2.1. First order PDEs


# Chapter 2. $1$-dimensional waves

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

### Introduction

Consider PDE $$au_t+bu_x=0. \label{eq-2.1.1}$$ 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: $$\frac{dt}{a}=\frac{dx}{b}. \label{eq-2.1.2}$$

Remark 1.

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

### Constant coefficients

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 $C$: $$u= \phi \bigl( \frac{t}{a}-\frac{x}{b}\bigr) \label{eq-2.1.3}$$ 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 $$u=f\bigl( x-ct\bigr)\qquad\text{with } c=b/a. \label{eq-2.1.4}$$ It is a solution of IVP \left\{\begin{aligned} &au_t+bu_x=0,\\ &u(x,0)=f(x). \end{aligned} \right. \label{eq-2.1.5} 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.

Visual examples (animation)

### Variable coefficients

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.

1. 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).
2. 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.

### Right-hand expression

Consider the same equation albeit with the right-hand expression $$au_t+bu_x=f, \qquad f=f(x,t,u). \label{eq-2.1.6}$$ 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 $$\frac{dt}{a}=\frac{dx}{b}=\frac{du}{f}. \label{eq-2.1.7}$$

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 $u=x(t-1 +e^{-t})$.

### Linear and semilinear equations

Definition 2.

1. 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.

2. 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.

### Quasilinear equations

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, $$g'(y)t +1\ne 0. \label{eq-2.1.8}$$

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 $$u=g(x-ut). \label{eq-2.1.9}$$ 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

### IBVP

Consider IBVP (initial-boundary value problem) for constant coefficient 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}

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) \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} 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 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}

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).