$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ ##[1.4. Origin of some equations](id:sect-1.4) > 1. [Wave equation](#sect-1.4.1) > 2. [Diffusion equation](#sect-1.4.2) > 2. [Laplace equation](#sect-1.4.3) ###[Wave equation](id:sect-1.4.1) **[Example 1.](id:example-1.4.1)** Consider a string as a curve $y=u(x,t)$ 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 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 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 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.](id:example-1.4.2)** 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 $Tds$ 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 $D$ between $x=x\_1$ and equals due to [(A1.1.1)](../ChapterA/A.1.html#mjx-eqn-eq-A.1.1) \begin{equation\*} -\int\_{L} T \mathbf{n}\cdot \nabla u\, ds= \iint \nabla \cdot (T\nabla u) \, dxdy \end{equation\*} According to Newton 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\_{[x\_1,x\_2]} T\Delta u\, dx \end{equation\*} as $\nabla \cdot (T\nabla u)=T \Delta u$. Since this equality holds for any segment, the integrands coincide: \begin{equation} \rho u\_{tt}= T\Delta u. \label{eq-1.4.2} \end{equation} **[Example 3.](id:example-1.4.3)** 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 $\rho \frac{d\ }{dt} \mathbf{v})$ (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 $\rho \mathbf{n}\cdot \mathbf{v}$. We need to add $p=p(\rho)$. 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 $c=\sqrt{p'(\rho\_0)}$. ###[Diffusion equation](id:sect-1.4.2) **[Example 4.](id:example-1.4.4)** Let $ u$ be a concentration of parfume in the still air. Consider some volume $V$, then the quantity of the parfume 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 parfume through the surface element $dS$ in the direction of the normal $\mathbf{n}$ for time $dt$ equals $-k\nabla u dSdt$ where $k$ is a *diffusion coefficient* and therefore the flow of the parfume into $V$ from outside for time $dt$ equals \begin{equation\*} \iint \_S (-k\nabla u)\, dS\times dt= \iiint\_V \nabla \cdot (k\nabla u)\,dxdydz\times dt \end{equation\*} due to [(A1.1.2)](../ChapterA/A.1.html#mjx-eqn-eq-A.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.](id:example-1.4.5)** Consider heat propagation. Let $T$ be a temperature. Then the heat energy contained in the volume $V$ equals $\iiint \_V Q(T)\,dxdydz$ and 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 $-k\nabla T dSdt$ 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. ###[Laplace equation](id:sect-1.4.3) **[Example 6.](id:example-1.4.6)** 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 \begin{equation} \Delta u=0. \label{eq-1.4.13} \end{equation} **[Example 7.](id:example-1.4.7)** In the theory of complex variables one studies holomorphic function $f(z)$ satisfying a Cauchy-Riemann equation $\partial\_{\bar{z}}f=0$. Here $z=x+iy$, $f=u(x,y)+iv(x,y)$ 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 (\ref{eq-1.4.13}). ________ [$\Leftarrow$](./S1.3.html)  [$\Uparrow$](../contents.html)  [$\Rightarrow$](./S1.P.html)