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2 Heat Equation

2.1 Derivation

Ref: Strauss, Section 1.3.

Below we provide two derivations of theheat equation, u t¡kuxx= 0k >0:(2.1) This equation is also known as thediffusion equation.

2.1.1 Diffusion

Consider a liquid in which a dye is being diffused through the liquid. The dye will move from higher concentration to lower concentration. Letu(x;t) be the concentration (mass per unit length) of the dye at positionxin the pipe at timet. The total mass of dye in the pipe fromx0tox1at timetis given by

M(t) =Z

x1 x

0u(x;t)dx:

Therefore,

dM dt =Z x1 x 0u t(x;t)dx:

By Fick"s Law,

dM dt = flow in¡flow out =kux(x1;t)¡kux(x0;t); wherek >0 is a proportionality constant. That is, the flow rate is proportional to the concentration gradient. Therefore, Z x1 x 0u t(x;t)dx=kux(x1;t)¡kux(x0;t):

Now differentiating with respect tox1, we have

u t(x1;t) =kuxx(x1;t): Or, u t=kuxx:

This is known as thediffusion equation.

2.1.2 Heat Flow

We now give an alternate derivation of (2.1) from the study of heat flow. LetDbe a region inRn. Letx= [x1;:::;xn]Tbe a vector inRn. Letu(x;t) be the temperature at pointx, 1 timet, and letH(t) be the total amount of heat (in calories) contained inD. Letcbe the specific heat of the material and½its density (mass per unit volume). Then

H(t) =Z

D c½u(x;t)dx:

Therefore, the change in heat is given by

dH dt =Z D c½u t(x;t)dx: Fourier"s Law says that heat flows from hot to cold regions at a rate· >0 proportional to the temperature gradient. The only way heat will leaveDis through the boundary. That is,dH dt =Z @D

·ru¢ndS:

where@Dis the boundary ofD,nis the outward unit normal vector to@DanddSis the surface measure over@D. Therefore, we have Z D c½u t(x;t)dx=Z @D

·ru¢ndS:

Recall that for a vector fieldF, the Divergence Theorem says Z @D

F¢ndS=Z

D r ¢F dx: (Ref: See Strauss, Appendix A.3.) Therefore, we have Z D c½u t(x;t)dx=Z D r ¢(·ru)dx: This leads us to the partial differential equation c½u t=r ¢(·ru): Ifc;½and·are constants, we are led to the heat equation u t=kΔu; wherek=·=c½ >0 and Δu=Pn i=1uxixi.

2.2 Heat Equation on an Interval inR

2.2.1 Separation of Variables

Consider the initial/boundary value problem on an intervalIinR, 8< :u t=kuxxx2 I;t >0 u(x;0) =Á(x)x2 I usatisfies certain BCs.(2.2) In practice, the most common boundary conditions are the following: 2 1.

Dirichlet (I= (0;l)) :u(0;t) = 0 =u(l;t).

2.

Neumann (I= (0;l)) :ux(0;t) = 0 =ux(l;t).

3. Robin (I= (0;l)) :ux(0;t)¡a0u(0;t) = 0 andux(l;t) +alu(l;t) = 0. 4. Periodic (I= (¡l;l)):u(¡l;t) =u(l;t) andux(¡l;t) =ux(l;t). We will give specific examples below where we consider some of these boundary condi- tions. First, however, we present the technique ofseparation of variables. This technique involves looking for a solution of a particular form. In particular, we look for a solution of the form u(x;t) =X(x)T(t) for functionsX,Tto be determined. Suppose we can find a solution of (2.2) of this form. Plugging a functionu=XTinto the heat equation, we arrive at the equation XT

0¡kX00T= 0:

Dividing this equation bykXT, we have

T 0 kT =X00 X for some constant¸. Therefore, if there exists a solutionu(x;t) =X(x)T(t) of the heat equation, thenTandXmust satisfy the equations T 0 kT X 00 X for some constant¸. In addition, in order foruto satisfy our boundary conditions, we need our functionXto satisfy our boundary conditions. That is, we need to find functionsX and scalars¸such that(¡X00(x) =¸X(x)x2 I

Xsatisfies our BCs.(2.3)

This problem is known as aneigenvalue problem. In particular, a constant¸which satisfies (2.3) for some functionX, not identically zero, is called aneigenvalueof¡@2xfor the given boundary conditions. The functionXis called aneigenfunctionwith associated eigenvalue¸. Therefore, in order to find a solution of (2.2) of the formu(x;t) =X(x)T(t) our first goal is to find all solutions of our eigenvalue problem (2.3). Let"s look at some examples below.

