Qualitative and Numerical Methods for Differential Equations

Almost every differential equation you meet outside a textbook has no solution formula. This topic gives you the two things you can always do anyway: read the qualitative behaviour of solutions straight off the equation (slope fields, phase lines, stability), and produce numerical values step by step (Euler's method). Both are pure applications of the central idea of differential calculus — near a point, a differentiable function looks like its tangent line.

First-order equations and slope fields

A first-order ODE in explicit form is $\dfrac{dy}{dx}=F(x,y)$. A function $y=g(x)$ is a solution on an interval $I$ if $g$ is differentiable on $I$ and $g'(x)=F\left(x,g(x)\right)$ for every $x$ in $I$. An initial value problem (IVP) attaches a condition $y(x_0)=y_0$.

Slope field (direction field). The equation hands you the slope of the solution through every point before you know any solution: at the point $(a,b)$ any solution passing through has slope $F(a,b)$. Draw a short segment of that slope at each point of a grid; the picture is the slope field. A solution curve is any curve that is tangent to the field at every one of its points, so you sketch solutions by "flowing with the segments".

Isocline. The isocline of level $k$ is the curve $F(x,y)=k$; every field segment on it has slope $k$. The level-$0$ isocline (the nullcline) shows where solutions have horizontal tangents, hence where a solution can have a local max or min. An isocline is not usually a solution: it is a solution only in the special case that it is itself a curve whose own slope equals $k$ everywhere on it.

Existence and uniqueness (Picard–Lindelöf). If $F$ and its $y$-partial derivative $F_y$ are both continuous on an open rectangle containing $(x_0,y_0)$, then the IVP $y'=F(x,y)$, $y(x_0)=y_0$ has exactly one solution on some open interval around $x_0$. Two consequences you use constantly: exactly one solution curve passes through each point of that rectangle, and distinct solution curves never touch or cross there.

Both hypotheses and the word "some" matter.

• Continuity of $F$ alone gives existence but not uniqueness: $y'=y^{2/3}$, $y(0)=0$ has $F_y=\dfrac{2}{3}y^{-1/3}$, discontinuous at $y=0$, and both $y=0$ and $y=\dfrac{x^{3}}{27}$ solve it.
• Uniqueness is local in $x$: $y'=y^{2}$, $y(0)=1$ has the unique solution $y=\dfrac{1}{1-x}$, which exists only for $x<1$.

Autonomous equations and the phase line

An ODE is autonomous when the independent variable does not appear on the right: $\dfrac{dy}{dt}=f(y)$. Two structural facts follow immediately.

1. The slope field depends only on the height $y$, so it is unchanged by horizontal shifts. Hence if $y(t)$ is a solution, so is $y(t-c)$ for every constant $c$: solution curves are horizontal translates of one another.
2. If $f(c)=0$, the constant function $y(t)=c$ for all $t$ is a solution. Such a $c$ is an equilibrium (also: critical point, stationary solution).

Phase line. Draw a $y$-axis, mark every root of $f$, and on each open interval between consecutive roots draw an arrow pointing up where $f(y)>0$ and down where $f(y)<0$. That one picture classifies every long-run behaviour.

Linearisation (derivative) test for stability. Suppose $f$ is continuously differentiable near $c$ and $f(c)=0$. Then

$\begin{cases} f'(c)<0 & \Rightarrow\ c \text{ is asymptotically stable},\\ f'(c)>0 & \Rightarrow\ c \text{ is unstable},\\ f'(c)=0 & \Rightarrow\ \text{no conclusion — use the sign chart of } f. \end{cases}$

Monotonicity theorem. Assume $f$ is continuously differentiable. A solution that is not an equilibrium can never reach an equilibrium value (uniqueness would force it to be that constant solution), so for its whole life it stays inside one open interval on which $f$ has no zero: between two consecutive equilibria, or above the largest one, or below the smallest one, or the entire line if $f$ never vanishes. On such an interval $f$ has one fixed sign, so every non-equilibrium solution of a scalar autonomous equation is strictly monotone — it can never oscillate. If it stays bounded it converges to an equilibrium as $t\to\infty$. If it is unbounded, beware: it may run off to $+\infty$ or $-\infty$ in finite time (as $y'=y^{2}$ does), so it need not exist for all $t$.

