Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 1.3

1. Solve $u_x^2+yu_y-u=0$ with initial condition $\displaystyle u(x,1)=\frac{x^2}4+1$.
Solution
Let $F:\mathbb{R}^5\to\mathbb{R}$ be defined by
$F(x,y,z,p,q)=p^2+yq-z\quad\text{for}~(x,y,z,p,q)\in\mathbb{R}^5.$
Consider the characteristic curve
$x=x(t),\quad y=y(t),\quad z=z(t)=u(x(t),y(t)),\quad p=p(t)=u_x(x(t),y(t)),\quad q=q(t)=u_y(x(t),y(t))$.
Then we have the following ordinary differential equations
$\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=F_p(x,y,z,p,q)=2p,\\&\frac{\mathrm dy}{\mathrm dt}=F_q(x,y,z,p,q)=y,\\&\frac{\mathrm dz}{\mathrm dt}=pF_p(x,y,z,p,q)+qF_q(x,y,z,p,q)=2p^2+yq,\\&\frac{\mathrm dp}{\mathrm dt}=-F_x(x,y,z,p,q)-pF_z(x,y,z,p,q)=p,\\&\frac{\mathrm dq}{\mathrm dt}=-F_y(x,y,z,p,q)-qF_z(x,y,z,p,q)=0\end{aligned}\right.$
with the initial condition
$\displaystyle x(0)=s,\quad y(0)=1,\quad z(0)=u(x(0),y(0))=\frac{s^2}4+1$.
From $F(x(0),y(0),z(0),p(0),q(0))=0$, we have
$\displaystyle p^2(0)+q(0)-\frac{s^2}4-1=0$.
By the strip condition, we have
$\displaystyle\frac s2=\frac{\mathrm dz(0)}{\mathrm ds}=p(0)\frac{\mathrm dx(0)}{\mathrm ds}+q(0)\frac{\mathrm dy(0)}{\mathrm ds}=p(0)\cdot1+q(0)\cdot0$.
Thus, $p(0)=s/2$ and $q(0)=1$. Now from $\displaystyle\frac{\mathrm dp}{\mathrm dt}=p$ and $\displaystyle\frac{\mathrm dq}{\mathrm dt}=0$, we obtain
$\displaystyle p(t)=p(0)e^t=\frac{s}2e^t,\quad q(t)=q(0)=1.$
Hence, the first three equations become
$\displaystyle\frac{\mathrm dx}{\mathrm dt}=se^t,\quad\frac{\mathrm dy}{\mathrm dt}=y,\quad\frac{\mathrm dz}{\mathrm dt}=\frac{s^2}2e^{2t}+y,$
which implies $x(t)=x(0)+s(e^t-1)=se^t$ and $y(t)=y(0)e^t=e^t$, and
$\displaystyle z(t)=z(0)+\frac{s^2}4(e^{2t}-1)+(e^t-1)=\frac{s^2}4e^{2t}+e^t=\left(\frac{se^t}2\right)^2+e^t.$
It is easy to get $t=\ln(y)$ and $s=x/y$, and we obtain
$\displaystyle u(x,y)=\frac{x^2}4+y\quad\text{for}~x\in\mathbb R~\text{and}~y>0.$
Here we note that $y=e^t>0$. It is remarkable that this solution can be extended to $\mathbb R^2$. Function $u(x,y)=\frac{x^2}4+|y|$ for $(x,y)\in\mathbb R^2$ is a solution in function space $\mathcal C^1(\mathbb R\times\mathbb R_*)\cap\mathcal C^0(\mathbb R^2)$, where $\mathbb R_*=\mathbb R-\{0\}$.


2. Solve $u=xu_x+yu_y+(u_x^2+u_y^2)/2$ with initial condition $u(x,0)=(1-x^2)/2$. [This is an example of Clairaut's equation $u=xp+yq+f(p,q)$.]
Solution

Let $F:\mathbb{R}^5\to\mathbb{R}$ be defined by

$F(x,y,z,p,q)=z-xp-yq-(p^2+q^2)/2\quad\text{for}~(x,y,z,p,q)\in\mathbb R^5.$

Consider the characteristic curve

$x=x(t),\quad y=y(t),\quad z=z(t)=u(x(t),y(t)),\quad p=p(t)=u_x(x(t),y(t)),\quad q=q(t)=u_y(x(t),y(t))$.

