HYPERBOLIC EQUATIONS

Perhaps the hardest evolutionary PDE problems to approximate are hyperbolic. Reasons for this are beyond scope of the course, but you'll be getting an appreciation of the difficulty in the homework assignments. Why start with the most difficult PDE's? Because they serve as a good venue to illustrate some of the basic numberical-analytical concepts. What we will do is solve some hyperbolic problems and avoid most of the complicated aspects.

In what follows we'll denote $x$ the spatial variable, which can be the whole real line or a closed subinterval of the real line. We will denote time as $t$ and assume $t\ge 0$. The dynamic variable is $U=U(x,t)$

Three arquetypical problems that are Hyperbolic, are

$$\begin {array}{ll} U_t+A(x,U,t)U_x=0 & \mbox {Advection or O... ...re $f$ doesn't depend on $U_x$, $U_{xx}$, etc...} \end{array}$$

Let's take a look at the scalar One-way Wave Equation: Take $U=U(x,t)\in {\mathbb{R}}^1$

The Simple Advection Equation

$$U_t+aU_x=0$$ (121)

$a$, constant

$$U(x,0)=U_0(x)$$

has solution $U_0(x-at)$, that is,

$$U(x,t)=U_0(x-at)$$ (122)

See Figure 19 for an illustration. Note that the wave travels to the right as time increases and the shape is retained.

Figure 19: Graphical representation of the solution to the simple advection or one way wave equation with constant speed $a>0$

What we glean from solution:

1. requires differentiability of $U$, but (123) does not, makes sense $\ldots$ this introduces the concept of ``weak'' solutions (e.g. shocks). Hyperbolic problems admit solutions with discontinuities .
2. See solution is a copy of $U_0$ except displaced by $a t$ (to the right if $a>0$, to the left if $a<0$).
3. Solution only depends on characteristic variable $\xi=x-at$.

Solution to a more general advection equation

$$\left\{\begin{array}{ll} U_t+aU_x+bU=f(t,x) & x\in {\mathbb{R}}^1, t>0 U(0,x)=U_0(x) & \end{array} \right.$$ (123)

$$a, b$$    constants

$$\left\{\begin{array}{ll} \tau=t\\ \xi =x-at \end{array}\right.\Rightarrow \left\{\begin{array}{ll} t=\tau\\ x=\xi+a\tau \end{array}\right.$$

and define $\tilde {U}(t,\xi)=U(t,x)$ (same function in both coordinate systems).

$$\therefore \frac{\partial \tilde{U}}... ...{\partial x}= \frac{\partial U}{dt}+a\frac{\partial U}{dx}=-bU+f(t,\xi +a\tau)$$

$\Rightarrow$ We get an ODE:

$$\frac{\partial \tilde{U}}{\partial \tau}=-bU+f(\tau,\xi +a\tau)$$ which we can solve:

$$\tilde{U}(t,\xi) = U_0(\xi)e^{-b\tau}+\int^{\tau}_0f(\sigma,\xi+a\sigma) e^{-b(\tau-\sigma)}d^\sigma$$

$$\therefore\quad U(t,x) = U_0(x-at)e^{-bt}+\int^t_0f(s,x-a(t-s))e^{-b(t-s)}ds$$

What we learn from this solution:

• Solution is found by a change of coordinate system, or more precisely, a change of reference frame.
• Along special lines in space-time, the function $U$ is an ODE. These lines are called characteristics. In this instance the lines are straight, but they are not in general, if the speed $a$ is not constant in space or time.
• The resulting ODE above shows that if $b>0$ that $\tilde U$ dissipates and grows if $b<0$. Note that if $b$ and $f$ is the derivative of another function that we get a conservation law for $\tilde U$, and that $\tilde U$ is conserved if $f=0$.

Equations with Variable Coefficients Now assume that the speed $a=a(x,t)$. Then

$$\left\{\begin{array}{c} U_t+a(t,x)U_x=0 U(0,x)=U_0(x) \end{array} \right.$$ (124)

then

$$\displaystyle \frac{\partial \tilde{U}}{\partial \tau} = \frac{\partial t}{\partial \tau}U_t+\frac{\partial x} {\partial \tau}U_x=0$$

$$= U_t+aU_x=0$$

$$\therefore\quad \frac{dx}{d\tau}=a(t,x)=a(\tau, x)$$

$\therefore$ (125) is equivalent to

$$\left\{\begin{array}{ll} \displaystyle \frac{d\tilde{u}}{\partial... ...\\ \\ \displaystyle \frac{dx}{d\tau}=a(\tau, x) & x(0)=\xi \end{array}\right.$$