Concavity of solutions. By the chain rule, $y''=\dfrac{d}{dt}f(y)=f'(y)\,y'=f'(y)f(y)$. Along a non-equilibrium solution $f(y)$ keeps one sign, so the concavity flips exactly where $f'(y)$ flips. Precisely: if the solution passes through a level $c$ with $f(c)\ne 0$ and $f'$ changes sign at $c$, the curve has an inflection point there. The sign change is essential — $f'(c)=0$ alone is not enough. For $f(y)=y^{3}+1$ we have $f'(y)=3y^{2}$, which vanishes at $c=0$ without changing sign, and $f(0)=1\ne 0$; a solution crossing the level $0$ has $y''=0$ there but is concave up on both sides, so there is no inflection.

The logistic equation

$\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right),\qquad k>0,\ M>0 .$

Here $k$ is the intrinsic growth rate and $M$ the carrying capacity. The right side $f(P)=kP\left(1-\dfrac{P}{M}\right)$ is a downward parabola with roots $P=0$ and $P=M$ and vertex at $P=\dfrac{M}{2}$, where $f\left(\dfrac{M}{2}\right)=\dfrac{kM}{4}$ is the largest possible growth rate.

Since $f'(P)=k\left(1-\dfrac{2P}{M}\right)$, we get $f'(0)=k>0$ (so $P=0$ is unstable) and $f'(M)=-k<0$ (so $P=M$ is asymptotically stable).

Separating variables and using partial fractions gives the explicit solution

$P(t)=\frac{M}{1+Ae^{-kt}},\qquad A=\frac{M-P_0}{P_0}\quad (P_0\ne 0),$

and indeed $P(0)=\dfrac{M}{1+\frac{M-P_0}{P_0}}=\dfrac{MP_0}{P_0+M-P_0}=P_0$, while $\lim_{t\to\infty}P(t)=M$ for every $P_0>0$ — exactly what the phase line predicted.

Euler's method

Derivation. Suppose you know the exact point $(x_n,y_n)$ on a solution. The linear approximation (tangent line) at that point is $L(x)=y_n+y'(x_n)(x-x_n)$, and the ODE hands you the slope: $y'(x_n)=F(x_n,y_n)$. Stepping forward a distance $h$ and calling the result $y_{n+1}$:

$x_{n+1}=x_n+h,\qquad y_{n+1}=y_n+h\,F\left(x_n,y_n\right)$

Geometrically you walk along the slope field in short straight segments, re-reading the slope at the start of each segment.

Worked example A — an Euler table. Approximate $y(0.8)$ for $y'=x+y$, $y(0)=1$, with $h=0.2$.

So $y(0.8)\approx 2.3472$. The exact solution is $y=2e^{x}-x-1$ (check: $y'=2e^{x}-1$ and $x+y=x+2e^{x}-x-1=2e^{x}-1$), giving $y(0.8)=2e^{0.8}-1.8\approx 2.65108$. The error is $\approx 0.30388$. Redoing the computation with $h=0.1$ (eight steps) gives $y_8=2.48717762$ and error $\approx 0.16390$ — the error dropped by a factor of about $1.85$, i.e. roughly in half when $h$ was halved.

Error analysis. Taylor's theorem with remainder gives, for a twice-differentiable exact solution,

$y(x_n+h)=y(x_n)+h\,y'(x_n)+\frac{h^{2}}{2}y''(\xi),\qquad \xi \text{ between } x_n \text{ and } x_n+h,$

so one step started from an exact value costs a local truncation error of size $\dfrac{h^{2}}{2}\left|y''(\xi)\right|=O\left(h^{2}\right)$. Covering a fixed interval $[a,b]$ takes $N=\dfrac{b-a}{h}$ steps, and $N\cdot O\left(h^{2}\right)=O(h)$: Euler's method is first-order accurate. Halving $h$ roughly halves the global error (and doubles the work).

