How to solve the coupled differential equation
$begingroup$
I have the following differential equations.
$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$
$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.
Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?
thanks
ordinary-differential-equations numerical-methods
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
$endgroup$
add a comment |
$begingroup$
I have the following differential equations.
$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$
$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.
Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?
thanks
ordinary-differential-equations numerical-methods
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
$endgroup$
$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51
$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52
$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01
$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00
$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00
add a comment |
$begingroup$
I have the following differential equations.
$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$
$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.
Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?
thanks
ordinary-differential-equations numerical-methods
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
$endgroup$
I have the following differential equations.
$$ tau Vfrac{partial r}{partial x} + Wfrac{partial ^2r}{partial x^2} + [r - Cu(1-r^2)][1-r^2]=0,$$
$$ Vfrac{partial u}{partial x} + Dfrac{partial^2 u}{partial x^2} - frac{V}{2}frac{partial r}{partial x}=0,$$ with far field boundary condition as $u_{x=infty} = 0.6.$ and $u_{x=0} = 0$. Here, $tau, V, W, D, C$ are dimensionless constants.
Now, I would like to solve the second equation subject to boundary conditions. But the last term in the second equation confuses me on how to proceed. Can anyone help me out with this ?
thanks
ordinary-differential-equations numerical-methods
ordinary-differential-equations numerical-methods
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
edited Dec 18 '18 at 13:03
newstudent
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
asked Dec 18 '18 at 12:38
newstudentnewstudent
366
366
$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51
$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52
$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01
$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00
$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00
add a comment |
$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51
$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52
$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01
$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00
$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00
$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51
$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51
$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52
$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52
$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01
$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
$endgroup$
– rafa11111
Dec 18 '18 at 13:01
$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00
$begingroup$
You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
$endgroup$
– user121049
Dec 19 '18 at 8:00
$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00
$begingroup$
You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
$endgroup$
– user121049
Dec 20 '18 at 9:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.
Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter
Let $a = frac{V}{D}$
$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$
The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form
$$ u_p(x) = p(x) + q(x)e^{-ax} $$
Then
begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}
or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$
You can integrate to find
begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}
Therefore
$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
Matching the boundary conditions, we find
$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
$endgroup$
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.
Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter
Let $a = frac{V}{D}$
$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$
The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form
$$ u_p(x) = p(x) + q(x)e^{-ax} $$
Then
begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}
or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$
You can integrate to find
begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}
Therefore
$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
Matching the boundary conditions, we find
$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
$endgroup$
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
add a comment |
$begingroup$
Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.
Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter
Let $a = frac{V}{D}$
$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$
The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form
$$ u_p(x) = p(x) + q(x)e^{-ax} $$
Then
begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}
or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$
You can integrate to find
begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}
Therefore
$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
Matching the boundary conditions, we find
$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
$endgroup$
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
add a comment |
$begingroup$
Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.
Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter
Let $a = frac{V}{D}$
$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$
The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form
$$ u_p(x) = p(x) + q(x)e^{-ax} $$
Then
begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}
or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$
You can integrate to find
begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}
Therefore
$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
Matching the boundary conditions, we find
$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
$endgroup$
Disclaimer: I've just realized you can't actually solve for $r(x)$ in the first equation, which makes my answer kind of useless. I'll leave it up anyway, to learn from my mistakes.
Suppose you've already solved the first equation, and $r(x)$ is known, then the second equation is linear and non-homogeneous, and can easily be solved using variations of parameter
Let $a = frac{V}{D}$
$$ frac{d^2u}{dx^2} + afrac{du}{dx} = frac{a}{2}frac{dr}{dx} $$
The homogeneous equation has solution $u_h = c_1 + c_2e^{-ax}$, so guess a particular solution of the form
$$ u_p(x) = p(x) + q(x)e^{-ax} $$
Then
begin{align} p'(x) + q'(x)e^{-ax} &= 0 \ -aq'(x)e^{-ax} &= frac{a}{2}r'(x) end{align}
or $p'(x) = frac{1}{2}r'(x)$ and $q'(x) = -frac{1}{2}r'(x)e^{-ax}$
You can integrate to find
begin{align} p(x) &= frac{1}{2}r(x) \ q(x) &=-frac12int r'(x) e^{ax} = -frac12 r(x)e^{ax} dx + frac{a}{2}int r(x)e^{ax} dx end{align}
Therefore
$$ u_p(x) = frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
Matching the boundary conditions, we find
$$ u(x) = u_{infty}(1-e^{-ax}) + frac{a}{2}e^{-ax}int_0^x r(t)e^{at} dt $$
where $u_infty$ is some constant depending on the behavior of $r(x)$ at ${xtoinfty}$
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
edited Dec 20 '18 at 4:03
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
answered Dec 19 '18 at 7:48
DylanDylan
14.1k31127
14.1k31127
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
add a comment |
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
There is a $u(x)$ in the first equation so how do you solve it for $r(x)$?
$endgroup$
– user121049
Dec 19 '18 at 20:37
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
$begingroup$
@user121049 Wow I didn't even notice. Thanks for pointing it out.
$endgroup$
– Dylan
Dec 20 '18 at 3:57
add a comment |
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$begingroup$
I would rather say that the last term in the first equation is the most difficult to deal with. Is $x$ the only independent variable? Why are those partial differential equations instead of ordinary equations? There are a handful numerical methods that would help you to solve these equations if they are ordinary.
$endgroup$
– rafa11111
Dec 18 '18 at 12:51
$begingroup$
@rafa11111 Ok. let us say these are ordinary equations. I would like to solve it analytically.
$endgroup$
– newstudent
Dec 18 '18 at 12:52
$begingroup$
If you are sure these are ordinary equations, please edit the question to address so (change the equation, the tag, etc.). I don't think there is an analytical solution to these equations, mainly due to the nonlinear term in the first equation. If $C$ is a small number, you can use a perturbation method, that provides an approximate analytical solution.
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– rafa11111
Dec 18 '18 at 13:01
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You can integrate the second equation once then substitute for $r(x)$ in the first to give an equation in $u(x)$ alone. I think your journey will end there!
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– user121049
Dec 19 '18 at 8:00
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You need some more boundary conditions on $r$. The only hope I see is to assume $r$ stays small so you can set $1-r^2approx 1$ and linearize the first equation. The two equations are then easy to solve. Then check $r$ really stays small.
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– user121049
Dec 20 '18 at 9:00