Biomedical Engineering - Mathematical and Numerical Methods in Engineering

Completed notes of the course - Mathematical Methods

Complete course

Mathematical Methods Course held by Prof. Giovanni CiprianiYear 2020-2021 Notes by Alessandro M. IppolitiAugustin-Louis Cauchy 1789-1857. 2 Contents 1 Overview 51.1 Euclidian Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Function Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.6 Dierential Operators . . . . . . . . . . . . . . . . . . . . . . . 9 1.7 Surfaces in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.8 Surface Integrals and the Divergence Theorem . . . . . . . . . 111.8.1 Derivation of Archimedes Principle . . . . . . . . . . . 12 2 Fourier Series and Convergence 152.1 Conditions for Convergence . . . . . . . . . . . . . . . . . . . 16 2.2 The Norm and Uniform Convegence . . . . . . . . . . . . . . . 17 2.3 Gibbs Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . 18 3 Properties of L2 19 3.1 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . 19 3.2 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . 23 3.4 Bilinear Functionals and the Lax-Milgram Theorem . . . . . . 23 3.4.1 Applications of the Lax-Milgram Theorem . . . . . . . 24 4 Transport Equations 274.1 Derivation of the transport Equation from a Conservation Law 27 4.1.1 Construction of the Solution to the T.Eq in the DriftCase . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.1.2 Construction of the Solution to the T.Eq in the DriftCase in the Presence of Sources . . . . . . . . . . . . . 30 4.1.3 Construction of the Solution to the T.Eq in the DriftCase with Extinction . . . . . . . . . . . . . . . . . . . 31 3 4 CONTENTS 5 Characteristic Curves 335.1 The Green Light Problem . . . . . . . . . . . . . . . . . . . . 35 5.2 Trac-Jam Ahead Problem . . . . . . . . . . . . . . . . . . . 36 5.3 Burger's Equation . . . . . . . . . . . . . . . . . . . . . . . . . 41 5.4 The Lax Condition . . . . . . . . . . . . . . . . . . . . . . . . 42 6 The Heat Equation 436.1 Heat Equation inR1 . . . . . . . . . . . . . . . . . . . . . . . 43 6.2 Fourier's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 6.3 Heat Equation in Euclidian Spaces . . . . . . . . . . . . . . . 47 6.4 Properties of the Solution to the Heat Equation . . . . . . . . 50 6.5 The Heat Equation in Arbitrary Domains . . . . . . . . . . . 51 6.6 The Maximum Principle for the Heat Equation (General Case) 53 6.7 Using the Maximum Principle to Estimate Solutions . . . . . . 56 6.8 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . 58 7 Poisson's Equation 637.0.1 Laplace's Equation . . . . . . . . . . . . . . . . . . . . 63 7.1 Electrostatic Model . . . . . . . . . . . . . . . . . . . . . . . . 64 7.2 Boundary Conditions for Laplace's Equation . . . . . . . . . . 67 7.3 The Mean Value Property of Harmonic Functions . . . . . . . 707.3.1 The Maximum Principle for Harmonic Functions . . . 74 7.3.2 Method of Separation of variables for the Laplace Equa-tion with D.B.C. on Rectangles . . . . . . . . . . . . . 77 8 The Wave Equation 818.1 Conservation of Energy for the Wave Equation . . . . . . . . . 84 8.2 D'Alembert's Formula . . . . . . . . . . . . . . . . . . . . . . 868.2.1 Method of Separation of Variables for the Heat andWave Equation . . . . . . . . . . . . . . . . . . . . . . 89 8.3 Weyl's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 92 9 Exam Cheatsheet 95 Chapter 1 Overview 1.1 Euclidian Spaces An example of aEuclidian Spaceis then-th dimensional intersection ofR, Rn =n times z}|{ RR:::R These spaces are characterized by the existence of anOrthonormal Base, Base= ek n k=1 where ek:=0 @0;0; :::;1 |{z} nth1 ;0;0; :::;01 A as such, any vectorx2Rn can be written as Rn 3x= (x 1; x 2; :::; x n) = x 1e 1+ x 2e 2+ :::+x ne n TheNormof a vectorx2Rn in this space is given by the formula, kxk:=v u u tn X k=1x 2 k This particular norm is also called theEuclidian Lengthof the vector. Similarly, we can de ne aEuclidian Distancebetween two vectorsx;y2 Rn as, 5 6 CHAPTER 1. OVERVIEW kxyk:=v u u tn X k=1( x k y k)2 Finally, we de neScalar ProductorDot Productof two vectorsx;y2 Rn as, x
y:=n X k=1x ky k and, as such, we can see that the norm of a vector can be written in terms of its scalar product with itself, kxk=px
x Two vectors are said to beOrthogonalif their dot product is null, x?y()x
y= 0 1.2 Sets In order to de ne various characteristics of Euclidian Spaces we must rst de ne the mathematical constructs known asBalls, ˆOpen BallB(x; R) :=n y2Rn :kxyk< Ro ˆClosed Ball B(x; R) :=n y2Rn :kxyk=< Ro They are the points whose euclidian distance from a speci c element x2Rn is less than a quantityR. De nition 1.1(Open Set).AnOpen SetARn can be de ned as follows: A is the set in which8x2A;9r >0 :B(x; r)A De nition 1.2(Limit).Given a functionf:ARn !Rm and two elementsx2Aand l2Rm we can write that, lim x!x 0f (x) = l!lim x!x 0 f(x) l = 0 where the limit on the right-hand side of the equation is to be understood as, 8 >0;9 >0 :kxx 0k < =) f(x) l <  1.3. PARTIAL DERIVATIVES 7 De nition 1.3(Continuous Function).A functionf:ARn !Rm is said to beContinuousinx 0if lim x!x 0f (x) =f(x 0) 1.3 Partial Derivatives This course will deal mainly withOrdinary Dierential Equations(ODEs) andPartial Dierential Equations(PDEs). As such we will have to keep in mind all the ordianry rules of dierentiation, such as theLeibnitz Rule for Scalar Functionsand the so-calledChain Rule: De nition 1.4(Leibnitz Rule for Scalar Functions).Given the product of two scalar functionsf g:ARn !Rwe can write that, @xk( f g) =@ k( f g) = (@ kf )g+f(@ kg ) De nition 1.5(Chain Rule).Given the open spacesARn ,BRm and a vector functiong:A!B, g= g1 ; g2 ; :::; gm
where the functionsgk :A!R; k= 1;2; :::; nare calledCoordinatesof g, we have that, ifg(A)B, the functionfg:A!Cis meaningful and we have that, @k( f(g)) (x) =m X j=1( @ kf ) (g(x)) @kgj
(x) 1.4 Function Spaces De nition 1.6(Function Space).AFunction Spaceis a set whose ele- ments are functions of euclidian variables. Given Rn open, we de ne the folowing function spaces: ˆC( ) :=n f: !Rwhich are continuouso ˆC1 ( ) :=n f: !Radmitting1st order continuous partial derivateso ˆC1 ( ) :=n f: !Radmitting1order continuous partial derivateso It is easy to see thatC1 ( )C1 ( )C( ). 8 CHAPTER 1. OVERVIEW 1.5 Vector Calculus This course will deal mainly withOrdinary Dierential Equations(ODEs) andPartial Dierential Equations(PDEs). As such we will have to keep in mind all the ordinary rules of dierentiation, such as theLeibnitz Rule for Scalar Functionsand the so-calledChain Rule. Before we do that it might help to remind the student of certain de nitions. De nition 1.7(Scalar Function).Given Rn aScalar Function, s(x) =s(x 1; x 2;


