11.1.5. Example Problem

Given:

$\frac{dy}{dx} = x + y, \quad y(0) = 1$

Find: $y$ at $x = 0.1$ using Heun’s Method with $h = 0.1$

This chunk introduces the problem we'll solve using Heun's Method. We have a first-order ordinary differential equation given by $\frac{dy}{dx} = x + y$, and an initial condition stating that when $x = 0$, the value of $y$ is $1$. Our goal is to find the value of $y$ at a new point, specifically when $x = 0.1$, and we will use a step size of $h = 0.1$ to compute this.

Think of this scenario like tracking the growth of a plant. Initially, at time 0 (when we start observing), the plant is 1 unit high. We want to predict how tall the plant will be (y) after 10% of a time period has passed (0.1 units), and we have a rule that dictates its growth based on both its current height and time.

Step 1:

$f(x_n, y_n) = f(0, 1) = 0 + 1 = 1$

$y^* = 1 + 0.1 \times 1 = 1.1$

This step involves using Euler's method to get an initial estimate of the function's value at the next step. We first calculate the function value at the initial conditions, which is given by $f(0, 1) = 0 + 1 = 1$. Then, we substitute this value into the predictor formula to get our predicted value: $y^* = y_n + h \cdot f(x_n, y_n)$, which gives us $y^* = 1 + 0.1 \cdot 1 = 1.1$.

Imagine you have a plant that grows based on its current height. You estimate its height after a short time step (0.1 days) by simply adding its growth during that time (1 unit based on your rule). Hence, if you estimate its height was 1 unit initially, you predict it to be 1.1 units after 0.1 days.

Step 2:

$f(x_n, y^*) = f(0.1, 1.1) = 0.1 + 1.1 = 1.2$

$y_{n+1} = 1 + \frac{1 + 1.2}{2} \times 0.1 = 1 + 0.05 \times 2.2 = 1.11$

In this step, we refine our prediction using the corrector formula. First, we find the value of the function with the predicted point: $f(0.1, 1.1)$, which calculates to $1.2$. Next, we average the slopes we found: the initial slope (1) and this new slope (1.2). Using this average, we can find the corrected value at the new x, computing the formula $y_{n+1} = y_n + \frac{1}{2} (f(x_n, y_n) + f(x_{n+1}, y^*)) \times h$, leading us to find that $y(0.1) \approx 1.11$.

If you plant your seed and wait another moment to see how it grows, you might notice after accounting for the recent growth (the new height being the average of initial and predicted), the plant isn't just 1.1 units but closer to 1.11 units. This process makes sure you get a more accurate estimate of the growth.

Result: $y(0.1) \approx 1.11$

The result indicates our final computed value of the function at $x = 0.1$ is approximately 1.11. This means based on our calculations using Heun's method, we've successfully approximated the behavior of the original differential equation over the specified interval.

Basically, after applying our plant growth model for a short burst of time, we estimate that after 0.1 days, our plant has grown to approximately 1.11 units tall—showing the predictive power of our calculations and methods.

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