Then we have the following ordinary differential equations

$\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=F_p(x,y,z,p,q)=-x-p,\\&\frac{\mathrm dy}{\mathrm dt}=F_q(x,y,z,p,q)=-y-q\\&\frac{\mathrm dz}{\mathrm dt}=pF_p(x,y,z,p,q)+qF_q(x,y,z,p,q)=p(-x-p)+q(-y-q)=-xp-yq-p^2-q^2,\\&\frac{\mathrm dp}{\mathrm dt}=-F_x(x,y,z,p,q)-pF_z(x,y,z,p,q)=-(-p)-p\cdot1=0,\\&\frac{\mathrm dq}{\mathrm dt}=-F_y(x,y,z,p,q)-qF_z(x,y,z,p,q)=-(-q)-q\cdot1=0\end{aligned}\right.$

with initial condition

$x(0)=s,\quad y(0)=0,\quad z(0)=u(x(0),y(0))=(1-s^2)/2$.

From $F(x(0),y(0),z(0),p(0),q(0))=0$, we have

$\displaystyle\frac{1-s^2}2-sp(0)-\frac{p^2(0)+q^2(0)}2=0$.

By the strip condition, we get

$\displaystyle-s=\frac{\mathrm dz(0)}{\mathrm ds}=p(0)\frac{\mathrm dx(0)}{\mathrm ds}+q(0)\frac{\mathrm dy(0)}{\mathrm ds}=p(0)\cdot1+q(0)\cdot0=p(0)$.

Thus, $p(0)=-s$ and $q(0)=\pm1$. Now from $\displaystyle\frac{\mathrm dp}{\mathrm dt}=\frac{\mathrm dq}{\mathrm dt}=0$, we obtain

$p(t)=p(0)=-s$ and $q(t)=q(0)=\pm1$.

Hence, the first three equations become

$\displaystyle\frac{\mathrm dx}{\mathrm dt}=-x+s,\quad\frac{\mathrm dy}{\mathrm dt}=-y\mp1,\quad\frac{\mathrm dz}{\mathrm dt}=sx\mp y-s^2-1$,

which implies $x(t)=s+(x(0)-s)e^{-t}=s$, $y(t)=\mp1+(y(0)\pm1)e^{-t}=\pm(e^{-t}-1)$, and

$\displaystyle\frac{\mathrm dz}{\mathrm dt}=s^2-(e^{-t}-1)-s^2-1=-e^{-t}$.

Thus, we can solve

$\displaystyle z=z(t)=z(0)+(e^{-t}-1)=e^{-t}-\frac{1+s^2}2$.

It is easy to get $t=-\ln(1\pm y)$ and $s=x$, and we obtain

$\displaystyle u(x,y)=1\pm y-\frac{1+x^2}2=\pm y-\frac{1-x^2}2\quad\text{for}~(x,y)\in\mathbb R^2$.



3. Consider $u=u_x^2+u_y^2$ with the initial condition $u(x,0)=ax^2$. For what positive constants $a$ is there a solution? Is it unique? Find all solutions.
Solution

When $a\in(0,1/4]$, there is a solution. More precisely, the partial differential equation admits a unique solution ($a=1/4$) and two solutions ($a\in(0,1/4)$), respectively.

Let $F:\mathbb{R}^5\to\mathbb{R}$ be defined by

$F(x,y,z,p,q)=p^2+q^2-z\quad\text{for}~(x,y,z,p,q)\in\mathbb{R}^5.$

Consider the characteristic curve

$x=x(t),\quad y=y(t),\quad z=z(t)=u(x(t),y(t)),\quad p=p(t)=u_x(x(t),y(t)),\quad q=q(t)=u_y(x(t),y(t))$.

Then we have the following ordinary differential equations

$\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=F_p(x,y,z,p,q)=2p,\\&\frac{\mathrm dx}{\mathrm dt}=F_q(x,y,z,p,q)=2q,\\&\frac{\mathrm dz}{\mathrm dt}=pF_p(x,y,z,p,q)+qF_q(x,y,z,p,q)=p\cdot2p+q\cdot2q=2(p^2+q^2)=2z,\\&\frac{\mathrm dp}{\mathrm dt}=-F_x(x,y,z,p,q)-pF_z(x,y,z,p,q)=p,\\&\frac{\mathrm dq}{\mathrm dt}=-F_y(x,y,z,p,q)-qF_z(x,y,z,p,q)=q\end{aligned}\right.$

with the initial condition

$x(0)=s,\quad y(0)=0,\quad z(0)=u(x(0),y(0))=as^2$.

From $F(x(0),y(0),z(0),p(0),q(0))=0$, we have $p^2(0)+q^2(0)=z(0)=as^2$. By the strip condition, we get

$\displaystyle2as=\frac{\mathrm dz(0)}{\mathrm ds}=p(0)\frac{\mathrm dx(0)}{\mathrm ds}+q(0)\frac{\mathrm dy(0)}{\mathrm ds}=p(0)\cdot1+q(0)\cdot0=p(0)$.

