$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

In this Chapter we first consider first order PDE and then move to $1$-dimensional wave equation which we analyze by the *method of characteristics*.

Consider PDE
\begin{equation} au_t+bu_x=0.
\label{eq-2.1.1}
\end{equation}
Note that the left-hand expression is a *derivative of $u$ along vector field $\ell=(a,b)$*. Consider an *integral lines* of this vector field:
\begin{equation}
\frac{dt}{a}=\frac{dx}{b}.
\label{eq-2.1.2}
\end{equation}

**Remark 1.**

- Recall from ODE course that an
*integral line*of the vector field is a line, tangent to it in each point. - Often it is called
*directional derivative*but also often then $\ell$ is*normalized*, replaced by the unit vector of the same direction $\ell^0=\ell/|\ell|$.

If $a$ and $b$ are constant then integral curves are just straight lines $t/a -x/b=C$ where $C$ is a constant along integral curves and it labels them (at least as long as we consider the whole plane $(x,t)$). Therefore $u$ depends only on $C$: \begin{equation} u= \phi \bigl( \frac{t}{a}-\frac{x}{b}\bigr) \label{eq-2.1.3} \end{equation} where $\phi$ is an arbitrary function.

This is a *general solution* of our equation.

Consider initial value condition $u|_{t=0}=f(x)$. It allows us define $\phi$: $\phi(-x/b)=f(x)\implies \phi (x)= f(-bx)$. Plugging in $u$ we get \begin{equation} u=f\bigl( x-ct\bigr)\qquad\text{with } c=b/a. \label{eq-2.1.4} \end{equation} It is a solution of IVP \begin{equation} \left\{\begin{aligned} &au_t+bu_x=0,\\ &u(x,0)=f(x). \end{aligned} \right. \label{eq-2.1.5} \end{equation} Obviously we need to assume that $a\ne 0$.

Also we can rewrite general solution in the form $u(x,t)=f(x-ct)$ where now $f(x)$ is another arbitrary function.

**Definition 1.**
Solutions $u=\chi(x-ct)$ are *running waves* where $c$ is a *propagation speed*.

If $a$ and/or $b$ are not constant these integral lines are curves.

**Example 1.**
Consider equation $u_t+tu_x=0$. Then equation of the integral curve is $\frac{dt}{1}=\frac{dx}{t}$ or equivalently $tdt-dx=0$ which solves as $x-\frac{1}{2}t^2=C$ and therefore $u=\phi (x-\frac{1}{2}t^2)$ is a general solution to this equation.

One can see easily that $u=f(x-\frac{1}{2}t^2)$ is a solution of IVP.

**Example 2.**

- Consider the same equation but let us consider IVP as $x=0$: $u(0,t)=g(t)$. However it is not a good problem: first, some integral curves intersect line $x=0$ more than once and if in different points of intersection of the same curve initial values are different we get a contradiction (therefore problem is not solvable for $g$ which are not even functions).
- On the other hand, if we consider even function $g$ (or equivalently impose initial condition only for $t>0$) then $u$ is not defined on the curves which are not intersecting $x=0$ (which means that $u$ is not defined for $x> \frac{1}{2}t^2$).

In [1] of this example both solvability and uniqueness are broken;
in [2] only uniqueness is broken. But each integral line intersects $\{(x,t)\colon t=0\}$ exactly once, so IVP of Example 1 is *well-posed*.

