2.1: Linear Equations
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let us begin with the linear homogeneous equation
a1(x,y)ux+a2(x,y)uy=0.
Assume there is a C1-solution z=u(x,y). This function defines a surface S which has at P=(x,y,u(x,y)) the normal
N=1√1+|∇u|2(−ux,−uy,1)
and the tangential plane defined by
ζ−z=ux(x,y)(ξ−x)+uy(x,y)(η−y).
Set p=ux(x,y), q=uy(x,y) and z=u(x,y). The tuple (x,y,z,p,q) is called surface element and the tuple (x,y,z) support of the surface element. The tangential plane is defined by the surface element. On the other hand, differential equation (???)
a1(x,y)p+a2(x,y)q=0
defines at each support (x,y,z) a bundle of planes if we consider all (p,q) satisfying this equation. For fixed (x,y), this family of planes Π(λ)=Π(λ;x,y) is defined by a one parameter family of ascents p(λ)=p(λ;x,y), q(λ)=q(λ;x,y).
The envelope of these planes is a line since
a1(x,y)p(λ)+a2(x,y)q(λ)=0,
which implies that the normal N(λ) on Π(λ) is perpendicular on (a1,a2,0).
Consider a curve x(τ)=(x(τ),y(τ),z(τ)) on S, let
Tx0 be the tangential plane at x0=(x(τ0),y(τ0),z(τ0)) of S
and consider on Tx0 the line
L: l(σ)=x0+σx′(τ0), σ∈R1,
see Figure 2.1.1.
Figure 2.1.1: Curve on a surface
We assume L coincides with the envelope, which is a line here, of the family of planes Π(λ) at (x,y,z). Assume that Tx0=Π(λ0) and consider two planes
Π(λ0): z−z0=(x−x0)p(λ0)+(y−y0)q(λ0)Π(λ0+h): z−z0=(x−x0)p(λ0+h)+(y−y0)q(λ0+h).
At the intersection l(σ) we have
(x−x0)p(λ0)+(y−y0)q(λ0)=(x−x0)p(λ0+h)+(y−y0)q(λ0+h).
Thus,
x′(τ0)p′(λ0)+y′(τ0)q′(λ0)=0.
From the differential equation
a1(x(τ0),y(τ0))p(λ)+a2(x(τ0),y(τ0))q(λ)=0
it follows
a1p′(λ0)+a2q′(λ0)=0.
Consequently
(x′(τ),y′(τ))=x′(τ)a1(x(τ,y(τ))(a1(x(τ),y(τ)),a2(x(τ),y(τ)),
since τ0 was an arbitrary parameter. Here we assume that x′(τ)≠0 and a1(x(τ),y(τ))≠0.
Then we introduce a new parameter t by the inverse of τ=τ(t), where
t(τ)=∫ττ0x′(s)a1(x(s),y(s)) ds.
It follows x′(t)=a1(x,y), y′(t)=a2(x,y). We denote x(τ(t)) by x(t) again.
Now we consider the initial value problem
x′(t)=a1(x,y), y′(t)=a2(x,y), x(0)=x0, y(0)=y0.
From the theory of ordinary differential equations it follows (Theorem of Picard-Lindelöf) that there is a unique solution in a neighbourhood of t=0 provided the functions a1, a2 are in C1. From this definition of the curves (x(t),y(t)) is follows that
the field of directions (a1(x0,y0),a2(x0,y0)) defines the slope of these curves at (x(0),y(0)).
Definition. The differential equations in (???) are called characteristic equations or characteristic system and solutions of the associated initial value problem are called characteristic curves.
Definition. A function ϕ(x,y) is said to be an integral of the characteristic system if ϕ(x(t),y(t))=const. for each characteristic curve. The constant depends on the characteristic curve considered.
Proposition 2.1. Assume ϕ∈C1 is an integral, then u=ϕ(x,y) is a solution of (???).
Proof. Consider for given (x0,y0) the above initial value problem (???). Since ϕ(x(t),y(t))=const. it follows
ϕxx′+ϕyy′=0
for $ |t|<t_0$, t0>0 and sufficiently small.
Thus
$$
\phi_x(x_0,y_0)a_1(x_0,y_0)+\phi_y(x_0,y_0)a_2(x_0,y_0)=0.
\]
◻
Remark. If ϕ(x,y) is a solution of equation (???) then also H(ϕ(x,y)), where H(s) is a given C1-function.
Examples
Example 2.1.1:
Consider
a1ux+a2uy=0,
where a1, a2 are constants. The system of characteristic equations is
x′=a1, y′=a2.
Thus the characteristic curves are parallel straight lines defined by
x=a1t+A, y=a2t+B,
where A, B are arbitrary constants. From these equations it follows that
ϕ(x,y):=a2x−a1y
is constant along each characteristic curve. Consequently, see Proposition 2.1, u=a2x−a1y is a solution of the differential equation. From an exercise it follows that
u=H(a2x−a1y),
where H(s) is an arbitrary C1-function, is also a solution. Since u is constant when a2x−a1y is constant, equation (???) defines cylinder surfaces which are generated by parallel straight lines which are parallel to the (x,y)-plane, see Figure 2.1.2.
Figure 2.1.2: Cylinder surfaces
Example 2.1.2:
Consider the differential equation
xux+yuy=0.
The characteristic equations are
x′=x, y′=y,
and the characteristic curves are given by
x=Aet, y=Bet,
where A, B are arbitrary constants. Then an integral is y/x, x≠0, and for a given C1-function the function u=H(x/y) is a solution of the differential equation. If y/x=const., then u is constant. Suppose that H′(s)>0, for example, then u defines right helicoids (in German: Wendelflächen), see Figure 2.1.3.
Figure 2.1.3: Right helicoid, a2<x2+y2<R2 (Museo Ideale Leonardo da Vinci, Italy)
Example 2.1.3:
Consider the differential equation
yux−xuy=0.
The associated characteristic system is
x′=y, y′=−x.
If follows
x′x+yy′=0,
or, equivalently,
ddt(x2+y2)=0,
which implies that x2+y2=const. along each characteristic. Thus, rotationally symmetric surfaces defined by u=H(x2+y2), where H′≠0, are solutions of the differential equation.
Example 2.1.4:
The associated characteristic equations to
ayux+bxuy=0,
where a, b are positive constants,
are given by
x′=ay, y′=bx.
It follows bxx′−ayy′=0, or equivalently,
ddt(bx2−ay2)=0.
Solutions of the differential equation are u=H(bx2−ay2), which define surfaces which have a hyperbola as the intersection with planes parallel to the (x,y)-plane. Here H(s) is an arbitrary C1-function, H′(s)≠0.
Contributors and Attributions
Integrated by Justin Marshall.