Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 2.2

1. Reduce to canonical form:
   1. $u_{xx}+5u_{xy}+6u_{yy}=0$
   2. $x^2u_{xx}-y^2u_{yy}=0$
Solution
1. This equation can be written as $au_{xx}+bu_{xy}+cu_{yy}=0$, where $a=1$, $b=5$ and $c=6$. Then by equation (10), we have

   $\displaystyle\frac{\mathrm dy}{\mathrm dx}=\frac{b\pm\sqrt{b^2-4ac}}{2a}=\frac{5\pm\sqrt{25-24}}{2}=\frac{5\pm1}2=2$ or $3$,

   which implies $y=2x+C_1$ and $y=3x+C_2$ are the characteristics, and the PDE is hyperbolic on the whole plane $\mathbb R^2$. Let $\mu(x,y)=y-2x$ and $\eta(x,y)=y-3x$. Then we define $\tilde u:\mathbb R^2\to\mathbb R$ by $\tilde u(\mu,\eta)=u(x,y)=u(\mu-\eta,3\mu-2\eta)$ for $(\mu,\eta)\in\mathbb R^2$. It is easy to verify that

   $\begin{aligned}\tilde u_{\mu\eta}(\mu,\eta)&=\partial_\mu\partial_\eta\tilde u(\mu,\eta)=\partial_\mu\left[u_x(\mu-\eta,3\mu-2\eta)\cdot(-1)+u_y(\mu-\eta,3\mu-2\eta)\cdot(-2)\right]\\&=\left[-u_{xx}(\mu-\eta,3\mu-2\eta)-3u_{xy}(\mu-\eta,3\mu-2\eta)\right]+[-2u_{yx}(\mu-\eta,3\mu-2\eta)-6u_{yy}(\mu-\eta,3\mu-2\eta)]\\&=0.\end{aligned}$

   This gives $\tilde u_{\mu\eta}(\mu,\eta)=0$.
2. This equation can be written as $a(x,y)u_{xx}+b(x,y)u_{xy}+c(x,y)u_{yy}=0$, where $a(x,y)=x^2$, $b(x,y)=0$ and $c(x,y)=-y^2$. Then by equation (10), we have

   $\displaystyle\frac{\mathrm dy}{\mathrm dx}=\frac{b\pm\sqrt{b^2-4ac}}{2a}=\frac{0\pm\sqrt{0+4x^2y^2}}{2x^2}=\pm\frac{y}x$,

   which implies $y=C_1x$ and $xy=C_2$. Let $\mu(x,y)=y/x$ and $\eta(x,y)=xy$. Then we define $\tilde u(\eta,\mu)=u(x,y)$ for $(x,y)$ and $(\mu,\eta)$ in some suitable domain. A direct computation gives

   $\begin{aligned} &u_x(x,y)=\tilde u_\mu(\mu,\eta)\mu_x+\tilde u_\eta(\mu,\eta)\eta_x=-\frac y{x^2}\tilde u_\mu(\mu,\eta)+y\tilde u_\eta(\mu,\eta),\\&u_y(x,y)=\tilde u_\mu(\mu,\eta)\mu_y+\tilde u_\eta(\mu,\eta)\eta_y=\frac1x\tilde u_\mu(\mu,\eta)+x\tilde u_\eta(\mu,\eta).\end{aligned}$

   Moreover, we get the second order partial derivatives

   $\begin{aligned} &u_{xx}(x,y)=\frac{y^2}{x^4}\tilde u_{\mu\mu}(\mu,\eta)-\frac{2y^2}{x^2}\tilde u_{\mu\eta}(\mu,\eta)+y^2\tilde u_{\eta\eta}(\mu,\eta)+\frac{2y}{x^3}\tilde u_\mu(\mu,\eta),\\&u_{yy}(x,y)=\frac1{x^2}\tilde u_{\mu\mu}(\mu,\eta)+2\tilde u_{\mu\eta}(\mu,\eta)+x^2\tilde u_{\eta\eta}(\mu,\eta)\end{aligned}$

   Thus, the equation becomes

   $\begin{aligned}0=x^2u_{xx}(x,y)-y^2u_{yy}(x,y)&=\left(\frac{y}x\right)^2\tilde u_{\mu\mu}(\mu,\eta)-2y^2\tilde u_{\mu\eta}(\mu,\eta)+(xy)^2\tilde u_{\eta\eta}(\mu,\eta)+\frac{2y}x\tilde u_\mu(\mu,\eta)\\&\quad-(\frac yx)^2\tilde u_{\mu\mu}(\mu,\eta)-2y^2\tilde u_{\mu\eta}(\mu,\eta)-(xy)^2\tilde u_{\eta\eta}(\mu,\eta)\\&=-4y^2\tilde u_{\mu\eta}(\mu,\eta)+\frac{2y}x\tilde u_\mu(\mu,\eta)=-4\mu\eta\tilde u_{\mu\eta}(\mu,\eta)+2\mu\tilde u_\mu(\mu,\eta),\end{aligned}$

   which means $\displaystyle\tilde u_{\mu\eta}(\mu,\eta)=\frac1{2\eta}\tilde u_\mu(\mu,\eta)$.


