Solving a non-linear ODE system with ParametricNDsolveValue

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enter image description here



I'm trying to solve a non linear ODE numerically with ParametricNDSolve, but as far as I got is shown below. My problem is to set the find root of these set of equations since I have three region, I need to match the solution between them. There might some error in my code I don't know how to match the three regions.
What I know is this: x'[0]==0, x[R]==0, x'[R]==0.



Here is my code: 



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps1 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[a] == x0, x'[ϵ] == 0, x, x', r, ϵ, a, n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps2 = ParametricNDSolveValue[
 x''[r] + 2 x'[r] == c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), 
 x[a] == x0, x'[a] == x[a], x, x', r, a, b, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps3 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[R] == 0, x'[R] == 0, x, x', r, b, R, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30]





differential-equations equation-solving parametric-functions







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  • How do you want to combine these three solutions of three different equations? Or is this one solution of an equation with discontinuous coefficients?
    – Alex Trounev
    4 hours ago










  • It is one equations has three regions see the image.
    – user60416
    4 hours ago










  • And why in conditions $R<b$
    – Alex Trounev
    4 hours ago










  • it is R>b. I fixed it
    – user60416
    4 hours ago














up vote
3
down vote

favorite












enter image description here



I'm trying to solve a non linear ODE numerically with ParametricNDSolve, but as far as I got is shown below. My problem is to set the find root of these set of equations since I have three region, I need to match the solution between them. There might some error in my code I don't know how to match the three regions.
What I know is this: x'[0]==0, x[R]==0, x'[R]==0.



Here is my code: 



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps1 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[a] == x0, x'[ϵ] == 0, x, x', r, ϵ, a, n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps2 = ParametricNDSolveValue[
 x''[r] + 2 x'[r] == c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), 
 x[a] == x0, x'[a] == x[a], x, x', r, a, b, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps3 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[R] == 0, x'[R] == 0, x, x', r, b, R, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30]





differential-equations equation-solving parametric-functions







share|improve this question









New contributor




user60416 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • How do you want to combine these three solutions of three different equations? Or is this one solution of an equation with discontinuous coefficients?
    – Alex Trounev
    4 hours ago










  • It is one equations has three regions see the image.
    – user60416
    4 hours ago










  • And why in conditions $R<b$
    – Alex Trounev
    4 hours ago










  • it is R>b. I fixed it
    – user60416
    4 hours ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











enter image description here



I'm trying to solve a non linear ODE numerically with ParametricNDSolve, but as far as I got is shown below. My problem is to set the find root of these set of equations since I have three region, I need to match the solution between them. There might some error in my code I don't know how to match the three regions.
What I know is this: x'[0]==0, x[R]==0, x'[R]==0.



Here is my code: 



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps1 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[a] == x0, x'[ϵ] == 0, x, x', r, ϵ, a, n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps2 = ParametricNDSolveValue[
 x''[r] + 2 x'[r] == c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), 
 x[a] == x0, x'[a] == x[a], x, x', r, a, b, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps3 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[R] == 0, x'[R] == 0, x, x', r, b, R, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30]





differential-equations equation-solving parametric-functions







share|improve this question









New contributor




user60416 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











enter image description here



I'm trying to solve a non linear ODE numerically with ParametricNDSolve, but as far as I got is shown below. My problem is to set the find root of these set of equations since I have three region, I need to match the solution between them. There might some error in my code I don't know how to match the three regions.
What I know is this: x'[0]==0, x[R]==0, x'[R]==0.



Here is my code: 



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps1 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[a] == x0, x'[ϵ] == 0, x, x', r, ϵ, a, n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps2 = ParametricNDSolveValue[
 x''[r] + 2 x'[r] == c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), 
 x[a] == x0, x'[a] == x[a], x, x', r, a, b, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30];
ps3 = ParametricNDSolveValue[x''[r] + 2 x'[r] == c n0 Exp[-x[r]], 
 x[R] == 0, x'[R] == 0, x, x', r, b, R, n0, 
 Method -> "StiffnessSwitching", WorkingPrecision -> 30]





differential-equations equation-solving parametric-functions




differential-equations equation-solving parametric-functions






share|improve this question









New contributor




user60416 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question









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edited 3 hours ago









Henrik Schumacher

39.9k255119




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asked 4 hours ago









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user60416 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.











