BVP Problems

Defines an BVP problem. Documentation Page: https://docs.sciml.ai/DiffEqDocs/stable/types/bvp_types/

Mathematical Specification of a BVP Problem

To define a BVP Problem, you simply need to give the function $f$ and the initial condition $u_0$ which define an ODE:

\[\frac{du}{dt} = f(u,p,t)\]

along with an implicit function bc which defines the residual equation, where

\[bc(u,p,t) = 0\]

is the manifold on which the solution must live. A common form for this is the two-point BVProblem where the manifold defines the solution at two points:

\[u(t_0) = a u(t_f) = b\]

Problem Type

Constructors

TwoPointBVProblem{isinplace}(f, bc, u0, tspan, p=NullParameters(); kwargs...)
BVProblem{isinplace}(f, bc, u0, tspan, p=NullParameters(); kwargs...)

or if we have an initial guess function initialGuess(p, t) for the given BVP, we can pass the initial guess to the problem constructors:

TwoPointBVProblem{isinplace}(f, bc, initialGuess, tspan, p=NullParameters(); kwargs...)
BVProblem{isinplace}(f, bc, initialGuess, tspan, p=NullParameters(); kwargs...)

For any BVP problem type, bc must be inplace if f is inplace. Otherwise it must be out-of-place.

If the bvp is a StandardBVProblem (also known as a Multi-Point BV Problem) it must define either of the following functions

bc!(residual, u, p, t)
residual = bc(u, p, t)

where residual computed from the current u. u is an array of solution values where u[i] is at time t[i], while p are the parameters. For a TwoPointBVProblem, t = tspan. For the more general BVProblem, u can be all of the internal time points, and for shooting type methods u=sol the ODE solution. Note that all features of the ODESolution are present in this form. In both cases, the size of the residual matches the size of the initial condition.

If the bvp is a TwoPointBVProblem then bc must be a Tuple (bca, bcb) and each of them must define either of the following functions:

begin
    bca!(resid_a, u_a, p)
    bcb!(resid_b, u_b, p)
end
begin
    resid_a = bca(u_a, p)
    resid_b = bcb(u_b, p)
end

where resid_a and resid_b are the residuals at the two endpoints, u_a and u_b are the solution values at the two endpoints, and p are the parameters.

Parameters are optional, and if not given, then a NullParameters() singleton will be used which will throw nice errors if you try to index non-existent parameters. Any extra keyword arguments are passed on to the solvers. For example, if you set a callback in the problem, then that callback will be added in every solve call.

Fields

• f: The function for the ODE.
• bc: The boundary condition function.
• u0: The initial condition. Either the initial condition for the ODE as an initial value problem, or a Vector of values for $u(t_i)$ for collocation methods.
• tspan: The timespan for the problem.
• p: The parameters for the problem. Defaults to NullParameters
• kwargs: The keyword arguments passed onto the solves.

Special Keyword Arguments

• nlls: Specify that the BVP is a nonlinear least squares problem. Use Val(true) or Val(false) for type stability. By default this is automatically inferred based on the size of the input and outputs, however this is type unstable for any array type that doesn't store array size as part of type information. If we can't reliably infer this, we set it to Nothing. Downstreams solvers must be setup to deal with this case.

Defines a second order BVP problem. Documentation Page: https://docs.sciml.ai/DiffEqDocs/stable/types/bvp_types/

Mathematical Specification of a second order BVP Problem

To define a second order BVP Problem, you simply need to give the function $f$ and the initial condition $u_0$ which define an ODE:

\[\frac{ddu}{dt} = f(du,u,p,t)\]

along with an implicit function bc which defines the residual equation, where

\[bc(du,u,p,t) = 0\]

is the manifold on which the solution must live. A common form for this is the two-point SecondOrderBVProblem where the manifold defines the solution at two points:

\[g(u(t_0),u'(t_0)) = 0 g(u(t_f),u'(t_f)) = 0\]

Problem Type

Constructors

TwoPointSecondOrderBVProblem{isinplace}(f, bc, u0, tspan, p=NullParameters(); kwargs...)
SecondOrderBVProblem{isinplace}(f, bc, u0, tspan, p=NullParameters(); kwargs...)

or if we have an initial guess function initialGuess(p, t) for the given BVP, we can pass the initial guess to the problem constructors:

TwoPointSecondOrderBVProblem{isinplace}(f, bc, initialGuess, tspan, p=NullParameters(); kwargs...)
SecondOrderBVProblem{isinplace}(f, bc, initialGuess, tspan, p=NullParameters(); kwargs...)