2. Find the general solution:
   1. $u_{xx}-2u_{xy}\sin x-u_{yy}\cos^2x-u_y\cos x=0$
   2. $y^2u_{xx}-2yu_{xy}+u_{yy}=u_x+6y$
Solution

1. The principal part of equation can be denoted by the principal symbol

   $\sigma(\xi)=\sigma(x,y;\xi)=\xi_1^2+(-2\sin x)\xi_1\xi_2+(-\cos^2x)\xi_2^2$.

   Then by (10), we have

   $\displaystyle\frac{\mathrm dy}{\mathrm dx}=\frac{-2\sin x\pm\sqrt{4\sin^2x-4\cdot1\cdot(-\cos^2x)}}{2}=-\sin x\pm1$,

   which gives $y=\cos x\pm x+C$. Let $\mu(x,y)=y-\cos x+x$ and $\eta(x,y)=y-\cos x-x$. Then we define $\tilde u(\mu,\eta)=u(x,y)$. A simple computation gives

   $\begin{aligned} &u_x(x,y)=\tilde u_\mu(\mu,\eta)\mu_x+\tilde u_\eta(\mu,\eta)\eta_x=(\sin x+1)\tilde u_\mu(\mu,\eta)+(\sin x-1)\tilde u_\eta(\mu,\eta),\\&u_y(x,y)=\tilde u_\mu(\mu,\eta)\mu_y+\tilde u_\eta(\mu,\eta)\eta_y=\tilde u_\mu(\mu,\eta)+\tilde u_\eta(\mu,\eta).\end{aligned}$

   Moreover, we get the second order partial derivaitves

   $\begin{aligned} &u_{xx}(x,y)=(\sin x+1)^2\tilde u_{\mu\mu}(\mu,\eta)-(2\cos^2x)\tilde u_{\mu\eta}(\mu,\eta)+(\sin x-1)^2\tilde u_{\eta\eta}(\mu,\eta)+(\cos x)\tilde u_\mu(\mu,\eta)+(\cos x)\tilde u_\eta(\mu,\eta),\\&u_{xy}(x,y)=(\sin x+1)\tilde u_{\mu\mu}(\mu,\eta)+(2\sin x)\tilde u_{\mu\eta}(\mu,\eta)+(\sin x-1)\tilde u_{\eta\eta}(\mu,\eta),\\&u_{yy}(x,y)=\tilde u_{\mu\mu}(\mu,\eta)+2\tilde u_{\mu\eta}(\mu,\eta)+\tilde u_{\eta\eta}(\mu,\eta).\end{aligned}$

   Therefore, the partial differential equation becomes

   $\tilde u_{\mu\eta}(\mu,\eta)=0$,

   which can be solved by $\tilde u(\mu,\eta)=F(\mu)+G(\eta)$ for some $F,G\in\mathcal C^2(\mathbb R)$. Therefore, we obtain

   $u(x,y)=F(y-\cos x+x)+G(y-\cos x-x)$.

2. The principal part of equation can be denoted by the principal symbol

   $\sigma(\xi)=\sigma(x,y;\xi)=y^2\xi_1^2-2y\xi_1\xi_2+\xi_2^2$.

   Then by (10), we have

   $\displaystyle\frac{\mathrm dy}{\mathrm dx}=\frac{-2y\pm\sqrt{4y^2-4\cdot y^2\cdot1}}{2y^2}=-\frac1y$,

   which gives $2x+y^2=C$. Let $\mu(x,y)=2x+y^2$ and $\eta(x,y)=y$. Then we define $\tilde u(\mu,\eta)=u(x,y)$. A simple computation gives

   $\begin{aligned} &u_x(x,y)=\tilde u_\mu(\mu,\eta)\mu_x+\tilde u_\eta(\mu,\eta)\eta_x=2\tilde u_\mu(\mu,\eta),\\&u_y(x,y)=\tilde u_\mu(\mu,\eta)\mu_y+\tilde u_\eta(\mu,\eta)\eta_y=2y\tilde u_\mu(\mu,\eta)+\tilde u_\eta(\mu,\eta).\end{aligned}$

   Moreover, we get the second order partial derivatives

   $\begin{aligned} &u_{xx}(x,y)=4\tilde u_{\mu\mu}(\mu,\eta),\\&u_{xy}(x,y)=8y\tilde u_{\mu\mu}(\mu,\eta)+2\tilde u_{\mu\eta}(\mu,\eta),\\&u_{yy}(x,y)=4y^2\tilde u_{\mu\mu}(\mu,\eta)+4y\tilde u_{\mu\eta}(\mu,\eta)+\tilde u_{\eta\eta}(\mu,\eta)+2\tilde u_\mu(\mu,\eta).\end{aligned}$

