8a. Laplace operator

[math] \newcommand{\mathds}{\mathbb}[/math]

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

Welcome to theoretical physics. We will discuss in this chapter some applications of the theory of complex functions developed so far, to some questions from physics, along with some more theory, needed in order to reach to these applications. We will be mainly interested in two fundamental equations of physics, namely the wave equation, and the heat equation. We will have a look as well into basic quantum mechanics.

All this sounds very exciting, doesn't it. However, in practice, technically speaking, we are not exactly ready yet for all these things. But the temptation remains high to talk about them, a bit like physicists do. So, what to do. Ask the cat of course, who says: \begin{cat} The more you know, the better that is. \end{cat} Thanks cat. So, we will talk indeed about all this, a bit like physicists do, by taking some freedom with surfing on difficult mathematics. In short, this will be a physics chapter, and for a more rigorous treatement of all this, which will be more general too, do not worry, that will come, later in this book, once we will know more calculus.

So, let us start, with some mathematics. We have:

Definition

A function [math]f:X\to\mathbb C[/math] with [math]X\subset\mathbb C[/math] is called differentiable in the real sense if the following limits, called its partial derivatives,

[[math]] \frac{df}{dx}(z)=\lim_{t\to0}\frac{f(z+t)-f(z)}{t} [[/math]]

[[math]] \frac{df}{dy}(z)=\lim_{t\to0}\frac{f(z+it)-f(z)}{t} [[/math]]

computed by using [math]t\in\mathbb R[/math], [math]t\to0[/math], exist at any [math]z\in X[/math], and are continuous.

We have already met this notion, in chapter 6, in a purely mathematical context, when taking about holomorphic functions. Our observation there was that a holomorphic function is differentiable in the above real sense, and also that there are further functions, such as the conjugate [math]f(z)=\bar{z}[/math] or the modulus [math]f(z)=|z|[/math], which are still differentiable in the above real sense, but are not holomorphic. But more on this later.

We will need as well the following definition, complementing Definition 8.2:

Definition

A function [math]f:X\to\mathbb C[/math] is called doubly differentiable if its partial derivatives are both differentiable. These double derivatives are denoted as follows:

[[math]] \frac{d^2f}{dx^2}=\frac{d}{dx}\left(\frac{df}{dx}\right) \quad,\quad \frac{d^2f}{dxdy}=\frac{d}{dx}\left(\frac{df}{dy}\right) [[/math]]

[[math]] \frac{d^2f}{dydx}=\frac{d}{dy}\left(\frac{df}{dx}\right) \quad,\quad \frac{d^2f}{dy^2}=\frac{d}{dy}\left(\frac{df}{dy}\right) [[/math]]

We can in fact talk, in a similar way, about order [math]k[/math] partial derivatives, for any [math]k\in\mathbb N[/math].

As before with Definition 8.2, there are many things that can be said here, mathematically speaking, but we will leave this discussion for later, starting with chapter 9 below, when systematically investigating functions of several variables. In fact, the material in this chapter will be, among others, an introduction to that. However, we will need:

Theorem

The double derivatives satisfy the formula

[[math]] \frac{d^2f}{dxdy}=\frac{d^2f}{dydx} [[/math]]

called Clairaut formula.

Show Proof

This is something very standard, the idea being as follows:

(1) Before pulling out a formal proof, as an intuitive justification for our formula, let us consider a product of power functions, [math]f(z)=x^py^q[/math]. We have then:

[[math]] \frac{d^2f}{dxdy}=\frac{d}{dx}\left(\frac{dx^py^q}{dy}\right)=\frac{d}{dx}\left(qx^py^{q-1}\right)=pqx^{p-1}y^{q-1} [[/math]]

[[math]] \frac{d^2f}{dydx}=\frac{d}{dy}\left(\frac{dx^py^q}{dx}\right)=\frac{d}{dy}\left(px^{p-1}y^q\right)=pqx^{p-1}y^{q-1} [[/math]]

Next, let us consider a linear combination of power functions, [math]f(z)=\sum_{pq}c_{pq}x^py^q[/math], which can be finite or not. We have then, by using the above computation:

[[math]] \frac{d^2f}{dxdy}=\frac{d^2f}{dydx}=\sum_{pq}c_{pq}pqx^{p-1}y^{q-1} [[/math]]

Thus, we can see that our commutation formula for derivatives holds indeed, and this due to the fact that the functions in [math]x[/math] and [math]y[/math] commute. Of course, all this does not prove our formula, in general. But exercise for you, to have this idea fully working.