For any BVP problem type, bc must be inplace if f is inplace. Otherwise it must be out-of-place.

If the bvp is a StandardSecondOrderBVProblem (also known as a Multi-Point BV Problem) it must define either of the following functions

bc!(residual, du, u, p, t)
residual = bc(du, u, p, t)

where residual computed from the current u. u is an array of solution values where u[i] is at time t[i], while p are the parameters. For a TwoPointBVProblem, t = tspan. For the more general BVProblem, u can be all of the internal time points, and for shooting type methods u=sol the ODE solution. Note that all features of the ODESolution are present in this form. In both cases, the size of the residual matches the size of the initial condition.

If the bvp is a TwoPointSecondOrderBVProblem then bc must be a Tuple (bca, bcb) and each of them must define either of the following functions:

begin
    bca!(resid_a, du_a, u_a, p)
    bcb!(resid_b, du_b, u_b, p)
end
begin
    resid_a = bca(du_a, u_a, p)
    resid_b = bcb(du_b, u_b, p)
end

where resid_a and resid_b are the residuals at the two endpoints, u_a and u_b are the solution values at the two endpoints, du_a and du_b are the derivative of solution values at the two endpoints, and p are the parameters.

Parameters are optional, and if not given, then a NullParameters() singleton will be used which will throw nice errors if you try to index non-existent parameters. Any extra keyword arguments are passed on to the solvers. For example, if you set a callback in the problem, then that callback will be added in every solve call.

Fields

• f: The function for the ODE.
• bc: The boundary condition function.
• u0: The initial condition. Either the initial condition for the ODE as an initial value problem, or a Vector of values for $u(t_i)$ for collocation methods.
• tspan: The timespan for the problem.
• p: The parameters for the problem. Defaults to NullParameters
• kwargs: The keyword arguments passed onto the solves.

Holds information on what variables to alias when solving an BVP. Conforms to the AbstractAliasSpecifier interface. BVPAliasSpecifier(;alias_p = nothing, alias_f = nothing, alias_u0 = nothing, alias_du0 = nothing, alias_tstops = nothing, alias = nothing)

When a keyword argument is nothing, the default behaviour of the solver is used.

Keywords

• alias_p::Union{Bool, Nothing}
• alias_f::Union{Bool, Nothing}
• alias_u0::Union{Bool, Nothing}: alias the u0 array. Defaults to false .
• alias_du0::Union{Bool, Nothing}: alias the du0 array for DAEs. Defaults to false.
• alias_tstops::Union{Bool, Nothing}: alias the tstops array
• alias::Union{Bool, Nothing}: sets all fields of the BVPAliasSpecifier to alias

prob_bvp_linear_1

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}y_1\]

with boundary condition

\[y_1(0)=1, y_1(1)=0\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = \frac{\exp(-t/\sqrt{\lambda}) - \exp((t-2)/\sqrt{\lambda})}{1-\exp(-2/\sqrt{\lambda})}\]

\[y_2(t)=y_1'(t)\]

References

Reference

prob_bvp_linear_2

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}y_2\]

with boundary condition

\[y_1(0)=1, y_1(1)=0\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = \frac{1-\exp((t-1)/\lambda)}{1-\exp(-1/\lambda)}\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_3

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1, y_2)\]

where

\[f(t, y_1, y_2) = -(2+\cos(\pi t))y_2 + y_1 -(1+\lambda \pi^2)\cos(\pi t) - (2+\cos(\pi t))\pi\sin(\pi t)\]

with boundary condition

\[y_1(-1)=-1, y_1(1)=-1\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_4

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y2+(1+\lambda)y1\]

with boundary condition

\[y_1(-1)=1+\exp(-2), y_1(1)=1+\exp(-2(1+\lambda))\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \exp(t-1)+\exp(-(1+\lambda)(1+t)/\lambda)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_5