  • How do you want to combine these three solutions of three different equations? Or is this one solution of an equation with discontinuous coefficients?
    – Alex Trounev
    4 hours ago










  • It is one equations has three regions see the image.
    – user60416
    4 hours ago










  • And why in conditions $R<b$
    – Alex Trounev
    4 hours ago










  • it is R>b. I fixed it
    – user60416
    4 hours ago
















  • How do you want to combine these three solutions of three different equations? Or is this one solution of an equation with discontinuous coefficients?
    – Alex Trounev
    4 hours ago










  • It is one equations has three regions see the image.
    – user60416
    4 hours ago










  • And why in conditions $R<b$
    – Alex Trounev
    4 hours ago










  • it is R>b. I fixed it
    – user60416
    4 hours ago















How do you want to combine these three solutions of three different equations? Or is this one solution of an equation with discontinuous coefficients?
– Alex Trounev
4 hours ago




How do you want to combine these three solutions of three different equations? Or is this one solution of an equation with discontinuous coefficients?
– Alex Trounev
4 hours ago












It is one equations has three regions see the image.
– user60416
4 hours ago




It is one equations has three regions see the image.
– user60416
4 hours ago












And why in conditions $R<b$
– Alex Trounev
4 hours ago




And why in conditions $R<b$
– Alex Trounev
4 hours ago












it is R>b. I fixed it
– user60416
4 hours ago




it is R>b. I fixed it
– user60416
4 hours ago










2 Answers
2






active

oldest

votes

















up vote
4
down vote













This problem has a solution. It is given below



c = 0.72;
 h = 300;
 a = 15;
 b = 17;
 R = 25;
 f[r_] := Piecewise[0, 0 <= r <= a, (3 h)/(a^3 - b^3), 
 a < r <= b, 0, b < r <= R]


ps = ParametricNDSolveValue[x''[r] + 2 x'[r] == 
 c n0 Exp[-x[r]] + f[r], x'[0] == 0, x[0] == x0, 
 x, r, 0, R, n0, x0];

 n = 
 FindRoot[ps[n0, x0][R] == 0, ps[n0, x0]'[R] == 0, n0, -.2, x0, 
 1]

 n0 -> 9.19855*10^-8, x0 -> 0.585175

Plot[
 Evaluate[Table[ps[n0, 1][r], n0, -.2, 2, .1]], r, 0, R, 
 PlotRange -> All],
 Plot[ps[n[[1, 2]], n[[2, 2]]][r], r, 0, R]


fig1






share|improve this answer






















  • Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
    – Henrik Schumacher
    13 mins ago

















up vote
3
down vote













As Alex Trounev said, this is a second-order ODE with discontinuous right-hand side. You can use Piecewise to set up the forcing term:



rhs = Piecewise[
 c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), a <= r < b
 ,
 c n0 Exp[-x[r]]
 ]



$$begincases frac3 ha^3-b^3+c ,textn0, e^-x(r) & aleq r<b \ c ,textn0 , e^-x(r) & textTrue endcases$$




Notice that for convenience, I used c n0 Exp[-x[r]] as the default term. The full equation can be set up as



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps = ParametricNDSolveValue[
 
 x''[r] + 2 x'[r] == rhs,
 x[R] == 0,
 x'[R] == 0
 ,
 x,
 r, ϵ, R,
 n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30
 ];


Solving it for a given parameter



f = ps[0.00001];


Plotting the result:



Plot[f[r], r, ϵ, R]


enter image description here



Something must be wrong in your model: The solution blows up heavily towards $r = 0$ so that x'[0] == 0 cannot be expected. Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2.






share|improve this answer
















  • 1




    "Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
    – J. M. is somewhat okay.♦
    3 hours ago











  • @Henrik Schumacher Look at the solution from the other side at x=0
    – Alex Trounev
    17 mins ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote













This problem has a solution. It is given below



c = 0.72;
 h = 300;
 a = 15;
 b = 17;
 R = 25;
 f[r_] := Piecewise[0, 0 <= r <= a, (3 h)/(a^3 - b^3), 
 a < r <= b, 0, b < r <= R]


ps = ParametricNDSolveValue[x''[r] + 2 x'[r] == 
 c n0 Exp[-x[r]] + f[r], x'[0] == 0, x[0] == x0, 
 x, r, 0, R, n0, x0];

 n = 
 FindRoot[ps[n0, x0][R] == 0, ps[n0, x0]'[R] == 0, n0, -.2, x0, 
 1]

 n0 -> 9.19855*10^-8, x0 -> 0.585175

Plot[
 Evaluate[Table[ps[n0, 1][r], n0, -.2, 2, .1]], r, 0, R, 
 PlotRange -> All],
 Plot[ps[n[[1, 2]], n[[2, 2]]][r], r, 0, R]


fig1






share|improve this answer






















  • Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
    – Henrik Schumacher
    13 mins ago