Thus, $p(0)=2as$ and $q(0)=\pm|s|\sqrt{a-4a^2}\in\mathbb R$, provided $a\in(0,1/4]$. Now from $\displaystyle\frac{\mathrm dp}{\mathrm dt}=p$ and $\displaystyle\frac{\mathrm dq}{\mathrm dt}=q$, we obtain

$p(t)=p(0)e^t=2ase^t,\quad q(t)=q(0)e^t=\pm|s|e^t\sqrt{a-4a^2}$.

Hence, the first three equations become

$\displaystyle\frac{\mathrm dx}{\mathrm dt}=4ase^t,\quad\frac{\mathrm dy}{\mathrm dt}=\pm2|s|e^t\sqrt{a-4a^2},\quad\frac{\mathrm dz}{\mathrm dt}=2z$,

which implies

$\begin{aligned} &x(t)=x(0)+4as(e^t-1)=s(1+4ae^t-4a),\\&y(t)=y(0)\pm2|s|(e^t-1)\sqrt{a-4a^2}=\pm2|s|(e^t-1)\sqrt{a-4a^2},\\&z(t)=z(0)e^{2t}=as^2e^{2t}.\end{aligned}$

• When $a=1/4$, it is clear that $x=se^t$ and $y=0$, and we obtain

  $\displaystyle u(x,y)=\frac{x^2}4\quad\text{for}~(x,y)\in\mathbb R^2$.

• Suppose $a\in(0,1/4)$. From $x=s+4as(e^t-1)$, we have $\displaystyle\frac{x-s}{4a}=s(e^t-1)$. On the other hand, $y^2=4s^2(e^t-1)^2(a-4a^2)$. Combining them, we get $\displaystyle 4a^2y^2=(x-s)^2(a-4a^2)$, and hence $\displaystyle s=x\pm\frac{2y\sqrt a}{\sqrt{1-4a}}$. From $x=s(1-4a+4ae^t)$, we get

  $\displaystyle t=\ln\left(1+\frac{x-s}{4as}\right)=\ln\left(\frac{2x\sqrt{a-4a^2}\pm y(4a-1)}{2x\sqrt{a-4a^2}\pm4ya}\right)$.

  Therefore, we obtain

  $\displaystyle u(x,y)=a\left(x\pm\frac{(4a-1)y}{2\sqrt{a-4a^2}}\right)^2=\left(x\sqrt a\mp\frac{y\sqrt{1-4a}}2\right)^2\quad\text{for}~(x,y)\in\mathbb R^2$.

  Here we have used the fact that

  $\displaystyle s^2e^{2t}=(se^t)^2=\left[s\cdot\left(1+\frac{x/s-1}{4a}\right)\right]^2=\left(s+\frac{x-s}{4a}\right)^2=\left(x\pm\frac{2y\sqrt a}{\sqrt{1-4a}}\mp\frac{y}{2\sqrt{a-4a^2}}\right)^2$



4. To generalize (26) to $n$ variables, let $u(x)=u(x_1,\dots,x_n)$ and $p_j=\partial u/\partial x_j$ for $j=1,\dots,n$. Then consider

   $F(x,u,p)=0$.

   1. Find the characteristic equations [i.e., generlizations of (29), (30)].
   2. Find the conditions on the initial values of $p$ [i.e., generalization of (31), (32)].
Solution

1. Let $x=x(t)=(x_1(t),\dots,x_n(t))$ be the characteristic curve. Define $z(s)=u(x(t))$, $p(t)=(p_1(t),\dots,p_n(t))=\nabla u(x(t))$. Differentiating $p_i$ with respect to $t$ yields\begin{align}\label{1}\frac{\mathrm dp_i}{\mathrm dt}=\frac{\mathrm d}{\mathrm dt}[u_{x_i}(x(t))]=\sum_{j=1}^nu_{x_ix_j}(x(t))\frac{\mathrm dx_j}{\mathrm dt}.\end{align}On the other hand, we also differentiate $F(x,u,\nabla u)=0$ with respect to $x_i$:

   $\displaystyle F_{x_i}(x,u,\nabla u)+F_u(x,u,\nabla u)u_{x_i}+\sum_{i=j}^nF_{p_i}(x,u,\nabla u)u_{x_jx_i}=0$,

   which implies\begin{align}\label{2}F_{x_i}(x(t),z(t),p(t))+p_i(t)F_u(x(t),z(t),p(t))+\sum_{j=1}^nF_{p_i}(x(t),z(t),p(t))u_{x_jx_i}(x(t))=0.\end{align}Suppose $\displaystyle\frac{\mathrm dx_i}{\mathrm dt}=F_{p_i}(x(t),z(t),p(t))$ for $i=1,\dots,n$. Then by \eqref{1} and \eqref{2}, we obtain

   $\displaystyle\frac{\mathrm dp_i}{\mathrm dt}=\sum_{j=1}^nu_{x_ix_j}(x(t))F_{p_j}(x(t),z(t),p(t))=-F_{x_i}(x(t),z(t),p(t))-p_i(t)F_u(x(t),z(t),p(t))$.