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1, y_2)\]

where

\[f(t, y_1, y_2)=ty_2+y_1-(1+\lambda\pi^2)\cos(\pi t)+\pi t\sin(\pi t)\]

with boundary condition

\[y_1(-1)=-1, y_1(1)=-1\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_6

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_2)\]

where

\[f(t, y_2)=ty_2 - \lambda\pi^2\cos(\pi t)-\pi t\sin(\pi t)\]

with boundary condition

\[y_1(-1)=-2, y_1(1)=0\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t) + \erf(t/\sqrt{2\lambda})/\erf(1/\sqrt{2\lambda})\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_7

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1, y_2)\]

where

\[f(t, y_1, y_2)=ty_2+y_1-(1+\lambda\pi^2)\cos(\pi t)+\pi t\sin(\pi t)\]

with boundary condition

\[y_1(-1)=-1, y_1(1)=1\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t) + t + \frac{t\erf(t/\sqrt{2\lambda}) + \sqrt{2\lambda/\pi}\exp(-t^2/2\lambda)}{}\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_8

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}y_2\]

with boundary condition

\[y_1(0)=1, y_1(1)=2\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = (2-\exp(-1/\lambda)-\exp(-t/\lambda))/(1-\exp(-1/\lambda))\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_9

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda+t^2}f(t, y_1, y_2)\]

where

\[f(t, y_1, y_2)=-4ty_2 - 2y_1\]

with boundary condition

\[y_1(-1)=1/(1+\lambda), y_1(1)=1/(1+\lambda)\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = 1/(\lambda+t^2)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_10

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_2)\]

where

\[f(t, y_2)=-ty_2\]

with boundary condition

\[y_1(-1)=0, y_1(1)=2\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = 1+\erf(t/\sqrt{2\lambda})/\erf(1/\sqrt{2\lambda})\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_11

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1)\]

where

\[f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)\]

with boundary condition

\[y_1(-1)=0, y_1(1)=2\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_12

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1)\]

where

\[f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)\]

with boundary condition

\[y_1(-1)=-1, y_1(1)=0\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t)+\frac{\exp((t+1)/\sqrt{\lambda})-\exp((-t-1))/\sqrt{\lambda}}{\exp(2/\sqrt{\lambda})-\exp(-2/\sqrt{\lambda})}\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_13

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1)\]

where

\[f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)\]

with boundary condition

\[y_1(-1)=0, y_1(1)=-1\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t)+\exp(-(t+1)/\sqrt{\lambda})\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_14

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1)\]

where

\[f(t, y_1)=y_1-(1+\lambda\pi^2)\cos(\pi t)\]

with boundary condition

\[y_1(-1)=\exp(-2/\sqrt{\lambda}, y_1(1)=\exp(-2/\sqrt{\lambda})\]

Solution

The analytical solution for $t \in [-1, 1]$ is

\[y_1(t) = \cos(\pi t)+\exp((t-1)/\sqrt{\lambda})+\exp(-(t+1)/\sqrt{\lambda})\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_15

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(t, y_1)\]

where

\[f(t, y_1)=ty_1\]

with boundary condition

\[y_1(-1)=1, y_1(1)=1\]

Solution

No analytical solution

References

Reference

prob_bvp_linear_16

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda^2}f(y_1)\]

where

\[f(t, y_1)=-π^2y_1/4\]

with boundary condition

\[y_1(0)=0, y_1(1)=\sin(\pi/(2*\lambda))\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = \sin(\pi t/2\lambda) when 1/\lambda is odd\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_17

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = f(y_1)\]

where

\[f(t, y_1)=-3\lambda y_1/(\lambda+t^2)^2\]

with boundary condition

\[y_1(-0.1)=-0.1/\sqrt{\lambda+0.01}, y_1(0.1)=0.1/\sqrt{\lambda+0.01}\]

Solution

The analytical solution for $t \in [-0.1, 0.1]$ is

\[y_1(t) = t/\sqrt{\lambda+t^2}\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_linear_18

Linear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}y_2\]

where

\[f(y_2)=-y_1\]

with boundary condition

\[y_1(0)=1, y_1(1)=0.1/\sqrt{\lambda+0.01}\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = \exp(-t/\lambda)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_nonlinear_1

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}y_2\]

where

\[f(y_2)=-y_1\]

with boundary condition

\[y_1(0)=1, y_1(1)=0.1/\sqrt{\lambda+0.01}\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = \exp(-t/\lambda)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_nonlinear_2

