07 - Characteristics and Solution Geometry of First-Order Semi-Linear PDEs

Domain of Definition

Region in $(x,y)$-plane in which the soluition is uniquely defined by the data

Solution Surface

Let $z=f(x,y)$ be the solution of the First Order Semi-Linear PDE then

Solution surface is

$$\Sigma :=\{(x,y,z):z=f(x,y)\}=\mathrm{graph}(f)$$

Characteristic Equations

Using the First Order Semi-Linear PDE form the Characteristic Equations are

$$\begin{array}{rl}\dot{x}& =P(x,y)\\ \dot{y}& =Q(x,y)\\ \dot{z}& =R(x,y,z)\end{array}$$

Characteristic Curve / Characteristic

Let $\Gamma$ be a curve with tangent $\mathbf{t}$

If $\Gamma =(x(t),y(t),z(t))$ in terms of parameter $t$ where $x,y,z$ satisfy the Characteristic Equations

Characteristic Projection / Characteristic Trace

Given Characteristic $(x(t),y(t),z(t))$ then the curve

$$(x(t),y(t))$$

which lies in the $(x,y)-$plane is the characteristic projection or characteristic trace

Properties of Characteristics

Using the First Order Semi-Linear PDE

Suppose

1. $P(x,y)$ and $Q(x,y)$ are Lipschitz continuous in $x$ and $y$
2. $R(x,y,z)$ is continuous and satisfies a Lipschitz condition in $z$

Then

1. Through every point $(x_0,y_0)\in {\mathbb{R}}^2$ there is a unique characteristic projection
2. Through every $(x_0,y_0,z_0)\in {\mathbb{R}}^3$ there is a unique characteristic
3. If

• $1)$ $f$ is a solution of the First Order Semi-Linear PDE
• $2)$ $\Gamma$ is a characteristic through a point $p$ contained in solution surface $\Sigma =graph(f)$

Then we have that the whole characteristic is contained in $\Sigma$

Note that characteristic projections can never intersect (only for semilinear equations)

Statements $(2)$ and $(3)$ apply for quasilinear PDEs (provided the Lipschitz Conditions)

Proof

1. Since $P$ and $Q$ don’t depend on $z$ then the first two equations

$$\dot{x}=P(x,y),\dot{y}=Q(x,y)$$

is a Plane Autonomous System so it has a unique trajectory by Picard’s Theorem
Hence trajectory is simply the characteristics projections we obtain

2. By $(1)$ we have unique solution $(x(t),y(x))$ for $t$ in some interval $I_1$ to

$$\dot{x}=P(x,y),\dot{y}=Q(x,y)$$

with initial conditions $x(0)=x_0$ and $y(0)=y_0$

So given $(x(t),y(t))$ we can set $F(t,z):=R(x(t),y(t),z)$ so we get ODE

$$\dot{z}(t)=F(t,z(t))$$

So as $R$ satisfies a Lipschitz condition in $z$ then we can use Picard’s Theorem
Hence we get a unique solution $z(t)$ with $z(0)=z_0$
Thus we find a unique characteristic through $(x_0,y_0,z_0)$

3. Let $f$ be a solution of the PDE
   Let $\Sigma =\mathrm{graph}(f)=\{(x,y,f(x,y))\}$ be corresponding solution surface

Suppose $x(t),y(t),z(t)$ is a characteristic through a point $p\in \Sigma$ then
By shifting time we can assume that $p=(x,y,z)(0)$ hence $z(0)=f(x(0),y(0))$
To prove that the whole curve stays in $\Sigma$ then we need that

$$w(t):=z(t)-f(x(t),y(t))$$

remains equal to $0$

Differentiate $w(t)$ and use characteristic equations then

$$\begin{array}{rl}\dot{w}& =\dot{z}-\left({\partial }_{x}f(x,y)\dot{x}+{\partial }_{y}f(x,y)\dot{y}\right)=R(x,y,z)-(P(x,y){\partial }_{x}f(x,y)+Q(x,y){\partial }_{y}f(x,y))\\ & =R(x,y,z)-R(x,y,f(x,y))\end{array}$$

As $w(0)=0$ then we get

$$|w(t)=\left|{\int }_0^t\dot{w}(s)ds\right|\le \left|{\int }_0^t|R(x,y,z)-R(x,y,f(x,y))|ds\right|\le L\left|{\int }_0^t|w(s)|ds\right|$$

using the Lipschitz condition $|R(x,y,z)-R(x,y,\tilde{z})|\le L|z-\tilde{z}|$
Then applying Gronwall’s Inequality we get that

$$|w(t)|\le 0\cdot e^{L|t|}$$

Hence $w(t)=0$ for all $t$ so $z(t)=f(x(t),y(t))$ for all $t$