03. First Order Semi-Linear PDEs - Method of Characteristics

3.1 The Problem

First Order Semi-Linear PDE

$$P(x,y)\frac{\partial z}{\partial x}+Q(x,y)\frac{\partial z}{\partial y}=R(x,y,z(x,y))$$

for some unknown function $z=z(x,y)$

Generally assumed

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

Note that PDE is quasi-linear if $P,Q$ depend on $z$

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Domain of Definition

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

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Solution Surface

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

Solution surface is

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

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3.2 The big idea: characteristics

Characteristic Equations

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

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

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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

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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

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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 $t$

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3.2.1 Examples of Characteristics

Refer to page $52$ of lecture notes

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3.3 The Cauchy Problem

Method of Characteristics

We have First Order Semi-Linear PDE with boundary data

Suppose we are seeking solution $z$ along curve ${\gamma }_0=({\gamma }_1,{\gamma }_2)$ (data curve) in the $(x,y)$-plane

So we have $z(x,y)=g(x,y)$ for some function $g$ for all points on data curve ${\gamma }_0$
By setting ${\gamma }_3=g({\gamma }_1,{\gamma }_2)$ we have curve $\gamma =({\gamma }_1,{\gamma }_2,{\gamma }_3)$ which is called the initial curve
where the initial curve $\gamma$ must be in the solution surface $\Sigma$

So we have the following steps (method of characteristics)

1. Parametrise $\gamma$ over some interval $I$

$$\gamma (s)=({\gamma }_1(s),{\gamma }_2(s),{\gamma }_3(s))\text{ for }s\in I$$

Assume that each component of $\gamma$ is continuously differentiable

2. Determine the solutions $(x(t,s),y(t,s),z(t,s))$ of characteristic equations

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

with initial data $x(0,s)={\gamma }_1(s)$, $y(0,s)={\gamma }_2(s)$, $z(0,s)={\gamma }_3(s)$ for $s\in I$

3. This gives us solution surface $\Sigma$ in parametric form

$$\Sigma =\{(x(t,s),y(t,s),z(t,s))\}$$

for all $s\in I$
For each $s$, consider the maximal set of $t$‘s which solves all three characteristic equations

4. Eliminate the parameters $s,t$ and write $\Sigma$ as a graph to read off the solution
   Note that there is a restriction on the data for method to work

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3.4 Examples

Refer to page $54$ of lecture notes

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3.5 Domain of Definition

Finding the Domain of Definition

If the initial curve is bounded then

The domain of definition is usually bounded by the projections of the characteristics through end points of the initial curve

See lecture notes page $55$ for an example

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3.6 Cauchy Data

Cauchy Problem

Combination of PDE with boundary data to give a unique solution at least locally

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Cauchy Data

We have First Order Semi-Linear PDE with boundary data then

The data is Cauchy Data if

$$P(x,y)y_s-Q(x,y)x_s\ne 0\text{ on the data curve}$$

In other words

$$P({\gamma }_1(s),{\gamma }_2(s)){\gamma }_2^{\prime }(s)-Q({\gamma }_1(s),{\gamma }_2(s)){\gamma }_1^{\prime }(s)\ne 0$$

for all $s$ in the interval over which we parametrise the data curve ${\gamma }_0$

Note that if you reach a point where it is equal $0$ then you can split the data into two different cases

Proof

We require that $J(s,0)\ne 0$ on initial curve ${\gamma }_0$ to get a unique solution close
As

$$J(s,0)=\det \left(\begin{array}{cc}{x}_t& y_t\\ x_s& y_s\end{array}\right)=\det \left(\begin{array}{cc}P& Q\\ x_s& y_s\end{array}\right)=P{y}_s-Q{x}_s$$

then we just evaluate it at points $(x(s,0),y(s,0))$ on data curve ${\gamma }_0=({\gamma }_1(s),{\gamma }_2(s))$

Geometric Interpretation

Data is Cauchy if

• tangent vector ${\gamma }_0^{\prime }(s)=(x_s(s,0),y_s(s,0))$ along the data curve

and

• tangent vector $(P,Q)=(\dot{x},\dot{y})=(x_t(s,0),y_{t(s,0)})$ along the characteristic projection

are never parallel through the same point

Extreme Case

Data fails to be Cauchy if the data curve is a characteristic projection

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3.7 Discontinuities in the First Derivatives

Discontinuities in the First Derivatives in Characteristic Projections

The only curves that can have discontinuity in the $xy$-plane is the characteristic projections

Proof

Suppose that $z(x,y)$ is a solution of PDE which is

1. continuously differentiable away from curve ${\alpha }_0=({\alpha }_1,{\alpha }_2)$ in the $xy-$plane
2. continuous across ${\alpha }_0$ but discontinuous in first order partial derivatives as we cross ${\alpha }_0$

Use subscript $\pm$ to denote solution on either side of $\gamma$
Denote jumps in partial derivative by

$$\begin{array}{rl}[z_x{]}_{-}^{+}& =z_x^{+}-z_x^{-}\\ [z_y{]}_{-}^{+}& =z_y^{+}-z_y^{-}\end{array}$$

As $z$ is continuous across ${\alpha }_0$ then

$$z^{+}({\alpha }_1(s),{\alpha }_2(s))-z^{-}({\alpha }_1(s),{\alpha }_2(s))=0$$

By differentiating then

$$[z_x{]}_{-}^{+}{\alpha }_1^{\prime }+[z_y{]}_{-}^{+}{\alpha }_2^{\prime }=0$$

But we also have $z^{+}$ and $z^{-}$ as solutions to PDE so

$$\begin{array}{rl}P{z}_x^{+}+Q{z}_y^{+}& =R(x,y,z^{+})\\ P{z}_x^{-}+Q{z}_y^{-}& =R(x,y,z^{-})\end{array}$$

As $z^{+}=z^{-}$ on $\alpha$ then the right hand sides agree , so by subtraction we get

$$0=P[z_x{]}_{-}^{+}+Q[z_y{]}_{-}^{+}\text{ on }{\alpha }_0$$

So vector $j=([z_x{]}_{-}^{+},[z_y{]}_{-}^{+})$ of the jumps in first derivatives solves homogenous system of

$$\left(\begin{array}{cc}P& Q\\ {\alpha }_1^{\prime }& {\alpha }_2^{\prime }\end{array}\right)\cdot j=0$$

So for there to be a non-zero jump then matrix must be singular

As the first row is tangent to a characteristic projection
and the second row is the tangent to curve ${\alpha }_0$ across which the derivatives jump then

$${\alpha }_0\text{ is a characteristic projection}$$

Hence the only curves in $xy-$plane that jump must be characteristic projections

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3.8 General Solution

General Solution of of an PDE

Expect the most general solution of a $n$-order ODE is to have $n$-arbitrary constants
Hence we expect the most general solution of a PDE of order $n$ to have $n$ arbitrary functions

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