April 2022

April 28, 2022April 28, 2022Michael LiuLeave a comment

Even, odd, and periodic functions

The concepts of oddness, evenness, and periodicity are closely related to the Fourier series.

Definition A function ${\phi(x)}$ that is defined for ${-\infty<x<\infty}$ is called periodic if there is a number ${p>0}$ such that

$\displaystyle \phi(x)=\phi(x+p) \hspace{0.5cm} \mathrm{for\hspace{3pt}all}\hspace{3pt} x \ \ \ \ \ (1)$

p is called a period of ${\phi(X)}$ . A periodic function has the following properties:

If ${\phi(x)}$ has period ${p}$ , then ${\phi(x+np)=\phi(x)}$ for all ${x}$ and for all integers ${n}$ .
The sum of two functions of period ${p}$ has period ${p}$ .
If ${\phi(x)}$ has period ${p}$ , then ${\int_{a}^{a+p}\phi(x)}$ does not depend on ${a}$ .

For a function defined on the interval ${-l<x<l}$ , its periodic extension is

$\displaystyle \phi_{per}(x)=\phi(x-2lm)\hspace{3pt} \mathrm{for}\hspace{3pt} -l+2lm<x<l+2lm \ \ \ \ \ (2)$

for all integers ${m}$ . This definition does not specify what the periodic extension is at the endpoints ${x = l + 2lm}$ . In fact, the function ${\phi(x)}$ may have jump discontinuities at the endpoints if the the one-sided limits ${\phi(l^-)}$ and ${\phi(-l^+)}$ both exist but not equal.

An even function is a function the satisfy the equation

$\displaystyle \phi(x)=\phi(-x) \ \ \ \ \ (3)$

This mean that the graph ${y=\phi(x)}$ is symmetric with respect to the y axis.

An odd function is a function the satisfy the equation

$\displaystyle \phi(-x)=-\phi(x) \ \ \ \ \ (4)$

This mean that the graph ${y=\phi(x)}$ is symmetric with respect to the origin.

A monomial ${x^n}$ is an even function if ${n}$ is even and is an odd function if n is odd. The functions cos ${x}$ , cosh ${x}$ , and any function of ${x^2}$ are even functions. The functions sin ${x}$ , tan ${x}$ , and sinh ${x}$ are odd functions. In fact, the products of functions follow the usual rules: even × even = even, odd × odd = even, odd × even = odd. The sum of two odd functions is again odd, and the sum of two evens is even.

But the sum of an even and an odd function can be anything. Proof: Let ${f(x)}$ be any function at all defined on ${(-l,l)}$ . Let ${\phi(x)=\frac{1}{2}[\phi(x)+\phi(-x)]}$ and ${\psi(x)=\frac{1}{2}[\psi(x)-\psi(-x)]}$ . Then we easily check that ${f(x)=\phi(x)+\psi(x)}$ , that ${\phi(x)}$ is even and that ${\psi(x)}$ is odd. The functions ${\phi}$ and ${\psi}$ are called the even and odd parts of ${f}$ , respectively. If ${p(x)}$ is any polynomial, its even part is the sum of its even terms, and its odd part is the sum of its odd terms.

Integration and differentiation change the parity (evenness or oddness) of a function. That is, if ${\phi(x)}$ is even, then both ${\frac{d\phi}{dx}}$ and ${\int_0^x\phi(s)ds}$ are odd. If ${\phi(x)}$ is odd, then its derivative and integral are even.

The graph of an odd function ${\phi(x)}$ must pass through the origin since ${\phi(0) = 0}$ follows directly from (4) by putting ${x = 0}$ . The graph of an even function ${\phi(x)}$ must cross the ${y}$ axis horizontally, ${\phi'(x)=0}$ , since the derivative is odd (provided the derivative exists).

The concepts of oddness, evenness, and periodicity have the following relationships with the boundary conditions:

${u(0,t)=u(l,t)=0}$ : Dirichlet BCs corresponding to the odd extension
${u_x(0,t)=u_x(l,t)=0}$ : Nuemann BCs corresponding to the even extension
${u(l,t)=u_(-l,t), u_x(l,t)=u_x=(-l,t)}$ : Periodic BCs corresponding to the periodic extension

PDE notes: heat equation

April 26, 2022June 1, 2022Michael LiuLeave a comment

1. Heat equation

1.1. Introduction

The heat equation corresponding to no sources and constant thermal properties is given as

$\displaystyle u_t = ku_{xx} \ \ \ \ \ (1)$

Equation (1) describes how heat energy spreads out. Other physical quantities besides temperature smooth out in much the same manner, satisfying the same partial differential equation (1). For this reason, (1) is also called the diffusion equation.