   Differenitaing $z=u(x(t))$ with respec to $t$ gives

   $\displaystyle\frac{\mathrm dz}{\mathrm dt}=\sum_{i=1}^nu_{x_i}(x(t))\frac{\mathrm dx_i}{\mathrm dt}=\sum_{i=1}^np_i(x(t))F_{p_i}(x(t),z(t),p(t))$.

2. Suppose $\Gamma$ is the noncharacteric curve parameterized by $(x,z)=(f(s),g(s))=(f_1(s),\dots,f_n(s),g(s))$. Then $h(s)=(h_1(s),\dots,h_n(s))=\nabla u(f(s))$ satisfies

   $F(f(s),g(s),h(s))=0$.

   In addition, from $g(s)=u(f(s))$, we have the strip condition:

   $\displaystyle g'(s)=\nabla u(f(s))\cdot\nabla f(s)=h(s)\cdot\nabla f(s)=\sum_{i=1}^nh_i(f(s))f_i'(s)$.



5. Show that the family of spheres $S_a$ given by $(x-a)^2+y^2+z^2=1$ has as its envelope $\mathscr E$ the unit cylinder $y^2+z^2=1$.
Solution
Let $w(x,y,z;a)=(x-a)^2+y^2+z^2-1$. Then the evelope $\mathscr E$ is the intersection
$\left\{\begin{aligned} &w(x,y,z;a)=(x-a)^2+y^2+z^2-1=0,\\&\partial _a(x,y,z;a)=-2(x-a)=0\end{aligned}\right.$
From the second equation, we get $x=a$, and hence $y^2+z^2=1$.


6. Consider Clairaut's equation $u=xu_x+yu_y+f(u_x,u_y)$.
   1. Verify that $u(x,y;a,b)=ax+by+f(a,b)$ is a complete integral.
   2. Use envelopes to generate a solution of $u=xu_x+yu_y+(u_x^2+u_y^2)/2$ that is not linear in $x$ and $y$.
Solution
1. Since $u(x,y;a,b)=ax+by+f(a,b)$, it is easy to get $u_x(x,y;a,b)=a$ and $u_y(x,y;a,b)=b$. Thus, $u=u(x,y;a,b)$ satisfies the partial differential equation:

   $xu_x(x,y;a,b)+yu_y(x,y;a,b)+f(u_x(x,y;a,b),u_y(x,y;a,b))=x\cdot a+b\cdot y+f(a,b)=u(x,y;a,b)$.

   Moreover, the independence of the parameters follows from the invertiblity of the matrix

   $\begin{bmatrix}u_{xa}(x,y;a,b)&u_{ya}(x,y;a,b)\\u_{xb}(x,y;a,b)&u_{yb}(x,y;a,b)\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$.

   This shows $u(x,y;a,b)=ax+by+f(a,b)$ is a complete integral.
2. From a., we consider the two-parameter solution $u(x,y;a,b)=ax+by+(a^2+b^2)/2$. To find the envelope, we set $b=w(a)$ for some $w\in\mathcal C^1$, which gives a one-parameter solution

   $u(x,y;a):=u(x,y;a,w(a))=ax+w(a)y+(a^2+(w(a))^2)/2$.

   For the sake of simplicity, we choose $w(a)=a$ for $a\in\mathbb R$. Then the envelope $\mathscr E$ can be determined by

   $\left\{\begin{aligned} &u(x,y;a)=0,\\&\partial_au(x,y;a)=0.\end{aligned}\right.$

   The second equation yields $x+1\cdot y+(a+a\cdot1)=0$, which gives $\displaystyle a=-\frac{x+y}2$. Plugging into the first equation, we obtain

   $\displaystyle u(x,y)=-\frac{x+y}2\cdot x-\frac{x+y}2\cdot y+\frac12\left[\left(\frac{x+y}2\right)^2+\left(\frac{x+y}2\right)^2\right]=-\frac{(x+y)^2}4\quad\text{for}~(x,y)\in\mathbb R^2$.

   Note that this solution is not linear in $x$ and $y$.


7. The Hamilton-Jacobi equation $u_t+H(\nabla_xu)=0$, where $H$ depends only on $\nabla_xu=(u_{x_1},u_{x_2},\dots,u_{x_n})$, arises frequently in physics, such as geometric optics (cf. the eikonal equation). Here we assume $n=1$.
   1. Verify that condition (34) fails for $u(x,t;a,b)=ax-tH(a)+b$, yet this is a complete integral.
   2. Use envelopes to generate a solution of $u_t+u_x^2=0$ that is not linear in $x$ and $t$.
Solution
1. Since $u(x,t;a,b)=ax-tH(a)+b$, it is easy to get $\nabla u_x(x,t;a,b)=a$ and $u_t(x,t;a,b)=-H(a)$. Thus, $u=u(x,t;a,b)$ satisfies the partial differential equation:

   $u_t(x,t;a,b)+H(\nabla u_x(x,y;a,b))=-H(a)+H(a)=0$.

