ode

Ordinary differential equations

Since R2023b

Description

An ode object defines a system of ordinary differential equations or differential algebraic equations to solve.

You can solve initial value problems of the form $y\text{'}=f\left(t,y\right)$ or problems that involve a mass matrix, $M\left(t,y\right)y\text{'}=f\left(t,y\right)$.

Define aspects of the problem using properties of the ode object, such as ODEFcn, InitialTime, and InitialValue. You can select a specific solver to use, or let MATLAB® choose an appropriate solver based on properties of the equations. After you create an ode object, you can solve the equations using the solve or solutionFcn object functions.

Creation

Description

example

F = ode creates an ode object with default properties.

example

F = ode(Name=Value) specifies one or more property values using name-value arguments. For example, you can specify the equations to be solved, the initial time for integration, and the value of the solution at the initial time using the ODEFcn, InitialTime, and InitialValue properties.

Properties

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Problem Definition

Equations to solve, specified as a function handle that defines the system of differential equations to be integrated. The function handle can be an anonymous function or a handle to a named function file.

The function dydt = ODEFcn(t,y), for a scalar t and a column vector y, must return a column vector dydt that corresponds to $f\left(t,y\right)$. ODEFcn must accept at least two input arguments, t and y, even if one of the arguments is not used in the function. To supply parameter values, define ODEFcn as a function that accepts three inputs, dydt = ODEFcn(t,y,p), and then use the Parameters property to store parameter values (such as a struct or cell array).

For example, to solve a single equation such as $y\text{'}=5y-3$, you can use the anonymous function @(t,y) 5*y-3.

For a system of equations, the output of ODEFcn is a vector. Each element in the vector is the computed value of the right side of one equation. For example, consider this system of two equations.

$\begin{array}{l}y{\text{'}}_{1}={y}_{1}+2{y}_{2}\\ y{\text{'}}_{2}=3{y}_{1}+2{y}_{2}\end{array}$

An anonymous function that defines these equations is @(t,y) [y(1)+2*y(2); 3*y(1)+2*y(2)].

Example: F.ODEFcn = @myFcn specifies a handle to a named function file myFcn.m containing the equations.

Example: F.ODEFcn = @(t,y) [y(2); -y(1)] specifies an anonymous function that defines a system of two equations.

Example: F.ODEFcn = @(t,y,p) 5*y*p(1)-3*p(2) specifies an anonymous function for a single equation that uses two parameters.

Data Types: function_handle

Initial time for integration, specified as a real scalar. The value of InitialTime is the beginning of the integration interval where the initial conditions specified in InitialValue are applied by the solver before beginning integration steps.

Example: F.InitialTime = 10;

Data Types: single | double

Value of solution at InitialTime, specified as a scalar or vector. InitialValue must be a vector with the same length as the output of ODEFcn so that an initial value is specified for each equation defined in ODEFcn. The value of InitialTime is the beginning of the integration interval where the initial conditions specified in InitialValue are applied by the solver before beginning integration steps.

Example: F.InitialValue = 1;

Example: F.InitialValue = [1 2 3];

Data Types: single | double

Equation parameters, specified as an array of any size or data type. The values you store in Parameters can be supplied to any of the functions used for the ODEFcn, Jacobian, MassMatrix, or EventDefinition properties by specifying an extra input argument in the function.

For instance, you can supply parameter values stored in the Parameters property to ODEFcn by specifying ODEFcn as a function handle that accepts three inputs, dydt = ODEFcn(t,y,p). For example, F.ODEFcn = @(t,y,p) 5*y*p(1)-3*p(2) specifies an anonymous function for a single equation that uses two parameters. If you then specify F.Parameters = [2 3], then ODEFcn uses the parameter values p(1) = 2 and p(2) = 3 whenever the solver calls the function.

Example: F.Parameters = [0.1 0.5] specifies two parameter values.

Jacobian matrix, specified as an odeJacobian object, matrix, or handle to a function that evaluates the Jacobian. The Jacobian is a matrix of partial derivatives of the functions that define the system of differential equations.

