# 2.1: Linear Equations

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- 2133

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Let us begin with the linear homogeneous equation

\begin{equation}

\label{homtwo}

a_1(x,y)u_x+a_2(x,y)u_y=0.

\end{equation}

Assume there is a \(C^1\)-solution \(z=u(x,y)\). This function defines a surface \(S\) which has at \(P=(x,y,u(x,y))\) the normal

$$

{\bf N}=\frac{1}{\sqrt{1+|\nabla u|^2}}(-u_x,-u_y,1)

$$

and the tangential plane defined by

$$

\zeta-z=u_x(x,y)(\xi-x)+u_y(x,y)(\eta-y).

$$

Set \(p=u_x(x,y)\), \(q=u_y(x,y)\) 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 (\ref{homtwo})

$$

a_1(x,y)p+a_2(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 \(\Pi(\lambda)=\Pi(\lambda;x,y)\) is defined by a one parameter family of ascents \(p(\lambda)=p(\lambda;x,y)\), \(q(\lambda)=q(\lambda;x,y)\).

The envelope of these planes is a line since

$$

a_1(x,y)p(\lambda)+a_2(x,y)q(\lambda)=0,

$$

which implies that the normal \({\bf N}(\lambda)\) on \(\Pi(\lambda)\) is perpendicular on \((a_1,a_2,0)\).

Consider a curve \({\bf x}(\tau)=(x(\tau),y(\tau),z(\tau))\) on \(\mathcal S\), let

\(T_{{\bf x}_0}\) be the tangential plane at \({\bf x}_0=(x(\tau_0),y(\tau_0),z(\tau_0))\) of \(\mathcal S\)

and consider on \(T_{{\bf x}_0}\) the line

$$

L:\ \ l(\sigma)={\bf x}_0+\sigma{\bf x}'(\tau_0),\ \ \sigma\in\mathbb{R}^1,

$$

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 \(\Pi(\lambda)\) at \((x,y,z)\). Assume that \(T_{{\bf x}_0}=\Pi(\lambda_0)\) and consider two planes

\begin{eqnarray*}

\Pi(\lambda_0):\ \ \ z-z_0&=&(x-x_0)p(\lambda_0)+(y-y_0)q(\lambda_0)\\

\Pi(\lambda_0+h):\ \ \ z-z_0&=&(x-x_0)p(\lambda_0+h)+(y-y_0)q(\lambda_0+h).

\end{eqnarray*}

At the intersection \(l(\sigma)\) we have

$$

(x-x_0)p(\lambda_0)+(y-y_0)q(\lambda_0)=(x-x_0)p(\lambda_0+h)+(y-y_0)q(\lambda_0+h).

$$

Thus,

$$

x'(\tau_0)p'(\lambda_0)+y'(\tau_0)q'(\lambda_0)=0.

$$

From the differential equation

$$

a_1(x(\tau_0),y(\tau_0))p(\lambda)+a_2(x(\tau_0),y(\tau_0))q(\lambda)=0

$$

it follows

$$

a_1p'(\lambda_0)+a_2q'(\lambda_0)=0.

$$

Consequently

$$

(x'(\tau),y'(\tau))=\frac{x'(\tau)}{a_1(x(\tau,y(\tau))}(a_1(x(\tau),y(\tau)),a_2(x(\tau),y(\tau)),

$$

since \(\tau_0\) was an arbitrary parameter. Here we assume that \(x'(\tau)\not=0\) and \(a_1(x(\tau),y(\tau))\not=0\).

Then we introduce a new parameter \(t\) by the inverse of \(\tau=\tau(t)\), where

$$

t(\tau)=\int_{\tau_0}^\tau \frac{x'(s)}{a_1(x(s),y(s))}\ ds.

$$

It follows \(x'(t)=a_1(x,y),\ y'(t)=a_2(x,y)\). We denote \({\bf x}(\tau(t))\) by \({\bf x}(t)\) again.