   Despite the matrix $\begin{bmatrix}u_{xa}(x,t;a,b)&u_{ta}(x,y;a,b)\\u_{xb}(x,t;a,b)&u_{tb}(x,t;a,b)\end{bmatrix}=\begin{bmatrix}1&-H'(a)\\0&0\end{bmatrix}$ is non invertible, $a$ and $b$ are independent paramters because $\begin{bmatrix}u_a(x,t;a,b)&u_{xa}(x,t;a,b)&u_{ta}(x,t;a,b)\\u_b(x,t;a,b)&u_{xb}(x,t;a,b)&u_{tb}(x,t;a,b)\end{bmatrix}=\begin{bmatrix}x-tH'(a)&1&-H'(a)\\1&0&0\end{bmatrix}$ has rank $2$. This shows $u=u(x,t;a,b)$ is a complete integral.
2. From a., we consider the two-parameter solution $u(x,t;a,b)=ax-ta^2+b$. To find the envelope, we set $b=w(a)$ for some $w\in\mathcal C^1$, which gives a one-parameter solution $u(x,t;a):=u(x,t;a,w(a))=ax-ta^2+w(a)$. For the sake of simplicity, we choose $w(a)=a$ for $a\in\mathbb R$. Then the envelope $\mathscr E$ can be determined by

   $\left\{\begin{aligned} &u(x,y;a)=0,\\&\partial_au(x,y;a)=0.\end{aligned}\right.$

   The second equation $x-2ta+1=0$, which gives $a=(x+1)/(2t)$. Plugging into the first equation, we obtain

   $\displaystyle u(x,y)=\frac{x+1}{2t}\cdot x-t\cdot\frac{(x+1)^2}{4t^2}+\frac{x+1}{2t}=\frac{(x+1)^2}{4t}\quad\text{for}~(x,t)\in\mathbb R^2$.

   Note that this solution is not linear in $x$ and $t$.


8. To describe the wave front produced by an initial disturbance at a point, consider (36) with $\Gamma$ begin given by $f\equiv g\equiv h\equiv 0$. Describe the solution(s).
Solution

Let $F:\mathbb R^5\to\mathbb R$ be defined by

$F(x,y,z,p,q)=c^2(p^2+q^2)-1\quad\text{for}~(x,y,z,p,q)\in\mathbb R^5$,

where $c>0$ is a constant. Consider the characteristic curve

$x=x(t),\quad y=y(t),\quad z=u(x(t),y(t)),\quad p=p(t)=u_x(x(t),y(t)),\quad q=q(t)=u_y(x(t),y(t))$.

Then we have the following ordinary differential equations

$\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=F_p(x,y,z,p,q)=2c^2p,\\&\frac{\mathrm dy}{\mathrm dt}=F_q(x,y,z,p,q)=2c^2q,\\&\frac{\mathrm dz}{\mathrm dt}=pF_p(x,y,z,p,q)+qF_q(x,y,z,p,q)=p\cdot2c^2p+q\cdot2c^2q=2,\\&\frac{\mathrm dp}{\mathrm dt}=-F_x(x,y,z,p,q)-pF_z(x,y,z,p,q)=0,\\&\frac{\mathrm dq}{\mathrm dt}=-F_y(x,y,z,p,q)-qF_z(x,y,z,p,q)=0\end{aligned}\right.$

with the initial condition

$x(0)=f(s)\equiv0,\quad y(0)=g(s)\equiv0,\quad z(0)=u(x(0),y(0))=h(s)\equiv0.$

From $F(x(0),y(0),z(0),p(0),q(0))=0$, we have $c^2(p^2(0)+q^2(0))=1$. On the other hand, the strip condition gives nothing but $0=0$. Thus, $p(0)=c^{-1}\cos s$ and $q(0)=c^{-1}\sin s$ for $s\in[0,2\pi)$. Now from $\displaystyle\frac{\mathrm dp}{\mathrm dt}=\frac{\mathrm dq}{\mathrm dt}=0$, we obtain

$p(t)=p(0)=c^{-1}\cos s,\quad q(t)=q(0)=c^{-1}\sin s$.

