1.4. Origin of some equations

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1.4. Origin of some equations

  1. Wave equation
  2. Diffusion equation
  3. Laplace equation

Wave equation

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 $\nabla \cdot (T\nabla u)=T\nabla \cdot \nabla = T \Delta u$. 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.

Diffusion equation

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.

Laplace equation

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 $\Delta v=0$.


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