Backward Euler

Backward Euler#

This example solves the first-order ordinary differential equation using the backward Euler method.

\[ \frac{dy}{dt} = \sin^2(t) \cdot y \]

with initial condition \(y(0) = 2\) over the time interval \([0,5]\).

Mathematical Background#

The backward Euler is an implicit method for solving first-order differential equations where we are evaluating the function at the next point in time \(y_{i+1}\).

\[ y_{i+1} = y_i + h \cdot f(t_{i+1}, y_{i+1}) \]

where:

  • \(h\) is the step-size

  • \(t_i\) is the current time

  • \(y_i\) is the solution at time \(t_i\)

  • \(y_{i+1}\) is the solution at the next time step \(t_{i+1}\)

  • \(f(t_{i+1}, y_{i+1})\) is the function at time step \(t_{i+1}\) and \(y_{i+1}\)

Implementation Details#

Note that \(f(t_{i+1}, y_{i+1})\) is not known due to \(y_{i+1}\) being on both the left and right hand side. Therefore, we can solve for \(y_{i+1}\) by finding the root: $\(y_{i+1} - y_i - h \cdot f(t_{i+1}, y_{i+1}) = 0\)$

In the C++ example, we used the fixed-point iteration method for rootfinding due to it’s simpler nature.

Results#


Backward Euler solutions y(t) per time-step t.
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This example is implemented in: