Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 1.2

1. Solve the initial value problem $a(u)u_x+u_y=0$ with $u(x,0)=h(x)$, and show the solution becomes singular for some $y>0$ unless $a(h(s))$ is a nondecreasing function of $s$.
Solution
Consider the characteristic curve $x=x(t)$, $y=y(t)$, $z=z(t)=u(x(t),y(t))$. We have the following ordinary differential equation
$\displaystyle\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=a(z),\\&\frac{\mathrm dy}{\mathrm dt}=1,\\&\frac{\mathrm dz}{\mathrm dt}=0,\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned} &x(0)=x_0,\\&y(0)=0,\\&z(0)=u(x(0),y(0))=h(x_0).\end{aligned}\right.$
We can solve these equations to obtain
$\left\{\begin{aligned} &x(t)=x_0+a(h(x_0))t,\\&y(t)=t,\\&z(t)=h(x_0).\end{aligned}\right.$
By $z(t)=u(x(t),y(t))$, we have
$u(x_0+a(h(x_0))t,t)=h(x_0)$.
By taking $t=y$, $x_0=x-a(u(x,y))y$, we can deduce that
$u(x,y)=h(x-a(u(x,y))y)$.
Now, for $x_1\neq x_0$, we consider another characteristic curve $x=x(s)$, $y=y(s)$, $z(s)=u(x(s),y(s))$. Then, we can obtain
$\left\{\begin{aligned}&x(s)=x_1+a(h(x_1))s,\\&y(s)=s,\\&z(s)=h(x_1).\end{aligned}\right.$
Without loss of generality, we may assume that $x_1>x_0$.
• Suppose $a\circ h$ is not a nondecreasing function of. Then there exists $s_1>s_2$ such that $a(h(s_1))<a(h(s_2))$, which implies $h(s_1)\neq h(s_2)$. Thus, we choose $x_0=s_2$, $x_1=s_1$, and

  $\displaystyle t=s=y^*:=\frac{s_1-s_2}{a\left(h\left(s_2\right)\right)-a\left(h\left(s_1\right)\right)}>0$,

  which implies that $\left(x\left(t\right),y\left(t\right)\right)=\left(x\left(s\right),y\left(s\right)\right)$, and hence

  $h\left(s_2\right)=z\left(t\right)=u\left(x\left(t\right),y\left(t\right)\right)=u\left(x\left(s\right),y\left(s\right)\right)=z\left(s\right)=h\left(s_1\right)$,

  which leads a singularity.
• Suppose $a\circ h$ is a nondecreasing functinon. When $t=s$, we can observe that

  $x\left(s\right)=x_1+a\left(h\left(x_1\right)\right)s>x_0+a\left(h\left(x_0\right)\right)t=x\left(t\right)$,

  which means that two characteristic curve cannot have any intersect points unless they are same. Hence the solution has no singularity.


2. Consider (17) with the intial condition

   $u\left(x,0\right)=h\left(x\right)=\begin{cases}u_0&\text{for }x\leq0\\u_0\left(1-x\right)&\text{for }0<x<1\\0&\text{for }x\geq1,\end{cases}$