Since the heat equation in (1) has one time derivative, we must be given one initial condition (IC) (usually at t = 0), the initial temperature. It is possible that the initial temperature is not constant, but depends on x. Thus, we must be given the initial temperature distribution,

$\displaystyle u(x,0)= f(x) \ \ \ \ \ (2)$

We also need to know that happens at the two boundaries, ${x = 0}$ and ${x = L}$ . Without knowing this information, we cannot predict the future. Two conditions are needed corresponding to the second spatial derivatives present in (1), usually one condition at each end. We call them boundary conditions. We deal with mainly four types of boundary conditions: Dirichlet, Neumann, Robin, and Periodic.

1.2. Separation of Variables (SoV)

When the PDE and the BCs are linear and homogeneous, we use a technique called the method of separation of variables to find the analytic solutions.

We introduce the heat operator,

$\displaystyle L(u)=u_t-ku_{xx} \ \ \ \ \ (3)$

The heat operator is a linear operator that satisfies the linearity property ${L(c_1u_1+c_2u_2)=c_1L(u_1)+c_2L(u_2)}$ .

A linear equation for the unknown ${u}$ is of the form

$\displaystyle L(u)=f \ \ \ \ \ (4)$

where ${L}$ is a linear operator and ${f}$ is known. The heat equation is a linear PDE:

$\displaystyle L(u)=u_t-ku_{xx}=f(x,t) \ \ \ \ \ (5)$

${f(x,t)}$ is a forcing term. If ${f=0}$ , then ${L(u)=0}$ and the PDE becomes

$\displaystyle u_t-ku_{xx}=0 \ \ \ \ \ (6)$

Equation (6) is a linear homogeneous equation.

We now propose to study heat equation with zero temperatures at finite ends:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t=ku_{xx} \ \ \ \ \ (7)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm} u(0,t)=0 \ \ \ \ \ (8)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm} u(L,t)=0 \ \ \ \ \ (9)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (10)$

defined for ${0\leq x \leq L}$ and ${t \geq 0 }$ . This problem consists of linear homogeneous partial differential equations with linear homogeneous boundary conditions.

Solution. We attempt to determine the product solution in the form of

$\displaystyle u(x,t)=G(t)\phi(x) \ \ \ \ \ (11)$

First, we substitute equation (11) into equation (7),

$\displaystyle \phi(x)G'(t)=kG(t)\phi^{''}(x) \ \ \ \ \ (12)$

We can “separate variables” by divide both sides of equation (12) by ${kG(t)\phi(x)}$ :

$\displaystyle \frac{1}{kG}G'=\frac{1}{\phi}\phi^{''} \ \ \ \ \ (13)$

Now the variables are “separated” in a sense that the left hand side is only a function of ${t}$ and the right hand side is only a function of ${x}$ . Next, we can claim that both sides of (13) must equal the same constant:

$\displaystyle \frac{1}{kG}G'=\frac{1}{\phi}\phi^{''}=-\lambda \ \ \ \ \ (14)$

where ${\lambda}$ is a constant known as the separation constant.

Equation (14) yields 2 ordinary differential equations,

$\displaystyle G'=-\lambda kG \ \ \ \ \ (15)$

and

$\displaystyle \phi''=-\lambda\phi \ \ \ \ \ (16)$

We now see that the advantage of the product method is that it transform a PDE, which we do not have to solve, into two ODEs. The boundary conditions impose two conditions on the ${x}$ -dependent ODE. The time-dependent ODE, equation (15), has no additional conditions. We can find a general solution for equation (15) quite easily,

$\displaystyle G(t)=ce^{-\lambda kt} \ \ \ \ \ (17)$

where ${c}$ is an arbitrary multiplicative constant. There can be three cases for ${\lambda}$ : ${\lambda>0}$ , ${\lambda=0}$ , ${\lambda<0}$ . We will determine all allowable values of ${\lambda}$ next.

The ODE in (16) has two boundary conditions:

$\displaystyle \phi(0)=0 \ \ \ \ \ (18)$

$\displaystyle \phi(L)=0 \ \ \ \ \ (19)$

We call this a boundary value problem for ordinary differential equations. We note that ${\phi(x)=0}$ satisfy the ODE (16) and both BCs (18) and (19) regardless of the value of the separation constant. Thus, ${\phi(x)=0}$ is a trivial solution of our PDE. It corresponds to ${u(x,t)\equiv0}$ since ${u(x,t)=\phi(x)G(t)}$ . However, we want to see if there are other nontrivial solutions. We will show that there are certain values of ${\lambda}$ , called eigenvalues of the boundary value problem, for which there are nontrivial solutions, ${\phi(x)}$ . The non-trivial solutions are called eigenfunctions corresponding to the eigenvalues ${\lambda}$

Now we solve equation (16). For linear and homogeneous second-order ODE, two independent solutions are usually obtained in the form of ${\phi=e^{rx}}$ . Substituting this result into equation (16) yields

$\displaystyle \phi''=r^2e^{rx} \ \ \ \ \ (20)$

and this implies that

$\displaystyle r^2=-\lambda \ \ \ \ \ (21)$

We consider the three cases of ${\lambda}$ separately.