x n) : !R is a function whosedomainis a vector space and whosevalueis a scalar eld. An example of this inR3 would be, s(x) = 3x2 y+z3 which evaluated at the point, x= (3;3;1) returns, s(x) = 81 + 1 = 82 which is a scalar value. De nition 1.8(Vector Function).Given Rn aVector Function, v(x) = (v 1( x); v 2( x); v 3( x)) is a function whosedomainis a vector space and whosevalueis a vector eld. An example inR3 would be, v(x) =0 @2 x+y2 y+z3 xe 5y1 A which evaluated at the same pointxas before returns, v(x) = (15;4;3e 15 ) which is a vector. 1.6. DIFFERENTIAL OPERATORS 9 1.6 Dierential Operators On both scalar and vector functions we can apply the followingDierential Operators. De nition 1.9(Gradient).TheGradient Operatorrof a function f(x) :Rm !Rn , r:Rm !Rm n is an operator whose value, when applied to a dierentiable function (ei- ther scalar of vectorial) is given by, (rs)(x) = @ s@ x 1; @ s@ x 2;


;@ s@ x n (rv)(x) =J=0 B B @@ v 1@ x 1@ v 1@ x 2;


@ v 1@ x n @ v2@ x 1@ v 2@ x 2


@ v 2@ x n


@ vn@ x 1@ v n@ x 2


@ v n@ x n1 C C A So the gradient of a scalar function is avector, and the gradient of a vec- tor function is atensor. The matrixJgiven by1:9is called theJacobian of the vector eld. This operator applied to our test functions 1:5 and 1:5 returns, (rs)(x) = 6xy;3x2 ;1
= (54;27;1) (rv)(x) =J=0 @2 2 y0 0 1 3z2 e 5y 5xe 5y 01 A=0 @2 ;6;0 0;1;3 e 15 ;15e 15 ;01 A De nition 1.10(Divergence Operator).TheDivergence Operatorr
is avector operatorwhose value, when applied to a vector function, is given by, (r
v) (x) =@ v 1( x)@ x 1+ @ v 2( x)@ x 2+


+@ v n( x)@ x n So that the divergence of a vector function returns a scalar value. Applying this operator to our test function we get, (r
v) (x) = 2 + 1 + 0 = 3 10 CHAPTER 1. OVERVIEW De nition 1.11(Curl Operator).TheCurl Operatorris avector operatorwhose value, when applied to a vector function inR3 written in Cartesian coordinates, is given by, (rv) (x) = i j k @@ x @@ y @@ z v1( x)v 2( x)v 3( x) = i @ v3@ y @ v 2@ z  +j @ v1@ z @ v 3@ x  +k @ v2@ x @ v 1@ y  So that the rotor of a vector function returns a vector. Applying this operator to our test function we get, (rv) =i (5)xe 5y 3z2
+j 0e 5y
+k(02y) = (15e 15 3;e 15 ;6) De nition 1.12(Laplacian Operator).TheLaplacian Operatorr2 = r
ris an operator whose value, when applied to a dierentiable function (either scalar of vectorial) is given by, r2 s(x) =r
(rs) (x) =@ 2 s@ x 2 1+ @ 2 s@ x 2 2+
+@ 2 s@ x 2 n=n X i=1@ 2 s@ x 2 i For vector functions the following equation holds, r2 v(x) =r(r
v)(x)r(rv)(x) and in Cartesian coordinates this reduces to, r2 v(x) = (r2 v1( x);r2 v2( x);


;r2 vn( x)) Moreover, functions whose laplacian is equal to zero are calledHarmon- ics. Applying this operator to our test functions we get, r2 s(x) = 6y+ 0 + 2 = 20 r2 v(x) = (2;6z;25e 5y ) = (2;6;25e 15 ) These de nitions give rise to a series of identities, the rst of which 1:12 we have already seen. Some other identities are, r
(rf(x)) = 0 1.7. SURFACES IN R 3 11 r(rf(x)) = 0 Given a vector functiong(x) :ARn !Rm such that, g(x) = (g 1( x); g 2( x);


; g m( x)) the expressionfg:A!Cis meaningful and the value of its gradient is given by the so-calledChain Rule r(fg)(x) = (f0 g)rg Similarly, the value of the gradient applied to the product of a scalar and a vector function is given byLeibniz's rule for scalar functions, r(sv)(x) =s(x)(r
v)(x) + (rs)(x)
v(x) 1.7 Surfaces in R3 De nition 1.13.ASurfaceR3 is a subset ofR3 for which there exists a parametrization ,  u: R2 !R3 such that,  u2C1 ;R3
(1.1) u( ) =  which is read as "is the Image of ." 1.8 Surface Integrals and the Divergence The-orem Given a surface  we have that,Z f d=Z f ( u(x; y))k@ 1u (x; y)^@ 2u (x; y)k |{z} J acobian of ud xdy(1.2) 12 CHAPTER 1. OVERVIEW Theorem 1.14(Divergence Theorem).Given R3 open, bounded with a smooth boundary surface@ , aVector Field,  v2C1 ;R3
and anOutward Normal Vector Field,  u:@ !R3 we have that,Z @  v
 ud=Z r
 vdxdydz(1.3) where the last equation is nothing but the generalization of theFirst Fundamental Theorem of Calculus(FFTC) in 3D. 1.8.1 Derivation of Archimedes Principle A rst application of this theorem can be seen in the derivation of Archimedes Principle: Take the vector function Findicating the force acting on a set of points in submerged in a liquid. We know that, according to the principle,  F=gV( ) RR(1;0] In this case,V( ) is the volume of space occupied by the points in . To prove this, letPbe the pressure at a certain depthz, P(x; y; z) :RR(1;0]!R P(x; y; z) =gz so that the quantity, dF j= P
n jd S represents the force acting on an in nitesimal set of points dSin the directionj. The minus sign in front is due to the fact that the force is directed inward with respect to the normal vectorn. The total force acting on the body is thus given by the formula, 1.8. SURFACE INTEGRALS AND THE DIVERGENCE THEOREM 13 F=Z @ P
ndS By separating the normal vectornin its spacial componentsn x; n y; n z we can divide the formula for the total force in three separate integrals, Fx=Z @ P
n xd =Z @ (P;0;0)
nd=Z r
(P;0;0)dxdydz =Z ( @ xP ) dxdydz= 0 where we have used the divergence theorem in the second-to-last step. Similarly, for the other components of u, Fy=Z @ P
n yd =Z @ (0; P;0)
nd=Z r
(0; P;0)dxdydz =Z ( @ yP ) dxdydz= 0 Fz=Z @ P
n zd =Z @ (0;0; P)
nd=Z r
(0;0; P)dxdydz =Z ( @ zP ) dxdydz=gZ 1 d xdydz=gV( ) We have thus proven that, F= 0 + 0gV( ) =gV( ) 14 CHAPTER 1. OVERVIEW Chapter 2 Fourier Series and Convergence A function can be represented as a superposition of other trigonometric func- tions. One such example is theFourier Series: De nition 2.1.Letu(x) :R!C,2T-periodic withT >0, u(x+ 2T) =u(x) We de ne the quantity!, !=T We call Fourier series ofu(x)the function, Su( x) =a 02 +1 X k=1a kcos( k!x) +b ksin( k!x) wherea kand b kare the Fourier Coecientsof the series and are given by the formulae, ak=1T Z T Tu (x) cos(k!x)dx k2 bk=1T Z T Tu (x) sin(k!x) dx k2N; k=>1 The existence of the Fourier series is well de ned, meaning thata k; b k2 C if, ZT Tj u(x)jdx > < > > :( @ t ) (x; t) +v m 12  m (@ x ) = 0 0 m (@ t ) (x; t) = 0 >  m (x;0) = 0( x)(5.1) To solve this problem we will introduce the concepts ofCharacteristics andShock-Wavesthat will allow us to see how the solution behaves in Space-Time. De nition 5.1.TheCharacteristics(x(t); t) :R[0;+1): are curves in space-time along which(x; t)isconstant, and are thus level lines for. (x(t); t) =(x(0);0) = 0( x 0) It follows from dierentiating this equation that, d(x(t); t)d t= ( @ t ) (x(t); t) + (@ x ) (x(t); t)x0 (t) = 08t >0 which gives us theDierential Equation of Characteristics, x0 (t) =( @ t ) (x(t); t)( @ x ) (x(t); t)(5.2) Now, by subtracting the T.Eq to the C.Eq we get, [(@ t (x(t); t)) +q0 ((x(t); t)) (@ x (x(t); t))][(@ t (x(t); t)) + (@ x (x(t); t))x0 (t)] = 0 (@ x ) (x(t); t) [x0 (t)q0 ((x(t); t))] = 0 which, wherever (@ t ) (x(t); t)6 = 0 gives us, x0 (t) =q0 ((x(t); t)) =q0 ( 0( x 0)) Finally by integrating we get, x(t) =x(0) +q0 ( 0( x 0)) t8t0 (5.3) This formula for the characteristic curves tells us that, ˆThe characteristics arestraight linesin Space-Time. ˆTheirvelocitydepends on their starting pointx 0. ˆDierent characteristics maycross pathsdue to their dierent ve- locities.This fact causes a problem in the uniqueness of the solution to the transport equation. 