ODE solvers (ode45) in MATLAB - Time & Space Complexity

Time Complexity: ODE solvers (ode45)

O(n)

Understanding Time Complexity

When solving ordinary differential equations (ODEs) with ode45, it is important to understand how the time to solve grows as the problem size or time span increases.

We want to know how the number of calculations changes when we ask ode45 to solve over longer times or with more complex equations.

Scenario Under Consideration

Analyze the time complexity of the following ode45 usage.

    function dydt = simpleODE(t, y)
        dydt = -2 * y + sin(t);
    end

    tspan = [0 10];
    y0 = 1;
    [t, y] = ode45(@simpleODE, tspan, y0);
    

This code solves a simple ODE from time 0 to 10 starting at y=1 using ode45.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: ode45 repeatedly evaluates the ODE function at many time points.
• How many times: The number of function evaluations depends on the solver's step size and error control, which adapts to the problem.

How Execution Grows With Input

As the time span or complexity of the ODE increases, ode45 takes more steps, so the number of operations grows roughly in proportion to the length of the time interval.

Input Size (time span length)	Approx. Operations (function calls)
10	~100 to 200
100	~1000 to 2000
1000	~10,000 to 20,000

Pattern observation: Doubling the time span roughly doubles the number of steps and function evaluations.

Final Time Complexity

Time Complexity: O(n)

This means the time to solve grows roughly linearly with the length of the time interval or the number of steps needed.

Common Mistake

[X] Wrong: "ode45 always takes the same number of steps regardless of the time span or equation complexity."

[OK] Correct: ode45 adapts its step size based on the problem and interval length, so more time or harder equations mean more steps and more work.

Interview Connect

Understanding how numerical solvers like ode45 scale helps you explain performance in simulations and scientific computing tasks, showing you grasp practical algorithm behavior.

Self-Check

"What if we changed the solver from ode45 to ode15s for stiff problems? How would the time complexity change?"

Practice

1. What does the MATLAB function ode45 primarily do?

easy

A. It solves ordinary differential equations numerically.

B. It plots graphs of functions automatically.

C. It calculates the derivative of a function symbolically.

D. It integrates functions analytically.

Solution

1. Step 1: Understand the purpose of ode45

   The ode45 function in MATLAB is designed to solve ordinary differential equations (ODEs) using a numerical method called Runge-Kutta.

2. Step 2: Compare options with ode45's function

   Options A, C, and D describe plotting, symbolic differentiation, and analytical integration, which are not what ode45 does.

3. Final Answer:

   It solves ordinary differential equations numerically. -> Option A

4. Quick Check:

   ode45 solves ODEs numerically = B [OK]

Hint: Remember ode45 solves ODEs numerically, not symbolically or graphing [OK]

Common Mistakes:

• Confusing ode45 with plotting functions
• Thinking ode45 does symbolic math
• Assuming ode45 integrates analytically

2. Which of the following is the correct syntax to call ode45 for solving dy/dt = -2y with initial value y(0)=1 over time interval [0 5]?

easy

A. [t,y] = ode45(@(y,t) -2*y, [0 5], 1);

B. [t,y] = ode45(@(t,y) -2*y, [0 5], 1);

C. [t,y] = ode45(@(t,y) 2*y, 0:5, 1);

D. [t,y] = ode45(@(t,y) -2*y, 0, 5, 1);

Solution

1. Step 1: Check function handle and argument order

   The function handle for ode45 must be @(t,y) for time t and variable y. [t,y] = ode45(@(t,y) -2*y, [0 5], 1); uses @(t,y) correctly with the expression -2*y.

2. Step 2: Verify time span and initial value format

   Time span must be a vector [0 5], and initial value is a scalar 1. [t,y] = ode45(@(t,y) -2*y, [0 5], 1); matches this format exactly.

3. Final Answer:

   [t,y] = ode45(@(t,y) -2*y, [0 5], 1); -> Option B

4. Quick Check:

   Correct syntax: @(t,y), time vector, initial value = A [OK]

Hint: Use @(t,y) for function and time vector in brackets [OK]

Common Mistakes:

• Swapping t and y in function handle
• Using colon operator instead of vector for time span
• Passing time points as separate arguments

3. Consider the code:

f = @(t,y) t - y;
[t,y] = ode45(f, [0 2], 1);
disp(y(end));

What is the output approximately?

medium

A. 1.3

B. 1.0

C. 0.5

D. 2.0

Solution

1. Step 1: Understand the ODE and initial condition

   The ODE is dy/dt = t - y with y(0) = 1, solved over [0 2]. This is a linear ODE where y tends to follow t but is pulled down by -y.

2. Step 2: Estimate solution behavior at t=2

   Numerical solution with ode45 typically gives y around 1.3 at t=2 for this ODE and initial value.

3. Final Answer:

   1.3 -> Option A

4. Quick Check:

   ode45 output y(end) ≈ 1.3 for dy/dt = t - y [OK]

Hint: Recall ode45 returns solution array; last value is y at final time [OK]

Common Mistakes:

• Assuming output equals initial value
• Confusing function input order
• Ignoring numerical approximation

4. Identify the error in this code snippet:

f = @(y,t) y^2 - t;
[t,y] = ode45(f, [0 1], 0);
plot(t,y);

medium

A. Time span must be a column vector.

B. Initial value cannot be zero.

C. Function handle arguments are in wrong order.

D. Plot command is missing labels.

Solution

1. Step 1: Check function handle argument order

   ODE functions for ode45 must have arguments in order (t,y), but here it is @(y,t), which is incorrect.

2. Step 2: Verify other parts of the code

   Initial value zero is allowed, time span as [0 1] is valid, and plot labels are optional, so these are not errors.

3. Final Answer:

   Function handle arguments are in wrong order. -> Option C

4. Quick Check:

   Function handle must be @(t,y), not @(y,t) [OK]

Hint: Always use @(t,y) for ode45 function handles [OK]

Common Mistakes:

• Swapping t and y in function handle
• Thinking initial value can't be zero
• Believing time span must be column vector

5. You want to solve the system:

dx/dt = y
dy/dt = -x

with initial conditions x(0)=0, y(0)=1 over [0 10]. Which MATLAB code correctly uses ode45 to solve this system?

hard

A. f = @(t,u) [u(1); u(2)];
[t,u] = ode45(f, [0 10], [0 1]);

B. f = @(t,x,y) [y; -x];
[t,u] = ode45(f, [0 10], [0 1]);

C. f = @(u,t) [u(2); -u(1)];
[t,u] = ode45(f, [0 10], [0 1]);

D. f = @(t,u) [u(2); -u(1)];
[t,u] = ode45(f, [0 10], [0 1]);

Solution

1. Step 1: Define function handle for system of ODEs

   The system has two variables x and y, combined as vector u where u(1)=x and u(2)=y. The function must be @(t,u) returning [y; -x], which is [u(2); -u(1)].

2. Step 2: Check initial conditions and call to ode45

   Initial conditions are [0 1] for x=0 and y=1. Time span is [0 10]. f = @(t,u) [u(2); -u(1)]; [t,u] = ode45(f, [0 10], [0 1]); matches this exactly.

3. Final Answer:

   f = @(t,u) [u(2); -u(1)]; [t,u] = ode45(f, [0 10], [0 1]); -> Option D

4. Quick Check:

   System ODEs use @(t,u) with vector u = D [OK]

Hint: Use vector input @(t,u) and return derivatives as vector [OK]

Common Mistakes:

• Using separate arguments for variables in function handle
• Swapping order of arguments in function handle
• Returning wrong derivative vector