Consider the same equation albeit with the right-hand expression \begin{equation} au_t+bu_x=f, \qquad f=f(x,t,u). \label{eq-2.1.6} \end{equation} Then as $\frac{dt}{a}=\frac{dx}{b}$ we have $du = u_t dt + u_xdx = (au_t+bu_x) \frac{dt}{a}=f \frac{dt}{a}$ and therefore we expand our ordinary equation (\ref{eq-2.1.2}) to \begin{equation} \frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}. \label{eq-2.1.7} \end{equation}

**Example 3.** Consider problem $u_t+u_x=x$. Then
$\frac{dx}{1}=\frac{dt}{1}=\frac{du}{x}$. Then $x-t=C$ and $u-\frac{1}{2}x^2=D$ and we get $u-\frac{1}{2}x^2 = \phi (x-t)$ as relation between $C$ and $D$ both of which are constants along integral curves. Here $\phi$ is an arbitrary function. So $u=\frac{1}{2}x^2 + \phi (x-t)$ is a general solution. Imposing initial condition $u|_{t=0}=0$ (sure, we could impose another condition) we have $\phi(x)=-\frac{1}{2}x^2$ and plugging into $u$ we get
$u(x,t)=\frac{1}{2}x^2-\frac{1}{2}(x-t)^2= xt - \frac{1}{2}t^2$.

**Example 4.** Consider $u_t+ xu_x = x t$. Then $\frac{dt}{1}=\frac{dx}{x}=\frac{du}{xt}$. Solving the first equation
$t-\ln(x)=-\ln(C)\implies x =Ce^t$ we get integral curves.
Now we have
\begin{multline*}
\frac{du}{xt}=dt \implies du= x t dt= Cte^t dt \\
\implies u=C(t-1)e^t +D =
x(t-1)+D
\end{multline*}
where $D$ must be constant along integral curves and therefore
$D=\phi (xe^{-t})$ with an arbitrary function $\phi$. So
$u=x(t-1)+\phi (xe^{-t})$ is a general solution of this equation.

Imposing initial condition $u|_{t=0}=0$ (sure, we could impose another condition) we have $\phi(x)=x$ and then $u=x(t-1 +e^{-t})$.

**Definition 2.**

If $a=a(x,t)$ and $b=b(x,t)$ equation is

*semilinear*.

In this case we first define integral curves which do not depend on $u$ and then find $u$ as a solution of ODE along these curves.Furthermore if $f$ is a linear function of $u$: $f=c(x,t)u + g(x,t)$ original equation is

*linear*.

In this case the last ODE is also linear.

**Example 5.** Consider $u_t+ xu_x = u$. Then $\frac{dt}{1}=\frac{dx}{x}=\frac{du}{u}$. Solving the first
equation $t-\ln(x)=-\ln(C)\implies x =Ce^t$ we get integral curves.
Now we have
\begin{equation*}
\frac{du}{u}=dt \implies \ln(u)= t+\ln(D) \implies u=De^t=\phi (xe^{-t})e^t
\end{equation*}
which is a general solution of this equation.

Imposing initial condition $u|_{t=0}=x^2$ (sure, we could impose another condition) we have $\phi(x)= x^2$ and then $u=x^2 e^{-t}$.

**Example 6.** Consider $u_t+ xu_x = -u^2$. Then $\frac{dt}{1}=\frac{dx}{x}=-\frac{du}{u^2}$. Solving the first equation $x =Ce^t$ we get integral curves.
Now we have
\begin{equation*}
-\frac{du}{u^2}=dt \implies u^{-1}= t+ D \implies u=(t+ \phi(xe^{-t}))^{-1}.
\end{equation*}
which is a general solution of this equation.

Imposing initial condition $u|_{t=0}=-1$ we get $\phi = -1$ and then $u= (t-1)^{-1}$. This solution ``blows up" as $t=1$ (no surprise, we got it, solving *non-linear* ODE.

**Definition 3.** If $a$ and/or $b$ depend on $u$ this is *quasininear* equation.

For such equations integral curves depend on the solution which can lead to breaking of solution. Indeed, while equations (\ref{eq-2.1.7}) define curves in 3-dimensional space of $(t,x,u)$, which do not intersect, their projections on $(t,x)$-plane can intersect and then $u$ becomes a multivalued function, which is not allowed.