5.1. THE GREEN LIGHT PROBLEM 35 5.1 The Green Light Problem Lets take problem 5:1 and apply the following initial distribution, 0( x) =( m8 x 0 This distribution models the behavior of cars at a green light placed at x= 0, since car density is maximal before the green light and zero right after it. The general Ch.Eq is, x(t) =x(0) +q0 ( 0( x 0)) t8t0 and we can see that, q((x; t)) =v m 1 m + (5.4) q0 ((x; t)) =v m 12  m +(5.5) q0 ( 0( x 0)) =( v m8 x 0(5.6) so that, depending on the starting point, we have two sets of Ch. lines given by 5:3, x(t) =( x0 v mt 8x 0< 0 x0+ v mt 8x 0> 0(5.7) We can divide space-time is 3 regions:ˆR, in which the Ch. lines havev m< 0 ˆT, in which the Ch. lines havev m> 0 ˆS, in which there are no Ch. lines Now lets ask ourselves what exactly happens inS? To analyze this we will introduceRarefaction Waves, Theorem 5.2.Suppose thatq: [ 1;  2] !R2C2 ([ 1;  2]) , and suppose that q0 isinvertiblewith inverse(q0 ) 1 . Then the solution of the trac equation is given by, 36 CHAPTER 5. CHARACTERISTIC CURVES (x; t) =r xt  x; t2R(0;+1) wherer(y)is the inverse function ofq0 (y), r(y) = (q0 (y)) 1 and the Ch. lines are given by the formula, x(t) =mt t0 (5.8) Equation 5:8 tells us that inSthe Ch. lines fan out in Space-Time, creating what we called Rarefaction Waves in our solution. Proof.Assume that(x; t) =r xt
. Then, by dierentiating, (@ t ) =r0 xt  xt 2 (@ x ) =r0 xt  1t Rewriting the T.Eq. we get, (@ t ) +q0 () (@ x ) (x; t) =1t r 0 xt  h xt q0 r xt i = 0 meaning that, q0 r xt  =xt or, q0 (r(y)) =y8y2R The functionris the inverse of the functionq0 for all values ofyand is thus its inverse.5.2 Trac-Jam Ahead Problem Lets take the same problem 5:1 and apply a dierent initial distribution, 0( x) =( 18  m8 x 0 5.2. TRAFFIC-JAM AHEAD PROBLEM 37Figure 5.1: William John Macquorn Rankine 1820-1872. 38 CHAPTER 5. CHARACTERISTIC CURVES This distribution models the behavior of cars at a trac jam starting at x= 0, since car density is small before the trac jam and maximal right after it. Without rewriting equations 5:4, 5:5 and 5:6 we have that with this initial density the Ch. lines given by 5:3 become, x(t) =( x0+34 v mt 8x 0< 0 x0 v mt 8x 0> 0 Because of the dierent slope of the two sets of curves depending on their starting point, the Ch. lines will at some point meet. This creates a problem as we have said that along one of these lines the solution must be equal to a constant. But if the lines meet we have one point associated with two dierent values of the function, leading to a problem in determining a unique solution to our problem. To solve this we will have to consider solutions that presentjump discontinuitiesin Space-Time, in particular along a line calledShock-Line. Suppose that(x; t) has discontinuities of the jump variety along a curve parametrized by the function, t!s(t) so that, lim y!s (t) (y; t) = (s(t); t)6 =+ (s(t); t) = lim y!s+ (t) (y; t) Theorem 5.3(Existence of Shock-Waves).The Shock-Linet!s(t)2 R[0;+1]is determined by the following condition, known as theRankine- Hugoniot Condition, s0 (t) =q (+ (s(t); t))q( (s(t); t)) + (s(t); t)) (s(t); t))(5.9) Which is the ratio between the jump in current and the jump in density along the Shock-Lines. Proof.Lets take the conservation equation, dd tZ x2 x1 (x; t)dx=q((x 1; t ))q((x 2; t )) and lets split up the integral in two parts,dd tZ s(t) x1 (x; t)dx+Z x2 s(t) (x; t)dx=q((x 1; t ))q((x 2; t )) 5.3. BURGER'S EQUATION 39 Zs(t) x1( @ t (x; t)) (x; t)dx+ (s(t); t)s0 (t)+Z x2 s(t)( @ t (x; t)) (x; t)dx+ (s(t); t)s0 (t) =q((x 1; t ))q((x 2; t )) Zx2 x1( @ t (x; t)) (x; t)dx+  (s(t); t)+ (s(t); t) s0 (t) =q((x 1; t ))q((x 2; t )) Now lets letx 1! s (t) andx 2! s+ (t) to obtain the nal result (as we approach the limits both integrals go to 0),  (s(t); t)+ (s(t); t) s0 (t) =q( (s(t); t))q( (s(t); t)) s0 (t) =q (+ (s(t); t))q( (s(t); t)) + (s(t); t)) (s(t); t))Lets now use this equation to solve our problem. In our case the function (x; t) had a discontinuity atx= 0, meaning that, q(+ (s(t); t)) = 0 q( (s(t); t)) =764 v m m + (s(t); t) = m  (s(t); t) =18  m giving us, s0 (t) =18 v m and since the shock-wave starts at (x; t) = 0 by integrating we get the nal equation for the shock-line, s(t) =18 v mt (5.10) The uniqueness if the solution is assured as the shock-line separates space- time in two distinct parts with their own Ch. lines. 40 CHAPTER 5. CHARACTERISTIC CURVESFigure 5.2: Peter Lax 1926 - . 5.3. BURGER'S EQUATION 41 5.3 Burger's Equation Let's now try to compare the two solution methods by taking a trac-jam ahead problem with the following initial distribution and current function, 0( x) =( 08x : 0( x) = 0x 0 u(x;0) =u 0( x)(6.7) where the functionu 0( x)2L1 (Rn )\L1 (Rn ) is the initial temperature distribution of our body, and is both integrable and bounded. Let's now assume that the solution to our initial value problem is of the form, u(x; t) =Z Rnk (t; x; y)u 0( y)dy x2Rn ; t >0 (6.8) where the function, k(t; x; y) = (4Dt) n2 e j xyj24 Dt (6.9) is called theHeat Kernel ofRn . To demonstrate that this is actually a solution lets start by calculating the time derivative of the heat kernel, (@ tk ) (t; x; y) = (4Dt) n2 e j xyj24 Dt j xyj24 Dt2! n2 (4 Dt) n2 1 (4D)e j xyj24 Dt = (4Dt) n2 e j xyj24 Dt" jxyj24 Dt2n2 t# =k(t; x; y)" jxyj24 Dt2n2 t# The laplacian isn't as immediate. First we must calculate the gradient in the variablex, 48 CHAPTER 6. THE HEAT EQUATION (r xk ) (t; x; y) = (4Dt) n2 e j xyj24 Dt 14 Dt rxj xyj2 =k(t; x; y)x y2 Dt By the Leibniz rule, r
 ~ Ef = r
~ E f+~ E(rf) so that, r2 xk
(t; x; y) =r x
(r xk )(t; x; y) =x y2 Dt( r xk ) (t; x; y)k(t; x; y)12 Dtr x
(xy) =x y2 Dtk (t; x; y)x y2 Dt k(t; x; y)12 Dtn =k(t; x; y)" jxyj2(2 Dt)2n2 Dt# (thencomes from the fact that we are working onRn .) By substituting these results into the heat equation we get, (@ tk ) (x; t)D r2 xk
(t; x; y) =k(t; x; y)" jxyj24 Dt2n2 t# D" k(t; x; y) jxyj2(2 Dt)2n2 Dt!# = 0 Therefore the heat kernel 6:9 is a solution to the general heat equation 6:5 but is not yet a solution to our initial values problem 6:7. Let's now take the integral function, u(x; t) =Z Rnk (t; x; y)u 0( y) dy(6.10) Because both the time derivative and the laplacian operator act only on the kernelk(t; x; y), the functionu(x; t) is still a solution to the heat equation, (@ tu ) (x; t)D r2 xu
(x; t) =Z Rnu 0( y) (@ tk ) (x; t)D r2 xk (t; x; y)
