Two quantities that change together are described by a system such as x′ = a·x + b·y, y′ = c·x + d·y. At each point (x, y) it names a direction and a speed, and Euler's method steps both quantities at once. A solution is a curve (x(t), y(t)) in the phase plane, and the plane has no axis for time: time is how far along the curve you have gone.
Core ideas
Three ideas, and the first one decides how to read every picture on this course. The other two are the arithmetic that fills the picture in.
The right-hand sides a·x + b·y and c·x + d·y are functions of the point (x, y), and nothing else. At (1, 1) the system x′ = x, y′ = −y names the arrow (1, −1): x is growing at rate 1 and y is shrinking at rate 1. The collection of all those arrows is a vector field, and the system is a rule for drawing it.
A pair of functions x(t) and y(t) solves the system when, at every t, the velocity (x′, y′) is the arrow at the point (x(t), y(t)). Plot the points (x(t), y(t)) as t runs and you get a curve in the plane that is tangent to the field everywhere it goes. That plane, with the unknowns for axes, is the phase plane.
From the point (x, y) with step h the next point is (x + h·x′, y + h·y′), each rate read from the arrow at the old point. Both coordinates move before either rate is read again. With rational coefficients every step is a pair of fractions, and the lab prints them exactly.
The material
The definition, the first system worked out, what the phase plane leaves out, and the one exact fact that shows Euler's polygon is only a recipe.
Two equations x′ = a·x + b·y and y′ = c·x + d·y, where a, b, c and d are numbers. Written with a matrix it is x′ = A·x, where x is the pair (x, y) and A has the rows a b and c d.
A solution is a pair of functions x(t), y(t) that satisfy both equations. The phase plane is the plane with the unknowns x and y as its axes, and the orbit of a solution is the curve it traces.
One equation y′ = f(t, y) gave a field of slopes on a plane with time across the bottom. A system has two unknowns and no time on either axis, and the field is a field of arrows instead. The equations here have no t on the right-hand side, so the arrow at a point is the same whenever the solution gets there. That is why the picture can leave time out: two solutions that meet the same point leave it in the same direction.
The commonest misreading is to treat the horizontal axis as time. In the phase plane the horizontal axis is x. A solution that starts at (1, 1) and reaches (5/4, 3/4) has moved right and down in space, and the picture does not say when. To recover the timing you go back to the formulas, or to the step count, because time is carried by the step and not by an axis.
Start at (1, 1) with h = 1/4. The arrow at (1, 1) is (1, −1), so the first step lands at (1 + 1/4, 1 − 1/4), which is (5/4, 3/4).
The arrow there is (5/4, −3/4), giving (5/4 + 5/16, 3/4 − 3/16), which is (25/16, 9/16). A third step gives (125/64, 27/64). The curve heads along the x direction away from the origin while y dies away, which is the picture of a saddle.
The first preset in the lab is this system, and it is set to eight steps rather than three, so the point it prints is the eighth. Drag the step control down to three and the third row above appears in the last-point tile. Notice that the arithmetic was the same one-line rule each time: x is multiplied by 1 + h and y by 1 − h.
Every solution of x′ = x, y′ = −y keeps the product x·y constant.
By the product rule (x·y)′ = x′·y + x·y′.
Substituting the equations gives x·y − x·y = 0, so the product never changes. The solution through (1, 1) therefore lies on the curve x·y = 1. The Euler points above do not: (125/64)·(27/64) = 3375/4096, which is (15/16)³. Each step multiplies the product by (1 + h)·(1 − h) = 15/16, so the polygon slips inside the curve.
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That is the difference between the method and the solution, shown with two fractions. Each Euler step is computed exactly, and the polygon is still not the curve. Sixteen steps of h = 1/16 would stay much closer, and the lab lets you try. What it cannot do is make the product exactly one.
Take x′ = y, y′ = −x from (1, 0). The arrow at (1, 0) is (0, −1), so with h = 1/4 the first step goes to (1, −1/4).
The true solutions here are circles about the origin, and the squared distance from the origin is 1 at the start. After the step it is 1 + 1/16 = 17/16. The lab prints that ratio, and it is the same at every step: Euler's polygon on a rotation spirals outwards by an exact factor of 1 + h², because each step moves along the tangent, and a tangent always leaves the circle. “Euler on an Oscillator” proved that factor for x′ = v, v′ = −x; this is the same system with y written for v, seen in the phase plane.
Interactive
Trace, determinant, eigenvalues and exact Euler steps
Method
Read the arrow, take the step on both coordinates, and then ask what the path is doing in the plane.
At the point (x, y) the arrow is (a·x + b·y, c·x + d·y). For x′ = x, y′ = −y it is (x, −y), so at (2, 1) it is (2, −1).
With h = 1/4 the arrow (2, −1) becomes the displacement (1/2, −1/4). Both components are scaled by the same h.
The new point is (2 + 1/2, 1 − 1/4), which is (5/2, 3/4). Read the next arrow from this new point and repeat.
Say where the point is heading in the plane: away from the origin along one axis and toward it along the other, round and round, or in toward the origin. Do not say how x varies with t unless you also say that you have left the phase plane.
Where the true solution keeps a quantity (the product x·y, or the squared distance on a rotation), compute that quantity at the Euler point and compare. The gap is the method's error, and it is exact.
The system is the first preset. The arrow at each point is (x, −y), and the step multiplies x by 1 + h = 5/4 and y by 1 − h = 3/4.
The path heads away from the origin along the x-axis while its y dies away, and no start on the y-axis ever leaves it. That mixture of going in along one direction and out along another is what a saddle is, and the trace and determinant tiles, 0 and −1, are the first look at the two numbers the rest of this course sorts by.
The product of the steps' coordinates is 3375/4096, not 1, so this polygon is inside the true curve x·y = 1. That is not a slip in the arithmetic. It is what one step size does to a curve that bends.
You should be able to name the arrow at a point, take an Euler step on both unknowns as exact fractions, say that the orbit is a curve in the plane with no time axis, and say what quantity the true solution keeps that the polygon does not.
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