$J=\frac{\partial f}{\partial y}=\left[\begin{array}{ccc}\frac{\partial {f}_{1}}{\partial {y}_{1}}& \frac{\partial {f}_{1}}{\partial {y}_{2}}& \cdots \\ \frac{\partial {f}_{2}}{\partial {y}_{1}}& \frac{\partial {f}_{2}}{\partial {y}_{2}}& \cdots \\ ⋮& ⋮& \end{array}\right]$

For stiff ODE problems, providing information about the Jacobian matrix is critical for reliability and efficiency of the solver. If you do not provide the Jacobian, then the ODE solver approximates it numerically using finite differences.

For large systems of equations where it is not feasible to provide the entire analytic Jacobian, you can specify the sparsity pattern of the Jacobian matrix instead. The solver uses the sparsity pattern to calculate a sparse Jacobian.

You can specify the value of the Jacobian property as:

• An odeJacobian object, which can represent either the Jacobian matrix or its sparsity pattern.

• A constant matrix with calculated values for $\frac{\partial f}{\partial y}$.

• A handle to a function that computes the matrix elements and that accepts two input arguments, dfdy = Fjac(t,y). To give the function access to parameter values in the Parameters property, specify a third input argument in the function definition, dfdy = Fjac(t,y,p).

If you specify a matrix or function handle, then MATLAB converts it to an odeJacobian object.

Example: F.Jacobian = @Fjac specifies the function Fjac that evaluates the Jacobian matrix.

Example: F.Jacobian = [0 1; -2 1] specifies a constant Jacobian matrix.

Example: F.Jacobian = odeJacobian(SparsityPattern=S) specifies the Jacobian sparsity pattern using sparse matrix S.

Events to detect, specified as an odeEvent object. Create an odeEvent object to define expressions that trigger an event when they cross zero. You can specify the direction of the zero crossing and what to do when an event triggers, including the use of a callback function.

Mass matrix, specified as an odeMassMatrix object, matrix, or handle to a function that evaluates the mass matrix.

ode objects can represent problems of the form $M\left(t,y\right)\text{\hspace{0.17em}}y\text{'}=f\left(t,y\right)$, where $M\left(t,y\right)$ is a mass matrix that can be full or sparse. The mass matrix encodes linear combinations of derivatives on the left side of the equation.

• When the mass matrix is nonsingular, the equation simplifies to $y\text{'}={M}^{-1}\text{\hspace{0.17em}}f\left(t,y\right)$ and the ODE has a solution for any initial value. However, it is often more convenient and natural to express the model in terms of the mass matrix directly using $M\left(t,y\right)\text{\hspace{0.17em}}y\text{'}=f\left(t,y\right)$, and avoiding the computation of the matrix inverse reduces the storage and execution time needed to solve the problem.

• When the mass matrix is singular, then the problem is a system of differential algebraic equations (DAEs). A DAE has a solution only when the initial values are consistent; that is, you must specify the initial slope $y{\text{'}}_{0}$ using the InitialSlope property such that the initial conditions are all consistent, $M\left({t}_{0},{y}_{0}\right)y{\text{'}}_{0}=f\left({t}_{0},{y}_{0}\right)$. If the specified initial conditions are not consistent with the InitialTime, InitialValue, and MassMatrix properties, then the solver treats them as guesses and attempts to compute consistent values for the initial slopes that are close to the guesses before continuing to solve the problem. For more information, see Solve Differential Algebraic Equations (DAEs).

You can specify the value of the MassMatrix property as:

• An odeMassMatrix object, which represents the mass matrix and its associated properties. You can specify whether the mass matrix is singular or has state dependence.

• A constant matrix with calculated values for $M\left(t,y\right)$.

• A handle to a function that computes the matrix elements and that accepts two input arguments, M = Mass(t,y). To give the function access to parameter values stored in the Parameters property, specify a third input argument in the function definition, M = Mass(t,y,p).

If you specify a matrix or function handle, then MATLAB converts it to an odeMassMatrix object.

Example: F.MassMatrix = @Mass specifies the function Mass that evaluates the mass matrix.

Example: F.MassMatrix = [1 0; -2 1] specifies a constant mass matrix.

Example: F.MassMatrix = odeMassMatrix(MassMatrix=@Mass,StateDependence="strong",SparsityPattern=S) specifies a state-dependent mass matrix and the sparsity pattern when the mass matrix multiplies a vector.

