the inverse function $y(x)$ in general (for arbitrary values of $a$ and $c_1$).
A fortiori no closed form for $t=intfrac{dx}{y(x)}$.
Of course, the solutions of equation $(1)$ exist but in general they cannot be expressed with a finite number of standard functions.
As a consequence, in practice solve the ODE $(1)$ with numerical calculus. It should be a waste of time to try to solve it analytically in general.
Nevertheless, it might be possible to solve the problem analytically is some particular cases, depending on the initial conditions of the ODE and/or with particular values of the parameter $a$.
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
$endgroup$
1
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
1
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
|
show 2 more comments
$begingroup$
$$frac{d^2x}{dt^2}=−frac{1}{a x + frac{dx}{dt}} tag 1$$
The usual change of function to reduce the order of this autonomous ODE is
$frac{dx}{dt}=y(x) quad;quad frac{d^2x}{dt^2}=frac{dy}{dx}frac{dx}{dt}=yfrac{dy}{dx}$
$$yfrac{dy}{dx}=−frac{1}{a x + y}$$
Consider the inverse function $x(y)$
$$frac{dx}{dy}=−(a x + y)y$$
This is a first order linear ODE with respect to the function $x(y)$. The solution is :
$$x(y)=-frac{y}{a}+sqrt{frac{pi}{2a^3}}e^{-ay^2/2}text{erfi}
left(sqrt{frac{a}{2}}:yright)+c_1 e^{-ay^2/2}$$
Unfortunately there is no closed form for the inverse function $y(x)$ in general (for arbitrary values of $a$ and $c_1$).
A fortiori no closed form for $t=intfrac{dx}{y(x)}$.
Of course, the solutions of equation $(1)$ exist but in general they cannot be expressed with a finite number of standard functions.
As a consequence, in practice solve the ODE $(1)$ with numerical calculus. It should be a waste of time to try to solve it analytically in general.
Nevertheless, it might be possible to solve the problem analytically is some particular cases, depending on the initial conditions of the ODE and/or with particular values of the parameter $a$.
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
$endgroup$
$$frac{d^2x}{dt^2}=−frac{1}{a x + frac{dx}{dt}} tag 1$$
The usual change of function to reduce the order of this autonomous ODE is
$frac{dx}{dt}=y(x) quad;quad frac{d^2x}{dt^2}=frac{dy}{dx}frac{dx}{dt}=yfrac{dy}{dx}$
$$yfrac{dy}{dx}=−frac{1}{a x + y}$$
Consider the inverse function $x(y)$
$$frac{dx}{dy}=−(a x + y)y$$
This is a first order linear ODE with respect to the function $x(y)$. The solution is :
$$x(y)=-frac{y}{a}+sqrt{frac{pi}{2a^3}}e^{-ay^2/2}text{erfi}
left(sqrt{frac{a}{2}}:yright)+c_1 e^{-ay^2/2}$$
Unfortunately there is no closed form for the inverse function $y(x)$ in general (for arbitrary values of $a$ and $c_1$).
A fortiori no closed form for $t=intfrac{dx}{y(x)}$.
Of course, the solutions of equation $(1)$ exist but in general they cannot be expressed with a finite number of standard functions.
As a consequence, in practice solve the ODE $(1)$ with numerical calculus. It should be a waste of time to try to solve it analytically in general.
Nevertheless, it might be possible to solve the problem analytically is some particular cases, depending on the initial conditions of the ODE and/or with particular values of the parameter $a$.
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
edited Jan 6 at 21:39
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
answered Jan 6 at 18:31
JJacquelinJJacquelin
43k21751
43k21751
1
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
1
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
|
show 2 more comments
1
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
1
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
1
1
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
should be $y^{-1}dx/dy=-(ax+y)$
$endgroup$
– user254433
Jan 6 at 20:22
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
You are right. Thank you for your remark. I corrected my calculus. Finally the conclusion remains the same.
$endgroup$
– JJacquelin
Jan 6 at 21:41
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
@JJacquelin I am much obliged to you for your help.
$endgroup$
– SOQEH
Jan 7 at 0:17
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
$begingroup$
I don't think $x=-y/a$ is a solution
$endgroup$
– Dylan
Jan 7 at 9:16
1
1
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
$begingroup$
Hi Jean ! You probably noticed that, assuming $a>0$ and small values of $y$, we have $x(y)=-frac{y^3}{3}+frac{a y^5}{15}+Oleft(y^7right)$ and then we could (may be !) get ... something. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 11:15
|
show 2 more comments
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