If ${\lambda<0}$ , the roots of the characteristic polynomials are ${r=\pm\sqrt{-\lambda}}$ , so the solutions are ${e^{\sqrt{-\lambda}x}}$ and ${e^{-\sqrt{-\lambda}x}}$ . The general solution is

$\displaystyle \phi(x)=c_1e^{\sqrt{-\lambda}x}+c_2e^{-\sqrt{-\lambda}x} \ \ \ \ \ (22)$

We use the hyperbolic functions,

$\displaystyle \phi(x)=c_3\mathrm{cosh}(\sqrt{-\lambda}x)+c_4\mathrm{sinh}(\sqrt{-\lambda}x) \ \ \ \ \ (23)$

We consider the BC from (18). Since ${\mathrm{cosh}(0)=1}$ , ${\phi(0)=0}$ implies that ${c_3=0}$ . So we have ${\phi(x)=c_4\mathrm{sinh}(\sqrt{-\lambda}x)}$ . Now we consider ${\phi(L)=0}$ , which implies that ${c_4\mathrm{sinh}(\sqrt{-\lambda}L)=0}$ . Since ${\sqrt{-\lambda}L>0}$ , ${\mathrm{sinh}(\sqrt{-\lambda}L)>0}$ . It follows that ${c_4=0}$ . This implies that ${\phi(x)=0}$ . The only solution of (16) for ${\lambda<0}$ that solves the homogeneous boundary condition is the trivial solution. Thus there are no negative eigenvalues.

We now consider the case of ${\lambda=0}$ . The general solution is

$\displaystyle \phi(x)=c_1+c_2x \ \ \ \ \ (24)$

Apply ${\phi(0)=0}$ and we arrive at ${c_1=0}$ . Apply ${\phi(L)=0}$ , we have ${c_2x=0}$ which implies that ${c_2=0}$ . This implies that ${\phi(x)=0}$ . The only solution of (16) for ${\lambda=0}$ that solves the homogeneous boundary condition is the trivial solution. Thus ${\lambda=0}$ is not an eigenvalue for this problem.

We now consider the case of ${\lambda>0}$ . The roots of the characteristic polynomials are ${r=\pm\sqrt{\lambda}i}$ , so the solutions are ${e^{\sqrt{\lambda}ix}}$ and ${e^{-\sqrt{\lambda}ix}}$ . The general solution is

$\displaystyle \phi(x)=c_1e^{\sqrt{\lambda}ix}+c_2e^{-\sqrt{\lambda}ix} \ \ \ \ \ (25)$

Since ${\mathrm{cos}(\sqrt{\lambda}x)}$ and ${\mathrm{sin}(\sqrt{\lambda}x)}$ are each linear combinations of ${e^{\sqrt{\lambda}ix}}$ and ${e^{-\sqrt{\lambda}ix}}$ , the general solution can also be expresed in the form of

$\displaystyle \phi(x)=c_3\mathrm{cos}(\sqrt{\lambda}x)+c_4\mathrm{sin}(\sqrt{\lambda}x) \ \ \ \ \ (26)$

For the BC ${\phi(0)=0}$ , we have ${\phi(0)=c_3\mathrm{cos}(\sqrt{\lambda}x)}$ which implies that ${c_3=0}$ . Consider the BC ${\phi(L)=0}$ , we have ${c_4\mathrm{sin}(\sqrt{\lambda}x)=0}$ . If we let ${c_4=0}$ , then we arrive at ${u(x,t)=0}$ , a trivial solution. To find the nontrivial solutions, the eigenvalue ${\lambda}$ much satisfy

$\displaystyle \mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (27)$

${\sqrt{\lambda}x}$ must be the zero of the sine function. Thus,

$\displaystyle \sqrt{\lambda}L=n\pi \ \ \ \ \ (28)$

The eigenvalues are

$\displaystyle \lambda=(\frac{n\pi}{L})^2, n=0,1,2,3... \ \ \ \ \ (29)$

The eigenfunctions corresponding to the eigenvalues are

$\displaystyle \phi(x)=c_4\mathrm{sin}(\frac{n\pi x}{L}), n=0,1,2,3... \ \ \ \ \ (30)$

Now we know that the product solution consists of ${\phi(x)=\\c_4\mathrm{sin}(\frac{n\pi x}{L})}$ and ${G(t)=e^{-(\frac{n\pi}{L})^2 kt}}$ where we determined from the boundary conditions ${\phi(0)=0}$ and ${\phi(L)=0}$ . We call this type of boundary conditions the Dirichlet boundary conditions. Thus, the product solutions of the heat equation with homogeneous Dirichlet boundary conditions are

$\displaystyle u(x,t)= B\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt}, n=0,1,2,3... \ \ \ \ \ (31)$

where B is an arbitrary constant.

