BVProblemLibrary.flat_earthConstant
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

BVProblemLibrary.flat_earth_dragConstant
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

BVProblemLibrary.flat_moonConstant
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

BVProblemLibrary.measlesConstant
measles

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

BVProblemLibrary.prob_bvp_linear_1Constant
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

BVProblemLibrary.prob_bvp_linear_10Constant
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

BVProblemLibrary.prob_bvp_linear_11Constant
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

BVProblemLibrary.prob_bvp_linear_12Constant
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

BVProblemLibrary.prob_bvp_linear_13Constant
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

BVProblemLibrary.prob_bvp_linear_14Constant
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

BVProblemLibrary.prob_bvp_linear_15Constant
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

BVProblemLibrary.prob_bvp_linear_16Constant
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

BVProblemLibrary.prob_bvp_linear_17Constant
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

BVProblemLibrary.prob_bvp_linear_18Constant
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

BVProblemLibrary.prob_bvp_linear_2Constant
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

BVProblemLibrary.prob_bvp_linear_3Constant
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

BVProblemLibrary.prob_bvp_linear_4Constant
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

BVProblemLibrary.prob_bvp_linear_5Constant
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

BVProblemLibrary.prob_bvp_linear_6Constant
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

BVProblemLibrary.prob_bvp_linear_7Constant
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

BVProblemLibrary.prob_bvp_linear_8Constant
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

BVProblemLibrary.prob_bvp_linear_9Constant
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

BVProblemLibrary.prob_bvp_nonlinear_1Constant
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

BVProblemLibrary.prob_bvp_nonlinear_10Constant
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

BVProblemLibrary.prob_bvp_nonlinear_11Constant
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

BVProblemLibrary.prob_bvp_nonlinear_12Constant
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

BVProblemLibrary.prob_bvp_nonlinear_13Constant
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

BVProblemLibrary.prob_bvp_nonlinear_14Constant
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

BVProblemLibrary.prob_bvp_nonlinear_15Constant
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

BVProblemLibrary.prob_bvp_nonlinear_2Constant
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

BVProblemLibrary.prob_bvp_nonlinear_3Constant
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

BVProblemLibrary.prob_bvp_nonlinear_4Constant
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

BVProblemLibrary.prob_bvp_nonlinear_5Constant
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

BVProblemLibrary.prob_bvp_nonlinear_6Constant
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

BVProblemLibrary.prob_bvp_nonlinear_7Constant
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

BVProblemLibrary.prob_bvp_nonlinear_8Constant
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

BVProblemLibrary.prob_bvp_nonlinear_9Constant
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