Hence, the first three equations become

$\displaystyle\frac{\mathrm dx}{\mathrm dt}=2c\cos s,\quad\frac{\mathrm dy}{\mathrm dt}=2c\sin s,\quad\frac{\mathrm dz}{\mathrm dt}=2$,

which implies

$x(t)=x(0)+2ct\cos s=2ct\cos s,\quad y(t)=y(0)+2ct\sin s=2ct\sin s,\quad z(t)=z(0)+2t=2t$.

It is easy to get $t=\pm\sqrt{x^2+y^2}/(2c)$ and $s=\tan^{-1}(y/x)$, and we obtain

$\displaystyle u(x,y)=\pm\frac{\sqrt{x^2+y^2}}c\quad\text{for}~(x,y)\in\mathbb R^2$.



9. Instead of assuming the propagation speed is constant, consider $c=c(x,y)$ in (36) and derive the characteristic equations. In the special case $c=|x|$ with initial condition $u(x,0)=0$, find the solution to be

   $\displaystyle u(x,y)=-\log\frac{\sqrt{x^2+y^2}+y}{x}\quad\text{for}~x>0.$

Solution

Let $F:\mathbb R^5\to\mathbb R$ be defined by

$F(x,y,z,p,q)=c^2(x,y)(p^2+q^2)-1\quad\text{for}~(x,y,z,p,q)\in\mathbb R^5$.

Consider the characteristic curve

$x=x(t),\quad y=y(t),\quad z=z(t)=u(x(t),y(t)),\quad p=p(t)=u_x(x(t),y(t)),\quad q=q(t)=u_y(x(t),y(t))$.

Then we have the following ordinary differential equations

$\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=F_p(x,y,z,p,q)=2c^2(x,y)p,\\&\frac{\mathrm dy}{\mathrm dt}=F_q(x,y,z,p,q)=2c^2(x,y)q,\\&\frac{\mathrm dz}{\mathrm dt}=pF_p(x,y,z,p,q)+qF_q(x,y,z,p,q)=p\cdot2c^2(x,y)p+q\cdot2c^2(x,y)q=2,\\&\frac{\mathrm dp}{\mathrm dt}=-F_x(x,y,z,p,q)-pF_z(x,y,z,p,q)=-2c(x,y)c_x(x,y)(p^2+q^2)=-\frac{2c_x(x,y)}{c(x,y)},\\&\frac{\mathrm dq}{\mathrm dt}=-F_y(x,y,z,p,q)-qF_z(x,y,z,p,q)=-2c(x,y)c_y(x,y)(p^2+q^2)=-\frac{2c_y(x,y)}{c(x,y)}.\end{aligned}\right.$

Now we suppose that $c(x,y)=|x|$. Then the five ordinary differential equations become

$\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=2x^2p,\\&\frac{\mathrm dy}{\mathrm dt}=2x^2q,\\&\frac{\mathrm dz}{\mathrm dt}=2,\\&\frac{\mathrm dp}{\mathrm dt}=-\frac{2\text{sgn}(x)}{|x|}=-\frac2x,\\&\frac{\mathrm dq}{\mathrm dt}=0.\end{aligned}\right.$

In addition, the noncharacteristic curve can be parameterized by

$x(0)=s,\quad y(0)=0,\quad z(0)=u(x(0),y(0))=0$.

From $F(x(0),y(0),z(0),p(0),q(0))=0$, we have $s^2(p^2(0)+q^2(0))=1$. By the strip condition, we get

$\displaystyle0=\frac{\mathrm dz(0)}{\mathrm ds}=p(0)\frac{\mathrm dx(0)}{\mathrm ds}+q(0)\frac{\mathrm dy(0)}{\mathrm dt}=p(0)\cdot1+q(0)\cdot0$.

Thus, $p(0)=0$ and $q(0)=\pm1/s$. From $\displaystyle\frac{\mathrm dq}{\mathrm dt}=0$, we get $q(t)=q(0)=\pm1/s$. Using the first and fourth equations, we have

$\displaystyle\frac{\mathrm dx}{2x^2p}=-\frac{x\,\mathrm dp}{2}\quad\text{or}\quad\frac{\mathrm dx}{x^3}=-p\,\mathrm dp$.

Thus, we obtain $\displaystyle\frac1{2x^2(0)}-\frac1{2x^2(t)}=\frac{p^2(0)-p^2(t)}2$. By the initial condition, we get $\displaystyle p^2(t)=\frac1{x^2(t)}-\frac1{s^2}=\frac{s^2-x^2(t)}{s^2x^2(t)}$, which implies $\displaystyle p(t)=\pm\frac{\sqrt{s^2-x^2(t)}}{sx(t)}$. Hence, the first equation becomes

$\displaystyle\frac{\mathrm dx}{\mathrm dt}=\pm\frac{2x\sqrt{s^2-x^2}}s$,

which can be integrated by separation of variables:

$\displaystyle-\tanh^{-1}\left(\frac{\sqrt{s^2-x^2(t)}}s\right)=\int_0^t\!\frac s{x(t)\sqrt{s^2-x^2(t)}}\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt=\pm\int_0^t\!2\,\mathrm dt=\pm2t$.