Now we consider the initial value problem

\begin{equation}

\label{chartwo}

x'(t)=a_1(x,y),\ \ y'(t)=a_2(x,y),\ \ x(0)=x_0,\ y(0)=y_0.

\end{equation}

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 \(a_1,\ a_2\) are in \(C^1\). From this definition of the curves \((x(t),y(t))\) is follows that

the field of directions \((a_1(x_0,y_0),a_2(x_0,y_0))\) defines the slope of these curves at \((x(0),y(0))\).

**Definition.** The differential equations in (\ref{chartwo}) are called *characteristic equations* or characteristic system and solutions of the associated initial value problem are called *characteristic curves*.

**Definition.** A function \(\phi(x,y)\) is said to be an *integral* of the characteristic system if \(\phi(x(t),y(t))=const.\) for each characteristic curve. The constant depends on the characteristic curve considered.

**Proposition 2.1.** *Assume \(\phi\in C^1\) is an integral, then \(u=\phi(x,y)\) is a solution of *(\ref{homtwo})*.*

*Proof.* Consider for given \((x_0,y_0)\) the above initial value problem (\ref{chartwo}). Since \(\phi(x(t),y(t))=const.\) it follows

$$

\phi_xx'+\phi_yy'=0

$$

for $ |t|<t_0$, \(t_0>0\) 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.

$$

\(\Box\)

**Remark.** If \(\phi(x,y)\) is a solution of equation (\ref{homtwo}) then also \(H(\phi(x,y))\), where \(H(s)\) is a given \(C^1\)-function.

### Examples

Example 2.1.1:

Consider

$$

a_1u_x+a_2u_y=0,

$$

where \(a_1,\ a_2\) are constants. The system of characteristic equations is

$$

x'=a_1,\ y'=a_2.

$$

Thus the characteristic curves are parallel straight lines defined by

$$

x=a_1t+A,\ y=a_2t+B,

$$

where \(A,\ B\) are arbitrary constants. From these equations it follows that

$$

\phi(x,y):=a_2x-a_1y

$$

is constant along each characteristic curve. Consequently, see Proposition 2.1, \(u=a_2x-a_1y\) is a solution of the differential equation. From an exercise it follows that

\begin{equation}

\label{cyl}

u=H(a_2x-a_1y),

\end{equation}

where \(H(s)\) is an arbitrary \(C^1\)-function, is also a solution. Since \(u\) is constant when \(a_2x-a_1y\) is constant, equation (\ref{cyl}) 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

$$

xu_x+yu_y=0.

$$

The characteristic equations are

$$

x'=x,\ y'=y,

$$

and the characteristic curves are given by

$$

x=Ae^t,\ y=Be^t,

$$

where \(A,\ B\) are arbitrary constants. Then an integral is \(y/x\), \(x\not=0\), and for a given \(C^1\)-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, \(a^2<x^2+y^2<R^2\) (Museo Ideale Leonardo da Vinci, Italy)

Example 2.1.3:

Consider the differential equation

$$

yu_x-xu_y=0.

$$

The associated characteristic system is

$$

x'=y,\ y'=-x.

$$

If follows

$$

x'x+yy'=0,

$$

or, equivalently,

$$

\frac{d}{dt}(x^2+y^2)=0,

$$

which implies that \(x^2+y^2=const.\) along each characteristic. Thus, rotationally symmetric surfaces defined by \(u=H(x^2+y^2)\), where \(H'\not=0\), are solutions of the differential equation.

Example 2.1.4:

The associated characteristic equations to

$$

ayu_x+bxu_y=0,

$$

where \(a,\ b\) are positive constants,

are given by

$$

x'=ay,\ y'=bx.

$$

It follows \(bxx'-ayy'=0\), or equivalently,

$$

\frac{d}{dt}(bx^2-ay^2)=0.

$$

Solutions of the differential equation are \(u=H(bx^2-ay^2)\), which define surfaces which have a hyperbola as the intersection with planes parallel to the \((x,y)\)-plane. Here \(H(s)\) is an arbitrary \(C^1\)-function, \(H'(s)\not=0\).