$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$

**Example 1.**
Consider a string as a curve $y=u(x,t)$ (so, it's shape depends on time $t$) with a tension $T$ and with a linear density $\rho$. We assume that $|u_x|\ll 1$.

Observe that at point $x$ the part of the string to the left from $x$ pulls it up with a force $-F(x):=-Tu_x$. Indeed, the force $T$ is directed along the curve and the slope of angle $\theta$ between the tangent to the curve and the horizontal line is $u_x$; so $\sin(\theta)=u_x/\sqrt{1+u_x^2}$ which under our assumption we can replace by $u_x$. On the other hand, at point $x$ the part of the string to the right from $x$ pulls it up with a force $F(x):=Tu_x$. Therefore the total $y$-component of the force applied to the segment of the string between $J=[x_1,x_2]$ equals \begin{equation*} F(x_2)-F(x_1)=\int_{J} \partial F(x)\, dx= \int_{J} Tu_{xx}\, dx. \end{equation*} According to Newton's law it must be equal to $\int_{J} \rho u_{tt}\, dx$ where $\rho dx$ is the mass and $u_{tt}$ is the acceleration of the infinitesimal segment $[x,x+dx]$: \begin{equation*} \int_{J} \rho u_{tt}\, dx=\int_{J} Tu_{xx}\, dx. \end{equation*} Since this equality holds for any segment $J$, the integrands coincide: \begin{equation} \rho u_{tt}= Tu_{xx}. \label{eq-1.4.1} \end{equation}

**Example 2.**
Consider a membrane as a surface $z=u(x,y,t)$ with a tension $T$ and with a surface density $\rho$. We assume that $|u_x|,|u_{yy}|\ll 1$.

Consider a domain $D$ on the plane, its boundary $L$ and a small segment of the length $ds$ of this boundary. Then the outer domain pulls this segment up with the force $-T\mathbf{n}\cdot \nabla u \,ds$ where $\mathbf{n}$ is the inner unit normal to this segment. Indeed, the total force is $T\,ds$ but it pulls along the surface and the slope of the surface in the direction of $\mathbf{n}$ is $\approx \mathbf{n}\cdot \nabla u$.

Therefore the total $z$-component of force applied to $D$ due to Gauss formula in dimension 2 to (A1.1.1) equals \begin{equation*} -\int_{L} T \mathbf{n}\cdot \nabla u\, ds= \iint_D \nabla \cdot (T\nabla u) \, dxdy. \end{equation*} According to Newton's law it must be equal to $\iint_D \rho u_{tt}\, dxdy$ where $\rho dxdy$ is the mass and $u_{tt}$ is the acceleration of the element of the area: \begin{equation*} \iint_D \rho u_{tt}\, dxdy=\iint_{D} T\Delta u\, dx \end{equation*} because $\nabla \cdot (T\nabla u)=T\nabla \cdot \nabla u= T \Delta u$. Since this equality holds for any domain, the integrands coincide: \begin{equation} \rho u_{tt}= T\Delta u. \label{eq-1.4.2} \end{equation}

**Example 3.**
Consider a gas and let $\mathbf{v}$ be its velocity and $\rho$ its density. Then \begin{align}
&\rho \mathbf{v}_t + \rho (\mathbf{v}\cdot \nabla ) \mathbf{v} = -\nabla p,
\label{eq-1.4.3}\\
&\rho_t + \nabla \cdot (\rho\mathbf{v})=0
\label{eq-1.4.4}
\end{align}
where $p$ is the pressure. Indeed, in (\ref{eq-1.4.3}) the left-hand expression is $\rho \dfrac{d\mathbf{v}}{dt} $ (the mass per unit of the volume multiplied by acceleration) and the right hand expression is the force of the pressure; no other forces are considered.

Further, (\ref{eq-1.4.4}) is *continuity equation* which means the mass conservation since the flow of the mass through the surface element $dS$ in the direction of the normal $\mathbf{n}$ for time $dt$ equals $\rho \mathbf{n}\cdot \mathbf{v}$.

**Remark 1.**
According to the chain rule
\begin{gather*}
\dfrac{du}{dt}=\dfrac{\partial u}{dt}+ (\nabla u)\cdot \dfrac{d\mathbf{x} }{dt}=
u_t + (\nabla u)\cdot \mathbf{v}
\end{gather*}
is a *derivative of $u$ along trajectory* which does not coincide with the *partial derivative* $u_t$; $\mathbf{v}\cdot \nabla u$ is called *convection term*. However in the linearization with $|\mathbf{v}|\ll 1$ it is negligible.

**Remark 2.**
Consider any domain $\Omega$ with a border $\Sigma$. The flow of the gas inwards for time $dt$ equals
\begin{gather*}
\iint_\Sigma \rho \boldsymbol{v}\cdot\boldsymbol{n}\,dS dt= -\iiint_\Omega \nabla\cdot(\rho \boldsymbol{v})\,dxdydz\times dt
\end{gather*}
again due to Gauss formula (A1.1.1).

This equals to the increment of the mass in $V$
\begin{gather*}
\partial_t \iiint_\Omega \rho\,dxdydz \times dt=\iiint_\Omega \rho_{tt} \,dxdydz\times dt.
\end{gather*}
Therefore
\begin{gather*}
-\iiint_\Omega \nabla\cdot(\rho \boldsymbol{v})\,dxdydz = \iiint_\Omega \rho_{tt} \,dxdydz
\end{gather*}
Since this equality holds for *any* domain $\Omega$ we can drop integral and arrive to (\ref{eq-1.4.4}).

