09 - Cauchy Data and Discontinuities of First Order Semi-Linear PDEs

Cauchy Problem

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

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

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