Example 1.

(Dirichlet Boundary Conditions) Find all solutions to the eigenvalue problem

½¡X00=¸X0< x < l

X(0) = 0 =X(l):(2.4)

3 Any positive eigenvalues?First, we check if we have any positive eigenvalues. That is, we check if there exists any¸=¯2>0. Our eigenvalue problem (2.4) becomes

½X00+¯2X= 0 0< x < l

X(0) = 0 =X(l):

The solutions of this ODE are given by

X(x) =Ccos(¯x) +Dsin(¯x):

The boundary condition

X(0) = 0 =)C= 0:

The boundary condition

X(l) = 0 =)sin(¯l) = 0 =)¯=n¼

l n= 1;2;:::: Therefore, we have a sequence of positive eigenvalues n=³n¼ l 2 with corresponding eigenfunctions X n(x) =Dnsin³n¼ l x´ Is zero an eigenvalue?Next, we look to see if zero is an eigenvalue. If zero is an eigenvalue, our eigenvalue problem (2.4) becomes

½X00= 0 0< x < l

X(0) = 0 =X(l):

The general solution of the ODE is given by

X(x) =C+Dx:

The boundary condition

X(0) = 0 =)C= 0:

The boundary condition

X(l) = 0 =)D= 0:

Therefore, the only solution of the eigenvalue problem for¸= 0 isX(x) = 0. By definition, the zero function is not an eigenfunction. Therefore,¸= 0 is not an eigenvalue. Any negative eigenvalues?Last, we check for negative eigenvalues. That is, we look for an eigenvalue¸=¡°2. In this case, our eigenvalue problem (2.4) becomes

½X00¡°2X= 0 0< x < l

X(0) = 0 =X(l):

4

The solutions of this ODE are given by

X(x) =Ccosh(°x) +Dsinh(°x):

The boundary condition

X(0) = 0 =)C= 0:

The boundary condition

X(l) = 0 =)D= 0:

Therefore, there are no negative eigenvalues.

Consequently,allthe solutions of (2.4) are given by n=³n¼ l 2X n(x) =Dnsin³n¼ l x´ n= 1;2;::::

Example 2.

(Periodic Boundary Conditions) Find all solutions to the eigenvalue problem

½¡X00=¸X¡l < x < l

X(¡l) =X(l); X0(¡l) =X0(l):(2.5)

Any positive eigenvalues?First, we check if we have any positive eigenvalues. That is, we check if there exists any¸=¯2>0. Our eigenvalue problem (2.5) becomes

½X00+¯2X= 0¡l < x < l

X(¡l) =X(l); X0(¡l) =X0(l):

The solutions of this ODE are given by

X(x) =Ccos(¯x) +Dsin(¯x):

The boundary condition

X(¡l) =X(l) =)Dsin(¯l) = 0 =)D= 0 or¯=n¼ l

The boundary condition

X

0(¡l) =X0(l) =)C¯sin(¯l) = 0 =)C= 0 or¯=n¼

l Therefore, we have a sequence of positive eigenvalues n=³n¼ l 2 with corresponding eigenfunctions X n(x) =Cncos³n¼ l x´ +Dnsin³n¼ l x´ 5 Is zero an eigenvalue?Next, we look to see if zero is an eigenvalue. If zero is an eigenvalue, our eigenvalue problem (2.5) becomes

½X00= 0¡l < x < l

X(¡l) =X(l); X0(¡l) =X0(l):

The general solution of the ODE is given by

X(x) =C+Dx:

The boundary condition

X(¡l) =X(l) =)D= 0:

The boundary condition

X

0(¡l) =X0(l) is automatically satisfied ifD= 0:

Therefore,¸= 0 is an eigenvalue with corresponding eigenfunction X

0(x) =C0:

Any negative eigenvalues?Last, we check for negative eigenvalues. That is, we look for an eigenvalue¸=¡°2. In this case, our eigenvalue problem (2.5) becomes

½X00¡°2X= 0¡l < x < l

X(¡l) =X(l); X0(¡l) =X0(l):

The solutions of this ODE are given by

X(x) =Ccosh(°x) +Dsinh(°x):

The boundary condition

X(¡l) =X(l) =)Dsinh(°l) = 0 =)D= 0:

The boundary condition

X

0(¡l) =X0(l) =)C°sinh(°l) = 0 =)C= 0:

Therefore, there are no negative eigenvalues.