(2) Getting now to more standard techniques, given a point in the complex plane, [math]z=a+ib[/math], consider the following functions, depending on [math]h,k\in\mathbb R[/math] small:

[[math]] u(h,k)=f(a+h,b+k)-f(a+h,b) [[/math]]

[[math]] v(h,k)=f(a+h,b+k)-f(a,b+k) [[/math]]

[[math]] w(h,k)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b) [[/math]]

By the mean value theorem, for [math]h,k\neq0[/math] we can find [math]\alpha,\beta\in\mathbb R[/math] such that:

[[math]] \begin{eqnarray*} w(h,k) &=&u(h,k)-u(0,k)\\ &=&h\cdot\frac{d}{dx}\,u(\alpha h,k)\\ &=&h\left(\frac{d}{dx}\,f(a+\alpha h,b+k)-\frac{d}{dx}\,f(a+\alpha h,b)\right)\\ &=&hk\cdot\frac{d}{dy}\cdot\frac{d}{dx}\,f(a+\alpha h,b+\beta k) \end{eqnarray*} [[/math]]

Similarly, again for [math]h,k\neq0[/math], we can find [math]\gamma,\delta\in\mathbb R[/math] such that:

[[math]] \begin{eqnarray*} w(h,k) &=&v(h,k)-v(h,0)\\ &=&k\cdot\frac{d}{dy}\,v(h,\delta k)\\ &=&k\left(\frac{d}{dy}\,f(a+h,b+\delta k)-\frac{d}{dy}\,f(a,b+\delta k)\right)\\ &=&hk\cdot\frac{d}{dx}\cdot\frac{d}{dy}\,f(a+\gamma h,b+\delta k) \end{eqnarray*} [[/math]]

Now by dividing everything by [math]hk\neq0[/math], we conclude from this that the following equality holds, with the numbers [math]\alpha,\beta,\gamma,\delta\in\mathbb R[/math] being found as above:

[[math]] \frac{d}{dy}\cdot\frac{d}{dx}\,f(a+\alpha h,b+\beta k)=\frac{d}{dx}\cdot\frac{d}{dy}\,f(a+\gamma h,b+\delta k) [[/math]]

But with [math]h,k\to0[/math] we get from this the Clairaut formula, at [math]z=a+ib[/math], as desired.

■

Regarding the double derivatives in one of the variables, here our intuition from the one-variable case can help, and in fact there is no need for more, at least of this stage of the things. But, as already mentioned, we will be back to this, later in this book.

Getting now into physics, which will be our starting point for the considerations in this chapter, we will be interested in the propagation of waves and heat. Let us start with the waves. In analogy with what we saw in chapter 3, in one dimension, we have:

Theorem

The wave equation in the plane [math]\mathbb R^2[/math] is

[[math]] \ddot{\varphi}=v^2\Delta\varphi [[/math]]

where dots denote time derivatives, [math]\Delta[/math] is the Laplace operator, given by

[[math]] \Delta\varphi=\frac{d^2\varphi}{dx^2}+\frac{d^2\varphi}{dy^2} [[/math]]

and [math]v \gt 0[/math] is the propagation speed.

Show Proof

We have already met this equation in chapter 3, in one dimension, and in 2 dimensions the study is similar, by using a lattice model, as follows:

(1) In order to understand the propagation of waves in 2 dimensions, we can model the whole space [math]\mathbb R^2[/math] as a network of balls, with springs between them, as follows:

[[math]] \xymatrix@R=12pt@C=15pt{ &\ar@{~}[d]&\ar@{~}[d]&\ar@{~}[d]&\ar@{~}[d]\\ \ar@{~}[r]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\\ \ar@{~}[r]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\\ \ar@{~}[r]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\bullet\ar@{~}[r]\ar@{~}[d]&\\ &&&&&} [[/math]]

As before in chapter 3 one dimension, let us send now an impulse, and zoom on one ball. The situation here is as follows, with [math]l[/math] being the spring length:

[[math]] \xymatrix@R=20pt@C=20pt{ &\bullet_{\varphi(x,y+l)}\ar@{~}[d]&\\ \bullet_{\varphi(x-l,y)}\ar@{~}[r]&\bullet_{\varphi(x,y)}\ar@{~}[r]\ar@{~}[d]&\bullet_{\varphi(x+l,y)}\\ &\bullet_{\varphi(x,y-l)}} [[/math]]

We have two forces acting at [math](x,y)[/math]. First is the Newton motion force, mass times acceleration, which is as follows, with [math]m[/math] being the mass of each ball:

[[math]] F_n=m\cdot\ddot{\varphi}(x,y) [[/math]]