Nonnegative solution components, specified as a scalar index or vector of indices. Use the NonNegativeVariables property to specify which solutions the solver must keep nonnegative. If dydt = ODEFcn(t,y), then the indices correspond to elements in the vector y. For example, if you specify a value of 3, then the solver keeps the solution component y(3) nonnegative.

Note

NonNegativeVariables is not available for the ode23s solver. Additionally, for ode15s, ode23t, and ode23tb the property is not available for problems that have a mass matrix.

Example: F.NonNegativeVariables = [1 3] specifies that the first and third solution components must be kept nonnegative.

Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64

Initial value of dy/dt, specified as a vector. Use this property with the ode15s and ode23t solvers when solving DAEs. The specified vector is the initial slope $y{\text{'}}_{0}$ such that $M\left({t}_{0},{y}_{0}\right)y{\text{'}}_{0}=f\left({t}_{0},{y}_{0}\right)$. If the specified initial conditions are not consistent with the InitialTime, InitialValue, and MassMatrix properties, then the solver treats them as guesses and attempts to compute consistent values for the initial slopes that are close to the guesses before continuing to solve the problem. For more information, see Solve Differential Algebraic Equations (DAEs).

Data Types: single | double

Sensitivity analysis information, specified as an odeSensitivity object. The ODE solver performs sensitivity analysis only when the Sensitivity property is nonempty. The analysis is with respect to the Parameters property, which must be nonempty and numeric. See odeSensitivity for more information.

Example: F.Sensitivity = odeSensitivity performs sensitivity analysis on all parameters.

Example: F.Sensitivity = odeSensitivity(ParameterIndices=1:2) performs sensitivity analysis on the first two parameters.

Solver Control

Solver method, specified as one of the values in this table.

ValueDescription

"auto"

Automatically selects a solver based on available problem information and tolerances. The SelectedSolver property shows the currently chosen solver, and the solver choice can change when you update property values of the ode object.

"stiff"

Similar to "auto" but chooses only from the stiff solvers: "ode15s", "ode23s", "ode23t", "ode23tb", "cvodesStiff", "idas".

"nonstiff"

Similar to "auto" but chooses only from the nonstiff solvers: "ode23", "ode45", "ode78", "ode89", "ode113", "cvodesNonstiff".

"ode23"

Runge-Kutta (2,3) pair. See ode23.

"ode45"

Runge-Kutta (4,5) pair. See ode45.

"ode78"

Runge-Kutta 8(7) pair with a 7th-order continuous extension. See ode78.

"ode89"

Runge-Kutta 9(8) pair with an 8th-order continuous extension. See ode89.

"ode113"

Variable-step, variable-order (VSVO) solver of orders 1 to 13. See ode113.

"ode15s"

Variable-step, variable-order (VSVO) solver based on the numerical differentiation formulas (NDFs) of orders 1 to 5. See ode15s.

"ode23s"

Modified Rosenbrock formula of order 2. See ode23s.

"ode23t"

Trapezoidal rule using a “free” interpolant. See ode23t.

"ode23tb"

Implicit Runge-Kutta formula with a trapezoidal rule step as its first stage and a backward differentiation formula of order 2 as its second stage. See ode23tb.

"cvodesNonstiff"

Variable-step, variable-order (VSVO) solver using Adams-Moulton formulas, with the order varying between 1 and 12. Supports sensitivity analysis. See CVODE and CVODES.

"cvodesStiff"

Variable-step, variable-order (VSVO) solver using Backward Differentiation Formulas (BDFs) in fixed-leading coefficient form, with order varying between 1 and 5. Supports sensitivity analysis. See CVODE and CVODES.

"idas"

Variable-order, variable-coefficient solver using Backward Differentiation Formulas (BDFs) in fixed-leading coefficient form, with order varying between 1 and 5. Supports sensitivity analysis. See IDA and IDAS.

Selected solver, returned as the name of a solver. If the value of Solver is "auto", "stiff", or "nonstiff", then the ode object automatically chooses a solver and sets the value of SelectedSolver to the name of the chosen solver. If you manually select a solver, then SelectedSolver and Solver have the same value.