dy= 0 Moreover, equation 6:10 is a solution to the Cauchy problem, as it satis- es the initial conditions. What this means is that the solutionu(x; t) is a superposition ofGaussian Kernelsk(t; x; y), whose analytical expression 6.3. HEAT EQUATION IN EUCLIDIAN SPACES 49 depends on the type of domain, controlled by the initial temperature distri- butionu 0( x). To check that the solution meets the initial condition we must rst notice that, Z Rnk (t; x; y) dy= 18t0;8x2Rn This allows us to write the following equality, u0( x)u(x; t) =u 0( x)Z Rnk (t; x; y)u 0( y) dy =u 0( x)Z Rnk (t; x; y) dyZ Rnk (t; x; y)u 0( y) dy=Z Rnk (t; x; y) [u 0( x)u 0( y)] dy If we now rewrite the heat kernel as a function of a new variablezandt only, kt( z) = (4Dt) n2 e j zj24 Dt z2Rn where this new variablezis de ned as, z=xy(6.11) so that, k(t; x; y) =k t( xy) =k t( z) our previous equation becomes, u0( x)u(x; t) =Z Rnk t( z) [u 0( x)u 0( xz)] dz= (4Dt) n2 Z Rne j zj24 Dt [u 0( x)u 0( xz)] dz If we now introduce yet another variable, w=z(4 Dt)12 (6.12) We have that, in ann-dimensional space, z=w(4Dt)12 dz= (4Dt)n2 dw 50 CHAPTER 6. THE HEAT EQUATION and by substituting this into our equation we can write, u0( x)u(x; t) = (4Dt) n2 (4Dt)n2 Z Rne j wj2h u0( x)u 0( xp4 Dtw)i dw Finally, integrating both sides with respect toxand taking the modulus we obtain, Z Rnj u 0( x)u(x; t)jdx=n2 Z Z Rn Rn ej wj2 u0( x)u 0( xp4 Dtw) d wdx The integral of the modulus of two functions will certainly be smaller than the product of the modulus of the two functions integrated separately, Z Rnj u 0( x)u(x; t)jdx n2 Z Rne j wj2 dwZ Rn u 0( x)u 0( xp4 Dtw) d x By taking the limit ast!0 we can see that the modulus inside of the right hand side integral vanishes, lim t!0+k u 0( x)u(x; t)k=n2 Z Rne j wj2 dwlim t!0+Z Rn u 0( x)u 0( xp4 Dtw) d x= 0 The initial condition is met not in a point-wise way but in aL1 , or average, sense. Since it allows us to write the solution to many dierent Cauchy problems inRn the heat kernelk(t; x; y) is also called theFundamental Solutionof the heat equation 6:5 inRn . 6.4 Properties of the Solution to the HeatEquation Letu 0( x)2L1 (Rn )\L1 (Rn ) be the integrable and bounded initial temper- ature distribution andu(
; t)2L1 (Rn )\L1 (Rn ) be the solution to the heat equation with initial conditionu 0( x). Then, ˆPositivity :Ifu 0( x)0)u(x; t)08x;8t. This follows from the fact that, sincek(t; x; y)0 the solution u(x; t) =Z Rnk (t; x; y)u 0( y) dy0 must also be greater than zero. 6.5. THE HEAT EQUATION IN ARBITRARY DOMAINS 51 ˆMaximum Principle:If 0u 0( x)u max) 0u(x; t)u max. Similarly to the previous property we can write the following inequality, 0u(x; t) =Z Rnk t( xy)u 0( y) dyZ Rnk t( xy)u maxd y=u maxZ Rnk t( xy) dy =u maxZ Rnk t( z) dz=u max ˆSpeed of Propagation :The speed of propagation of the solution is1. Sincek(t; x; y)0 ifu 0 0 but also strictly positive at some pointx2Rn , the integral, Z Rnk (t; x; y)u 0( y) dy=u(x; t)>0 cannot vanish. At any momentt >0 therefore we must have a solution u(x; t)6 = 0. ˆConservation of Energy :The total amount of energy in the system does not change in the absence of sources. Let's consider a simpli ed case of a homogeneous material wherec V=  0= 1, QRn (t) =Z Rnu (x; t) dx=Z Rn Z Rnu 0( y)k(t; x; y) dy dx =Z Rnu 0( y) dyZ Rnk (t; x; y) dx=Z Rnu 0( y) dy=Q Rn (0) ˆSmoothingu2C1 (R(0;+1)). Even if the initial distribution u0( x) isn't smooth the overall solution always is thanks to the Gaussian term in the integral. 6.5 The Heat Equation in Arbitrary Domains Let's now consider the heat equation in an arbitrary domain, not necessarily an unbounded euclidean space. In order to be able to solve the heat equation in a unique way, we must as usual set the initial and boundary conditions. There are three kinds of boundary conditions that have a physical meaning, 52 CHAPTER 6. THE HEAT EQUATION ˆDirichlet Boundary Condition: We can de ne a function, k(x) =u(x; t)x2@ ; t >0 that describes the behavior of the solution at the boundary. ˆNeumann Boundary Condition: We can de ne a function, g(x) = (ru) (x; t)
~ n(x; t)x2@ ; t >0 that describes the rate at which heat is owing outwards from the boundary of the system. ˆRobin Boundary Condition: This condition imposes that, ~ j(x; t)
~ n(x; t) = (Uu(x; t)) the rate at which heat is owing outwards from the system is pro- portional with >0 to the dierence in temperature between the boundary of the system and the external temperatureU. Example of Diusion on a half line with Homogeneous D.B.C. Let's take a1Dhalf-line as our domain , = (0;+1)@ =f0g and try to nd a solution to the H.Eq,8 > < > :@ tu D@2 xu = 0x >0; t >0 u(x;0) =u 0( x)x >0 u(0; t) = 0t >0 It can be proven that the solution to this problem takes the form, u(x; t) =Z 1 0u 0( y) [k t( xy)k t( x+y)] dy where the term in brackets represent the heat kernel of this domain. Therefore the heat kernel function of a domain changes according to itsge- ometry. The knowledge of this function provides the solution to the H.Eq. 6.6. THE MAXIMUM PRINCIPLE FOR THE HEAT EQUATION (GENERAL CASE) 53 Example of Diusion on a half line with Homogeneous N.B.C. Let's take the same1Dhalf-line as before, = (0;+1)@ =f0g and try to nd a solution to the H.Eq,8 > < > :@ tu D@2 xu = 0x >0; t >0 u(x;0) =u 0( x)x >0 (@ xu ) (0; t) = 0t >0 In this case the solution takes the form, u(x; t) =Z 1 0u 0( y) [k t( xy) +k t( x+y)] dy And as we can see the heat kernel of the problem changes also according to the conditions we impose on the problem. Example of Diusion on a half line with Sources and Zero Initial ValueConsider now a problem similar to the rst example in the same domain, = (0;+1)@ =f0g with the added dierence that now the system contains a source term, along with vanishing initial conditions, (@tu D@2 xu =f(x; t)x >0; t >0 u(x;0) = 0x >0 It can be proven that the solution to this problem takes the form of a convolution integral, u(x; t) =Z 0d sZ 1 0f (y; t)k ts( xy) dy 6.6 The Maximum Principle for the Heat Equa-tion (General Case) In the previous section we have proven the maximum principle in a euclidean space (of any dimension) but are yet to see if it holds true in any general bounded region. 54 CHAPTER 6. THE HEAT EQUATION Theorem 6.2(Maximum Principle for the Heat Equation).Let Rn be a bounded region with boundary@ . We denote theSpace-Time Domain as, Q= (0;+1) and itsParabolic Boundaryby, @pQ = (  f0g)\(@ [0;+1]) Also set,  Q=Q\(@ Q) and letf(x; t)2C(Q)represent anexternal source. Assume that u(x; t)is a solution to the H.Eq, @tu (x; t)Dr2 u(x; t) =f(x; t) Then, iff(x; t)0onQ, max (x;t)2 Qu (x; t) =max (x;t)2@ pQu (x; t) (6.13) The solution assumes its maximum values on the parabolic boundary@ pQ .In particular, ifu(x; t)0on@ pQ thenu(x; t)0on the whole Q. What the theorem tells us is that if the source inside the system is non- positive (and therefore is absorbing heat), the solution cannot have maximal points inside the region. In the case of our1Dline bounded at (a; b) the solution will take its maximum value either ataorb. Proof.Let simplify the hypothesis by settingf(x; t) < > :@ tu @2 xu = 0 (x; t)2(0;1)(0;+1) u(x;0) =u 0( x)x2(0;1) u(0; t) =u(1; t) = 0t >0 We want toboundthe solution, which may have a very complex explicit form, with a more controllable known function, such as an exponential multi- plied by a polynomial. Let's assume we want to nd the parameters; 2R such that, 0u(x; t)x(1x)e t We would therefore like the solution be bounded by functions both at the lower bound, by the functiong(x; t) = 0 and at the higher bound, by the exponential. By the I.C. we have that, u(x;0) =x(1x)02(0;1) but, by the rst B.C., u(0; t) =u(1; t) = 08t >0 we must have that, u(x; t)08(x; t)2@ pQ By one of the corollaries of the maximum principle, u(x; t)08(x; t)Q 6.7. USING THE MAXIMUM PRINCIPLE TO ESTIMATE SOLUTIONS 57 We have thus proven that the function is bounded from below by 0with- out explicitly knowing the form of the solution. Now for the second part set, w(x; t) =x(1x)e t This function is a solution to the problem, @tw @2 xw = w(x; t)(2)e t =e t [2 x(1x)] which is a H.Eq with a source function. The functionw(x; t) also satis es the following B.C., (w(0; t) =w(1; t) = 0t >0 w(x;0) =x(1x) Lets construct now the new function, v(x; t) =w(x; t)u(x; t) This function solves the problem, @tv @2 xv = @tw @2 xw
