Why solve numerically? · Differential Equations

Intuition

Most differential equations have no formula for their solutions. Even one as short as y′ = x − y² has none in the functions of calculus. Yet the existence and uniqueness theorem says its initial value problem has exactly one solution, so there is something definite to find. A numerical method does not look for a formula. It computes approximate values of that one solution at a chain of points a small step apart, starting from the initial value and using nothing but the slope the equation gives.

A navigator with no satellite fix. From a known position, the speed and the heading give the position an hour later, and from that the position an hour after. Each estimate is built on the last, and a small error in one hour’s reckoning is carried into every hour after it.

The field of $y'=x-y^{2}$ and its solution from $y(0)=0$ , which no formula in elementary functions expresses. A numerical method computes the dots — approximate values at $x=0,\ 0.5,\ 1,\ \dots$ — and nothing in between.

A grid of points, and values on it

For the initial value problem $y'=f(x,y)$ , $y(x_{0})=y_{0}$ , choose a step size and the grid points . A numerical method computes numbers with , each from those before it; a one-step method computes from alone. The error at the -th point is , and reaching takes steps.

x_{n}=x_{0}+nh,\qquad y_{n}\approx y(x_{n})

What a method is, and is not

The answer is a table of numbers, not a function: values at the grid points and nothing in between.
It needs a solution to exist and to be the only one. Where the existence and uniqueness theorem does not apply, a method still produces numbers, and they may describe no solution, or only one of several.
Each step starts from the previous estimate, not from the true solution, so errors are carried forward and add up.
A smaller step means more points and, for a good method, a smaller error — and more work: halving $h$ doubles the number of steps.
Every one-step method rests on the integral form of the equation proved below: a step is an integral of the slope, and methods differ in how they approximate it.

A step is an integral of the slope

y(x_{n+1})=y(x_{n})+\int_{x_{n}}^{x_{n+1}}f\bigl(x,y(x)\bigr)\,dx

Let y be the solution on an interval holding the step. Because its derivative is f(x, y(x)) at every point there, integrating the equation over the step gives on the left the change in y, by the fundamental theorem of calculus, and on the right the integral of the slope along the solution. The integrand involves the unknown solution itself, so the integral cannot be computed exactly: every one-step method is a rule for approximating it from values the method already has.

Proof steps

y'(x)=f\bigl(x,y(x)\bigr)\quad\text{for}\ x_{n}\le x\le x_{n+1}

The solution satisfies the equation at every point of the step.

\int_{x_{n}}^{x_{n+1}}y'(x)\,dx=\int_{x_{n}}^{x_{n+1}}f\bigl(x,y(x)\bigr)\,dx

Integrate both sides over the step.

\int_{x_{n}}^{x_{n+1}}y'(x)\,dx=y(x_{n+1})-y(x_{n})

The fundamental theorem of calculus, since $y'=f(x,y(x))$ is continuous.

y(x_{n+1})=y(x_{n})+\int_{x_{n}}^{x_{n+1}}f\bigl(x,y(x)\bigr)\,dx

The value at the end of the step is the value at its start plus the integral of the slope.

\text{the integrand needs}\ y(x)\ \text{inside the step}

So it cannot be evaluated without the solution itself: a method approximates it.

Applications

Practice

Numbers at Grid Points

A numerical method replaces the unknown function by approximate values at the points x₀, x₀ + h, x₀ + 2h, …, computed one after another from the initial value.

x_{n}=x_{0}+nh,\qquad y_{n}\approx y(x_{n})

Try it

What does a numerical method produce for an initial value problem?