Solver-specific options, specified as a matlab.ode.options.* object. The ode object automatically populates the SolverOptions property with an options object appropriate for the selected solver. You can check available options for an existing ode object with the command properties(F.SolverOptions). Specify options for the ODE problem by changing property values of the matlab.ode.options.* object. For example, F.SolverOptions.OutputFcn = @odeplot specifies an output function that the solver calls after each successful time step.

When the Solver property is "auto", "stiff", or "nonstiff", the SolverOptions property is read-only.

This table summarizes the available options for each solver.

Absolute error tolerance, specified as a positive scalar or vector. This tolerance is a threshold below which the value of the solution becomes unimportant. If the solution |y| is less than AbsoluteTolerance, then the solver does not need to obtain any correct digits in |y|. For this reason, keep in mind the scale of the solution components when setting the value of AbsoluteTolerance.

If AbsoluteTolerance is a vector, then it must be the same length as the number of solution components. If AbsoluteTolerance is a scalar, then the value applies to all solution components.

At each step, the ODE solver estimates the local error e in the ith component of the solution. To be successful, the step must have an acceptable error, as determined by both the relative and absolute error tolerances:

|e(i)| <= max(RelativeTolerance*abs(y(i)),AbsoluteTolerance(i))

Data Types: single | double

Relative error tolerance, specified as a positive scalar. This tolerance measures the error relative to the magnitude of each solution component. The relative error tolerance controls the number of correct digits in all solution components, except those smaller than AbsoluteTolerance.

At each step, the ODE solver estimates the local error e in the ith component of the solution. To be successful, the step must have an acceptable error, as determined by both the relative and absolute error tolerances:

|e(i)| <= max(RelativeTolerance*abs(y(i)),AbsoluteTolerance(i))

Data Types: single | double

Object Functions

 solve Solve ODE over interval or at specified points solutionFcn Construct function that interpolates ODE solution

Examples

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Create an empty ode object, and then specify values for the ODEFcn and InitialValue properties.

F = ode;
F.ODEFcn = @(t,y) 2*t;
F.InitialValue = 0;

Check which solver is selected for the problem, and then solve the equation over the time range [0 10].

F.SelectedSolver
ans =
SolverID enumeration

ode45

sol = solve(F,0,10)
sol =
ODEResults with properties:

Time: [0 0.2500 0.5000 0.7500 1 1.2500 1.5000 1.7500 2 2.2500 2.5000 2.7500 3 3.2500 3.5000 3.7500 4 4.2500 4.5000 4.7500 5 5.2500 5.5000 5.7500 6 6.2500 6.5000 6.7500 7 7.2500 7.5000 7.7500 8 8.2500 8.5000 8.7500 9 9.2500 9.5000 9.7500 10]
Solution: [0 0.0625 0.2500 0.5625 1.0000 1.5625 2.2500 3.0625 4 5.0625 6.2500 7.5625 9 10.5625 12.2500 14.0625 16 18.0625 20.2500 22.5625 25 27.5625 30.2500 33.0625 36 39.0625 42.2500 45.5625 49 52.5625 56.2500 60.0625 64 68.0625 ... ] (1x41 double)

Plot the results.

plot(sol.Time,sol.Solution,"-o")

The Van der Pol oscillator equation is a second-order differential equation. The equation includes a parameter $\mu$, and the equation becomes stiff when the value of $\mu$ is large.

$\frac{{d}^{2}x}{d{t}^{2}}-\mu \left(1-{x}^{2}\right)\frac{dx}{dt}+x=0$

Using the substitutions ${\mathit{y}}_{1}=\mathit{x}$ and ${\mathit{y}}_{2}=\frac{\mathrm{dx}}{\mathrm{dt}}$ produces a system of two first-order equations.

$\begin{array}{l}\frac{d{y}_{1}}{dt}={y}_{2}\\ \frac{d{y}_{2}}{dt}=\mu \left(1-{y}_{1}^{2}\right){y}_{2}-{y}_{1}\end{array}$

The Jacobian matrix for these equations is the matrix of partial derivatives of each equation with respect to both ${\mathit{y}}_{1}$ and ${\mathit{y}}_{2}$.