up vote
4
down vote













This problem has a solution. It is given below



c = 0.72;
 h = 300;
 a = 15;
 b = 17;
 R = 25;
 f[r_] := Piecewise[0, 0 <= r <= a, (3 h)/(a^3 - b^3), 
 a < r <= b, 0, b < r <= R]


ps = ParametricNDSolveValue[x''[r] + 2 x'[r] == 
 c n0 Exp[-x[r]] + f[r], x'[0] == 0, x[0] == x0, 
 x, r, 0, R, n0, x0];

 n = 
 FindRoot[ps[n0, x0][R] == 0, ps[n0, x0]'[R] == 0, n0, -.2, x0, 
 1]

 n0 -> 9.19855*10^-8, x0 -> 0.585175

Plot[
 Evaluate[Table[ps[n0, 1][r], n0, -.2, 2, .1]], r, 0, R, 
 PlotRange -> All],
 Plot[ps[n[[1, 2]], n[[2, 2]]][r], r, 0, R]


fig1






share|improve this answer






















  • Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
    – Henrik Schumacher
    13 mins ago












up vote
4
down vote










up vote
4
down vote









This problem has a solution. It is given below



c = 0.72;
 h = 300;
 a = 15;
 b = 17;
 R = 25;
 f[r_] := Piecewise[0, 0 <= r <= a, (3 h)/(a^3 - b^3), 
 a < r <= b, 0, b < r <= R]


ps = ParametricNDSolveValue[x''[r] + 2 x'[r] == 
 c n0 Exp[-x[r]] + f[r], x'[0] == 0, x[0] == x0, 
 x, r, 0, R, n0, x0];

 n = 
 FindRoot[ps[n0, x0][R] == 0, ps[n0, x0]'[R] == 0, n0, -.2, x0, 
 1]

 n0 -> 9.19855*10^-8, x0 -> 0.585175

Plot[
 Evaluate[Table[ps[n0, 1][r], n0, -.2, 2, .1]], r, 0, R, 
 PlotRange -> All],
 Plot[ps[n[[1, 2]], n[[2, 2]]][r], r, 0, R]


fig1






share|improve this answer














This problem has a solution. It is given below



c = 0.72;
 h = 300;
 a = 15;
 b = 17;
 R = 25;
 f[r_] := Piecewise[0, 0 <= r <= a, (3 h)/(a^3 - b^3), 
 a < r <= b, 0, b < r <= R]


ps = ParametricNDSolveValue[x''[r] + 2 x'[r] == 
 c n0 Exp[-x[r]] + f[r], x'[0] == 0, x[0] == x0, 
 x, r, 0, R, n0, x0];

 n = 
 FindRoot[ps[n0, x0][R] == 0, ps[n0, x0]'[R] == 0, n0, -.2, x0, 
 1]

 n0 -> 9.19855*10^-8, x0 -> 0.585175

Plot[
 Evaluate[Table[ps[n0, 1][r], n0, -.2, 2, .1]], r, 0, R, 
 PlotRange -> All],
 Plot[ps[n[[1, 2]], n[[2, 2]]][r], r, 0, R]


fig1







share|improve this answer














share|improve this answer



share|improve this answer








edited 14 mins ago

























answered 3 hours ago









Alex Trounev

2,535312




2,535312











  • Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
    – Henrik Schumacher
    13 mins ago
















  • Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
    – Henrik Schumacher
    13 mins ago















Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
– Henrik Schumacher
13 mins ago




Ah, now I get it. The third boundary condition is used to determine the parameter n0. Good job (and of course +1)!
– Henrik Schumacher
13 mins ago










up vote
3
down vote













As Alex Trounev said, this is a second-order ODE with discontinuous right-hand side. You can use Piecewise to set up the forcing term:



rhs = Piecewise[
 c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), a <= r < b
 ,
 c n0 Exp[-x[r]]
 ]



$$begincases frac3 ha^3-b^3+c ,textn0, e^-x(r) & aleq r<b \ c ,textn0 , e^-x(r) & textTrue endcases$$




Notice that for convenience, I used c n0 Exp[-x[r]] as the default term. The full equation can be set up as



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps = ParametricNDSolveValue[
 
 x''[r] + 2 x'[r] == rhs,
 x[R] == 0,
 x'[R] == 0
 ,
 x,
 r, ϵ, R,
 n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30
 ];


Solving it for a given parameter



f = ps[0.00001];


Plotting the result:



Plot[f[r], r, ϵ, R]


enter image description here



Something must be wrong in your model: The solution blows up heavily towards $r = 0$ so that x'[0] == 0 cannot be expected. Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2.






share|improve this answer
















  • 1




    "Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
    – J. M. is somewhat okay.♦
    3 hours ago











