1. Calculate the Jacobians $\frac{\partial(u, v)}{\partial(x, y)}$ and $\frac{\partial(x, y)}{\partial(u, v)}$, and verify that $\frac{\partial(u, v)}{\partial(x, y)} \frac{\partial(x, y)}{\partial(u, v)}=1$, in each of the following cases:
    (i) $u=x+y, v=\frac{y}{x};\quad $(ii) $u=\frac{x^{2}}{y},  v=\frac{y^{2}}{x}$
    (i) $\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1&1\\-\frac{y}{x^2}&\frac1x\end{vmatrix}=\frac{x+y}{x^2}$,$\left.\begin{array}{ll}1=x_u+y_u&0=\frac{-y}{x^2}x_u+\frac{y_u}x\\0=x_v+y_v&1=-\frac y{x^2}x_v+\frac{y_v}x\end{array}\right\}$⇒$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\frac{x}{x+y}&\frac y{x+y}\\-\frac{x^2}{x+y}&x^2\over x+y\end{vmatrix}=\frac{x^2}{x+y}$⇒$\frac{\partial(u, v)}{\partial(x, y)}\frac{\partial(x, y)}{\partial(u, v)}=1$.
    (ii) $\frac{\partial(u, v)}{\partial(x, y)}=\begin{vmatrix}\frac{2x}y&-\frac{x^2}{y^2}\\-\frac{y^2}{x^2}&\frac{2y}x\end{vmatrix}=3$,$\left.\begin{array}{ll}1=x_u\frac{2x}y-y_u\frac{x^2}{y^2}&0 = \frac{y^2}{x^2}x_u + 2\frac yx y_u\\0= x_v\frac{2x}y - y_v\frac{x^2}{y^2}&1 = -\frac{y^2}{x^2}x_v + \frac {2y}x y_v\end{array}\right\}⇒\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}
    \frac{2 y}{3 x} & \frac{y^2}{3 x^2} \\
    \frac{x^2}{3 y^2} & \frac{2 x}{3 y} \\
    \end{vmatrix}=\frac13$. $\frac{\partial(u, v)}{\partial(x, y)}\frac{\partial(x, y)}{\partial(u, v)}=3\cdot\frac13=1$.
  2. The variables $u$ and $v$ are given by $u=x^{2}-x y$, $v=y^{2}+x y$ for all real $x$ and $y$. By finding an appropriate Jacobian matrix, calculate the partial derivatives $x_u$,$x_v$,$y_u$ and $y_v$ in terms of $x$ and $y$ only. State the values of $x$ and $y$ for which your results are valid.
    Jacobian matrix is $\begin{pmatrix}u_x&u_y\\v_x&v_y\end{pmatrix}$, where $u_x=2x-y,u_y=-x,v_x=y,v_y=2y+x$.
    By calculating inverse matrix, $x_u=\frac{x+2 y}{2 x^2+4 x y-2 y^2}$, $x_v=\frac{x}{2 x^2+4 x y-2 y^2}$, $y_u=-\frac{y}{2 x^2+4 x y-2 y^2}$, $y_v=\frac{2 x-y}{2 x^2+4 x y-2 y^2}$.
    Valid condition: $y\ne(1\pm\sqrt2) x$.
  3. Recall the definition of parabolic coordinates $(u,v)$ given by the relationships with the Cartesian coordinates $(x,y)$: $x=\frac{1}{2}\left(u^{2}-v^{2}\right), y = uv.$
    Show that Laplace’s equation in Cartesian coordinates, which is given by $\frac{\partial^{2} F}{\partial x^{2}}+\frac{\partial^{2} F}{\partial y^{2}}=0$ transforms into the same equation in parabolic coordinates.
    $\frac{\partial F}{\partial u}=u\frac{\partial F}{\partial x}+v\frac{\partial F}{\partial y}$, $\frac{\partial F}{\partial v}=-v\frac{\partial F}{\partial x}+u\frac{\partial F}{\partial y}$.
    $\frac{\partial^2F}{\partial u^2}=\frac{\partial F}{\partial x}+u^2\frac{\partial^2 F}{\partial x^2}+v^2\frac{\partial^2F}{\partial y^2}+2uv\frac{\partial^2F}{\partial x\partial y}$, $\frac{\partial^2F}{\partial v^2}=-\frac{\partial F}{\partial x}+v^2\frac{\partial^2 F}{\partial x^2}+u^2\frac{\partial^2 F}{\partial y^2}-2uv\frac{\partial^2F}{\partial x\partial y}$.