@tu @2 xu
where the last terms on the right must be equal to 0 as per the H.Eq. Therefore, 8 > < > :@ tv @2 xv =e t [2 x(1x)] =f(x; t) v(x;0) =w(x;0)u(x;0) =x(1x)x(1x) = (1)x(1x) v(0; t) =v(1; t) =w(1; t)u(1; t) = 0 If1 then, v(x;0)0x2(0;1) If 18 then, f(x; t)0x2(0;1); t >0 Applying the maximum principle to the functionv(x; t) we have that, 0v(x; t) =u(x; t)w(x; t) 58 CHAPTER 6. THE HEAT EQUATION or,u(x; t)w(x; t) and have thus proven the higher bound. 6.8 Separation of Variables There is another powerful method to obtain a solution, calledSeparation of Variables. The method consists in nding as many solutions to the problem, (@sv @2 yv = 0 (y; s)2(0; )[0;+1) v(0; s) =v(; s) = 0s2[0;+1) as we can of the form, v(y; s) =u(y)w(s) (in which the variables are in fact separated) where, u: (0; )!Rfunction of space only w: (0;+1)!Rfunction of time only By inserting this solution into the problem we have that, 0 =@ sv @2 yv =u(y)w0 (s)u00 (y)w(s) u00 (y)u (y)= w 0 (s)w (s) Since this is an equality between functions of independent variables, it holds true only if both sides are constant and equal to a certain value. Therefore, w0 (s) =w(s) u00 (y) =u(y) We have reduced the initial H.Eq into these two independent equations. We must distinguish three cases, 6.8. SEPARATION OF VARIABLES 59 Case 1:0< =2 If the value ofispositive, then the second equation has solutions, u(y) =Aey +Be y By imposing the B.C., v(0; s) =v(; s)()u(0) =u() = 0 meaning that, 0 =Ae 0 +Be 0 =A+B!A=B 0 =Ae +Be  =B e e 
which only happens ifB=A= 0 Case 2:= 0 If the value ofiszero, then the second equation has solutions, u(y) =A+By By imposing the same B.C., 0 =u(0) =A!A= 0 0 =u() =A+B!B= 0 we get the same null solution as before. Case 3:0> =2 If the value ofisnegative, then the second equation has solutions, u(y) =Acos(y) +Bsin(y) By once again imposing the same B.C., 0 =u(0) =A 0 =u() =Acos() +Bsin() =Bsin() which has a non-zero solution if, 2N f0g 60 CHAPTER 6. THE HEAT EQUATION We can now write a one-parameter family of functions that verify the spatial B.C., uk( y) =sin(ky)k1 By imposing the rst B.C. we have that, w0 (s) =k2 w(s) which has solutions, wk( s) =e k2 s s0 Overall we have a one parameter family of solutions of the H.Eq with homogeneous D.B.C. of the form, vk( y; s) =u k( y)w k( s) =e k2 s sin(ky)y2(0; ); s0 Since derivation is a linear operation, any linear combination of these functions, v(y; s) =N X k=1b kv k( y; s) will still be a solution to the H.Eq. This fact is very advantageous as we can x the values of the coecients in order to satisfy any generic initial condition set by the problem. Theorem 6.3.There9!solution of the H.Eq with D.B.C. on(0; )with Cauchy data, 8 > < > :@ sv @2 yv = 0 (y; s)2(0; )[0;+1) v(0; s) =v(; s) = 0s2[0;+1) v(y;0) =v 0( y)v 02 L2 (0; ) The solution to this problem is given by a Fourier series, v(y; s) =1 X k1b ke k2 s sin(ky) where,bk=2 Z  0v 0( y)sin(ky) dy k1 6.8. SEPARATION OF VARIABLES 61 are the Fourier coecients ofv 02 L2 (0; ), or, better yet, are the Fourier coecients of theodd extensionofv 0! (pi; ). (We are simply taking the Fourier coecients of the functionu 0( y), and, since the exponential goes !1ast!0, what we are left with is simply the Fourier series corresponding to the initial temperature distribution. Moreover the initial condition is veri ed in theL2 mean, lim s!0+k v 0( y)v(
; s)k L2 (0;)= 0 lim s!0+Z  0j v 0( y)v(y; s)j2 dy 62 CHAPTER 6. THE HEAT EQUATION Chapter 7 Poisson's Equation We will now consider a third class of equations of this course, after the transport and the heat equations, calledPoisson's Equations. The study of this class of PDEs grants us the ability to investigate the geometric properties of a particular space by deriving their solutions. As such, these equations do not involve the time variablet, unlike the previous two. Let Rn andu(x) : !R; u(x)2C2 ( ), these equations take on the form, r2 u(x) =f(x)x2 (7.1) wheref(x) : !Ris a known function, also called thedatum. It may be useful to recall that for a multivariable function, r2 u(x) =n X k=1@ 2 ku (x) 7.0.1 Laplace's Equation Laplace's Equationis a speci c case of Poisson's equation where the know term is equal to zero, r2 u(x) = 0x2 (7.2) Functions that satisfy these equations are calledHarmonicsand their shape is greatly dependent on the dimensionality of the problem. Example 1Forn= 1 we have that = (a; b)Rand the laplacian operator becomes, 63 64 CHAPTER 7. POISSON'S EQUATION r2 u(x) =u00 (x)x2(a; b) Therefore, Poisson's equation in the1Dcase is simply, u00 (x) = 0x2(a; b) (7.3) Solutions to this equation are of the type, u(x) =mx+q(7.4) wherem; q2Rare known coecients. Example 2More interesting solutions can be found in greater dimension spaces, such asn= 2. Consider the function, u(x; y) =xy(x; y)2R2 By calculating it's laplacian we get, r2 u(x; y) = @2 xu
(x; y) + @2 yu
(x; y) =@ x( y) +@ y( x) = 0 meaning thatu(x; y) is a harmonic in the spaceR2 . Another example of a harmonic function in this space could be, v(x; y) =x2 y2 so that, r2 v(x; y) = @2 xu
(x; y) + @2 yu
(x; y) =@ x(2 x) +@ y( 2y) = 22 = 0 7.1 Electrostatic Model A physical implementation of these kind of equations can be found in the study of electromagnetism. Let R3 be a section of space where there persists anElectric Vector Field, ~ E(x; y; z) : !R3 (7.5) generated by a certain distribution of chargesq(V) in . According to Gauss's law the uxof the electric vector eld across the boundary@ Vis proportional to the total amount of chargeq(V) contained in the regionV, 7.1. ELECTROSTATIC MODEL 65 (~ E; @ V) =4  q (V) (7.6) Let's create a function(x; y; z) : !Rthat describes thecharge densityin the region. The total charge inVcould be written as q(V) =Z  (x; y; z) dxdydz Using the divergence theorem the ux can be rewritten as a similar inte- gral, (~ E; @ V) =Z @ V~ E
dS=Z Vr
~ E(x; y; z) dxdydz so that Gauss's law becomes,Z Vr
~ E(x; y; z) dxdydz=4  Z  (x; y; z) dxdydz This identity has to hold for any regionV . Therefore, r
~ E(x; y; z) =4   (x; y; z)8x; y; z2 (7.7) which is is the local form of 7:6. Since the electric vector eld~ Eis conservativein , it must admit a potential functionU: !Rsuch that, rU(x; y; z) =~ E(x; y; z)x; y; z2 (7.8) which gives us, r
~ E(x; y; z) =r