We need to add We need to add $p=p(\rho, T)$ where $T$ is the temperature, but we assume $T$ is constant. Assuming that $\mathbf{v}$, $\rho-\rho_0$ and their first derivatives are small ($\rho_0=\const$) we arrive instead to
\begin{align}
&\rho_0 \mathbf{v}_t = -p'(\rho_0)\nabla \rho,
\label{eq-1.4.5}\\
&\rho_t + \rho_0\nabla \cdot \mathbf{v}=0
\label{eq-1.4.6}
\end{align}
and then applying $\nabla\cdot $ to (\ref{eq-1.4.5}) and $\partial_t $ to (\ref{eq-1.4.6}) we arrive to
\begin{equation}
\rho_{tt}=c^2\Delta \rho
\label{eq-1.4.7}
\end{equation}
with $c=\sqrt{p'(\rho_0)}$ is *the speed of sound*.

**Example 4.**
Let $ u$ be a concentration of perfume in the still air. Consider some volume $V$, then the quantity of the perfume in $V$ at time $t$ equals
$\iiint _V u \, dxdydz$ and its increment for time $dt$ equals
\begin{equation*}
\iiint _V u_t \, dxdydz\times dt.
\end{equation*}

On the other hand, the law of diffusion states that the flow of perfume through the surface element $dS$ in the direction of the normal $\mathbf{n}$ for time $dt$ equals $-k\nabla u\cdot \mathbf{n}\, dSdt$ where $k$ is a *diffusion coefficient* and therefore the flow of the perfume into $V$ from outside for time $dt$ equals
\begin{equation*}
-\iint _S k\nabla u\cdot \mathbf{n}\, dS\times dt=
\iiint_V \nabla \cdot (k\nabla u)\,dxdydz\times dt
\end{equation*}
due to Gauss formula (A1.1.2).
Therefore if there are neither sources nor sinks (negative sources) in $V$ these two expression must be equal
\begin{equation*}
\iiint _V u_t \, dxdydz= \iiint_V \nabla \cdot (k\nabla u)\,dxdydz
\end{equation*}
where we divided by $dt$. Since these equalities must hold for any volume the integrands must coincide and we arrive to *continuity equation*:
\begin{equation}
u_t = \nabla \cdot (k\nabla u).
\label{eq-1.4.8}
\end{equation}
If $k$ is constant we get
\begin{equation}
u_t = k\Delta u.
\label{eq-1.4.9}
\end{equation}

**Example 5.**
Consider heat propagation. Let $T$ be a temperature. \pause Then the heat energy contained in the volume $V$ equals $\iiint _V Q(T)\,dxdydz$ where $Q(T)$ is a *heat energy density*. On the other hand, the heat flow (the flow of the heat energy) through the surface element $dS$ in the direction of the normal $\mathbf{n}$ for time $dt$ equals $-k\nabla T\cdot \mathbf{n} \, dSdt$ where $k$ is a *thermoconductivity coefficient*. Applying the same arguments as above we arrive to
\begin{equation}
Q_t = \nabla \cdot (k\nabla T).
\label{eq-1.4.10}
\end{equation}
which we rewrite as
\begin{equation}
cT_t = \nabla \cdot (k\nabla T).
\label{eq-1.4.11}
\end{equation}
where $c= \frac{\partial Q}{\partial T}$ is a *thermocapacity coefficient*.

If both $c$ and $k$ are constant we get \begin{equation} c T_t = k\Delta T. \label{eq-1.4.12} \end{equation}

In the real life $c$ and $k$ depend on $T$. Further, $Q(T)$ has jumps at *phase transition temperature*. For example to melt an ice to a water (both at $0^\circ$) requires a lot of heat and to boil the water to a vapour (both at $100^\circ$) also requires a lot of heat.

**Example 6.**
Considering all examples above and assuming that unknown function does not depend on $t$ (and thus replacing corresponding derivatives by $0$), we arrive to the corresponding *stationary equations* the simplest of which is Laplace equation
\begin{equation}
\Delta u=0.
\label{eq-1.4.13}
\end{equation}

**Example 7.**
In the theory of complex variables one studies *complex-valued* holomorphic function $f(z)$ satisfying a Cauchy-Riemann equation $\partial_{\bar{z}}f=0$. Here $z=x+iy$, $f=u(x,y)+iv(x,y)$ with *real-valued* $u=u(x,y)$ and $v=v(x,y)$ and
$\partial_{\bar{z}}=\frac{1}{2}(\partial_x +i \partial_y)$; then this equation could be rewritten as
\begin{align}
& \partial_x u -\partial_y v=0,\label{eq-1.4.14}\\
& \partial_x v + \partial_y u=0,\label{eq-1.4.15}
\end{align}
which imply that both $u,v$ satisfy Laplace equation (\ref{eq-1.4.13}).

Indeed, differentiating the first equation by $x$ and the second by $y$ and adding we get $\Delta u=0$, and differentiating the second equation by $x$ and the first one by $y$ and subructing we get $\Delta v=0$.