1.3. Initial value problems

The principle of superposition can be extended to show that if ${u_1 , u_2 , u_3 , . . . , u_M}$ are solutions of a linear homogeneous problem, then any linear combination of these is also a solution, ${\sum_{n=1}^{M} c_nu_n}$ ,where ${c_n}$ are arbitrary constants. Since we know from the method of separation of variables that ${\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt}}$ is a solution of the heat equation (solving zero boundary conditions) for all positive ${n}$ , it follows that any linear combination of these solutions is also a solution of the linear homogeneous heat equation. Thus,

$\displaystyle u(t,x)=\sum_{n=1}^{M} B_n\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt} \ \ \ \ \ (32)$

For example, we wish to solve the following initial value problem:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (33)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u(0,t)=0 \ \ \ \ \ (34)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u(L,t)=0 \ \ \ \ \ (35)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = 3\mathrm{sin}(\frac{4\pi x}{L})+7\mathrm{sin}(\frac{9\pi x}{L}) \ \ \ \ \ (36)$

Solution. We can let ${u_1(x,0) = 3\mathrm{sin}(\frac{4\pi x}{L})}$ and ${u_2(x,0) = 7\mathrm{sin}(\frac{9\pi x}{L})}$ such that ${u(x,0)=u_1(x,0)+u_2(x,0)}$ . We can use the product solutions from (31) and we see that the product solution satisfy the initial conditions ${B\mathrm{sin}(\frac{n\pi x}{L})}$ . By picking ${B=3}$ and ${n=4}$ , we have satisfied the initial condition ${u_1(x,0)}$ . The solution we get is

$\displaystyle u_1(t,x)=3\mathrm{sin}(\frac{4\pi x}{L})e^{-(\frac{4\pi}{L})^2 kt} \ \ \ \ \ (37)$

By picking ${B=7}$ and ${n=9}$ , we have satisfied the initial condition ${u_2(x,0)}$ . The solution we get is

$\displaystyle u_1(t,x)=7\mathrm{sin}(\frac{9\pi x}{L})e^{-(\frac{9\pi}{L})^2 kt} \ \ \ \ \ (38)$

By superposition principle we know that the product solution of our problem is a linear combinations of ${u_1(x,t)}$ and ${u_2(x,t)}$ . Thus,

$\displaystyle u(t,x)=3\mathrm{sin}(\frac{4\pi x}{L})e^{-(\frac{4\pi}{L})^2 kt}+ 7\mathrm{sin}(\frac{9\pi x}{L})e^{-(\frac{9\pi}{L})^2 kt} \ \ \ \ \ (39)$

${\square}$

The IC in the previous problem is a finite sum of the sine functions. What should we do in the situation that the IC is not a finite linear combinations of the sine functions? Consider the following problem:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (40)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u(0,t)=0 \ \ \ \ \ (41)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u(L,t)=0 \ \ \ \ \ (42)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = 100 \ \ \ \ \ (43)$

Solution. From equation (32) we know that the solution is

$\displaystyle u(t,x)=\sum_{n=1}^{M} B_n\mathrm{sin}(\frac{n\pi x}{L})e^{-(\frac{n\pi}{L})^2 kt} \ \ \ \ \ (44)$

We want to apply the IC to our problem. First, from the previous problem, note that we can solve the heat equation if initially

$\displaystyle u(x,0)= f(x) = \sum_{n=1}^{M} B_n\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (45)$

Thus, we use the theory of Fourier series and claim that “any” (with restrictions) initial condition ${f(x)}$ can be written as an infinite linear combination of ${\mathrm{sin}(\frac{n\pi x}{L})}$ , known as a type of Fourier series:

$\displaystyle f(x) = \sum_{n=1}^{\infty} B_n\mathrm{sin}(\frac{n\pi x}{L}) \ \ \ \ \ (46)$

We want to determine the coefficient ${B_n}$ in equation (46). To do that, we use the fact the ${\mathrm{sin}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{sin}(\frac{n \pi x}{L})\mathrm{sin}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m \\ \end{array} \right. \ \ \ \ \ (47)$

Multiply (46) by ${\mathrm{sin}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx = B_{m}\int_{0}^{L}\mathrm{sin^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (48)$

Solving for ${A_m}$ yields,

$\displaystyle B_m=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{sin}(\frac{m \pi x}{L})\, dx \ \ \ \ \ (49)$