$J=\left[\begin{array}{cc}\frac{\partial {f}_{1}}{\partial {y}_{1}}& \frac{\partial {f}_{1}}{\partial {y}_{2}}\\ \frac{\partial {f}_{2}}{\partial {y}_{1}}& \frac{\partial {f}_{2}}{\partial {y}_{2}}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -2\mu {y}_{1}{y}_{2}-1& \mu \left(1-{y}_{1}^{2}\right)\end{array}\right]$

Solve the Van der Pol oscillator using $\mu =1000$ and initial values of [2; 0] by creating an ode object to represent the problem.

• Store the value of $\mu$ in the Parameters property.

• Specify the initial values in the InitialValue property.

• Specify the system of equations in the ODEFcn property, specifying three input arguments so that the value for $\mu$ is passed to the function.

• Specify a function that calculates the Jacobian matrix in the Jacobian property, specifying three input arguments so that the value for $\mu$ is passed to the function.

F = ode;
F.Parameters = 1000;
F.InitialValue = [2; 0];
F.ODEFcn = @(t,y,p) [y(2); p(1)*(1-y(1)^2)*y(2)-y(1)];
F.Jacobian = @(t,y,p) [0 1; -2*p(1)*y(1)*y(2)-1  p(1)*(1-y(1)^2)];

Display the ode object. The SelectedSolver property shows that the ode15s solver was automatically chosen for this problem.

F
F =
ode with properties:

Problem definition
ODEFcn: @(t,y,p)[y(2);p(1)*(1-y(1)^2)*y(2)-y(1)]
Parameters: 1000
InitialTime: 0
InitialValue: [2x1 double]
Jacobian: [1x1 odeJacobian]

Solver properties
AbsoluteTolerance: 1.0000e-06
RelativeTolerance: 1.0000e-03
Solver: auto
SelectedSolver: ode15s

Show all properties

Solve the system of equations over the time interval [0 3000] by using the solve method. Plot the first solution component.

S = solve(F,0,3000);
plot(S.Time,S.Solution(1,:),"-o")

Consider this system of first-order equations.

$\begin{array}{l}{y}_{1}^{\prime }={y}_{1}{y}_{3}-{y}_{2}\\ {y}_{2}^{\prime }={y}_{1}-1\\ 0={y}_{1}+{y}_{2}+{y}_{3}\end{array}$

The left side of the equations contain time derivatives, ${\mathit{y}}_{\mathit{n}}^{\prime }=\frac{{\mathrm{dy}}_{\mathit{n}}}{\mathrm{dt}}$. However, because the derivative for ${\mathit{y}}_{3}$ does not appear in the system, the equations define a system of differential algebraic equations. Rewriting the system in the form ${\mathrm{My}}^{\prime }=\mathit{f}\left(\mathit{t},\mathit{y}\right)$ shows a constant, singular mass matrix on the left side.

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]\underset{}{\overset{˙}{\left[\begin{array}{c}{y}_{1}^{\prime }\\ {y}_{2}^{\prime }\\ {y}_{3}^{\prime }\end{array}\right]}}=\left[\begin{array}{c}{y}_{1}{y}_{3}-{y}_{2}\\ {y}_{1}-1\\ {y}_{1}+{y}_{2}+{y}_{3}\end{array}\right]$

Solve the system of equations using the initial values [1 1 -2] by creating an ode object to represent the problem.

• Specify the initial values in the InitialValue property.

• Specify the system of equations as an anonymous function in the ODEFcn property.

• Use an odeMassMatrix object to specify the constant, singular mass matrix in the MassMatrix property.

F = ode;
F.InitialValue = [1 1 -2];
F.ODEFcn = @(t,y) [y(1)*y(3)-y(2);
y(1)-1;
y(1)+y(2)+y(3)];
F.MassMatrix = odeMassMatrix(MassMatrix=[1 0 0; 0 1 0; 0 0 0],Singular="yes");

Display the ode object. The SelectedSolver property shows that the ode15s solver was automatically chosen for this problem.