  • @Henrik Schumacher Look at the solution from the other side at x=0
    – Alex Trounev
    17 mins ago














up vote
3
down vote













As Alex Trounev said, this is a second-order ODE with discontinuous right-hand side. You can use Piecewise to set up the forcing term:



rhs = Piecewise[
 c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), a <= r < b
 ,
 c n0 Exp[-x[r]]
 ]



$$begincases frac3 ha^3-b^3+c ,textn0, e^-x(r) & aleq r<b \ c ,textn0 , e^-x(r) & textTrue endcases$$




Notice that for convenience, I used c n0 Exp[-x[r]] as the default term. The full equation can be set up as



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps = ParametricNDSolveValue[
 
 x''[r] + 2 x'[r] == rhs,
 x[R] == 0,
 x'[R] == 0
 ,
 x,
 r, ϵ, R,
 n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30
 ];


Solving it for a given parameter



f = ps[0.00001];


Plotting the result:



Plot[f[r], r, ϵ, R]


enter image description here



Something must be wrong in your model: The solution blows up heavily towards $r = 0$ so that x'[0] == 0 cannot be expected. Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2.






share|improve this answer
















  • 1




    "Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
    – J. M. is somewhat okay.♦
    3 hours ago











  • @Henrik Schumacher Look at the solution from the other side at x=0
    – Alex Trounev
    17 mins ago












up vote
3
down vote










up vote
3
down vote









As Alex Trounev said, this is a second-order ODE with discontinuous right-hand side. You can use Piecewise to set up the forcing term:



rhs = Piecewise[
 c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), a <= r < b
 ,
 c n0 Exp[-x[r]]
 ]



$$begincases frac3 ha^3-b^3+c ,textn0, e^-x(r) & aleq r<b \ c ,textn0 , e^-x(r) & textTrue endcases$$




Notice that for convenience, I used c n0 Exp[-x[r]] as the default term. The full equation can be set up as



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps = ParametricNDSolveValue[
 
 x''[r] + 2 x'[r] == rhs,
 x[R] == 0,
 x'[R] == 0
 ,
 x,
 r, ϵ, R,
 n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30
 ];


Solving it for a given parameter



f = ps[0.00001];


Plotting the result:



Plot[f[r], r, ϵ, R]


enter image description here



Something must be wrong in your model: The solution blows up heavily towards $r = 0$ so that x'[0] == 0 cannot be expected. Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2.






share|improve this answer












As Alex Trounev said, this is a second-order ODE with discontinuous right-hand side. You can use Piecewise to set up the forcing term:



rhs = Piecewise[
 c n0 Exp[-x[r]] + (3 h)/(a^3 - b^3), a <= r < b
 ,
 c n0 Exp[-x[r]]
 ]



$$begincases frac3 ha^3-b^3+c ,textn0, e^-x(r) & aleq r<b \ c ,textn0 , e^-x(r) & textTrue endcases$$




Notice that for convenience, I used c n0 Exp[-x[r]] as the default term. The full equation can be set up as



c = 0.72;
h = 300;
a = 15;
b = 17;
R = 25;
ϵ = $MachineEpsilon;

ps = ParametricNDSolveValue[
 
 x''[r] + 2 x'[r] == rhs,
 x[R] == 0,
 x'[R] == 0
 ,
 x,
 r, ϵ, R,
 n0,
 Method -> "StiffnessSwitching", WorkingPrecision -> 30
 ];


Solving it for a given parameter



f = ps[0.00001];


Plotting the result:



Plot[f[r], r, ϵ, R]


enter image description here



Something must be wrong in your model: The solution blows up heavily towards $r = 0$ so that x'[0] == 0 cannot be expected. Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2.







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answered 3 hours ago









Henrik Schumacher

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  • 1




    "Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
    – J. M. is somewhat okay.♦
    3 hours ago











  • @Henrik Schumacher Look at the solution from the other side at x=0
    – Alex Trounev
    17 mins ago












  • 1




    "Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
    – J. M. is somewhat okay.♦
    3 hours ago











  • @Henrik Schumacher Look at the solution from the other side at x=0
    – Alex Trounev
    17 mins ago







1




1




"Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
– J. M. is somewhat okay.♦
3 hours ago





"Actually, you cannot prescribe more than two boundary conditions for an ODE of order 2." - I'm actually surprised at how frequently people don't remember to check that the number of their conditions matches the order of their ODE...
– J. M. is somewhat okay.♦
3 hours ago













@Henrik Schumacher Look at the solution from the other side at x=0
– Alex Trounev
17 mins ago




@Henrik Schumacher Look at the solution from the other side at x=0
– Alex Trounev
17 mins ago










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