   where $u_0>0$. Show that a shock develops at a finite time and describe the weak solution.
Solution
It is well known that the solution has the implicit form $u\left(x,y\right)=h\left(x-u\left(x,y\right)y\right)$ and the family of the characteristic curve passing $\left(x,y\right)=\left(s,0\right)$ is $x=h\left(s\right)y$.
• If $s\geq1$, then the characteristic curve is a vertical line $x=s$.
• If $s\leq0$, then the characteristic curve is $x=u_0y+s$ whose slope is $\displaystyle\frac1{u_0}$.
• If $0<s<1$, then the characteristic curve is $x=u_0\left(1-s\right)y+s$ whose slope is $\displaystyle\frac1{u_0\left(1-s\right)}$.
Note that the characteristic curve passes the point $\displaystyle\left(1,\frac1{u_0}\right)$ for all $0<s<1$, and hence the solution has the singularity at this point.
To deal with the singularity, we need to consider the shock wave and introduce the curve
$\displaystyle\xi'\left(y\right)=\frac{G\left(u_r\right)-G\left(u_l\right)}{u_r-u_l}=\frac{u_0}2$
and the curve passes $\displaystyle\left(1,\frac1{u_0}\right)$. Thus, we can get $\displaystyle x=\xi\left(y\right)=\frac{1+u_0y}2$.
Therefore, we obtain the solution
$u\left(x,y\right)=\begin{cases}0,&\text{if}~x\geq1,\,2x-1>u_0y,\\\displaystyle\frac{u_0\left(1-x\right)}{1-u_0y},&\text{if}~0<x<1,~x>u_0y,\\u_0,&\text{if}~u_0y>x,~u_0y>2x-1.\end{cases}$
The figure of the solution can be shown as below.



3. Consider (17) with the initial condition

   $u\left(x,0\right)=h\left(x\right)=\begin{cases}0&\text{for }x<0\\u_0\left(x-1\right)&\text{for }x>0,\end{cases}$

   where $u_0>0$. There is a weak solution $u\left(x,y\right)$ that has a jump discontinuity along $x=\xi\left(y\right)$. Find this curve and describe the weak solution.
Solution
It is well known that the solution has the implicit form $u\left(x,y\right)=h\left(x-u\left(x,y\right)y\right)$ and the family of the characteristic curve passing $\left(x,y\right)=\left(s,0\right)$ is $x=h\left(s\right)y$.
• If $s<0$, then the characteristic curve is a vertical line $x=s$.
• If $s>0$, then the characteristic curve is a line $x=u_0\left(s-1\right)y+s$.
  • If $s>1$, then the slope of the charcteristic curve is $\displaystyle\frac1{u_0\left(s-1\right)}>0$.
  • If $s=1$, then the characteristic curve is $x=1$ and $u\left(1,y\right)=0$ for all $y>0$.
  • If $0<s<1$, then the slope of the characteristic curve is $\displaystyle\frac1{u_0\left(s-1\right)}<0$.
It is clear that $u$ has a singularity at origin. For $y>0$, we consider the ordinary differential equation
$\displaystyle\xi'\left(y\right)=\frac{G\left(u_r\right)-G\left(u_l\right)}{u_r-u_l}=\frac{u_r\left(\xi\left(y\right),y\right)}2$,
whose value is determined by the characteristic curve passing $\left(\xi\left(y\right),y\right)$. By taking $\displaystyle s=\frac{\xi\left(y\right)+u_0y}{1+u_0y}$, we obtain
$\displaystyle\xi'\left(y\right)=\frac{u_0\xi\left(y\right)-u_0}{2+2u_0y}$,
which can be written as
$\displaystyle\xi'\left(y\right)-\frac{u_0}{2+2u_0y}\xi\left(y\right)=-\frac{u_0}{2+2u_0y}$.
Multiplying it by the integral factor $\left(1+u_0y\right)^{-1/2}$ gives
$\displaystyle\frac{d}{dt}\left(\frac{\xi\left(y\right)}{\sqrt{1+u_0y}}\right)=-\frac{u_0}2\left(1+u_0y\right)^{-\frac32}$,
which can be solved by
$x=\xi\left(y\right)=1-\sqrt{1+u_0y}$.
Thus, the solution is
$\displaystyle u\left(x,y\right)=\begin{cases}0,&\text{if}~x<1-\sqrt{1+u_0y},\\\displaystyle\frac{u_0\left(x-1\right)}{1+u_0y},&\text{if}~x>1-\sqrt{1+u_0y}.\end{cases}$
The figure of the solution can be shown as below.