   Therefore, the partial differential equation becomes

   $\tilde u_{\eta\eta}(\mu,\eta)=6\eta$,

   which implies $\tilde u_\eta(\mu,\eta)=3\eta^2+F(\mu)$, and hence $\tilde u(\mu,\eta)=\eta^3+F(\mu)\eta+G(\mu)$ for some $F,G\in\mathcal C^2(\mathbb R)$. Therefore, we obtain

   $u(x,y)=y^3+F(2x+y^2)y+G(2x+y^2)$.



3. Show that the function

   $u(x,y)=\begin{cases}0&\text{if}~x\leq y\\(x-y)^2&\text{if}~x>y\end{cases}$

   satisfies $u_{xx}-u_{yy}=0$ for all $x\neq y$. Is $u\in C^1(\mathbb{R}^2)$? Where does $u$ fail to be $C^2$?
Solution
It is clear that
$u_x(x,y)=\begin{cases}0&\text{if}~x<y;\\2(x-y)&\text{if}~x>y,\end{cases}\quad\text{and}\quad u_y(x,y)=\begin{cases}0&\text{if}~x<y;\\-2(x-y)&\text{if}~x>y.\end{cases}$
Now we consider the case that $x=y$. Note that
$\begin{aligned} &\lim_{h\to0^+}\frac{u(a+h,a)-u(a,a)}h=\lim_{h\to0^+}\frac{h^2-0}h=0=\lim_{h\to0^-}\frac{0-0}h=\lim_{h\to0^-}\frac{u(a+h,a)-u(a,a)}h,\\&\lim_{h\to0^+}\frac{u(a,a+h)-u(a,a)}h=\lim_{h\to0^+}\frac{0-0}h=0=\lim_{h\to0^-}\frac{(-h)^2-0}h=\lim_{h\to0^-}\frac{u(a,a+h)-u(a,a)}h.\end{aligned}$
This implies $u_x(a,a)=u_y(a,a)=0$ for any $a\in\mathbb R$. Therefore, we have
$u_x(x,y)=\begin{cases}0&\text{if}~x\leq y;\\2(x-y)&\text{if}~x>y,\end{cases}\quad\text{and}\quad u_y(x,y)=\begin{cases}0&\text{if}~x\leq y;\\-2(x-y)&\text{if}~x>y.\end{cases}$
This shows $u\in\mathcal C^1(\mathbb R^2)$. Moreover, we compute the second order partial derivatives as follows. It is clear that
$u_{xx}(x,y)=\begin{cases}0&\text{if}~x<y;\\2&\text{if}~x>y,\end{cases}\quad\text{and}\quad u_{yy}(x,y)=\begin{cases}0&\text{if}~x<y;\\2&\text{if}~x>y.\end{cases}$
Now we consider the case that $x=y$. Note that
$\begin{aligned} &\lim_{h\to0^+}\frac{u_x(a+h,a)-u_x(a,a)}h=\lim_{h\to0^+}\frac{2h-0}h=2\neq0=\lim_{h\to0^-}\frac{0-0}h=\lim_{h\to0^-}\frac{u_x(a+h,a)-u_x(a,a)}h,\\&\lim_{h\to0^+}\frac{u_y(a,a+h)-u_y(a,a)}h=\lim_{h\to0^+}\frac{0-0}h=0\neq2=\lim_{h\to0^-}\frac{-2(-h)-0}h=\lim_{h\to0^-}\frac{u_y(a,a+h)-u_y(a,a)}h.\end{aligned}$
Therefore, $u_{xx}(a,a)$ and $u_{yy}(a,a)$ does not exist for any $a\in\mathbb R$, and $u\in\mathcal C^2(\{(x,y)\in\mathbb R^2:\,y\neq x\})$. $u$ is not $C^2$ on the set $\{(x,y)\in\mathbb R^2:\,y=x\}$. Clearly, $u_{xx}-u_{yy}=0$ for all $x\neq y$.


4. Show that the minimal surface equation $(1+u_y^2)u_{xx}-2u_xu_yu_{xy}+(1+u_x^2)u_{yy}=0$ is everywhere elliptic.
Solution
Let $F:\mathbb R^6\to\mathbb R$ be defined by
$F(z,p,q,r,s,t)=(1+q^2)r-2pqs+(1+q^2)t\quad\text{for}~(z,p,q,r,s,t)\in\mathbb R^6$.
Then let
$\begin{aligned} &a(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=F_r(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=1+u_y^2,\\&b(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=F_s(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=-2u_xu_y,\\&c(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=F_t(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=1+u_x^2.\end{aligned}$
Because $b^2-4ac=4u_x^2u_y^2-4(1+u_y^2)(1+u_x^2)=-4-4(u_x^2(x,y)+u_y^2(x,y))\leq-4<0$ for any $(x,y)\in\mathbb R^2$, the minimal surface equation is everywhere elliptic.