Now, we calculate the coefficient ${B_n}$ from equation (49) for ${f(x)=100}$ ,

$\displaystyle B_n=\frac{2}{L} \int_{0}^{L}100\mathrm{sin}(\frac{n \pi x}{L})\, dx \ \ \ \ \ (50)$

Solving for ${B_n}$ yields,

$\displaystyle \left\{ \begin{array}{lr} 0 & n \hspace{5pt} even\\ \frac{400}{n\pi} & n\hspace{5pt} odd \\ \end{array} \right. \ \ \ \ \ (51)$

Each succeeding term in the series is much smaller than the first. We can then approximate the infinite series by only the first term:

$\displaystyle u(t,x)\approx \frac{400}{\pi}\mathrm{sin}(\frac{\pi x}{L})e^{-(\frac{\pi}{L})^2 kt} \ \ \ \ \ (52)$

${\square}$

Let us summarize the method of separation of variables as it appears for the heat equation with homogeneous Dirichlet boundary conditions:

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (53)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u(0,t)=0 \ \ \ \ \ (54)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u(L,t)=0 \ \ \ \ \ (55)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (56)$

Make sure that you have a linear and homogeneous PDE with linear and homogeneous BC.
Temporarily ignore the nonzero IC.
Separate variables (determine differential equations implied by the assumption of product solutions) and introduce a separation constant.
Determine separation constants as the eigenvalues of a boundary value problem.
Solve other differential equations. Record all product solutions of the PDE obtainable by this method.
Apply the principle of superposition (for a linear combination of all product solutions).
Attempt to satisfy the initial condition.
Determine coefficients using the orthogonality of the eigenfunctions.

1.4. Other Boundary Value Problems

The following problem is defined for ${0\leq x \leq L}$ and ${t \geq 0 }$ .

$\displaystyle \mathrm{PDE: } \hspace{0.5cm} u_t = ku_{xx} \ \ \ \ \ (57)$

$\displaystyle \mathrm{BC1:} \hspace{0.5cm}u_x(0,t)=0 \ \ \ \ \ (58)$

$\displaystyle \mathrm{BC2:}\hspace{0.5cm}u_x(L,t)=0 \ \ \ \ \ (59)$

$\displaystyle \mathrm{IC:}\hspace{0.5cm} u(x,0) = f(x) \ \ \ \ \ (60)$

This is a heat conduction problem in a one-dimensional rod with constant thermal properties and no sources. The ends of the rod are insulated. Both the PDE and BCs are linear and homogeneous. We call this type of BC the Neumann boundary condition. We apply SoV.

Solution. The assumed product solutions are:

$\displaystyle u(x,t)=\phi(x)G(t) \ \ \ \ \ (61)$

Substitute into the PDE,

$\displaystyle \phi G'= kG\phi^{''}= -\lambda \ \ \ \ \ (62)$

where ${\lambda}$ is the separation constant. This implies that

$\displaystyle G^{'}=-\lambda kG \ \ \ \ \ (63)$

$\displaystyle \phi^{''}=-\lambda\phi \ \ \ \ \ (64)$

Solving (64),

$\displaystyle G(t)=e^{-\lambda kt} \ \ \ \ \ (65)$

The insulated BCs imply that the separated solutions must satisfy

$\displaystyle \phi^{'}(0)=0 \ \ \ \ \ (66)$

$\displaystyle \phi^{'}(L)=0 \ \ \ \ \ (67)$

We determine the separation constant ${\lambda}$ by finding those ${\lambda}$ for which nontrivial solutions exist.

For ${\lambda <0}$ , the general solution is

$\displaystyle \phi(x)=c_1e^{\sqrt{-\lambda}x}+c_2e^{-\sqrt{-\lambda}x} \ \ \ \ \ (68)$

We also need

$\displaystyle \phi^{'}(x)=\sqrt{-\lambda}(c_1e^{\sqrt{-\lambda}x}-c_2e^{-\sqrt{-\lambda}x}) \ \ \ \ \ (69)$

The BC ${\phi^{'}(0)=0}$ implies that ${c_1=c_2}$ , and ${\phi^{'}(L)=0}$ means that

$\displaystyle \sqrt{-\lambda}(c_1e^{\sqrt{-\lambda}L}-c_2e^{-\sqrt{-\lambda}L})=2\sqrt{-\lambda}c_1\mathrm{sinh}(\sqrt{-\lambda} L)=0 \ \ \ \ \ (70)$

Since ${\sqrt{-\lambda} L >0}$ , ${\mathrm{sinh}(\sqrt{-\lambda} L)>0}$ . This implies that ${c_1=0}$ . Since ${\phi(x)=0}$ is a trivial solution, there is no negative eigenvalues.