4. Consider (17) with initial condition

   $u\left(x,0\right)=h\left(x\right)=\begin{cases}u_l&\text{for }x<x_0\\u_r&\text{for }x>x_0,\end{cases}$

   where $u_l<u_r$. Find a continuous weak solution.
Solution
It is well known that the solution has the implicit form $u\left(x,y\right)=h\left(x-u\left(x,y\right)y\right)$ and the family of the characteristic curve passing $\left(x,y\right)=\left(s,0\right)$ is $x=h\left(s\right)y$.
• If $s>x_0$, then $x=u_ry+s$ whose slope is $\displaystyle\frac1{u_r}$.
• If $s<x_0$, then $x=u_ly+s$ whose slope is $\displaystyle\frac1{u_l}$.
Since $u_l<u_r$, the rarefaction wave of the solution is
$\displaystyle u\left(x,y\right)=\begin{cases}u_r,&\text{if}~x-x_0>u_ry,\\\displaystyle\frac{x-x_0}y,&\text{if}~u_ly<x-x_0<u_ry,\\u_l,&\text{if}~x-x_0<u_ly.\end{cases}$
Moreover, the figure of the solution can be shown as follows.
• Case 1. $0<u_l<u_r$.
  
• Case 2. $u_l<0<u_r$.
  
• Case 3. $u_l<u_r<0$.
  


5. Find a simple wave solution $u\left(x,y\right)=v\left(x/y\right)$ for (19) when $G\left(u\right)=u^4/4$. Use this to define a continuous weak solution of (19) for $y>0$ that satisfies (22) with $u_0<0$.
Solution
According to the condition that $G\left(u\right)=u^4/4$, the partial differential equation can be rewritten as $u^3u_x+u_y=0$. Since $u\left(x,y\right)=v\left(x/y\right)$, we have
$\displaystyle u_x\left(x,y\right)=\frac{v'\left(x/y\right)}y$,　$\displaystyle u_y\left(x,y\right)=-\frac{xv'\left(x/y\right)}{y^2}$.
Thus, the partial differential equation can be transformed into $v\left(x/y\right)=\left(x/y\right)^{1/3}$, and hence $u\left(x,y\right)=\left(x/y\right)^{1/3}$. Now, we consider the characteristic curve $x=x\left(t\right)$, $y=y\left(t\right)$, $z=z\left(t\right)=u\left(x\left(t\right),y\left(t\right)\right)$ satisfying
$\left\{\begin{aligned}&\displaystyle\frac{dx}{dt}=z^3,\\&\displaystyle\frac{dy}{dt}=1,\\&\displaystyle\frac{dz}{dt}=0\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned}&x\left(0\right)=x_0,\\&y\left(0\right)=0,\\&z\left(0\right)=u\left(x\left(0\right),y\left(0\right)\right)=h\left(x_0\right).\end{aligned}\right.$
We can solve it by $x\left(t\right)=h^3\left(x_0\right)t+x_0$, $y\left(t\right)=t$ and $z\left(t\right)=h\left(x_0\right)$. Thus, we obtain
$u\left(x,y\right)=h\left(x-u^3\left(x,y\right)y\right)$
• If $x_0>0$, then the characteristic curve is vertical line $x=x_0$.
• If $x_0<0$, then the characteristic curve is the line $x=u_0^3y+x_0$ whose slope is $\displaystyle\frac1{u_0^3}$.
Hence, the solution can be represented as
$\displaystyle u\left(x,y\right)=\begin{cases}0,&\text{if}~x>0,\\\displaystyle\left(\frac{x}y\right)^{1/3},&\text{if}~u_0^3y<x<0,\\u_0,&\text{if}~x<u_0^3y.\end{cases}$
The figure of the solution can be shown as below.