This gives $x(t)=\sqrt{s^2-s^2\tanh^2(\mp2t)}=|s|\text{sech}(\mp2t)$. On the other hand, the second equation becomes

$\displaystyle\frac{\mathrm dy}{\mathrm dt}=\pm2s\cdot\text{sech}^2(\mp2t)$,

which can be solved by

$\displaystyle y(t)=y(0)-s\tanh(\mp2t)=-s\tanh(\mp2t)$.

Finally, we have $z(t)=z(0)+2t=2t$. By $\tanh^2(\theta)+\text{sech}^2(\theta)=1$ for all $\theta\in\mathbb R$, we have $x^2+y^2=s^2$, which gives

$\displaystyle s=\sqrt{x^2+y^2}\quad\text{and}\quad t=\pm\frac12\tanh^{-1}\left(\frac y{x^2+y^2}\right)=\pm\frac14\ln\left(\frac{\sqrt{x^2+y^2}+y}{\sqrt{x^2+y^2}-y}\right)$.

Therefore, we obtain

$\begin{aligned}u(x,y)&=\pm\frac12\ln\left(\frac{\sqrt{x^2+y^2}+y}{\sqrt{x^2+y^2}-y}\right)=\pm\frac12\ln\left(\frac{(\sqrt{x^2+y^2}+y)^2}{(x^2+y^2)-y^2}\right)=\pm\ln\left(\frac{\sqrt{x^2+y^2}+y}{|x|}\right)\quad\text{for}~x\neq0,\,y\in\mathbb R\end{aligned}$.



10. Consider the eikonal equation in three dimensions

    $u_x^2+u_y^2+u_z^2=1$.

    1. Solve the initial value problem with $u=k=$constant on the plane $\alpha x+\beta y+z=0$.
    2. Find a complete integral.
Solution
1. Let $F:\mathbb R^7\to\mathbb R$ be defined by

   $F(x,y,z,u,p,q,r)=p^2+q^2+r^2-1\quad\text{for}~(x,y,z,u,p,q,r)\in\mathbb R^7$.

   Consider the characteristic curve

   $\begin{aligned} &x=x(t),\quad y=y(t),\quad z=z(t),\quad u=u(t)=u(x(t),y(t),z(t)),\\&p=p(t)=u_x(x(t),y(t),z(t)),\quad q=q(t)=u_y(x(t),y(t),z(t)),\quad r=r(t)=u_z(x(t),y(t),z(t)).\end{aligned}$

   Then we have the following ordinary differential equations

   $\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=F_p(x,y,z,u,p,q,r)=2p,\\&\frac{\mathrm dy}{\mathrm dt}=F_q(x,y,z,u,p,q,r)=2q,\\&\frac{\mathrm dz}{\mathrm dt}=F_r(x,y,z,u,p,q,r)=2r,\\&\frac{\mathrm du}{\mathrm dt}=pF_p(x,y,z,u,p,q,r)+qF_q(x,y,z,u,p,q,r)+rF_r(x,y,z,u,p,q,r)=2p^2+2q^2+2r^2=2,\\&\frac{\mathrm dp}{\mathrm dt}=-F_x(x,y,z,u,p,q,r)-pF_u(x,y,z,u,p,q,r)=0,\\&\frac{\mathrm dq}{\mathrm dt}=-F_y(x,y,z,u,p,q,r)-qF_u(x,y,z,u,p,q,r)=0,\\&\frac{\mathrm dr}{\mathrm dt}=-F_z(x,y,z,u,p,q,r)-rF_u(x,y,z,u,p,q,r)=0\end{aligned}\right.$

   with the initial condition

   $x(0)=s_1,\quad y(0)=s_2,\quad z(0)=-\alpha s_1-\beta s_2,\quad u(0)=u(x(0),y(0),z(0))=k$.

