Why Coupled Euler Computations Need Care
A coupled differential-equation system describes several changing quantities whose rates depend on one another. At each time, the rate of change of one quantity may depend on the current value of another. Forward Euler can approximate such a system, but only if every component is updated consistently.
The central bookkeeping rule is simple: calculate all rates from the same current state, then use those rates to calculate all next values. A table makes this rule visible. It also makes it easier to compare the computed points with the directions suggested by the system’s vector field.
Definition of a Coupled Initial-Value Problem
For two dependent variables \(x\) and \(y\), a coupled initial-value problem has the form
The independent variable is \(t\); the values \(x(t)\) and \(y(t)\) form the state of the system. The right-hand sides \(f\) and \(g\) specify the two rates at a given time and state. In vector notation, write the state as \(\mathbf{z}=(x,y)\) and the rate as \(\mathbf{F}(t,\mathbf{z})=(f(t,x,y),g(t,x,y))\).
These formulas are simultaneous updates: both right-hand sides use \(x_n\) and \(y_n\), not a mixture of old and newly computed values. The procedure is explicit: once the current state and step size are known, the next state is calculated directly.
Why the Update Formula Works
Suppose a differentiable solution has state \(\mathbf{z}(t)\), and its derivative is \(\mathbf{F}(t,\mathbf{z}(t))\). The definition of differentiability gives, for a small positive step \(h\),
Since \(\mathbf{z}'(t)=\mathbf{F}(t,\mathbf{z}(t))\), the first two terms give a tangent-line estimate of the state after time \(h\). Forward Euler keeps those terms and omits the remainder:
Taking components yields the two Euler update formulas. This justification explains both the method’s name and its limitation: each step follows the tangent direction at the start of the step, rather than tracking the changing direction throughout the interval.
Worked Example: A Coupled Oscillator
Worked Example: Tabulating an Oscillator Step by Step
Use forward Euler with step size \(h=0.5\) on the system \(x'=y,\ y'=-x\), starting from \((x_0,y_0)=(1,0)\) at \(t_0=0\). The rates at each row are \(x'=y_n\) and \(y'=-x_n\), evaluated using that row’s state.
| \(n\) | \(t_n\) | \(x_n\) | \(y_n\) | \(x'=y_n\) | \(y'=-x_n\) |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 0 | 0 | \(-1\) |
| 1 | 0.5 | 1 | \(-0.5\) | \(-0.5\) | \(-1\) |
| 2 | 1 | 0.75 | \(-1\) | \(-1\) | \(-0.75\) |
| 3 | 1.5 | 0.25 | \(-1.375\) | \(-1.375\) | \(-0.25\) |
| 4 | 2 | \(-0.4375\) | \(-1.5\) | \(-1.5\) | 0.4375 |
At \(n=0\), the rates are \(0\) and \(-1\). Thus \(x_1=1+0.5(0)=1\), while \(y_1=0+0.5(-1)=-0.5\).
At \(n=1\), the rates are \(-0.5\) and \(-1\). Therefore \(x_2=1+0.5(-0.5)=0.75\), and \(y_2=-0.5+0.5(-1)=-1\).
At \(n=2\), the rates are \(x'=y_2=-1\) and \(y'=-x_2=-0.75\). Therefore \(x_3=0.75+0.5(-1)=0.25\), and \(y_3=-1+0.5(-0.75)=-1.375\).
At \(n=3\), the rates are \(-1.375\) and \(-0.25\). Hence \(x_4=0.25+0.5(-1.375)=-0.4375\), and \(y_4=-1.375+0.5(-0.25)=-1.5\).
Check: Substituting the row-\(2\) state \((0.75,-1)\) into the system gives \(x'=-1\) and \(y'=-0.75\), as recorded. Applying the update formulas gives \((0.25,-1.375)\), which is the row-\(3\) state. The later rows likewise use each row’s displayed rates and the same step size.
Reading the Numerical Path Against the Vector Field
For a two-variable system, the vector field assigns the arrow \((f(t,x,y),g(t,x,y))\) to each state at a fixed time. The arrow indicates the direction in which the state initially moves. An Euler step follows that arrow for a distance scaled by \(h\). Connecting successive computed states with straight segments creates a polygonal numerical path.
For the oscillator in the example, the rate vector is \((y,-x)\). Its dot product with the state \((x,y)\) is
Thus the vector field is perpendicular to the radius from the origin: its arrows are tangent to circles centered at the origin. The exact solution, when started away from the origin, moves along a circle. The Euler polygon, however, need not stay on that circle.
To see the difference, take one Euler step from \((x,y)\) for this system. The next state is \((x+hy,y-hx)\), so its squared distance from the origin is
For every \(h\ne 0\) and every nonzero state, this is larger than \(x^2+y^2\). The numerical point moves outward even though the vector field is tangent to a circle. This is a qualitative comparison, not a claim that each individual component must increase: the signs of the two rates vary with the state.
