1.4 Isoclines

Consider an equation of the form $y^{\prime}=f(x,y)$ . The function $f(x,y)$ defines the slope of the solution curve at each point in the $xy$ -plane. Lines of constant slope, known as isoclines, satisfy

equation (1.10) (1.10)

f(x,y)=m,

where $m$ is a constant ( $m\in\mathbb{R}$ ). Isoclines can be used to sketch the solution curves.

Example 1.7.

Consider the following ODE:

y^{\prime}=y+2x.

By finding the isoclines, sketch some integral curves (i.e. solution curves) for this ODE (Exercise: show that the general solution is $\displaystyle{y=A\mathrm{e}^{x}-2-2x}$ , where $A$ is a constant).

Solution.

The isoclines correspond to

y+2x=m\Longrightarrow y=m-2x.

These are straight lines on which the slope of the solution curve is $m$ . The short line elements (or flow vectors) on the sketch indicate the corresponding direction field, i.e. the slope of the solution curves at each point in space. By connecting these, we can sketch the solution.

Example 1.8.

Find the isoclines and sketch some solution curves for the following ODE,

y^{\prime}=-\frac{x}{y}.

Check your solution curves by finding the exact solution.

Solution.

The isoclines are given by

-\frac{x}{y}=m\Longrightarrow y=-\frac{x}{m},

which are straight lines, of gradient $-1/m$ , passing through the origin.

Connecting up the flow vectors (see sketch) we obtain the solution curves. These look like concentric circles, centred at the origin. To find the exact solution, we can separate variables:

\int y\,\mathrm{d}y=-\int x\,\mathrm{d}x\Longrightarrow\frac{y^{2}}{2}=-\frac{% x^{2}}{2}+C,

where $C$ is a constant of integration. Rearranging,

y^{2}+x^{2}=2C,

which describes a circle, centred at the origin in the $xy$ -plane, of radius $\sqrt{2C}$ . This is consistent with the solution curves obtained from the isoclines.

Terminology: Consider an ODE of the form

y^{\prime}=\frac{g(x,y)}{h(x,y)},

with isoclines defined by

\frac{g(x,y)}{h(x,y)}=m.

Lines of zero slope (i.e. $m=0$ ) are defined by $g(x,y)=0$ , whilst lines of infinite slope (i.e. $1/m=0$ ) are defined by $h(x,y)=0$ . If

g(x,y)=h(x,y)=0

at any point in the $xy$ -plane, then the point is said to be singular. At this point, the slope is not single-valued: it lies on both the $m=0$ and $1/m=0$ isoclines. In Example 1.8, all of the isoclines intersect at the origin, which is the only singular point in this case. See handout for a more complicated example.