6. Consider (17) with (22) where $u_0<0$. In addition to the rarefaction solution described in the text, show that there is a weak solution with a shock along $x=u_0y/2$.
Solution
Let
$\displaystyle u\left(x,y\right)=\begin{cases}0,&\text{if}~2x>u_0y,\\u_0,&\text{if}~2x<u_0y.\end{cases}$
Fix $y\in\mathbb{R}$ and choose two real $a,b$ satisfying $2a<u_0y<2b$. We verify directly that $u$ is the weak solution, i.e., to prove the following equality holds:
$\displaystyle G\left(u\left(b,y\right)\right)-G\left(u\left(a,y\right)\right)+\frac{d}{dy}\int_a^bu\left(x,y\right)dx=0$.
A direct computation gives
$\displaystyle G\left(u\left(b,y\right)\right)-G\left(u\left(a,y\right)\right)=\frac12u^2\left(b,y\right)-\frac12u^2\left(a,y\right)=-\frac12u_0^2=-\frac{d}{dy}\int_a^{\frac{u_0y}2}u_0dx=-\frac{d}{dy}\int_a^bu\left(x,y\right)dx$,
as desired.


7. A reasonable model for low-density traffic is (25) with $dq/d\rho=c$, where $c$ is a constant.
   1. Show that $\rho$ is constant along the (characteristic) curves $x=ct+x_0$.
   2. If a car is alone on the highway, what does $\rho\left(x,0\right)$ look like? What does $\rho\left(x,t\right)$ look like?
   3. Explain why $c$ represents the free speed of the highway.
Solution
1. Consider $z\left(s\right)=\rho\left(x\left(s\right),y\left(s\right)\right)=\rho\left(cs+x_0,s\right)$. Then a direct computation gives

   $z'\left(s\right)=c\rho_x\left(cs+x_0,s\right)+\rho_t\left(cs+x_0,s\right)=0$.

   Thus, $\rho$ is constant along with the line $x=ct+x_0$.
2. Suppose the center of mass of the car is located at $x_0$ initially and the length of the car is $L$. Then, the density function is

   $\rho\left(x,0\right)=\begin{cases}\displaystyle\frac1L,&\displaystyle\text{if}~\left|x-x_0\right|\leq\frac{L}2,\\0,&\displaystyle\text{if}~\left|x-x_0\right|>\frac{L}2.\end{cases}$

   According to (a), we know that

   $\rho\left(x,t\right)=\begin{cases}\displaystyle\frac1L,&\displaystyle\text{if}~\left|x-ct-x_0\right|\leq\frac{L}2,\\0,&\displaystyle\text{if}~\left|x-ct-x_0\right|>\frac{L}2.\end{cases}$

3. Based on the results of (a) and (b), we find that the car moves from $x_0$ to $x_0+ct_0$ as $t=0$ becomes $t=t_0$, which implies the velocity of the car is $c$.


8. If $\rho_{\max}$ denotes the maximum density of cars on a highway (i.e., under bumper-to-bumper conditions), then a reasonable relation between $q$ and $\rho$ is given by $G\left(\rho\right)=c\rho\left(1-\rho/\rho_{\max}\right)$ where the constant $c$ is the free speed of the highway (cf. Exercise 7). Suppose the initial density is

   $\rho\left(x,0\right)=\begin{cases}\displaystyle\frac{1}{2}\rho_{\max}&\text{for }x<0\\\rho_{\max}&\text{for }x>0.\end{cases}$