5. Show that the Monge-Ampère equation $u_{xx}u_{yy}-u_{xy}^2=f(x)$ is elliptic for a solution $u$ exactly when $f(x)>0$. [In this case the graph of $u(x,y)$ is convex.]
Solution
Let $F:\mathbb R^4\to\mathbb R$ defined by
$F(x,r,s,t)=rt-s^2-f(x)\quad\text{for}~(x,r,s,t)\in\mathbb R^4$.
Then let
$\begin{aligned} &a(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=F_r(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=u_{yy},\\&b(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=F_s(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=-2u_{xy},\\&c(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=F_t(u,u_x,u_y,u_{xx},u_{xy},u_{yy})=u_{xx}.\end{aligned}$
For $x$ satisfying $f(x)>0$, we have $b^2-4ac=4u_{xy}^2-4u_{xx}u_{yy}=-4f(x)<0$. This means the Monge-Ampère equation is elliptic on the set $\{(x,y)\in\mathbb R^2\,:\,f(x)<0\}$.


6. Reduce to the form (22) and solve the initial value problem

   $\vec u_t+\begin{pmatrix}-4&-6\\3&5\end{pmatrix}\vec u_x=\begin{pmatrix}1\\-1\end{pmatrix},\quad\vec u(x,0)=\begin{pmatrix}x\\0\end{pmatrix}.$

Solution

Let $A=\begin{pmatrix}-4&-6\\3&5\end{pmatrix}$. Then the characteristic polynomial is

$\text{ch}_A(x)=\det(xI-A)=\begin{vmatrix}x+4&6\\-3&x-5\end{vmatrix}=(x+4)(x-5)+18=x^2-x-2=(x-2)(x+1)$.

Thus, the eigenvalues of matrix $A$ are $-1$ and $2$. Clearly, the associated eigenvectors are $\begin{pmatrix}-2\\1\end{pmatrix}$ and $\begin{pmatrix}1\\-1\end{pmatrix}$, respectively. Let $\Gamma=\begin{pmatrix}-2&1\\1&-1\end{pmatrix}$ and $\Lambda=\begin{pmatrix}-1&0\\0&2\end{pmatrix}$. Then $A=\Gamma^{-1}\Lambda\Gamma$. Define $\begin{pmatrix}v^{(1)}\\v^{(2)}\end{pmatrix}=\vec v=\Gamma^{-1}\vec u$. Then equation with its initial condition beceoms

$\vec v_t+\Lambda\vec v_x=\Gamma^{-1}\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}0\\1\end{pmatrix}\quad\text{with}\quad\vec v(x,0)=\Gamma^{-1}\vec u(x,0)=\begin{pmatrix}-x\\-x\end{pmatrix}$,

which means $\left\{\begin{aligned} &v_t^{(1)}-v_x^{(1)}=0\\&v^{(1)}(x,0)=-x\end{aligned}\right.$ and $\left\{\begin{aligned} &v_t^{(2)}+2v_x^{(2)}=1\\&v^{(2)}(x,0)=-x\end{aligned}\right.$. Fix $x_0\in\mathbb R$. Define

$z^{(1)}(s)=v^{(1)}(x_0-s,s)$ and $z^{(2)}(s)=v^{(2)}(x_0+2s,s)$ for $s\in\mathbb R$.

It is clear that

$\displaystyle\frac{\mathrm dz^{(1)}}{\mathrm ds}=v_t^{(1)}(x_0+s,-s)-v_x^{(1)}(x_0+s,s)=0,\quad\frac{\mathrm dz^{(2)}}{\mathrm ds}=v_t^{(2)}(x_0+2s,s)+2v_x^{(2)}(x_0+2s,s)=1$.

It follows that

$z^{(1)}(s)=z^{(1)}(0)=v^{(2)}(x_0,0)=-x_0$ and $z^{(2)}(s)=z^{(2)}(0)+s=v^{(2)}(x_0,0)+s=-x_0+s$.

Hence, we get

$v^{(1)}(x,t)=-x-t,\quad v^{(2)}(x,t)=-x+3t.$

Therefore, we obtain the solution

$\vec u=\Gamma\vec v=\begin{pmatrix}-2&1\\1&-1\end{pmatrix}\begin{pmatrix}-x-t\\-x+3t\end{pmatrix}=\begin{pmatrix}1+5t\\-4t\end{pmatrix}$.



