$\newcommand{\const}{\mathrm{const}}$

Consider PDE
\begin{equation}
au_t+bu_x=0.
\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}.
\end{equation}
*Remark.*
Recall from ODE cours that an *integral line* of the vector field is a line, tangent to it in each point.

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) \end{equation} wher $\phi$ 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 $\phi$: $\phi(x)=f(x)$. Plugging in $u$ we get \begin{equation} u=f\bigl( \frac{t}{a}-\frac{x}{b}\bigr). \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. \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 $\phi_1(x)=\phi(-xb^{-})$ is another arbitrary function.

*Definition*
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.*
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.*
Consider the same equation but let us consider IVP as $x=0$: $u(0,t)=g(tx)$. 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.

Consider equation \begin{equation} au_t+bu_x=f. \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 to \begin{equation} \frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}. \end{equation}

*Definition*
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* 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.* Consider
$u_t+ xu_x = x t$. Then $\frac{dt}{1}=\frac{dx}{x}=\frac{du}{tu}$. 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 +C_1 =\\
x(t-1)+C_1\end{multline*}
where $C_1$ must be constant along integral curves and therefore $C_1=\phi (xe^{-t})$ 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.* 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.* 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.

Consider IVP 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. \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.

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.
\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. \end{equation}

*Remark.*
Nonlinear equation
\begin{equation}
F(x_1,x_2,u,u_{x_1},u_{x_2})=0
\end{equation}
(we prefer such notations here) also could be solved through ODEs but it is much more complicated: one needs simultaneously find $x,u$ and $p_j=u_{x_j}$ along trajectories from the system of equations:
\begin{equation}
\left\{\begin{aligned}
&\frac{dx_j}{dt}=F_{p_j}, \\
&\frac{dp_j}{dt}=-F_{x_j}-F_u p_j ,\\
&\frac{du}{dt}=\sum_{j=1}^n F_{p_j}p_j
\end{aligned}\right.
\end{equation}
where in the right hand expressions we consider as function of $2n+1$ variables
$x_1,\ldots,x_n, u, p_1,\ldots, p_n$, $n=2$.

More

*Remark.*
Multidimensional equations (from linear to semilinear)
\begin{equation}
au_t + \sum_{j=1}^n b_j u_{x_j}=f(x_1,\ldots,x_n,t,u)
\end{equation}
and nonlinear
\begin{equation}
F(x_1,\ldots,x_n,t,u,u_{x_1},\ldots,u_{x_n},u_t)=0
\end{equation}
could be solved by the same methods.

For example, if $a=1$, $b_j=\const$ and $f=0$ the general solution is $u=\phi (x_1-b_1t,\ldots,x_n-b_nt)$ where $\phi$ is an arbitrary function of $n$ variables.