(rU(x; y; z)) =r2 U(x; y; z) Hence, the local formulation of Gauss's law in terms of the electric potential reads as follows, r2 U(x; y; z) =4   (x; y; z)x; y; z2 (7.9) which is a Poisson equation forU(x; y; z) given the charge density(x; y; z). By taking a region 0  in which there are no charges, (x; y; z) = 0x; y; z2 0 we can see that the electric potential also satis es Laplace's equation in this region, 66 CHAPTER 7. POISSON'S EQUATION r2 U(x; y; z) = 0x; y; z2 0 (7.10) or, in other words, the electric potentialU(x; y; z) is harmonic in the sub-region 0 where there are no charges. Electrostatic ExampleLet R3 and assume a chargeqis placed at the origin. The electric potential will be a functionU(x; y; z) :R3 n f0g !R of the form, U(x; y; z) =U(x) =4 q 1j xjx 6 = 0 (7.11) where we have set, jxj= 3 X k=1x 2 k! 12 =px 2 +y2 +z2 U(x) will be a harmonic function in the entire spaceR3 outside of the origin, r2 U(x) = 0x2R3 n f0g To verify this fact we can write, rU(x; y; z) =4 q r (jxj) 1 =4 q r 3 X k=1x 2 k! 12 =4 q  12  3 X k=1x 2 k! 32 3 X k=1r x2 k =4 q  12  3 X k=1x 2 k! 32 (2x 1; 2x 2; 2x 3) =4 q xj xj3 Using Leibniz's chain rule we can calculate the laplacian operator through the divergence operator, r2 U(x; y; z) =r
(rU) =r
 4 q xj xj3 =4 q r
xjxj 3
7.2. BOUNDARY CONDITIONS FOR LAPLACE'S EQUATION 67 =4 q rjxj 3
x+jxj 3 r
x
Now, for each of the variablesx kwe have that the gradient returns, rkj xj 3 =r k n X i=1x 2 i! 32 = 32  n X i=1x 2 i! 52 rk x2 k
=3x kj xj 5 So that the total gradient is,rjxj 3 =3xj xj5 The divergence of the vectorxis simply, r
x=r
n X i=1x i! =n X i=1@ i( x i) = n= 3 Therefore, in the region 0  the laplacian of the potentialU(x) is equal to, r2 U(x) =4 q  3xj xj5
x+3j xj3 =4 q  3j xj3 +3j xj3 = 0 (7.12) and the potential is in fact a harmonic function in 0 . 7.2 Boundary Conditions for Laplace's Equa-tion As with the previous problems, in order to obtain a unique solution we must rst set the boundary conditions, which are of the usual types Dirichlet Boundary Conditionsu(x) =g(x)x2@ whereg(x) :@ !Ris a given function. We are prescribing the behavior of the solution at the boundary. 68 CHAPTER 7. POISSON'S EQUATION Neumann Boundary Conditionsru(x)
~ n=h(x)x2@ whereh(x) :@ !Ris a given function. We are prescribing the behavior of the gradient component at the boundary. Robin Boundary Conditionsru(x)
~ n+u(x) =g(x)x2@ where >0 is a constant andg(x) :@ !Ris a given function. Robin's boundary conditions are simply the sum of Dirichlet and Neumann conditions. Theorem 7.1(Uniqueness of the Solution to the Poisson Equation under D/N/R B.C.).Suppose thatu 1( x)andu 2( x)are solutions to either of the following problems, (r2 uk( x) =f(x)x2@ uk( x) =g(x)x2@ ; k= 1;2 (r2 uk( x) =f(x)x2 ru k( x)
~ n=h(x)x2@ ; k= 1;2 (r2 uk( x) =f(x)x2 ru k( x)
~ n+u(x) = 0x2@ ; k= 1;2 Consider the function, w(x) =u 1( x)u 2( x) This function solves either of the following problems,(r2 w(x) = 0x2 w(x) = 0x2@ (r2 w(x) = 0x2 rw(x)
~ n= 0x2@ (r2 w(x) = 0x2 rw(x)
~ n+w(x) = 0x2@ 7.2. BOUNDARY CONDITIONS FOR LAPLACE'S EQUATION 69 According to Leibniz's rule,12 r w(x)2
=w(x)rw(x) Computing the divergence of the right hand side we get, r
(w(x)
rw(x)) =rw(x)
rw(x)+w(x)r
(rw)(x) =jrw(x)j2 +w(x)r2 w(x) but sincew(x)is harmonic, r2 w(x) = 0 and the result is just, r
(w(x)
rw(x)) =jrw(x)j2 We know that this value must be non-negative for all pointsx2 . This in turn means that the integral, 0Z j rw(x)j2 dV must also be non-negative. Using the divergence theorem, Z j rw(x)j2 dV=Z r
(w(x)rw(x)) dV=Z @ w (x)rw(x)
dS=Z @ w (x)rw(x)
~ ndS Depending on the type of boundary condition imposed on the initial prob- lem this integral can take on three dierent values, 0Z j rw(x)j2 =8 > < > :D:B :C: 0 N:B:C:0 R:B :C:R w2 (x) dV In case of D.B.C and N.B.C. the value of the integral is obviously zero as either the function itself or its gradient are null at the boundary, meaning that the solutionw(x)must take on a constant value across . In the case of R.B.C. we have that the resulting integral, which as we have originally stated is greater than zero, 0Z j rw(x)j2 dV=Z w 2 (x) dV 70 CHAPTER 7. POISSON'S EQUATION is equal to the product of a positive constantand a non-negative integral, all with a minus sign in front. The only way for this inequality to hold is if, Z j rw(x)j2 dV= 0 meaning that the functionw(x)remains constant over the interval in this case as well. Ifw(x) = 0on the boundary for all three problems we have that, w(x) =u 1( x)u 2( x) = 0 or alternatively, u1( x) =u 2( x) Poisson's equations, when coupled with boundary conditions admit a unique solution. 7.3 The Mean Value Property of HarmonicFunctions Let's start with an example inn= 1, = (a; b)R. As we have seen in a previous example harmonic functions in this space are of the form, u(x) =mx+q wherem; q2Rare constants. These being linear functions we have the following identity, u(a) +u(b)2 = u a+b2  (7.13) We are going to prove that this result holds8n. Theorem 7.2(Mean Value Property for Harmonic Functions).Letu(x)2 C2 ( )be a solution to, r2 u(x) = 0x2 Rn Then, for any ballB(x; R) we have the following identities, 7.3. THE MEAN VALUE PROPERTY OF HARMONIC FUNCTIONS 71 ˆThe value of the solutionu(x)can be calculated through the following volume integral, u(x) =1j B(x; R)jZ B(x;R)u (y) dy(7.14) which is simply the mean value ofu(x)over the ballB(x; R). ˆThe value of the solutionu(x)can be calculated through the following surface integral, u(x) =1j @ B(x; R)jZ @ B(x;R)u (y) dy(7.15) Although similar to the one above, this integral has a dierent meaning. We are saying that the value ofu(x)at the center of a generic ball B(x; R)coincides with the mean value of the function on the boundary of the ball. In words, what these properties are telling us is that harmonic functions arerigid.We cannot deform a harmonic function local ly without deforming it global ly if we stil l want for the function to remain harmonic. If we want to perturb the harmonic function we have to perturb it ina global way. It is exactly this property that, when properly used, allows us to reveal the geometry of the space. Proof.We will prove this forn= 2 only, so that is a simple planar region. Equation 7:15 tells us that inside the circleB(x; R), u(x) =12 RZ @ B(x;R)u (y) dy Let's consider the functiong(r) : (0; R]!R, g(r) =12 rZ @ B(x;r)u (y) dy8r2(0; R] (7.16) whose value coincides with the mean value ofu(x) on the circle@ B(x; r). Let's now perform an exchange of variables, @ B(x; r)3y=x+rz()z=y xr 2 @ B(0;1) dy=rdz 72 CHAPTER 7. POISSON'S EQUATION The variablezallows us to transform the integral in dyover the generic circle@ B(x; r) into an integral inrdzover the unit circle@ B(0;1) centered at the origin. Introducing this variable into equation 7:16 we get, g(r) =12 rZ @ B(x;r)u (y) dy=12 rZ @ B(0;1)u (x+rz)rdz=12 Z @ B(0;1)u (x+rz) dz Notably, through this substitution we have brought the dependence of the functiong(r) onrinside the integral. Now let's consider the functionv(z) :@ B(0;1)!R, v(z) =u(x+rz)jzj 1 The gradient of this function with respect tozis, rzv (z) =r zu (x+rz) =rru(x+rz) and using this to nd the value of the laplacian, r2 v(z) =r