7. Reduce the following systems to the form (22).
   1. $\left\{\begin{aligned} &u_t+v_x=u\\&v_t+u_x=v\end{aligned}\right.$
   2. $\left\{\begin{aligned} &u_t+v_x=u\\&v_t+u_x=0\end{aligned}\right.$
   Do they both decouple?
Solution
1. Let $\vec w=\begin{pmatrix}u\\v\end{pmatrix}$ and $A=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. The linear first order system of partial differential equations can be represented as $\vec w_t+A\vec w_x=\vec w$. It is clear that $A$ has two distinct eigenvalues $1$ and $-1$ and the associated eigenvectors $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\-1\end{pmatrix}$, respectively. Thus, we set $\Gamma=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ and $\Lambda=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and obtain

   $A=\Gamma\Lambda\Gamma^{-1}=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}^{-1}$.

   Define $\vec z=\Gamma^{-1}\vec w$. Then it is easy to verify that

   $\vec z_t+\Lambda\vec z_x=\Gamma^{-1}\vec w_t+\Lambda\Gamma^{-1}\vec w_x=\Gamma^{-1}(\vec w_t+A\vec w_x)=\Gamma^{-1}\vec w=\vec z,$

   which shows the system can be decoupled.
2. Let $\vec z=\Gamma^{-1}\vec w=\begin{pmatrix}1&1\\1&-1\end{pmatrix}^{-1}\begin{pmatrix}u\\v\end{pmatrix}$ and $\Lambda=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Then we find

   $\vec z_t+\Lambda \vec z_x=\Gamma^{-1}\vec w_t+\Lambda\Gamma^{-1}\vec w_x=\Gamma^{-1}(\vec w_t+A\vec w_x)=\Gamma^{-1}\begin{pmatrix}u\\0\end{pmatrix}$,

   which cannot be decounpled.
   
   Note that this system can be reduced in term of function $v$ as follows. Note that $v_{tt}=-u_{tx}$ by differentiating the second equation with respect to $t$. Then the first equation gives $u_{tx}+v_{xx}=u_x$, which implies

   $-v_{tt}+v_{xx}=-v_t$,

   which is hyperbolic.


8. Solve (26), (27) under the condition (29) [by solving (30), (31)].
Solution

Based on the reduction in the textbook, the equations (30) and (31) can be read as

$\left\{\begin{aligned} &v_t^{(1)}+\frac1{\sqrt{LC}}v_x^{(1)}=-\frac{RC+GL}{2LC}v^{(1)}=-\frac RLv^{(1)},\\&v^{(1)}(x,0)=\frac1{2\sqrt{LC}}[I_0(x)\sqrt L+V_0(x)\sqrt C],\end{aligned}\right.\quad\text{and}\quad\left\{\begin{aligned} &v_t^{(2)}-\frac1{\sqrt{LC}}v_x^{(2)}=-\frac{RC+GL}{2LC}v^{(2)}=-\frac RLv^{(2)},\\&v^{(2)}(x,0)=\frac1{2\sqrt{LC}}[I_0(x)\sqrt L-V_0(x)\sqrt C],\end{aligned}\right.$

where $G$, $C$, $R$ and $L$ are all positive constants with $RC=GL$.

Fix $x_0\in\mathbb R$. By the method of characteristic line, we let

$z^{(1)}(s)=v^{(1)}(x_0+s/\sqrt{LC},s)$ and $z^{(2)}(s)=v^{(2)}(x_0-s/\sqrt{LC},s)$ for $s\in\mathbb R$.

Then it is clear that

$\begin{aligned} &\frac{\mathrm dz^{(1)}}{\mathrm ds}=v_t^{(1)}(x_0+s/\sqrt{LC},s)+\frac1{\sqrt{LC}}v_x^{(1)}(x_0+s/\sqrt{LC},s)=-\frac RLv^{(1)}(x_0+s/\sqrt{LC},s)=-\frac RLz^{(1)},\\&\frac{\mathrm dz^{(2)}}{\mathrm ds}=v_t^{(2)}(x_0-s/\sqrt{LC},s)-\frac1{\sqrt{LC}}v_x^{(2)}(x_0-s/\sqrt{LC},s)=-\frac RLv^{(2)}(x_0-s/\sqrt{LC},s)=-\frac RLz^{(2)}.\end{aligned}$

Thus, we obtain

$\begin{aligned} &z^{(1)}(s)=z^{(1)}(0)\exp(-Rs/L)=\frac{I_0(x_0)\sqrt L+V_0(x_0)\sqrt C}{2\sqrt{LC}}\exp(-Rs/L),\\&z^{(2)}(s)=z^{(2)}(0)\exp(-Rs/L)=\frac{I_0(x_0)\sqrt L-V_0(x_0)\sqrt C}{2\sqrt{LC}}\exp(-Rs/L).\end{aligned}$