Nonlinear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}f(y_2)\]

where

\[f(y_2)=--y_2^2+1\]

with boundary condition

\[y_1(0)=1+\lambda\ln\cosh(-0.745/\lambda), y_1(1)=1+\lambda\ln\cosh(0.255/\lambda)\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = 1+\lambda\ln\cosh((t-0.745)/\lambda)\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_nonlinear_3

Nonlinear boundary value problem with analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}f(y, y_1)\]

where

\[f(y_1)=y_1+y_1^2-\exp(-2t/\sqrt{\lambda})\]

with boundary condition

\[y_1(0)=1, y_1(1)=\exp(-1/\sqrt{\lambda})\]

Solution

The analytical solution for $t \in [0, 1]$ is

\[y_1(t) = \exp(-t/\sqrt{\lambda})\]

\[y_2(t) = y_1'(t)\]

References

Reference

prob_bvp_nonlinear_4

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_2-y_1^2\]

with boundary condition

\[y_1(0)=0, y_1(1)=1/2\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_5

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}f(y_1)\]

where

\[f(y_1)=\lambda\sinh(\lambda z)\]

with boundary condition

\[y_1(0)=0, y_1(1)=1\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_6

This problem describes a shock wave in a one dimension nozzle flow.

The steady state Navier-Stokes equations generate a second order differential equations which can be reduced to a first order system described by nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = -\frac{1}{\lambda}f(y_1)\]

where

\[f(t, y_1, y_2)=(\frac{1+\gamma}{2}-\lambda A'(t))y_1y_2-\frac{y_2}{y_1}-\frac{A'(t)}{A(t)}(1-(\frac{\gamma-1}{2})y_1^2)\]

with boundary condition

\[y_1(0)=0.9129, y_1(1)=0.375\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_7

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_1y_2+y_1\]

with boundary condition

\[y_1(0)=-1/3, y_1(1)=1/3\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_8

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_1y_2+y_1\]

with boundary condition

\[y_1(0)=1, y_1(1)=-1/3\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_9

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_1y_2+y_1\]

with boundary condition

\[y_1(0)=1, y_1(1)=1/3\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_10

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_1y_2+y_1\]

with boundary condition

\[y_1(0)=1, y_1(1)=3/2\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_11

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_1y_2+y_1\]

with boundary condition

\[y_1(0)=0, y_1(1)=3/2\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_12

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}f(y_1, y_2)\]

where

\[f(y_1, y_2)=-y_1y_2+y_1\]

with boundary condition

\[y_1(0)=-7/6, y_1(1)=3/2\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_13

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = \sin(y_2)\]

\[\frac{dy_2}{dt} = y_3\]

\[\frac{dy_3}{dt} = -y_4/\lambda\]

\[\frac{dy_4}{dt} = f(y_1, y_2, y_3, y_4)\]

where

\[f(z, \theta, M, Q)=\frac{1}{\lambda}((z-1)\cos\theta-M\sec\theta)+\lambda Q\tan\theta\]

with boundary condition

\[y_1(0)=0, y_3(0)=0, y_1(1)=0, y_3(1)=0\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_14

This problem arises from fluid injection through one side of a long vertical channel

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = y_3\]

\[\frac{dy_3}{dt} = y_4\]

\[\frac{dy_4}{dt} = f(y_1, y_2, y_3, y_4)\]

where

\[f(y_1, y_2, y_3, y_4)=\lambda(y_2y_3-y_1y_4)\]

with boundary condition

\[y_1(0)=0, y_2(0)=0, y_1(1)=1, y_2(1)=0\]

Solution

No analytical solution

References

Reference

prob_bvp_nonlinear_15

This problem arises from fluid injection through one side of a long vertical channel

Nonlinear boundary value problem with no analytical solution, given by

\[\frac{dy_1}{dt} = y_2\]

\[\frac{dy_2}{dt} = \frac{1}{\lambda}y_1y_4-y_3y_2\]

\[\frac{dy_3}{dt} = y_4\]

\[\frac{dy_4}{dt} = y_5\]

\[\frac{dy_5}{dt} = y_6\]

\[\frac{dy_6}{dt} = \frac{1}{\lambda}(-y_3y_6-y_1y_2)\]

with boundary condition

\[y_1(0)=-1, y_3(0)=0, y_4(0)=0, y_1(1)=1, y_3(1)=0, y_4(1)=0\]

Solution

No analytical solution

References

Reference

flat_moon

This test problem is about the optimal-time launching of a satellite into orbit from flat moon without atmospheric drag.