F
F =
ode with properties:

Problem definition
ODEFcn: @(t,y)[y(1)*y(3)-y(2);y(1)-1;y(1)+y(2)+y(3)]
InitialTime: 0
InitialValue: [1 1 -2]
Jacobian: []
MassMatrix: [1x1 odeMassMatrix]

Solver properties
AbsoluteTolerance: 1.0000e-06
RelativeTolerance: 1.0000e-03
Solver: auto
SelectedSolver: ode15s

Show all properties

Solve the system of equations over the time interval [0 10] by using the solve method. Plot all three solution components.

S = solve(F,0,10);
plot(S.Time,S.Solution,"-o")
legend("y_1","y_2","y_3",Location="southeast")

Solve an ODE system with two equations and two parameters, and perform sensitivity analysis on the parameters.

Create an ode object to represent this system of equations.

$\begin{array}{l}\frac{{\mathrm{dy}}_{1}}{\mathrm{dt}}={\mathit{p}}_{1}{\mathit{y}}_{1}-{\mathit{y}}_{2}\\ \frac{{\mathrm{dy}}_{2}}{\mathrm{dt}}={-\mathit{p}}_{2}{\mathit{y}}_{2}\end{array}$

Specify the initial conditions as ${\mathit{y}}_{1}\left(0\right)=2$ and ${\mathit{y}}_{2}\left(0\right)=3$, and parameter values of ${\mathit{p}}_{1}=0.05$ and ${\mathit{p}}_{2}=1.5$. To enable sensitivity analysis of the parameters, set the Sensitivity property of the ode object to an odeSensitivity object.

p = [0.05 1.5];
F = ode(ODEFcn=@(t,y,p) [p(1)*y(1)-y(2); -p(2)*y(2)], ...
InitialValue=[2 3], ...
Parameters=p, ...
Sensitivity=odeSensitivity)
F =
ode with properties:

Problem definition
ODEFcn: @(t,y,p)[p(1)*y(1)-y(2);-p(2)*y(2)]
Parameters: [0.0500 1.5000]
InitialTime: 0
InitialValue: [2 3]
Sensitivity: [1x1 odeSensitivity]

Solver properties
AbsoluteTolerance: 1.0000e-06
RelativeTolerance: 1.0000e-03
Solver: auto
SelectedSolver: cvodesNonstiff

Show all properties

Because the equations are nonstiff and sensitivity analysis is enabled, the ode object automatically chooses the cvodesNonstiff solver for this problem.

Solve the ODE over the time interval [0 5], and plot the solution for each component.

S = solve(F,0,5)
S =
ODEResults with properties:

Time: [0 2.9540e-09 2.9543e-05 2.2466e-04 4.1978e-04 0.0024 0.0139 0.0255 0.0484 0.0713 0.1363 0.2014 0.2992 0.3970 0.4948 0.6661 0.8373 1.0085 1.1798 1.3510 1.5222 1.8047 2.0872 2.3697 2.6522 2.9347 3.2172 3.4997 3.7822 ... ] (1x34 double)
Solution: [2x34 double]
Sensitivity: [2x2x34 double]

plot(S.Time,S.Solution(1,:),"-o",S.Time,S.Solution(2,:),"-o")
legend("y1","y2")

The values in S.Sensitivity are partial derivatives of the equations with respect to the parameters. To examine the effects of the parameter values during the integration, plot the sensitivity values.

figure
hold on
plot(S.Time,squeeze(S.Sensitivity(1,1,:)),"-o")
plot(S.Time,squeeze(S.Sensitivity(1,2,:)),"-o")
plot(S.Time,squeeze(S.Sensitivity(2,1,:)),"-o")
plot(S.Time,squeeze(S.Sensitivity(2,2,:)),"-o")
legend("p1,eq1","p2,eq1","p1,eq2","p2,eq2")
hold off

Consider a ball thrown upward with an initial velocity $\mathrm{dy}/\mathrm{dt}$. The ball is subject to acceleration due to gravity aimed downward, so its acceleration is

$\frac{{d}^{2}y}{d{t}^{2}}=-g.$

Rewriting the equation as a first-order system of equations with the substitutions ${\mathit{y}}_{1}=\mathit{y}$ and ${\mathit{y}}_{2}=\frac{\mathrm{dy}}{\mathrm{dt}}$ yields

$\begin{array}{l}{y}_{1}^{\prime }={y}_{2}\\ {y}_{2}^{\prime }=-g.\end{array}$

Solve the equations for the position ${\mathit{y}}_{1}$ and velocity ${\mathit{y}}_{2}$ of the ball over time.

