$\renewcommand{\Re}{\operatorname{Re}}$
$\renewcommand{\Im}{\operatorname{Im}}$
$\newcommand{\erf}{\operatorname{erf}}$
$\newcommand{\dag}{\dagger}$
$\newcommand{\const}{\mathrm{const}}$
$\newcommand{\arcsinh}{\operatorname{arcsinh}}$
#[Chapter 2. $1$-dimensional waves](id:chapter-2)
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*.
##[2.1. First order PDEs](id:sect-2.1)
-----
> 1. [Introduction](#sect-2.1.1)
> 2. [Constant coefficients](#sect-2.1.2)
> 3. [Variable coefficients](#sect-2.1.3)
> 4. [Right-hand expression](#sect-2.1.4)
> 5. [Linear and semilinear equations](#sect-2.1.5)
> 6. [Quasilinear equations](#sect-2.1.6)
> 7. [IBVP](#sect-2.1.7)
> 8. [Problems](./S2.1.P.html)
###[Introduction](id:sect-2.1.1)
Consider PDE
\begin{equation} au\_t+bu\_x=0.
\label{eq-2.1.1}
\end{equation}
Note that the left-hand expression is a *directional derivative* of $u$ in the
direction $\ell=(a,b)$. 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.](id:remark-2.1.1)**
Recall from ODE cours that an *integral line* of the vector field is a line, tangent to it in each point.
###[Constant coefficients](id:sect-2.1.2)
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 $a\ne 0$.
If $a=1$ we can rewrite general solution in the form $u(x,t)=\phi\_1 (x-bt)$ where $\phi\_1(x)=\phi(-x/b)$ is another arbitrary function.
**[Definition 1.](id:def-2.1.1)**
Solutions $u=\chi(x-ct)$ are *running waves* where $c$ is a *propagation speed*.
###[Variable coefficients](id:sect-2.1.3)
If $a$ and/or $b$ are not constant these integral lines are curves.
**[Example 1.](id:example-2.1.1)**
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.](id:example-2.1.2)** 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 this example both *solvability* and *unicity* are broken.
###[Right-hand expression](id:sect-2.1.4)
Consider the same equation albeit with the right-hand expression
\begin{equation}
au\_t+bu\_x=f.
\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.](id:example-2.1.3)** 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 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.](id:example-2.1.4)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{equation\*}
\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{equation\*}
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 $u=x(t-1 +e^{-t})$.
###[Linear and semilinear equations](id:sect-2.1.5)
**[Definition 2.](id:definition-2.1.2)** 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.
**[Definition 3.](id:definition-2.1.3)**
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.](id:example-2.1.5)** 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.](id:example-2.1.6)** 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.
###[Quasilinear equations](id:sect-2.1.6)
**[Definition 4.](id:definition-2.1.4)** 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.
**[Example 7.](id:example-2.1.7)**
Consider Hopf 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 $u(x,0)=f(x)$. We take initial point $(y,0)$, find here $u=f(y)$, then $x-f(y)t =y$ (think why?) and we get $u=f(y)$ where $y=y(x,t)$ is a solution of equation $x=f(y)t +y$.
The trouble is that we can define $y$ for all $x$ only if
$\frac{\partial }{\partial y}\bigl(f(y)t +y\bigr)$ does not vanish. So,
$f'(y)t +1\ne 0$.
This is possible for all $t\>0$ if and only if $f'(y)\ge 0$ i.e. $f$ is a monotone non-decreasing function.
So, classical solution breaks if $f$ is not a monotone non-decreasing function. A proper understanding of the *global solution* for such equation goes well beyond our course.
**[Example 8.](id:example-2.1.8)**
Traffic flow is considered in [Appendix](./S2.1.A.html)
###[IBVP](id:sect-2.1.7)
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-2.1.8}
\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!â€“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.
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-2.1.9}
\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-2.1.10}
\end{equation}
______
[$\Leftarrow$](../Chapter1/S1.X.html) [$\Uparrow$](../contents.html) [$\Downarrow$](./S2.1.A.html) [$\downarrow$](./S2.1.P.html) [$\Rightarrow$](./S2.2.html)