For ${\lambda=0}$ , the solution is

$\displaystyle \phi(x)=c_1+c_2x \ \ \ \ \ (71)$

and

$\displaystyle \phi^{'}(x)=c_2 \ \ \ \ \ (72)$

Both BCs give ${c_2=0}$ . Thus, there are non trivial solutions of the BVP for ${\lambda=0}$ , namely,

$\displaystyle \phi(x)=c_1 \ \ \ \ \ (73)$

The time dependent part gives ${G(t)=e^{-\lambda kt}=1}$ . Thus, the resulting product solution of the PDE is

$\displaystyle u(x,t)=A \ \ \ \ \ (74)$

where ${A}$ is any constant.

For ${\lambda>0}$ , the solutions is in the form

$\displaystyle \phi(x)=e^{\pm\sqrt{\lambda}ix} \ \ \ \ \ (75)$

which gives

$\displaystyle \phi(x)=c_1\mathrm{cos}(\sqrt{\lambda} x)+c_2\mathrm{sin}(\sqrt{\lambda} x) \ \ \ \ \ (76)$

and

$\displaystyle \phi^{'}(x)=\sqrt{\lambda}(c_2\mathrm{cos}(\sqrt{\lambda} x)-c_1\mathrm{sin}(\sqrt{\lambda} x)) \ \ \ \ \ (77)$

The BC ${\phi^{'}(0)=0}$ implies that ${c_2\sqrt{\lambda}=0}$ , so ${c_2=0}$ . The BC ${\phi^{'}(L)=0}$ gives

$\displaystyle -\sqrt{\lambda}c_1\mathrm{sin}(\sqrt{\lambda}L)=0 \ \ \ \ \ (78)$

For nontrivial solutions, ${c_1\neq0}$ . This gives us ${\mathrm{sin}(\sqrt{\lambda}L)=0}$ .

$\displaystyle \lambda= (\frac{n\pi}{L})^2, n=0,1,2,3... \ \ \ \ \ (79)$

However, the corresponding eigenfunctions are cosines,

$\displaystyle \phi(x)=c_1\mathrm{cos}(\frac{n\pi x}{L}), n=0,1,2,3... \ \ \ \ \ (80)$

The product solution of the PDE is

$\displaystyle u(t,x)=Ae^{-(\frac{n\pi}{L})^2kt}\mathrm{cos}(\frac{n\pi x}{L}), n=0,1,2,3... \ \ \ \ \ (81)$

where A is a arbitrary multiplicative constant.

In order to satisfy the IC, we use the principle of superposition so that,

$\displaystyle u(t,x)=A_0+\sum_{n=1}^{\infty}A_{n}e^{-(\frac{n\pi}{L})^2kt}\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (82)$

The IE ${u(x,0)=f(x)}$ is satisfied if

$\displaystyle f(x)=A_0+\sum_{n=1}^{\infty}A_{n}\mathrm{cos}(\frac{n\pi x}{L}) \ \ \ \ \ (83)$

for ${0\leq x \leq L}$ .

To complete the solution, we need to determine the arbitrary coefficient ${A_0}$ and ${A_n}$ . We use the fact that ${\mathrm{cos}(\frac{n\pi x}{L})}$ satisfies the following orthogonality relation:

$\displaystyle \int_{0}^{L} \mathrm{cos}(\frac{n \pi x}{L})\mathrm{cos}(\frac{m \pi x}{L}) \,dx = \left\{ \begin{array}{lr} 0 & n \neq m \\ \frac{L}{2} & n=m\neq 0 \\ L & n=m=0 \end{array} \right. \ \ \ \ \ (84)$

for ${n}$ and ${m}$ nonnegative integers.

Multiply (83) by ${\mathrm{cos}(\frac{m \pi x}{L})}$ and integrating from ${0}$ to ${L}$ yields

$\displaystyle \int_{0}^{L} f(x)\mathrm{cos}(\frac{m \pi x}{L})\, dx = A_{m}\int_{0}^{L}\mathrm{cos^2}(\frac{m\pi x}{L})\, dx \ \ \ \ \ (85)$

Solving for ${A_m}$ yields,

$\displaystyle A_0= \frac{1}{L} \int_{0}^{L}f(x)\, dx \ \ \ \ \ (86)$

$\displaystyle A_m=\frac{2}{L} \int_{0}^{L}f(x)\mathrm{cos}(\frac{m \pi x}{L})\, dx , m\geq 1 \ \ \ \ \ (87)$

${\square}$

Percolation theory: percolation on a square lattice

April 11, 2022April 12, 2022Michael LiuLeave a comment

Figure 1. Percolation on a 100×100 square lattice at p=0.3, 0.5, 0,7 (in the order of top to bottom). The red squares are ones that can be reached from the top or bottom by going only on filled squares.