Define Equations and Initial Conditions

Create a function handle for the first-order system of equations that accepts two inputs for (t,y). Use the value $\mathit{g}=9.8$ for the acceleration due to gravity.

dydt = @(t,y) [y(2); -9.8];

Next, create a vector with the initial conditions. The ball starts at position ${\mathit{y}}_{1}=3$ at $\mathit{t}=0$ as it is thrown upward with initial velocity ${\mathit{y}}_{2}=20$.

y0 = [3 20];

Model Ball Bounces as Events

The ball initially travels upward until the force due to gravity causes it to change direction and head back down to the ground. If you solve the equations without more consideration, then the ball falls back downward forever without striking the ground. Instead, you can use an event function to detect when the position of the ball goes to zero where the ground is located. Because the solution component ${\mathit{y}}_{1}=\mathit{y}$ is the position of the ball, the event function tracks the value of ${\mathit{y}}_{1}$ so that an event triggers whenever ${\mathit{y}}_{1}=0$.

Create a function handle for the event function that accepts two inputs for (t,y).

bounceEvent = @(t,y) y(1);

When the ball strikes the ground, its direction changes again as it heads back upwards with a new (smaller) initial velocity. To model this situation, use a callback function along with the event function. When an event triggers, the ODE solver invokes the callback function. The callback function resets the position and initial velocity of the ball so that the integration can restart with the correct initial conditions. When an event occurs, the callback function sets the position ${\mathit{y}}_{1}=0$ and attenuates the velocity by a factor of $0.9$ while reversing its direction back upward. Define a callback function that performs these actions.

function [stop,y] = bounceResponse(t,y)
stop = false;
y(1) = 0;
y(2) = -0.9*y(2);
end

(The callback function is included as a local function at the end of the example.)

Create an odeEvent object to represent the bouncing ball events. Specify Direction="descending" so that only events where the position is decreasing are detected. Also, specify Response="callback" so that the solver invokes the callback function when an event occurs.

E = odeEvent(EventFcn=bounceEvent, ...
Direction="descending", ...
Response="callback", ...
CallbackFcn=@bounceResponse)
E =
odeEvent with properties:

EventFcn: @(t,y)y(1)
Direction: descending
Response: callback
CallbackFcn: @bounceResponse

Solve Equations

Create an ode object for the problem, specifying the equations dydt, initial conditions y0, and events E as property values.

F = ode(ODEFcn=dydt,InitialValue=y0,EventDefinition=E);

Integrate the equations over the time interval [0 30] by using the solve method. Specify Refine=8 to generate 8 points per step. The resulting object has properties for the time and solution, and because events are being tracked, the object also displays properties related to the events that triggered during the integration.

S = solve(F,0,30,Refine=8)
S =
ODEResults with properties:

Time: [0 0.0038 0.0075 0.0113 0.0151 0.0188 0.0226 0.0264 0.0301 0.0490 0.0678 0.0867 0.1055 0.1243 0.1432 0.1620 0.1809 0.2751 0.3692 0.4634 0.5576 0.6518 0.7460 0.8402 0.9344 1.3094 1.6844 2.0594 2.4344 2.8094 3.1844 ... ] (1x468 double)
Solution: [2x468 double]
EventTime: [4.2265 8.1607 11.7015 14.8882 17.7563 20.3375 22.6606 24.7514 26.6331 28.3267 29.8509]
EventSolution: [2x11 double]
EventIndex: [1 1 1 1 1 1 1 1 1 1 1]

Plot Results

Plot the position ${\mathit{y}}_{1}$ of the ball over time, marking the initial position with a green circle and events with red circles.

plot(S.Time,S.Solution(1,:),"--")
hold on
plot(S.EventTime,S.EventSolution(1,:),"ro")
plot(0,y0(1),"go")
hold off
ylim([0 25])
xlabel("Time")
ylabel("Position y_1")

Local Functions

function [stop,y] = bounceResponse(t,y)
stop = false;
y(1) = 0;
y(2) = -0.9*y(2);
end

Version History

Introduced in R2023b