(r zv (z)) =r
(r zu (x+rz)r) =r2 r2 u
(x+rz) but sinceu(x+rz) is harmonic, r2 v(z) =r2 u(x+rz) = 0 v(z) is also harmonic. For the last step let's derive our functiong(r), g0 (r) =r rg (r) =r r 12 Z @ B(0;1)u (x+rz) dz and assuming that the functions we are dealing with are smooth enough, we can bring the derivation (gradient) inside of the integral, =12 Z @ B(0;1)r ru (x+rz) dz=12 Z @ B(0;1)z ru(x+rz) dz Now we can multiply and divide byrto obtain, =12 rZ @ B(0;1)zr ru(x+rz) dz=12 rZ @ B(0;1)z r zv (z) dz= and by the divergence theorem, (since in this casen=z), 7.3. THE MEAN VALUE PROPERTY OF HARMONIC FUNCTIONS 73 =12 rZ B(0;1)r
(r zv (z)) dz=12 rZ B(0;1) r2 v
dz= 0 where we have use the fact thatv(z) is harmonic. Overall we have found that, g0 (r) = 08r2(0; R] (7.17) meaning thatg(r) is a constant function; more speci cally, this means that the value of the function will remain constant and equal to it's initial valueg(0) as the variablerspans the interval (0; R], g(0) =g(r) =g(R)r2(0; R] (7.18) By using it's initial de nition 7:16, g(r) =12 rZ @ B(x;r)u (y) dy8r2(0; R] Knowing thatu(x) is continuous, by themean value theorem for integralsthere must9x r2 @ B(x; t) such that, 2ru(x r) =Z @ B(x;r)u (y) dy(7.19) Comparing the right hand side integral to the de nition ofg(r) we get, g(r) =12 rZ @ B(x;r)u (y) dy=12 r(2 ru(x r)) = u(x r) r2(0; R] (7.20) The value of the constant functiong(r) is therefore determined by the value that our original functionu(x) takes at a speci c pointx ron the circle centered around the pointx. It is then easy to see that by taking the limit asr!0,x r! xand, lim r!0g (r) = lim r!0u (x r) = lim xr! xu (x r) = u(x) which becomes, u(x) =1j B(x; R)jZ @ B(x;R)u (y) dy(7.21) 74 CHAPTER 7. POISSON'S EQUATION 7.3.1 The Maximum Principle for Harmonic Functions It is easy to verify that a harmonic function belonging toR1 , u(x) =mx+q cannot have any local maxima/minima within the interval (a; b)2R1 in which it is de ned, and only attains these values at the boundary. While not as intuitively, this property actually holds for all harmonic functions belonging to a genericRn . Theorem 7.3(Maximum Principle for Harmonic Functions).Suppose  Rn is anopen setand thatu2C2 ( )such that, r2 u(x) = 0x2 Then, ifu(x)is not a constant function, it cannot have local maxima or minima in . Moreover, ifu2C2 ( )\C( [@ )then it attains its extreme values on@ . Otherwise stated, ifx 02 is a minimum/maximum point, thenu(x)has to be constant on the whole set . Proof.By contradiction, suppose thatx 02 is a minimum point, u(x)u(x 0) = m 08 x2 Since is open, there must9R 0> 0 such that a ballB(x 0; R 0)  having radiusR 0can be taken around the point x 0. Then, if u(x) is not a constant function, there9z2B(x 0; R 0) such that, u(z)> u(x 0) = m 0 Sinceu(x) is continuous inzthere must also9r >0 such that a second smaller ballB(z; r) having radiusrcan be taken around the pointzsuch that, m1= inf B(z;r)u (x)> u(x 0) = m 0 Applying the mean-value property for harmonic functions we have previ- ously allows us to write, m0= u(x 0) =1j B(x 0; R 0) jZ B(x 0;R 0)u (x) dx This integral can be split into two parts, one covering the smaller ball B(z; r) and one covering the remaining volumeB(x 0; R 0) nB(z; r), 7.3. THE MEAN VALUE PROPERTY OF HARMONIC FUNCTIONS 75 =1j B(x 0; R 0) j Z B(z;r)u (x) dx+Z B(x 0;R 0) nB(z;r)u (x) dx Since the minimum value thatu(x) attains on the smaller ball ism 1, we can bound the rst integral by writing Z B(z;r)u (x) dxZ B(z;r)m 1d x=m 1j B(z; r)j and, similarly for the second integral, Z B(x 0;R 0) nB(z;r)u (x) dxZ B(x 0;R 0) nB(z;r)m 0d x=m 0j B(x 0; R 0) nB(z; r)j so that, m0= u(x 0) 1j B(x 0; R 0) j[ m 1j B(z; r)j+m 0j B(x 0; R 0) nB(z; r)j] but we also know that, ifu(x 0) is to be a local minimum for the whole ballB(x 0; R 0), m1> m 0 so that the right-hand term can be further bounded by writing, m0= u(x 0) >1j B(x 0; R 0) j[ m 0j B(z; r)j+m 0j B(x 0; R 0) nB(z; r)j] =m 0j B(x 0; R 0) j[ jB(z; r)j+jB(x 0; R 0) nB(z; r)j] =m 0j B(x 0; R 0) jj B(x 0; R 0) j=m 0 But we have arrived at a contradiction since, comparing the rst and last terms of the inequality we have proven that, m0> m 0 Therefore the functionu(x) cannot have local minima at least within the ballB(x 0; R 0) and must be constant within this boundary. To prove that it must also be constant on the whole set we can repeat the very same line of reasoning choosing any pointy2 and using a "chain" of balls that will at some point includey. Finally we have proven that if the functionu(x) has a minimum inx 02 it must be constant over the entire set . 76 CHAPTER 7. POISSON'S EQUATION Corollary 1: Stability of the solution to the Laplace equation with D.B.C.Letg 1( x); g 2( x)2C(@ ) and letu 1( x); u 2( x)2C2 ( )\C( [@ ) be solutions to the following problem, (r2 u1( x) = 0x2 u1( x) =g 1( x)x2@ (r2 u2( x) = 0x2 u2( x) =g 2( x)x2@ Then we must have that, the maximum distance between the two solu- tionsu 1( x); u 2( x) on the whole set [@ must be found at the bound- ary@ and must be equal to the maximum distance between the functions g1( x); g 2( x), max [@ j u 1( x)u 2( x)j= max @ j g 1( x)g 2( x)j which can be written as, ku 1( x)u 2( x)k C( [@ )= kg 1( x)g 2( x)k C(@ ) Proof.To prove this result let's start by creating the function, w(x) =u 1( x)u 2( x)2C( )\C2 ( [@ ) which is a solution to the problem,(r2 w(x) =r2 u1( x) r2 u2( x) = 0x2 w(x) =g 1( x)g 2( x)x2@ Applying the maximum principle to this function we must have that, max [@ ( w(x)) = max @ ( g 1( x)g 2( x))max @ j g 1( x)g 2( x)j Applying the same principle to the opposite functionw(x) =u 2( x) u1( x) we must also have that, max [@ ( w(x)) = max @ ( g 2( x)g 1( x))max @ j g 1( x)g 2( x)j In conclusion, max [@ j w(x)j= max @ j u 1( x)u 2( x)j max @ j g 1( x)g 2( x)j 7.3. THE MEAN VALUE PROPERTY OF HARMONIC FUNCTIONS 77 Corollary 2: Comparison between two solutions of the Laplace Equations with D.B.C.Keeping the same notation as before suppose that, g1( x)g 2( x)8x2@ then we must have, u1( x)u 2( x)8x2 Proof.The proof goes as follows: Let us consider the function, w(x) =u 1( x)u 2( x) This function is a solution to the problem,(r2 w(x) =r2 u1( x) r2 u2( x) = 08x2 w=g 1( x)g 2( x)08x2@ Sincew(x) isnon-negativeon the boundary@ , by the maximum prin- ciple it must benon-negativeon the whole set , w(x) =u 1( x)u 2( x)08x2 meaning that, u1( x)u 2( x)8x2 7.3.2 Method of Separation of variables for the Laplace Equation with D.B.C. on Rectangles Supposew(x; y) :R2 !Ris a solution to the following problem, 8 > > > < > > > :r 2 w(x; y) = 0 (x; y)2 = [0; L][0; H] w(x;0) =g(x)x2[0; L] w(x; H) = 0x2[0; L] w(0; y) =w(L; y) = 0y2[0; H](7.22) whereg(x) : [0; L]! is a given boundary datum. The problem is de ned over a rectangular and we know the values ofw(x; y) to be equal to 0 on all sides of the rectangle except for one, where it's value is given by 78 CHAPTER 7. POISSON'S EQUATION the functiong(x). We now look