Consequently,allthe solutions of (2.5) are given by n=³n¼ l 2X n(x) =Cncos³n¼ l x´ +Dnsin³n¼ l x´ n= 1;2;:::

0= 0X0(x) =C0:

6 Now that we have done a couple of examples of solving eigenvalue problems, we return to using the method of separation of variables to solve (2.2). Recall that in order for a function of the formu(x;t) =X(x)T(t) to be a solution of the heat equation on an intervalI ½R which satisfies given boundary conditions, we needXto be a solution of the eigenvalue problem,½X00=¡¸X x2 I

Xsatisfies certain BCs

for some scalar¸andTto be a solution of the ODE

¡T0=k¸T:

We have given some examples above of how to solve the eigenvalue problem. Once we have solved the eigenvalue problem, we need to solve our equation forT. In particular, for any scalar¸, the solution of the ODE forTis given by

T(t) =Ae¡k¸t

for an arbitrary constantA. Therefore, for each eigenfunctionXnwith corresponding eigen- value¸n, we have a solutionTnsuch that the function u n(x;t) =Tn(t)Xn(x) is a solution of the heat equation on the intervalIwhich satisfies our boundary conditions. Note that we have not yet accounted for our initial conditionu(x;0) =Á(x). We will look at that next. First, we remark that iffungis a sequence of solutions of the heat equation onI which satisfy our boundary conditions, than anyfinitelinear combination of these solutions will also give us a solution. That is, u(x;t)´NX n=1u n(x;t) will be a solution of the heat equation onIwhich satisfies our boundary conditions, assuming eachunis such a solution. In fact, one can show that aninfiniteseries of the form u(x;t)´1X n=1u n(x;t) will also be a solution of the heat equation, under proper convergence assumptions of this series. We will omit discussion of this issue here.

2.2.2 Satisfying our Initial Conditions

We return to trying to satisfy our initial conditions. Assume we have found all solutions of our eigenvalue problem. We letfXngdenote our sequence of eigenfunctions andf¸ngdenote our sequence of eigenvalues. Then for each¸n, we have a solutionTnof our equation forT. Let u(x;t) =X nX n(x)Tn(t) =X nA nXn(x)e¡k¸nt: 7 Our goal is to chooseAnappropriately such that our initial condition is satisfied. In partic- ular, we need to chooseAnsuch that u(x;0) =X nA nXn(x) =Á(x): In order to findAnsatisfying this condition, we use the followingorthogonalityproperty of eigenfunctions. First, we make some definitions. For two real-valued functionsfandgdefined on Ω, hf;gi=Z f(x)g(x)dx is defined as theL2inner productoffandgon Ω. TheL2normoffon Ω is defined as jjfjj2

L2(Ω)=hf;fi=Z

jf(x)j2dx:

We say functionsfandgareorthogonalon Ω½Rnif

hf;gi=Z f(x)g(x)dx= 0:

We say boundary conditions aresymmetricif

[f0(x)g(x)¡f(x)g0(x)]jx=b x=a= 0 for all functionsfandgsatisfying the boundary conditions.

Lemma 3.

Consider the eigenvalue problem(2.3)with symmetric boundary conditions. If X n;Xmare two eigenfunctions of(2.3)with distinct eigenvalues, thenXnandXmare or- thogonal.

Proof.

LetI= [a;b].

nZ b a X n(x)Xm(x)dx=¡Z b a

X00n(x)Xm(x)dx

Z b a

X0n(x)X0m(x)dx¡X0n(x)Xm(x)jx=b

x=a =¡Z b a X n(x)X00m(x)dx+ [Xn(x)X0m(x)¡X0n(x)Xm(x)]jx=b x=a =¡¸mZ b a X n(x)Xm(x)dx; using the fact that the boundary conditions are symmetric. Therefore, (¸n¡¸m)Z b a X n(x)Xm(x)dx= 0; 8 but¸n6=¸mbecause the eigenvalues are assumed to be distinct. Therefore, Z b a X n(x)Xm(x)dx= 0; as claimed. We can use this lemma to find coefficientsAnsuch that X nA nXn(x) =Á(x): In particular, multiplying both sides of this equation byXmfor a fixedmand integrating overI, we have A mhXm;Xmi=hXm;Ái; which implies A m=hXm;Ái hXm;Xmi:

Example 4.

(Dirichlet Boundary Conditions) In the case of Dirichlet boundary conditions on the interval [0;l], we showed earlier that our eigenvalues and eigenfunctions are given by n=³n¼ l 2; X n(x) = sin³n¼ l x´ n= 1;2;::::

Our solutions forTnare given by

T n(t) =Ane¡k¸nt=Ane¡k(n¼=l)2t:

Now let

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