**Example 7.**
Consider Burgers equation $u_t+uu_x=0$ (which is an extremely simplified model of gas dynamics). We have $\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}$ and therefore $u=\const$ along integral curves and therefore integral curves are $x-ut=C$.

Consider initial problem $u(x,0)=g(x)$. We take initial point $(y,0)$, find here $u=g(y)$, then $x-g(y)t =y$ (because $x=y+ut$ and $u$ is constant along integral curves) and we get $u=g(y)$ where $y=y(x,t)$ is a solution of equation $x=g(y)t +y$.

The trouble is that we can define $y$ for all $x$ by *implicit function theorem* only if
$\frac{\partial }{\partial y}\bigl(y+ g(y)t\bigr)$ does not vanish. So,
\begin{equation}
g'(y)t +1\ne 0.
\label{eq-2.1.8}
\end{equation}

This is possible for all $t>0$ if and only if $g'(y)\ge 0$ i.e. $f$ is a monotone non-decreasing function.

Equivalently, $u$ is defined as an implicit function by \begin{equation} u=g(x-ut). \label{eq-2.1.9} \end{equation} Then the implicit function theorem could be applied iff $\frac{\partial }{\partial u} \bigl(u-g(x-ut)\bigr)= 1+ g(x-ut)t\ne 0$, which leads us to the previous conclusion.

Integral lines for Burgers equation if $f(x)=\tanh(x)$.

Integral lines for Burgers equation if $f(x)=-\tanh(x)$.

So, classical solution breaks for some $t>0$ if $g$ is not a monotone non-decreasing function. A proper understanding of the *global solution* for such equation goes well beyond our course. Some insight is provided by the analysis in Section 12.1.

**Example 8.**
Traffic flow is considered in Appendix 2.1.A

Consider IBVP (initial-boundary value problem) for constant coefficient equation \begin{equation} \left\{\begin{aligned} &u_t +cu_x=0, \qquad &&x>0,\ t>0,\\ &u|_{t=0}= f(x) \qquad &&x>0. \end{aligned}\right. \label{eq-2.1.10} \end{equation}

The general solution is $u=\phi(x-ct)$ and plugging into initial data we get $\phi(x)=f(x)$ (as $x>0$).

So, $u(x,t)= f(x-ct)$. Done! – Not so fast. $f$ is defined only for $x >0$ so $u$ is defined for $x-ct >0$ (or $x >ct$). It covers the whole quadrant if $c\le 0$ (so waves run to the left) and only in this case we are done.

On the other hand, if $c>0$ (waves run to the right) $u$ is not defined as $x< ct$
and to define it here we need a *boundary condition* at $x=0$. So we
get IBVP (initial-boundary value problem)
\begin{equation}
\left\{\begin{aligned}
&u_t +cu_x=0, \qquad &&x>0, t>0,\\
&u|_{t=0}= f(x) \qquad &&x>0,\\
&u|_{x=0}=g(t) \qquad &&t >0.
\end{aligned}\right.
\label{eq-2.1.11}
\end{equation}
Then we get $\phi(-ct)=g(t)$ as $t>0$ which implies $\phi(x)=g(-\frac{1}{c}x)$ as $x<0$ and then $u(x,t)=g(-\frac{1}{c}(x-ct))=g(t-\frac{1}{c}x)$ as $x< ct$.

So solution is \begin{equation} u=\left\{\begin{aligned} &f(x-ct)\qquad &&x> c t,\\ &g(t-\frac{1}{c}x)\qquad && x < ct. \end{aligned}\right. \label{eq-2.1.12} \end{equation}

**Remark 2.**
Unless $f(0)=g(0)$ this solution is discontinuous as $x=ct$. Therefore it is not a classical solution--derivatives do not exist. However it is still a *weak solution* (see Section 11.4 and *solution in the sense of distributions* (see Section 11.1).

$\Leftarrow$ $\Uparrow$ $\Downarrow$ $\downarrow$ $\Rightarrow$