Worked Example: A Conserved Sum in a Transfer System
Worked Example: Checking an Invariant Along Euler Steps
Consider \(x'=-x+y,\ y'=x-y\), with initial state \((2,0)\) at \(t=0\), and take \(h=0.25\). Each pair of rates is computed from the current row.
| \(n\) | \(t_n\) | \(x_n\) | \(y_n\) | \(x'=-x_n+y_n\) | \(y'=x_n-y_n\) |
|---|---|---|---|---|---|
| 0 | 0 | 2 | 0 | \(-2\) | 2 |
| 1 | 0.25 | 1.5 | 0.5 | \(-1\) | 1 |
| 2 | 0.5 | 1.25 | 0.75 | \(-0.5\) | 0.5 |
| 3 | 0.75 | 1.125 | 0.875 | \(-0.25\) | 0.25 |
From the initial state, the rates are \(-2\) and \(2\), so \(x_1=2+0.25(-2)=1.5\) and \(y_1=0+0.25(2)=0.5\). At \((1.5,0.5)\), the rates are \(-1\) and \(1\); the next state is \(x_2=1.5-0.25=1.25\), \(y_2=0.5+0.25=0.75\). At \((1.25,0.75)\), the rates are \(-0.5\) and \(0.5\); this gives \(x_3=1.25-0.125=1.125\), \(y_3=0.75+0.125=0.875\).
Check: The rate sum is zero at every state, because \((-x+y)+(x-y)=0\). Accordingly, the update changes \(x+y\) by \(0.25\) times zero. The table has \(x_n+y_n=2\) in every row: \(2\), \(1.5+0.5\), \(1.25+0.75\), and \(1.125+0.875\) all equal \(2\). The field arrows point along lines of constant sum, and the Euler points remain on the same line.
Worked Example: Including Time in the Rates
Worked Example: Evaluating a Nonautonomous Coupled System
Use \(h=0.5\) for \(x'=y+t,\ y'=x-t\), starting from \((x_0,y_0)=(1,0)\) at \(t_0=0\). Since the rates depend explicitly on time, each row must use its own \(t_n\) as well as its current state.
| \(n\) | \(t_n\) | \(x_n\) | \(y_n\) | \(x'=y_n+t_n\) | \(y'=x_n-t_n\) |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 0 | 0 | 1 |
| 1 | 0.5 | 1 | 0.5 | 1 | 0.5 |
| 2 | 1 | 1.5 | 0.75 | 1.75 | 0.5 |
| 3 | 1.5 | 2.375 | 1 | 2.5 | 0.875 |
At time \(0\), the rates are \(0+0=0\) and \(1-0=1\), giving \((x_1,y_1)=(1,0.5)\). At time \(0.5\), the rates are \(0.5+0.5=1\) and \(1-0.5=0.5\), giving \((x_2,y_2)=(1.5,0.75)\). At time \(1\), the rates are \(0.75+1=1.75\) and \(1.5-1=0.5\), so \(x_3=1.5+0.5(1.75)=2.375\) and \(y_3=0.75+0.5(0.5)=1\).
Check: At the final displayed state \((t_3,x_3,y_3)=(1.5,2.375,1)\), the system gives \(x'=1+1.5=2.5\) and \(y'=2.375-1.5=0.875\), matching the table. This row’s rates would be used to compute the next state if another step were required.
A Common Mistake: Updating One Component Too Early
In the transfer system \(x'=-x+y,\ y'=x-y\), start from \((2,0)\) with \(h=0.25\). The correct rates at the starting state are \(-2\) and \(2\), so the simultaneous update is
A tempting but incorrect calculation first updates \(x\), then uses the new \(x_1=1.5\) to recalculate the rate for \(y\) while keeping \(y_0=0\). That gives \(y'=1.5-0=1.5\) and the wrong value \(y_1=0+0.25(1.5)=0.375\). This mixes states from different time levels. It also changes the sum from \(2\) to \(1.5+0.375=1.875\), contradicting the system’s zero total rate.
A Reliable Table Workflow
For a coupled system, use the following sequence at every step:
Write \(t_n\) and every current component value before evaluating any rates.
Substitute \(t_n\) and the entire current state into every right-hand side.
For each variable, add \(h\) times its rate to its current value.
Substitute the new state into the system to verify the rates that should appear there.
Check whether the signs and direction of each update agree with the arrow at the old state.
This table workflow is useful because errors in coupled computations are often relational: a rate may have the wrong sign, may use the wrong component, or may be evaluated at the wrong time. Checking rows against the original equations helps detect all three. Comparing the polygonal path with the vector field then gives a separate qualitative check: each Euler segment should initially point in the direction of the rate vector at its starting point.
Check Your Understanding
Use the simultaneous-update rule and the system’s rate equations to answer these questions.
- For \(x'=y,\ y'=-x\), what rate pair is obtained at the state \((x,y)=(0.75,-1)\)?
- For \(x'=-x+y,\ y'=x-y\), why does every Euler step preserve \(x+y\), regardless of the step size?
- When the right-hand sides depend on \(t\), which time value should be used to calculate the rates for the step from \(t_n\) to \(t_{n+1}\)?
- Explain why updating one component and then using that new value to calculate another component’s rate does not give the forward Euler update.
- For the oscillator system \(x'=y,\ y'=-x\), why does the vector field point tangent to circles centered at the origin?