Therefore, we obtain

$\begin{aligned} &v^{(1)}(x,t)=\frac{I_0(x-t/\sqrt{LC})\sqrt L+V_0(x-t/\sqrt{LC})\sqrt C}{2\sqrt{LC}}\exp(-Rt/L),\\&v^{(2)}(x,t)=\frac{I_0(x+t/\sqrt{LC})\sqrt L-V_0(x+t/\sqrt{LC})\sqrt C}{2\sqrt{LC}}\exp(-Rt/L),\end{aligned}$

and the solution is

$\begin{aligned}\vec u&=\Gamma\vec v=\begin{pmatrix}\sqrt C&\sqrt C\\\sqrt L&-\sqrt L\end{pmatrix}\begin{pmatrix}\displaystyle\frac{I_0(x-t/\sqrt{LC})\sqrt L+V_0(x-t/\sqrt{LC})\sqrt C}{2\sqrt{LC}}\exp(-Rt/L)\\\displaystyle\frac{I_0(x+t/\sqrt{LC})\sqrt L-V_0(x+t/\sqrt{LC})\sqrt C}{2\sqrt{LC}}\exp(-Rt/L)\end{pmatrix}\\&=\frac{\exp(-Rt/L)}{2\sqrt{LC}}\begin{pmatrix}\sqrt C&\sqrt C\\\sqrt L&-\sqrt L\end{pmatrix}\begin{pmatrix}I_0(x-t/\sqrt{LC})\sqrt L+V_0(x-t/\sqrt{LC})\sqrt C\\I_0(x+t/\sqrt{LC})\sqrt L-V_0(x+t/\sqrt{LC})\sqrt C\end{pmatrix}\\&=\frac{\exp(-Rt/L)}2\begin{pmatrix}\displaystyle I_0(x-t/\sqrt{LC})+I_0(x+t/\sqrt{LC})+\frac{\sqrt C}{\sqrt L}V_0(x-t/\sqrt{LC})+\frac{\sqrt C}{\sqrt L}V_0(x+t/\sqrt{LC})\\\displaystyle V_0(x-t/\sqrt{LC})+V_0(x+t/\sqrt{LC})+\frac{\sqrt L}{\sqrt C}I_0(x-t/\sqrt{LC})-\frac{\sqrt L}{\sqrt C}I_0(x+t/\sqrt{LC})\end{pmatrix}.\end{aligned}$



9. Solve the wave equation $u_{xx}-u_{yy}=0$ with initial conditions $u(x,0)=g(x)$, $u_y(x,0)=h(x)$ by using a $3\times3$ system of first-order equations.
Solution

Let $\vec v=\begin{pmatrix}u\\u_x\\u_y\end{pmatrix}$. Then by the wave equation $u_{xx}-u_{yy}=0$, $\vec v$ satisfies

$\vec v_x+A\vec v_y=\begin{pmatrix}u_x\\0\\0\end{pmatrix}$,

where $A=\begin{pmatrix}0&0&0\\0&0&-1\\0&-1&0\end{pmatrix}$. It is easy to find the eigenvalues of $A$ are $1$, $-1$ and $0$ and the associated eigenvectors are $\begin{pmatrix}0\\1\\-1\end{pmatrix}$, $\begin{pmatrix}0\\1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\0\\0\end{pmatrix}$, respectively. Thus, we put

$\Gamma=\begin{pmatrix}0&0&1\\1&1&0\\-1&1&0\end{pmatrix},\quad\Lambda=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}$.

Note that $A=\Gamma\Lambda\Gamma^{-1}$. Then we define $\vec w=\Gamma^{-1}\vec v$ so $\vec w=\begin{pmatrix}w^{(1)}\\w^{(2)}\\w^{(3)}\end{pmatrix}$ satisfies

$\vec w_x+\Lambda \vec w_y=\Gamma^{-1}\vec v_x+\Lambda\Gamma^{-1}\vec v_y=\Gamma^{-1}(\vec v_x+A\vec v_y)=\Gamma^{-1}\begin{pmatrix}u_x\\0\\0\end{pmatrix}=\begin{pmatrix}0&1/2&-1/2\\0&1/2&1/2\\1&0&0\end{pmatrix}\begin{pmatrix}u_x\\0\\0\end{pmatrix}=\begin{pmatrix}0\\0\\u_x\end{pmatrix}$.

Moreover, the initial condition

$\vec w(x,0)=\Gamma^{-1}\vec v(x,0)=\begin{pmatrix}u(x,0)\\u_x(x,0)\\u_y(x,0)\end{pmatrix}=\begin{pmatrix}0&1/2&-1/2\\0&1/2&1/2\\1&0&0\end{pmatrix}\begin{pmatrix}g(x)\\g'(x)\\h(x)\end{pmatrix}=\begin{pmatrix}(g'(x)-h(x))/2\\(g'(x)+h(x))/2\\g(x)\end{pmatrix}$