And second is the Hooke force, displacement of the spring, times spring constant. Since we have four springs at [math](x,y)[/math], this is as follows, [math]k[/math] being the spring constant:

[[math]] \begin{eqnarray*} F_h &=&F_h^r-F_h^l+F_h^u-F_h^d\\ &=&k(\varphi(x+l,y)-\varphi(x,y))-k(\varphi(x,y)-\varphi(x-l,y))\\ &+&k(\varphi(x,y+l)-\varphi(x,y))-k(\varphi(x,y)-\varphi(x,y-l))\\ &=&k(\varphi(x+l,y)-2\varphi(x,y)+\varphi(x-l,y))\\ &+&k(\varphi(x,y+l)-2\varphi(x,y)+\varphi(x,y-l)) \end{eqnarray*} [[/math]]

We conclude that the equation of motion, in our model, is as follows:

[[math]] \begin{eqnarray*} m\cdot\ddot{\varphi}(x,y) &=&k(\varphi(x+l,y)-2\varphi(x,y)+\varphi(x-l,y))\\ &+&k(\varphi(x,y+l)-2\varphi(x,y)+\varphi(x,y-l)) \end{eqnarray*} [[/math]]

(2) Now let us take the limit of our model, as to reach to continuum. For this purpose we will assume that our system consists of [math]B^2 \gt \gt 0[/math] balls, having a total mass [math]M[/math], and spanning a total area [math]L^2[/math]. Thus, our previous infinitesimal parameters are as follows, with [math]K[/math] being the spring constant of the total system, taken to be equal to [math]k[/math]:

[[math]] m=\frac{M}{B^2}\quad,\quad k=K\quad,\quad l=\frac{L}{B} [[/math]]

With these changes, our equation of motion found in (1) reads:

[[math]] \begin{eqnarray*} \ddot{\varphi}(x,y) &=&\frac{KB^2}{M}(\varphi(x+l,y)-2\varphi(x,y)+\varphi(x-l,y))\\ &+&\frac{KB^2}{M}(\varphi(x,y+l)-2\varphi(x,y)+\varphi(x,y-l)) \end{eqnarray*} [[/math]]

Now observe that this equation can be written, more conveniently, as follows:

[[math]] \begin{eqnarray*} \ddot{\varphi}(x,y) &=&\frac{KL^2}{M}\times\frac{\varphi(x+l,y)-2\varphi(x,y)+\varphi(x-l,y)}{l^2}\\ &+&\frac{KL^2}{M}\times\frac{\varphi(x,y+l)-2\varphi(x,y)+\varphi(x,y-l)}{l^2} \end{eqnarray*} [[/math]]

With [math]N\to\infty[/math], and therefore [math]l\to0[/math], we obtain in this way:

[[math]] \ddot{\varphi}(x,y)=\frac{KL^2}{M}\left(\frac{d^2\varphi}{dx^2}+\frac{d^2\varphi}{dy^2}\right)(x,y) [[/math]]

Thus, we are led in this way to the following wave equation in two dimensions, with [math]v=\sqrt{K/M}\cdot L[/math] being the propagation speed of our wave:

[[math]] \ddot{\varphi}(x,y)=v^2\left(\frac{d^2\varphi}{dx^2}+\frac{d^2\varphi}{dy^2}\right)(x,y) [[/math]]

But we recognize at right the Laplace operator, and we are done. As before in 1D, there is of course some discussion to be made here, arguing that our spring model in (1) is indeed the correct one. But do not worry, experiments confirm our findings.

(3) Finally, for completness, let us mention that the same argument, namely a lattice model, carries on in arbitrary [math]N[/math] dimensions, and we obtain here the same wave equation as before, namely [math]\ddot{\varphi}=v^2\Delta\varphi[/math], with the following straightforward definition for the Laplacian, based on the obvious [math]N[/math]-dimensional analogues of Definition 8.2 and Definition 8.3:

[[math]] \Delta\varphi=\sum_{i=1}^N\frac{d^2\varphi}{dx_i^2} [[/math]]

This is quite nice, because at [math]N=1[/math] this Laplace operator is just the second derivative, so what we have here is a unification of what we did in chapter 3 and in (1,2) above. In addition, at [math]N=3[/math] we obtain in this way the wave equation in the case of main interest, namely the one of our real-life world. But more on this later in this book.

■

Regarding now heat diffusion, we have here a similar equation, as follows:

Theorem

Heat diffusion in [math]\mathbb R^2[/math] is described by the heat equation

[[math]] \dot{\varphi}=\alpha\Delta\varphi [[/math]]

where [math]\alpha \gt 0[/math] is the thermal diffusivity of the medium, and [math]\Delta[/math] is the Laplace operator.