In the previous post, we introduced the basic concepts of percolation theory. Here we define the terminologies formally. Let $E$ denote the edge set of a graph $\wedge$ . We take $\Omega = \prod_{e\in E} \{0,1\}$ as a sample space, points in which are represented as $\omega=(\omega(e):e\in \mathbb{E}^d)$ and are called configurations. A bond $e$ is open in the configuration $\omega$ if $\omega(e)=1$ and is closed if $\omega(e)=0$ , so configurations correspond to open subgraphs. We take $F$ to be the $\sigma-$ field of subsets of $\Omega$ generated by the cylinder sets. The associated probability measure is going to be the product measure with density $p$ on $(\Omega, F)$ ,

$P_p=\prod_{e\in E} \mu_e$

where $\mu_e$ is Bernoulli measure on {0,1}, given by $\mu_p(1)=p$ and $\mu_p(0)=1-p$ . We can take $\Omega$ to be the set of possible outcomes of our random subgraph and take $P_p$ to be describing its distribution. We can basically ignore the measure-theoretic details above but take note that we do use the notation $P_p$ to describe probabilities when the parameter used is $p$ .

Let us recall the definition critical probability, $p_H =$ sup $\{p:\theta(p)=0\} =$ inf $\{p:\theta(p)>0\}$ . We now present a basic result in probability theory.

Theorem 2.1 Suppose that ${X_n}$ is a sequence of independent random variables. Let A be an event in the $\sigma$ -field generated by ${X_n}$ . Suppose that the event A is independent of each finite subset of ${X_n}$ , then $\mathbb{P}(A)$ is 0 or 1 .

This is known as Kolmogorov’s 0-1 law. In percolation, we know that each site or bond has an independent probability $p$ . Let $E$ be an event that there is an infinite open cluster. The event $E$ is invariant under finite changes of sites or bonds. Thus, Kolmogorov’s 0-1 law implies that $\mathbb{P}$ (E) is either 0 or 1.

If $p<p_H$ , then

$P_p(E)\leq \sum_{x}P_p(|C_x|=\infty)=\sum_{x}\theta_x(p)=0$

If $p>p_H$ , then

$P_p(E)\geq P_p(|C_x|=\infty)=\theta_x(p) > 0$

for some site $x$ , implying that $P_p(E)=1$ . In this case, we say that percolation occurs.

Let’s sum up what we have shown so far with an analogy. Consider the case of water moving through a very large medium (so large that it will take infinitely long for water to reach the surface from an origin inside the medium). Let the probability that a pore is big enough for water to pass through be $p$ . If the medium is made of less porous material, the pores having small $p$ are less likely to be open for the water to flow through, so any open path that water travels through will likely be short. However, if it is made of more porous material with higher $p$ , there are more open channels, so water can travel through a longer path from the origin. It is obvious to see that as $p$ increases, it is likely that, eventually, water will travel infinitely far. We have observed that there is a critical probability $p_H$ representing a threshold for such an event to occur. If $p>p_H$ , then there is a positive probability that the path that water travels through is infinite. Kolmogorov’s 0-1 law guarantees that if $p>p_H$ , an infinite path must exist inside the medium, though in our case, water may not flow in that path (as there can be many disconnected open paths inside the medium and, base on our assumption, water only flow in one of them).

Figure 2. A sketch of the structure of a two-dimensional porous medium. The lines indicate open paths. Comparing both graphs, we see that water flows in a shorter path in a less porous medium with a lower value of p.

We see that on either side of the critical probability, the global behavior of the system is fundamentally different. And at critical probability, a sharp transition takes place which transforms the behavior of the system from one form to the other. Thus, the existence of a critical probability makes percolation a mathematically interesting and rich subject.

Next, we continue with our study of percolation on $\mathbb{Z}^2$ . First, we want to show that bond percolation on $\mathbb{Z}^2$ is non-trivial: $0<p_H<1$ . Let $\mu(\wedge;x)$ be the number of self-avoiding walks in $\wedge$ starting at $x$ . In graph theory, a self-avoiding walk is also known as a path. As $\mathbb{Z}^2$ is 4 regular, the number of self-avoiding walks of length $n$ starting at the origin 0 is $\mu_n=\mu_n(\mathbb{Z}^2)=\mu_n(\mathbb{Z}^2;0)\leq 4\times 3^{n-1}$ , since there are 4 choices for the first step and at most 3 choices in the later steps.

Theorem 2.2 For bond percolation in $\mathbb{Z}^2$ , we have $p_H \geq \frac{1}{3}$ .