Example)

$$U_t+xU_x=0$$

$$U(0,x)=\left\{\begin{array}{ll} 1 & 0\le x\le 1\\ 0 & \mbox{otherwise} \end{array}\right.$$

$$\Rightarrow \displaystyle \frac{d\tilde{U}}{d\tau}=0, \frac{dx}{d\tau}=x,\quad x(0)=\xi$$

the second of these can be integrated to give $x(t)=ce^{\tau}$
Thus,

$$x(0)=\xi=c\quad \therefore \quad x(\tau)=\xi e^{\tau}$$

$$\therefore \xi=xe^{-t}$$

$$\therefore \tilde{u}=\tilde{u}(\xi)$$

$$\tilde{u}(t,\xi)=u_0(\xi)$$

$$\therefore\quad u(t,x)=\tilde{u}(t,\xi)=u_0(\xi)=u_0(xe^{-t})$$

so we get, for $t>0$

$$U(t,x)= \left\{\begin{array}{ll} 1 & \mbox{if } 0\le x\le e^t\\ 0 & \mbox { otherwise.} \end{array}\right.$$

Quasi-linear (mildly nonlinear) Equations Consider

$$U_t + U U_x =0$$

with

$$U(0,x)\equiv \phi(x) = \left\{\begin{array}{ll} 2 & x < 0 \\ 2-x & 0\le x\le 1\\ 1 & x>1 \end{array}\right.$$

Since the speed $a(u) = U(x,t)$, the characteristics are straight lines emanating from $(\xi,0)$ with speed $a(\phi(\xi))=\phi(\xi)$. For $x<0$ the lines have speed $2$. for $x>1$ the lines have speed $1$. For $x \in [0,1]$ the lines have speed $2-x$ and these all intersect at $x=2$ and $t=1$. Thus, solution cannot exist for $t>1$. Actually, it does in what we call weak form. At $t=1$ we get wave breaking or a shock, i.e. the function no longer is single valued. To find the solution for $t<1$ note $U(x,t)=2$ for $x<2t$ and $U(x,t)=1$ for $x > t+1$. For $2t < x < t+1$ we get

$$x=(2-\xi) t + \xi$$

which in turns gives

$$\xi = \frac{x - 2 t}{1-t}$$

Thus the solution is

$$U(x,t) = \frac{2-x}{1-t}$$

for $2t < x < t+1$, and $t<1$.

Remarks: As we see from the above examples hyperbolic problems propagate signals or information -in the form of waves- with finite speed. An example of such information is the initial data. The direction in which the signal travels depends on the sign of the speed: as posed above, and for $t>0$, the signal will travel at speed $\vert a\vert$ and to the right if $a>0$ (remember that this speed may depend on $x$, $t$, even on $U$), and to the left if $a<0$ (see Figure 19).

Let's consider a system of hyperbolic equations:

$\left(\begin{array}{c} U\\ V \end{array}\right)_{t}+\left(\begin{array}{ll} a... ...\\ b & a\end{array}\right)\left(\begin{array}{l} U\\ V \end{array}\right)_x=0$

$$0\le x\le 1$$

the eigenvalues or eigenspeeds are $a+b, a-b$. Take $a,b>0$, constants.

If $0<b<a\Rightarrow$$\mbox {both characteristics travel to right.}\\ 0<a<b\Rightarrow \mbox {charactertistics travel in opposite direction}$

The situation is portrayed in Figure 20

Figure 20: Signal propagation direction, depending on the size of the eigenspeeds

Physical problems are often posed on a finite span in $x$. Assume this span is of length $l$. The hyperbolic problems considered above are well posed if initial data is specified and appropriate boundary conditions used, and all of these are consistent. Not only is information from the initial data advected but so is boundary data that is to the left (right) and before if $a>0$ (if $a<0$). One of the many difficulties associated with hyperbolic problems is in fact the issue of boundary conditions. Since we are always computing over finite domains, they will always require careful consideration. In what follows of this presentation we will not consider the hard boundary issues....even in your assignments these will be carefully avoided.

By way of example
Consider

$$U_t+aU_x =$$ 0

$$0 \leq x \leq 1 t >0$$

and

take $a>0\left\{\begin{array}{ll} U(0,x)=U_0(x)\\ U(t,0)=g(t) \end{array}\right.$

then solution is $\left\{\begin{array}{ll} U_0(x-at) & x-at>0\\ g(t-a^{-1}x) & x-at<0 \end{array}\right.$

Along $x-at=0$ there'll be a jump in the solution if $u_0(0)\ne g(0)$.

For $a<0$, roles are reversed. See Figure 21 (Convince yourself!)

Figure 21: Data specification for $a>0$ and $a<0$ case

Periodic Boundary Conditions: in this case we prescribe $u(t,x+l)=u(t,x)$, where $l$ is length of strip. These can add strong structure to solution.