From the first two equations, we have

$\left\{\begin{aligned} &w_x^{(1)}+w_y^{(1)}=0\\&w^{(1)}(x,0)=(g'(x)-h(x))/2\end{aligned}\right.\quad\text{and}\quad\left\{\begin{aligned} &w_x^{(2)}-w_y^{(2)}=0\\&w^{(2)}(x,0)=(g'(x)+h(x))/2\end{aligned}\right.$

By the method of characteristic lines, we obtain

$\displaystyle w^{(1)}(x,y)=\frac{g'(x-y)-h(x-y)}2,\quad w^{(2)}(x,y)=\frac{g'(x+y)+h(x+y)}2.$

By $\vec u=\Gamma\vec w$, we have

$u_x(x,y)+u_y(x,y)=w^{(1)}(x,y),\quad-u_x(x,y)+u_y(x,y)=w^{(2)}(x,y),$

which gives

$\begin{aligned} &u_x(x,y)=\frac{w^{(1)}(x,y)+w^{(2)}(x,y)}2=\frac{g'(x-y)+g'(x+y)-h(x-y)+h(x+y)}4,\\&u_y(x,y)=\frac{w^{(1)}(x,y)-w^{(2)}(x,y)}2=\frac{g'(x-y)-g'(x+y)-h(x-y)-h(x+y)}2.\end{aligned}$

Integrating $u_y$ with respect to $y$ yields

$\displaystyle u(x,y)=\frac{g(x+y)+g(x-y)}2+\frac12\int_{-y}^y\!h(x+\tau)\,\mathrm d\tau+s(x)$.

Then differentiating $u$ with respect to $x$, we obtain

$\begin{aligned}u_x(x,y)&=\frac{g'(x+y)+g'(x-y)}2+\int_{-y}^y\!h'(x+\tau)\,\mathrm d\tau+s'(x)\\&=\frac{g(x+y)+g(x-y)+h(x+y)-h(x-y)}2+s'(x),\end{aligned}$

which implies $s'(x)=0$. Therefore, we obtain the solution

$\displaystyle u(x,y)=\frac{g(x+y)+g(x-y)}2+\frac12\int_{-y}^y\!h(x+\tau)\,\mathrm d\tau+C$.

Due to the initial condition, we get $C=0$ and

$\displaystyle u(x,y)=\frac{g(x+y)+g(x-y)}2+\frac12\int_{-y}^y\!h(x+\tau)\,\mathrm d\tau$.



10. If $au_{xx}+2bu_{xy}+cu_{yy}=d$ is elliptic (i.e., $ac-b^2>0$), let $W=\sqrt{ac-b^2}$. Show that solutions $\mu$, $\eta$ of the Beltrami equations

    $\displaystyle\mu_x=\frac{b\eta_x+c\eta_y}{W}\quad\text{and}\quad\mu_y=-\frac{a\eta_x+b\eta_y}W$

    provide new coordinates transforming (7) to the form (14). [Note that $\displaystyle\mu(x,y)=\int_{\gamma}\!\mu_x\,\mathrm dx+\mu_y\,\mathrm dy$, where $\gamma$ is a path joining $(x,y)$ and a fixed point $p_0$; path independence is provided by the Beltrami equations.]
Solution

Define $\tilde u(\mu,\eta)=u(x,y)$, where $\mu=\mu(x,y)$ and $\eta=\eta(x,y)$ are the solution of Beltrami equations. Using the chain rule, we have

$\begin{aligned} &u_x(x,y)=\tilde u_\mu(\mu,\eta)\mu_x+\tilde u_\eta(\mu,\eta)\eta_x,\\&u_y(x,y)=\tilde u_\mu(\mu,\eta)\mu_y+\tilde u_\eta(\mu,\eta)\eta_y.\end{aligned}$

Moreover, we compute the second order partial derivatives:

$\begin{aligned} &u_{xx}(x,y)=\tilde u_{\mu\mu}(\mu,\eta)\mu_x^2+2\tilde u_{\mu\eta}(\mu,\eta)\mu_x\eta_x+\tilde u_{\eta\eta}(\mu,\eta)\eta_x^2+\tilde u_\mu(\mu,\eta)\mu_{xx}+\tilde u_{\eta}(\mu,\eta)\eta_{xx},\\&u_{xy}(x,y)=\tilde u_{\mu\mu}(\mu,\eta)\mu_x\mu_y+\tilde u_{\mu\eta}(\mu,\eta)(\mu_x\eta_y+\eta_x\mu_y)+\tilde u_{\eta\eta}(\mu,\eta)\eta_x\eta_y+\tilde u_\mu(\mu,\eta)\mu_{xy}+\tilde u_\eta(\mu,\eta)\eta_{xy},\\&u_{yy}(x,y)=\tilde u_{\mu\mu}(\mu,\eta)\mu_y^2+2\tilde u_{\mu\eta}(\mu,\eta)\mu_y\eta_y+\tilde u_{\eta\eta}(\mu,\eta)\eta_y^2+\tilde u_\mu(\mu,\eta)\mu_{yy}+\tilde u_\eta(\mu,\eta)\eta_{yy}.\end{aligned}$