    Adding, $\frac{\partial^2F}{\partial u^2}+\frac{\partial^2F}{\partial v^2}=(u^2+v^2)\left(\frac{\partial^2 F}{\partial x^2}+\frac{\partial^2 F}{\partial y^2}\right)$. Therefore, $\frac{\partial^2F}{\partial u^2}+\frac{\partial^2F}{\partial v^2}=0$.
  4. Given the partial differential equation $\frac{\partial^{2} z}{\partial x^{2}}-5 \frac{\partial^{2} z}{\partial x \partial y}+6 \frac{\partial^{2} z}{\partial y^{2}}=0$.
    make the change of variables $s = y + 2x$,$t = y + 3x$ and show that the PDE becomes $\frac{\partial^{2} z}{\partial s \partial t}=0$.
    Hence find the general solution, $z(x, y)$, to the original PDE.
    $\frac{\partial z}{\partial x}=2\frac{\partial z}{\partial s}+3\frac{\partial z}{\partial t}$,$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial s}+\frac{\partial z}{\partial t}$,
    $\frac{\partial^2 z}{\partial x^2}=4\frac{\partial^2z}{\partial s^2}+12\frac{\partial z}{\partial s\partial t}+9\frac{\partial^2 z}{\partial t^2}$
    $\frac{\partial^2 z}{\partial x\partial y}=2\frac{\partial^2 z}{\partial s^2}+5\frac{\partial^2 z}{\partial s\partial t}+3\frac{\partial^2 z}{\partial t^2}$
    $\frac{\partial^2 z}{\partial y^2}=\frac{\partial^2z}{\partial s^2}+2\frac{\partial z}{\partial s\partial t}+\frac{\partial^2 z}{\partial t^2}$
    The PDE becomes $\frac{\partial^{2} z}{\partial s \partial t}=0$. So $z=f(s)+g(t)=f(y+2x)+g(y+3x)$.
  5. Given the pair of equations (called the Cauchy-Riemann equations, which are fundamental to complex analysis)
    $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
    show that both $u$ and $v$ satisfy Laplace’s equation $\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0$
    If $x = r \cos \theta, y = r \sin \theta$ show that the Cauchy-Riemann equations become
    $\frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta}, \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta}$
    $\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x\partial y},  \frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y\partial x}$, by $\frac{\partial^2 v}{\partial x\partial y}=\frac{\partial^2 v}{\partial y\partial x}$, we find $u$ satisfy Laplace's equation.
    $\frac{\partial^2 v}{\partial x^2}=-\frac{\partial^2 u}{\partial x\partial y},  \frac{\partial^2 v}{\partial y^2}=\frac{\partial^2 u}{\partial y\partial x}$, by $\frac{\partial^2 u}{\partial x\partial y}=\frac{\partial^2 u}{\partial y\partial x}$, we find $v$ satisfy Laplace's equation.
    $\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cos\theta+\frac{\partial u}{\partial y}\sin\theta$, $\frac{\partial v}{\partial \theta}=-\frac{\partial v}{\partial x}r\sin\theta+\frac{\partial v}{\partial y}r\cos\theta$, by the Cauchy-Riemann equations, we have $\frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta}$.
    $\frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\cos\theta+\frac{\partial v}{\partial y}\sin\theta$, $\frac{\partial u}{\partial \theta}=-\frac{\partial u}{\partial x}r\sin\theta+\frac{\partial u}{\partial y}r\cos\theta$, by the Cauchy-Riemann equations, we have $\frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta}$.
  6. Use chain rule of Jacobian to show that, in spherical coordinates,
       \cos \varphi & -r \sin \theta \sin\varphi   & 0 \\
      \sin  \varphi & r \sin\theta  \cos\varphi & 0 \\
    0 & 0 & 1 \\
    \begin{array}{ccc}\sin\theta& r \cos\theta & 0 \\
    0 & 0 & 1 \\
    \cos\theta  & -r \sin\theta & 0 \\\end{array}\right)=\left(
    \sin \theta \cos \varphi  & r \cos \theta \cos\varphi & -r \sin\theta\sin\varphi \\
    \sin \theta \sin \varphi  & r \cos \theta \sin\varphi & r \sin\theta\cos \varphi \\
    \cos\theta & -r \sin \theta  & 0 \\\end{array}\right)$