   Find the shock curve and describe the weak solution. Interpret your result for the traffic flow.
Solution
Based the setting in Problem, the partial differential equation is read as $G'\left(\rho\right)\rho_x+\rho_t=0$. To solve the equation, we consider the characteristic curve $x=x\left(s\right)$, $t=t\left(s\right)$, $z=\rho\left(x\left(s\right),t\left(s\right)\right)$ satisfying the following ordinary differential equation
$\left\{\begin{aligned}&\displaystyle\frac{dx}{ds}=G'\left(z\right)=c\left(1-\frac{2z}{\rho_{\max}}\right),\\&\displaystyle\frac{dt}{ds}=1,\\&\displaystyle\frac{dz}{ds}=0\end{aligned}\right.$ with the initial condition $\left\{\begin{aligned}&x\left(0\right)=x_0,\\&t\left(0\right)=0,\\&z\left(0\right)=\rho\left(x_0,0\right).\end{aligned}\right.$
We can solve it by $\displaystyle x\left(s\right)=cs\left(1-\frac{2\rho\left(x_0,0\right)}{\rho_\max}\right)+x_0$, $t\left(s\right)=s$, and $z\left(s\right)=\rho\left(x_0,0\right)$.
• If $x_0<0$, then the characteristic curve is vertical line $x=x_0$.
• If $x_0>0$, then the characteristic curve is the line $x=-ct+x_0$ whose slope is $\displaystyle-\frac1c$.
In order to deal with the shock wave, we consider the ordinary differential equation
$\displaystyle\xi'\left(t\right)=\frac{G\left(\rho_r\right)-G\left(\rho_l\right)}{\rho_r-\rho_l}=\frac{-c\rho_\max/4}{\displaystyle\rho_\max-\frac12\rho_\max}=-\frac{c}2$
Thus, we get the shock curve $x=\xi\left(t\right)=-\frac{ct}2$ which passes $\left(0,0\right)$. Hence, the weak solution is
$\rho\left(x,t\right)=\begin{cases}\displaystyle\frac12\rho_\max,&\text{if}~x<-\frac{ct}2,\\\rho_\max,&\text{if}~x>-\frac{ct}2.\end{cases}$
The figure of the solution can be shown as below.

For the traffic flow, the region of the right side is crowded and the region of the left side is sparse. As time goes on, the crowded region extends to the left side.


9. Using $G\left(\rho\right)$ as in Exercise 8, describe the traffic flow after a long red light turns green at $t=0$; that is, the initial density is

   $\rho\left(x,0\right)=\begin{cases}\rho_{\max}&\text{for }x<0\\0&\text{for }x>0.\end{cases}$

   In particular, find the density at the green light, $\rho\left(0,t\right)$, while the light remains green.
Solution
According to Exercise 8, we can solve the ordinary differential equation with the initial condition $\left(x\left(0\right),t\left(0\right),z\left(0\right)\right)=\left(x_0,0,\rho\left(x_0,0\right)\right)$ by
$\displaystyle x\left(s\right)=cs\left(1-\frac{2\rho\left(x_0,0\right)}{\rho_\max}\right)+x_0$, $t\left(s\right)=s$, and $\displaystyle z\left(s\right)=\rho\left(x_0,0\right)$.
• If $x_0>0$, then the characteristic curve is the line $x\left(s\right)=ct+x_0$ whose slope is $\displaystyle\frac1c$.
• If $x_0<0$, then the characteristic curve is the line $x\left(s\right)=-ct+x_0$ whose slope is $\displaystyle-\frac1c$.
Since the solution is undefined on the middel region, we need to consider the rarefaction wave. Let $\rho\left(x,t\right)=v\left(x/t\right)$. Inserting it into the partial differential equaiton yields
$\displaystyle\frac{c}t\left(1-\frac{2v}{\rho_\max}\right)v'\left(\frac{x}t\right)-\frac{x}{t^2}v'\left(\frac{x}t\right)=0$,
which gives
$\displaystyle\rho\left(x,t\right)=v\left(\frac{x}t\right)=\frac{\left(ct-x\right)\rho_\max}{2ct}=\frac{\rho_\max}2\left(1-\frac{x}{ct}\right).$
Hence, we obtain the weak solution
$\displaystyle\rho\left(x,t\right)=\begin{cases}0,&\text{if}~x>ct,\\\displaystyle\frac{\rho_\max}2\left(1-\frac{x}{ct}\right),&\text{if}~-ct<x<ct,\\\rho_\max,&\text{if}~x<-ct.\end{cases}$
The figure of the solution is shown as below.

When the red light turns green, the cars move from crowded to sparse, which makes the densitiy of the cars chagnes. In addition, we note that $\rho\left(0,t\right)=\rho_\max/2$.