Given by

\[\frac{dz_1}{dt}=z_3t_f\]

\[\frac{dz_2}{dt}=z_4t_f\]

\[\frac{dz_3}{dt}=A\cos(z_5)t_f\]

\[\frac{dz_4}{dt}=(A\sin(z_5)-g)t_f\]

\[\frac{dz_5}{dt}=-z_6\cos(z_5)t_F\]

\[\frac{dz_6}{dt}=z_6^2\sin(z_5)t_f\]

\[\frac{dz_7}{dt}=0\]

with boundary condition

\[z_1(0)=0\]

\[z_2(0)=0\]

\[z_3(0)=0\]

\[z_4(0)=0\]

\[z_5(1)=h\]

\[z_6(1)=V_c\]

\[z_7(1)=0\]

Solution

No analytical solution

References

Reference

flat_earth

Launch of a satellite into circular orbit from a flat Earth where we assume a uniform gravitational field $g$.

Given by

\[\frac{dz_1}{dt}=z_3\frac{V_c}{h}\]

\[\frac{dz_2}{dt}=z_4\frac{V_c}{h}\]

\[\frac{dz_3}{dt}=acc\frac{1}{|V_c|\sqrt{1+z_6^2}}\]

\[\frac{dz_4}{dt}=acc\frac{1}{|V_c|\sqrt{1+z_6^2}}-frac{g}{V_c}\]

\[\frac{dz_5}{dt}=0\]

\[\frac{dz_6}{dt}=-z_5\frac{V_c}{h}\]

\[\frac{dz_7}{dt}=0\]

with boundary condition

\[z_1(0)=0\]

\[z_2(0)=0\]

\[z_3(0)=0\]

\[z_4(0)=0\]

\[z_5(1)=h\]

\[z_6(1)=V_c\]

\[z_7(1)=0\]

Solution

No analytical solution

References

Reference

flat_earth_drag

Launch into circular orbit from a flat Earth including athmosferic drag.

Given by

\[\frac{dz_1}{dt}=z_3\frac{V_c}{h}\]

\[\frac{dz_2}{dt}=z_4\frac{V_c}{h}\]

\[\frac{dz_3}{dt}=\frac{f}{V_c}(-\frac{z_6}{z_6^2+z_7^2}-V_c\eta\exp(-z_2\beta)z_3\sqrt{z_3^3+z_4^2})/m\]

\[\frac{dz_4}{dt}=\frac{f}{V_c}(-\frac{z_7}{z_6^2+z_7^2}-V_c\eta\exp(-z_2\beta)z_4\sqrt{z_3^3+z_4^2})/m - g_{accel}/V_c\]

\[\frac{dz_5}{dt}=-\eta\beta\exp(-z_2\beta)(z_6z_3+z_7z_4)\sqrt{z_3^3+z_4^2}\frac{V_c}{m}\]

\[\frac{dz_6}{dt}=\eta\exp(-z_2\beta)(z_6(2z_3^2+z_4^2)+z_7z_3z_4)V_c/\sqrt{z_3^2+z_4^2}/m\]

\[\frac{dz_7}{dt}=\eta\exp(-z_2\beta)(z_7(z_3^2+2z_4^2)+z_6z_3z_4)V_c/\sqrt{z_3^2+z_4^2}/m\]

with boundary condition

\[z_1(0)=0\]

\[z_2(0)=0\]

\[z_3(0)=0\]

\[z_4(0)=0\]

\[z_5(1)=h\]

\[z_6(1)=V_c\]

\[z_7(1)=0\]

Solution

No analytical solution

References

Reference

measles

This is an epidemiology model, about the spread of diseases.

Given by

\[\frac{dy_1}{dt}=\mu-\beta(t)y_1y_3\]

\[\frac{dy_2}{dt}=\beta(t)y_1y_3-y_2/\lambda\]

\[\frac{dy_3}{dt}=y_2/\lambda-y_3/\eta\]

with boundary condition

\[y(0)=y(1)\]

Solution

No analytical solution

References

Reference