   From $F(x(0),y(0),z(0),u(0),p(0),q(0),r(0))=0$, we have $p^2(0)+q^2(0)+r^2(0)=1$. By the strip condition, we have

   $\displaystyle\begin{aligned}&0=\frac{\mathrm dz(0)}{\mathrm ds_1}=p(0)\frac{\mathrm dx(0)}{\mathrm ds_1}+q(0)\frac{\mathrm dy(0)}{\mathrm ds_1}+r(0)\frac{\mathrm dz(0)}{\mathrm ds_1}=p(0)\cdot1+q(0)\cdot0+r(0)\cdot-\alpha=p(0)-\alpha r(0),\\&0=\frac{\mathrm dz(0)}{\mathrm ds_2}=p(0)\frac{\mathrm dx(0)}{\mathrm ds_2}+q(0)\frac{\mathrm dy(0)}{\mathrm ds_2}+r(0)\frac{\mathrm dz(0)}{\mathrm ds_2}=p(0)\cdot0+q(0)\cdot1+r(0)\cdot-\beta=q(0)-\beta r(0).\end{aligned}$

   Thus, we obtain

   $\displaystyle p(0)=\pm\frac{\alpha}{\sqrt{\alpha^2+\beta^2+1}},\quad q(0)=\pm\frac{\beta}{\sqrt{\alpha^2+\beta^2+1}},\quad r(0)=\pm\frac1{\sqrt{\alpha^2+\beta^2+1}}$.

   From $\displaystyle\frac{\mathrm dp}{\mathrm dt}=\frac{\mathrm dq}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}=0$, we get

   $\displaystyle p(t)=\pm\frac{\alpha}{\sqrt{\alpha^2+\beta^2+1}},\quad q(t)=\pm\frac{\beta}{\sqrt{\alpha^2+\beta^2+1}},\quad r(t)=\pm\frac1{\sqrt{\alpha^2+\beta^2+1}}$.

   Hence, the first four equations become

   $\left\{\begin{aligned} &\frac{\mathrm dx}{\mathrm dt}=\pm\frac{2\alpha}{\sqrt{\alpha^2+\beta^2+1}},\\&\frac{\mathrm dy}{\mathrm dt}=\pm\frac{2\beta}{\sqrt{\alpha^2+\beta^2+1}},\\&\frac{\mathrm dz}{\mathrm dt}=\pm\frac2{\sqrt{\alpha^2+\beta^2+1}},\\&\frac{\mathrm du}{\mathrm dt}=2.\end{aligned}\right.$

   It yields that

   $\begin{aligned} &x(t)=s_1\pm\frac{2t\alpha}{\sqrt{\alpha^2+\beta^2+1}},\quad y(t)=s_2\pm\frac{2t\beta}{\sqrt{\alpha^2+\beta^2+1}},\\&z(t)=-\alpha s_1-\beta s_2\pm\frac{2t}{\sqrt{\alpha^2+\beta^2+1}},\quad u(t)=2t+k.\end{aligned}$

   Here $(t,s_1,s_2)$ can be expressed in terms of $(x,y,z)$ by

   $\displaystyle t=\pm\frac{\alpha x+\beta y+z}{2\sqrt{\alpha^2+\beta^2+1}},\quad s_1=\frac{\beta^2x+x-\alpha\beta y-\alpha z}{\alpha^2+\beta^2+1},\quad s_2=\frac{\alpha^2y+y-\alpha\beta x-\beta z}{\alpha^2+\beta^2+1}$.

   Therefore, we obtain

   $\displaystyle u(x,y,z)=k\pm\frac{\alpha x+\beta y+z}{\sqrt{\alpha^2+\beta^2+1}}\quad\text{for}~(x,y,z)\in\mathbb R^3$.

2. Let $u(x,y,z;a,b,c)=x\cos a\sin b+y\sin a\sin b+z\cos b+c$. It is easy to get

   $u_x(x,y,z;a,b,c)=\cos a\sin b,\quad u_y(x,y,z;a,b,c)=\sin a\sin b,\quad u_z(x,y,z;a,b,c)=\cos b$

   and verify that $u_x^2(x,y,z;a,b,c)+u_y^2(x,y,z;a,b,c)+u_z^2(x,y,z;a,b,c)=1$, which means $u(x,y,z;a,b,c)$ is a solution. Moreover, the matrix

   $\begin{aligned} &\quad\begin{bmatrix}u_a(x,y,z;a,b,c)&u_{xa}(x,y,z;a,b,c)&u_{ya}(x,y,z;a,b,c)&u_{za}(x,y,z;a,b,c)\\u_b(x,y,z;a,b,c)&u_{xb}(x,y,z;a,b,c)&u_{yb}(x,y,z;a,b,c)&u_{zb}(x,y,z;a,b,c)\\u_c(x,y,z;a,b,c)&u_{xc}(x,y,z;a,b,c)&u_{yc}(x,y,z;a,b,c)&u_{zc}(x,y,z;a,b,c)\end{bmatrix}\\&=\begin{bmatrix}-x\sin a\sin b+y\cos a\sin b&\sin a\sin b&\cos a\sin b&0\\x\cos a\cos b+y\sin a\sin b-z\sin b&\cos a\cos b&\sin a\cos b&-\sin b\\1&0&0&0\end{bmatrix}\end{aligned}$

   is of rank $3$. Therefore, $u=u(x,y,z;a,b,c)$ is a complete integral.