Plugging into the elliptic partial differential equation, we obtain

$\begin{aligned}d&=au_{xx}(x,y)+2bu_{xy}(x,y)+cu_{yy}(x,y)\\&=a\left[\tilde u_{\mu\mu}(\mu,\eta)\mu_x^2+2\tilde u_{\mu\eta}(\mu,\eta)\mu_x\eta_x+\tilde u_{\eta\eta}(\mu,\eta)\eta_x^2+\tilde u_\mu(\mu,\eta)\mu_{xx}+\tilde u_{\eta}(\mu,\eta)\eta_{xx}\right]\\&\quad+2b\left[\tilde u_{\mu\mu}(\mu,\eta)\mu_x\mu_y+\tilde u_{\mu\eta}(\mu,\eta)(\mu_x\eta_y+\eta_x\mu_y)+\tilde u_{\eta\eta}(\mu,\eta)\eta_x\eta_y+\tilde u_\mu(\mu,\eta)\mu_{xy}+\tilde u_\eta(\mu,\eta)\eta_{xy}\right]\\&\quad+c[\tilde u_{\mu\mu}(\mu,\eta)\mu_y^2+2\tilde u_{\mu\eta}(\mu,\eta)\mu_y\eta_y+\tilde u_{\eta\eta}(\mu,\eta)\eta_y^2+\tilde u_\mu(\mu,\eta)\mu_{yy}+\tilde u_\eta(\mu,\eta)\eta_{yy}]\\&=(a\mu_x^2+2b\mu_x\mu_y+c\mu_y^2)\tilde u_{\mu\mu}(\mu,\eta)+(2a\mu_x\eta_x+2b\mu_x\eta_y+2b\mu_y\eta_x+2c\mu_y\eta_y)\tilde u_{\mu\eta}(\mu,\eta)\\&\quad+(a\eta_x^2+2b\eta_x\eta_y+c\eta_y^2)\tilde u_{\eta\eta}(\mu,\eta)+(a\mu_{xx}+2b\mu_{xy}+c\mu_{yy})\tilde u_\mu(\mu,\eta)+(a\eta_{xx}+2b\eta_{xy}+c\eta_{yy})\tilde u_\eta(\mu,\eta).\end{aligned}$

Now we need to compute the coefficients of $\tilde u_{\mu\mu}$, $\tilde u_{\mu\eta}$, $\tilde u_{\eta\eta}$, $\tilde u_\mu$ and $\tilde u_\eta$. A simple calculation gives

$\begin{aligned} &\begin{aligned}a\mu_x^2+2b\mu_x\mu_y+c\mu_y^2&=\frac{a(b\eta_x+c\eta_y)^2-2b(b\eta_x+c\eta_y)(a\eta_x+b\eta_y)+c(a\eta_x+b\eta_y)^2}{W^2}\\&=\frac{(ab^2-2ab^2+ca^2)\eta_x^2+(2abc-2abc-2b^3+2abc)\eta_x\eta_y+(ac^2-2b^2c+b^2c)\eta_y^2}{W^2}\\&=\frac{a(ac-b^2)\eta_x^2+2b(ac-b^2)\eta_x\eta_y+c(ac-b^2)\eta_y^2}{W^2}=a\eta_x^2+2b\eta_x\eta_y+c\eta_y^2,\end{aligned}\\&\begin{aligned}2a\mu_x\eta_x+2b\mu_x\eta_y+2b\mu_y\eta_x+2c\mu_y\eta_y&=\frac{2a(b\eta_x+c\eta_y)\eta_x+2b(b\eta_x+c\eta_y)\eta_y-2b(a\eta_x+b\eta_y)\eta_x-2c(a\eta_x+b\eta_y)\eta_y}W\\&=0\end{aligned}\\&\begin{aligned}a\mu_{xx}+2b\mu_{xy}+c\mu_{yy}&=\frac{a(b\eta_{xx}+c\eta_{xy})+2b(b\eta_{xy}+c\eta_{yy})-c(a\eta_{xy}+b\eta_{yy})}W=\frac{ab\eta_{xx}+2b^2\eta_{xy}+bc\eta_{yy}}{W}\\&=\frac bW(a\eta_{xx}+2b\eta_{xy}+c\eta_{yy}).\end{aligned}\end{aligned}$

In addition, by $\mu_{xy}=\mu_{yx}$, we can get $a\eta_{xx}+2b\eta_{xy}+c\eta_{yy}=0$. Therefore, we obtain

$\displaystyle\tilde u_{\mu\mu}(\mu,\eta)+\tilde u_{\eta\eta}(\mu,\eta)=\frac{d}{a\mu_x^2+2b\mu_x\mu_y+c\mu_y^2}.$