Show Proof

We have already met this equation in chapter 3, in one dimension, and in 2 dimensions the study is similar, by using a lattice model, as follows:

(1) In order to understand the propagation of heat in 2 dimensions, we can model the whole space [math]\mathbb R^2[/math] as a network as follows, with all lengths being [math]l \gt 0[/math]:

[[math]] \xymatrix@R=12pt@C=15pt{ &\ar@{-}[d]&\ar@{-}[d]&\ar@{-}[d]&\ar@{-}[d]\\ \ar@{-}[r]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\\ \ar@{-}[r]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\\ \ar@{-}[r]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\circ\ar@{-}[r]\ar@{-}[d]&\\ &&&&& } [[/math]]

We have to implement now the physical heat diffusion mechanism, namely “the rate of change of the temperature of the material at any given point must be proportional, with proportionality factor [math]\alpha \gt 0[/math], to the average difference of temperature between that given point and the surrounding material”. In practice, this leads to a condition as follows, expressing the change of the temperature [math]\varphi[/math], over a small period of time [math]\delta \gt 0[/math]:

[[math]] \varphi(x,y,t+\delta)=\varphi(x,y,t)+\frac{\alpha\delta}{l^2}\sum_{(x,y)\sim(u,v)}\left[\varphi(u,v,t)-\varphi(x,y,t)\right] [[/math]]

In fact, we can rewrite our equation as follows, making it clear that we have here an equation regarding the rate of change of temperature at [math]x[/math]:

[[math]] \frac{\varphi(x,y,t+\delta)-\varphi(x,y,t)}{\delta}=\frac{\alpha}{l^2}\sum_{(x,y)\sim(u,v)}\left[\varphi(u,v,t)-\varphi(x,y,t)\right] [[/math]]

(2) Now, let us do the math. In the context of our 2D model the neighbors of [math]x[/math] are the points [math](x\pm l,y\pm l)[/math], so the equation above takes the following form:

[[math]] \begin{eqnarray*} &&\frac{\varphi(x,y,t+\delta)-\varphi(x,y,t)}{\delta}\\ &=&\frac{\alpha}{l^2}\Big[(\varphi(x+l,y,t)-\varphi(x,y,t))+(\varphi(x-l,y,t)-\varphi(x,y,t))\Big]\\ &+&\frac{\alpha}{l^2}\Big[(\varphi(x,y+l,t)-\varphi(x,y,t))+(\varphi(x,y-l,t)-\varphi(x,y,t))\Big] \end{eqnarray*} [[/math]]

Now observe that we can write this equation as follows:

[[math]] \begin{eqnarray*} \frac{\varphi(x,y,t+\delta)-\varphi(x,y,t)}{\delta} &=&\alpha\cdot\frac{\varphi(x+l,y,t)-2\varphi(x,y,t)+\varphi(x-l,y,t)}{l^2}\\ &+&\alpha\cdot\frac{\varphi(x,y+l,t)-2\varphi(x,y,t)+\varphi(x,y-l,t)}{l^2} \end{eqnarray*} [[/math]]

As it was the case before when modeling the wave equation, we recognize on the right the usual approximation of the second derivative, coming from calculus. Thus, when taking the continuous limit of our model, [math]l\to 0[/math], we obtain the following equation:

[[math]] \frac{\varphi(x,y,t+\delta)-\varphi(x,y,t)}{\delta} =\alpha\left(\frac{d^2\varphi}{dx^2}+\frac{d^2\varphi}{dy^2}\right)(x,y,t) [[/math]]

Now with [math]t\to0[/math], we are led in this way to the heat equation, namely:

[[math]] \dot{\varphi}(x,y,t)=\alpha\cdot\Delta\varphi(x,y,t) [[/math]]

(3) Finally, for completness, let us mention that in arbitrary [math]N[/math] dimensions the same argument, namely a basic lattice model, carries over, and we obtain a similar equation, namely [math]\dot{\varphi}=\alpha\Delta\varphi[/math], with the following straightforward definition for the Laplacian, based on the obvious [math]N[/math]-dimensional analogues of Definition 8.2 and Definition 8.3:

[[math]] \Delta\varphi=\sum_{i=1}^N\frac{d^2\varphi}{dx_i^2} [[/math]]

Which is quite nice, unfiying what we did in chapter 3 in one dimension, and in the above in 2 dimensions. We will be back to this, later in this book.

■

General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].