Proof Let $F_n$ be the event that there is an open cluster $C_0$ of size $n$ starting at the origin 0 where each bond is open with probability $p$ . Then the probability that all bonds are open is $p^n$ . For every site $x \in C_0$ , there is at least one open path from from 0 to $x$ . Thus, we have

$P_p(F_n)\leq \mu_n p^n= \frac{4}{3}(3p^n)$

For $p<\frac{1}{3}, (3p^n)<1$ . Since ${|C_0|=\infty}\subseteq F_n$ $\forall$ $n$, let there be an infinite cluster from the origin with $p<\frac{1}{3}$ , we have

$P_p(|C_0|=\infty)=\lim_{n\rightarrow\infty}P_\textit{p}(F_n)\leq \lim_{n\rightarrow\infty} \frac{4}{3}(3p^n)=0$

This is the equivalent to saying if $p<\frac{1}{3}$ , $\theta(p)=0$ . By definition of $p_H$ , it follows that $p_H \geq \frac{1}{3}$ .

$\square$

We now consider an upper bound. We will make use of a key result from graph theory.

Lemma 2.3 (Grimmett) Let $C$ be a vertex set of a connected subgraph of $\mathbb{Z}^2$ . $|C|<\infty$ if and only if $\exists$ a simple cycle with $C$ in its interior.

No proof will be provided here, but by drawing some pictures we can convince ourselves that it is very believable. Looking at figures 3 and 4 will also be helpful.

Theorem 2.4 For bond percolation in $\mathbb{Z}^2$ , we have $p_H \leq \frac{2}{3}$ .

Proof The first thing we do is to introduce the dual $\wedge^*$ of a graph $\wedge$ drawn in the plane has a vertex for each face of $\wedge$ and an edge $e^*$ for each edge $e$ in $\wedge$ . When $\wedge=\mathbb{Z}^2$ , we take $\wedge^* = \mathbb{Z}^2+(\frac{1}{2},\frac{1}{2})$ , which is isomorphic to $\wedge$ . We call an bond in the dual graph $(\mathbb{Z}^2)^*$ open if and only if the corresponding bond in the original graph $\mathbb{Z}^2$ is closed. If the set of open bonds in $\mathbb{Z}^2$ are given by $P_p$ , then the distribution of the set of closed bonds in $(\mathbb{Z}^2)^*$ will also be given by $P_p$ . An \textit{open dual cycle} is a cycle in the dual graph consisting of dual bonds that are open.

Now that we have defined all the necessary terms. Suppose that $p>\frac{2}{3}$ , let $L_k$ be the line segment joining the origin to the point $(k,0)$ , and let $S$ be a dual cycle surrounding $L_k$ , which has a length $n$ . Then $S$ must contain a dual bond $e^*$ crossing the positive $x$ -axis at some coordinate between $(k+\frac{1}{2},0)$ and $(\frac{1}{2}(n-3),0)$ . This means that we have less than $\frac{n}{2}$ choices for $e^*$ . As the rest of S is a path of length $n-1$ in the dual lattice, we have shown that the number of cycles around the origin of length $n$ is at most $\frac{n}{2}\times 4(3^{n-1})$ . Let $A_n$ be the event that there exists an open dual cycle surrounding $L_k$ , since dual bonds are open with probability $1-p$ , we have

$\sum_{n=4}^{\infty} P_p(A_n) \leq \sum_{n=4}^{\infty}\frac{2n}{3}(3(1-p))^n$

Since $3(1-n)<1$ , the sum is convergent. We can choose a $N$ so that $\sum_{n\geq N}^{\infty}\frac{2n}{3}(3(1-p))^n <1$ . Now, let $E_1$ be the event that the $N$ bonds in the line segment $L_N$ are open. Let $E_2$ be the event that there are no open dual cycles surrounding $L_N$ . Since $\sum_{n=4}^{\infty} P_p(A_n)<1$ , $P_p(E_2)>0$ . If both $E_1$ and $E_2$ holds then $|C|=\infty$ by Lemma 2.3. Since $E_1$ and $E_2$ are independent, we have

$\theta(p)\geq P_p(E_1 \cap E_2) = P_p(E_1)P_p(E_2) = P_p(E_2)p^N >0$

Since $p>\frac{2}{3}$ , by definition of $p_H$ , it follows that $p_H \leq \frac{2}{3}$ .

$\square$

Figure 4. A finite open cluster at the origin, surrounded by an open dual cycle (corresponding to closed bonds in the original lattice)

We have shown that the critical probability $p_H$ for $\mathbb{Z}^2$ is in between $\frac{1}{3}$ and $\frac{2}{3}$ , so there exists a non-trivial critical phenomenon. On the basis of Monte Carlo simulations, it was suggested that the critical probability should be $\frac{1}{2}$ . In the next post, we will show that the critical probability for $\mathbb{Z}^2$ is in fact $\frac{1}{2}$ .