Error bound theorem. Suppose $F$ is continuous on a rectangle that contains both the exact solution curve and every Euler point $\left(x_n,y_n\right)$ over $[a,b]$, suppose $F$ satisfies a Lipschitz condition in its second variable there, $\left|F(x,u)-F(x,v)\right|\le L\left|u-v\right|$ with $L>0$, and suppose the exact solution satisfies $\left|y''\right|\le M$ there. Then for every step $x_n=a+nh$ in $[a,b]$,

$\left|y(x_n)-y_n\right|\le \frac{hM}{2L}\left(e^{L(x_n-a)}-1\right).$

Direction of the error. A tangent line lies below a concave-up curve and above a concave-down curve. So on a step where the exact solution is concave up, Euler undershoots; where it is concave down, Euler overshoots. This is exact for a single step from a true point and a reliable rule of thumb over many steps as long as the concavity does not change sign.

Worked example B — a phase line. Analyse $\dfrac{dy}{dt}=y^{2}-y^{3}=y^{2}(1-y)$.

Step 1 — equilibria. $y^{2}(1-y)=0$ gives $y=0$ (double root) and $y=1$.

Step 2 — sign chart. For $y<0$: $y^{2}>0$ and $1-y>0$, so $f>0$. For $0<y<1$: both factors positive, $f>0$. For $y>1$: $1-y<0$, so $f<0$.

Step 3 — phase line. Arrows point up on $(-\infty,0)$, up on $(0,1)$, down on $(1,\infty)$. Hence $y=0$ is semi-stable (attracting from below, repelling from above) and $y=1$ is asymptotically stable.

Step 4 — cross-check with the derivative test. $f'(y)=2y-3y^{2}$. Then $f'(1)=2-3=-1<0$, confirming that $y=1$ is stable. But $f'(0)=0$, so the test says nothing at $y=0$ — only the sign chart of Step 2 reveals the semi-stability.

Step 5 — shape of solutions. $y''=f'(y)f(y)=\left(2y-3y^{2}\right)\left(y^{2}-y^{3}\right)=y(2-3y)\cdot y^{2}(1-y)=y^{3}(2-3y)(1-y)$. For $0<y<1$ this is positive when $y<\dfrac{2}{3}$ and negative when $y>\dfrac{2}{3}$: a solution starting at $y(0)=0.1$ rises, is concave up until it passes the level $y=\dfrac{2}{3}$, then is concave down as it levels off at $y=1$.

Common mistakes

• Treating an isocline as a solution. For $y'=x-y$ the field segments are horizontal along $y=x$, so students draw $y=x$ as a solution curve. Wrong: the function $y=x$ has slope $1$, not $0$. Right: $y=x-1$ is a solution, because there its own slope $1$ matches $F=x-(x-1)=1$.
• "$f'(c)=0$, so $c$ is stable." Wrong — that is precisely the inconclusive case. Right: fall back on the sign of $f$ on both sides.
• Differentiating the wrong function for stability. The test uses $f'$, the derivative of the right-hand side with respect to $y$ — not $y''$, and not the second derivative of a solution.
• Reading equilibria off the wrong equation. For $y'=y^{2}-4$ the equilibria are $y=\pm 2$, from $f(y)=0$; $y=0$ is not one.
• Missing repeated roots. $f(y)=(y-1)^{2}(y+2)$ has an equilibrium at $y=1$ that is easy to skip and is semi-stable, not stable.
• Letting a solution cross an equilibrium. With $f$ continuously differentiable, uniqueness forbids it, and a non-equilibrium solution never even reaches an equilibrium value in finite time — it only approaches it as $t\to\infty$.
• Euler bookkeeping. Wrong: $y_{n+1}=y_n+F(x_n,y_n)$ (forgot $h$), or $y_{n+1}=y_n+hF(x_{n+1},y_n)$ (advanced $x$ too early). Right: compute the slope at the point you are standing on, then advance both $x$ and $y$.
• "Halving $h$ quarters the error." That is the local error, $O(h^{2})$. The accumulated error over a fixed interval is $O(h)$, so halving $h$ roughly halves it.
• "Euler always underestimates." Only when the solution is concave up. For $y'=3-2y$ with $y(0)=1$ the solution is concave down and Euler overestimates.