for a particular family of solutions to the Laplace equation of the type, w(x; y) =u(x)v(y)8(x; y)2 (7.23) whereu(x) : [0; L]!Randv(y) : [0; H]!R. In order for this function to be a solution to our original problem it must not only be a harmonic function but satisfy the boundary conditions as well. According to the rst equation of this system, r2 w(x; y) =u00 (x)v(y) +u(x)v00 (y) = 0 which can be rewritten as, u00 (x)u (x)= v 00 (y)v (y)(7.24) Since the two sides of the equation are functions of dierent independent variables, the only way this equation could hold8(x; y)2 is if both sides of the equation are equal to a constant negative value2 , with2R, u00 (x)u (x)= v 00 (y)v (y)= 2 We then split our Laplace equation in two separate second-order equations each dependent on a dierent variable, u00 (x) =2 u(x) (7.25) v00 (y) =2 v(y) (7.26) Assuming that these functions are not null over the entire , in order to satisfy the boundary conditions we must have that, u(0) =u(L) = 0 v(0) =v(H) = 0 The original problem has been split into two separate problems,(u00 (x) =2 u(x) u(0) =u(L) = 0(7.27) 7.3. THE MEAN VALUE PROPERTY OF HARMONIC FUNCTIONS 79 (v00 (y) =2 v(y) v(0) =v(H) = 0(7.28) The solution to the problem 7:27 in thexvariable is a linear combination of trigonometric functions, u(x) =Asin(x) +Bcos(x) (7.29) According to the rst boundary condition of the rst problem, u(0) =B= 0 Using this result in the second boundary condition, u(L) =Asin(L) +Bcos(L) =Asin(L) = 0 which holds only if, k= k2 (7.30) So that the solution to the rst problem is, uk( x) =Asin( kx ) (7.31) The problem 7:28 in theyvariable has become, v00 k( y) =2 kv (y) which has solutions, vk( y) =C e ky +De  ky Imposing the boundary condition we have that, v(H) =C e kH +De  kH = 0 which means that, D=C e2  kH so that, vk( y) =C e ky e2  kH e  ky
= 2C e kH e k( yH) e  k( yH)2  80 CHAPTER 7. POISSON'S EQUATION vk( y) = 2C e kH sinh( k( yH)) (7.32) For an appropriate value of the constantC, on which no hypothesis has been made, we can set the constant term in front of the hyperbolic sine equal to 1 without loss of generality. With this simpli cation the solution the original problem 7:22 inx; ycan be written as, wk( x; y) =u k( x)v k( y) =sin( kx )sinh( k( yH)) (7.33) Given that the Laplace equation is linear, any linear combinations of functionsw k( x; y) will still still be a solution, w(x; y) =1 X k=1a kw k( x; y) =1 X k=1a ksin ( kx )sinh( k( yH)) At this point we are very close to the nal form of the solution. the only thing left to do is impose the last boundary condition, which we have since now ignored, w(x;0) =g(x) With the solution written as a linear combination we are free to choose the coecientsa kin order to satisfy the requirement, w(x;0) =1 X k=1a kw k( x;0) =1 X k=1a ksin ( kx )sinh( kH ) =1 X k=1A ksin ( kx )g(x) As we can see, the coecientsA k= a ksinh ( kH ) are just theFourier coecientscorresponding to the functiong(x), g(x) =1 X k=1A ksin ( kx ) and can thus be calculated through the formula,Ak= a ksinh ( kH ) (7.34) Ak=2L Z L 0g (x)sin( kx ) dx(7.35) Chapter 8 The Wave Equation In this section we will present the problem relating to the wave equation, as well as some properties of it's solution. Imagine a horizontal string lying in a2Dplane. At rest, all the points of the string will be lying on thexaxis. If the string is pulled, at timet, the points of the strings will lie on a curve, t R2 which will vibrate in time. tis simply the graph of the function, u(
; t) :R!R which describes thevertical displacementat timet2Rof the point of the string that at rest was occupying the positionxon the horizontal axis, so that, t= f(x; u(x; t))2R2 :x2Rg In order to be able to perform our computations we will reparametrize t through its curviliear abscissa, t( y) = (y; u(y; t)) :R! t(8.1) A property of this parametrization in a Cartesian coordinate system is that, 0 (y) = (1; @ yu )! k 0 (y)k=q1 + j(@ yu )(y; t)j2 (8.2) Let's now introduce another function, t( x) : t! [0;+1) 81 82 CHAPTER 8. THE WAVE EQUATION This function describes themass densityat timet2Rof the points on t. If we suppose the string to be isolatedwe can use the law of conservation of mass for an intervalI= [x; x+
x]Ron thex-axis. According to this law, the mass originally placed within this intervalIat timet= 0 is at a later momenttplaced on it's correspondingarcon the curve , which we will call, t( I) t so that at any moment t( I) pro jects vertically ontoI. In integral form, the mass conservation law can therefore be written as, Z I 0( s)ds=Z t( I) t( s)ds Taking a discrete interval on thex-axisI= [x; x+
x] we notice that for t= 0 the curvilinear abscissa is just the coordinatexof the points on the string, while fort6 = 0 it is given by t( x), Zx+
x x 0( y) dy=Z x+
x x t( t( y)) d t=Z x+
x x t( t( y)) 0 td y=Z x+
x x t( y; u(y; t))q1 + j(@ yu )(y; t)j2 dy In order for this relation to hold for any value of (
x; t) the two integrands must be equal, meaning that, t( y; u(y; t)) = 0( y)q 1 + j(@ yu )(y; t)j2 The term at the denominator is strictly greater than 1 fort6 = 0, meaning that as the string vibrates the density of the points on the curve tdecreases to compensate for the stretching.In order for the string to start vibrating we must provide a some sort of forcing term. LetT(x) be theintensityof a force acting on a point t( x)2 twhich we can suppose to be tangentto tat any point, tan((x)) = (@ xu )(x; t) Since the points only move vertically, the horizontal component of this force, T(x)cos((x)) =T(x+
x)cos((x+
x)) = can be assumed to beconstantand equal tofor all intervals on the string. The vertical component of this force, 83 T(x)sin((x)) will be responsible for the vertical oscillations of the points on the vibrat- ing string. CallingF t( I) the vertical force acting at timeton the segment t( I), Ft( I) =T(x+
x)sin((x+
x)T(x)sin((x)) we can see that, by reworking the terms, =T(x+
x)cos((x+
x))tan((x+
x))T(x)cos((x))tan((x)) =[tan((x+
x))tan((x))] =tau[(@ xu )(x+
x; t)(@ xu )(x; t)] Then, thelinear vertical momentumof the segment t( I) can be calculated as the density at a point of the curve t( I) times the temporal derivative of the vertical displacement, p(I ; t) =Z t( I) t( t( y)) (@ tu ) (y; t) |{z} vertical velocityd t= =Z x+
x x t( y) (@ tu ) (y; t)q1 + j(@ yu )(y; t)j2 dy=Z x+
x x 0( y) (@ tu ) (y; t) dy According to Newton's law, the force acting on a segmentIof the string is related to its change in momentum through the formula, Ft( I) =dd tp (I ; t) meaning that, [(@ yu ) (x+
x; t)(@ yu ) (x; t)] =dd tZ x+
x x( @ tu ) (y; t) 0( y) dy Assuming that the functions are well-behaved with respect to the ex- change between integration and derivation order we can bring the partial derivative with respect to space inside the integral. Dividing by
xand taking the limit as
x!0 returns, lim
x!0( @ yu ) (x+
x; t)(@ yu ) (x; t)
x= lim
x!01
xZ x+
x x 0( y) @2 tu
(y; t) dy 84 CHAPTER 8. THE WAVE EQUATION which, according to thesecond fundamental theorem of calculus becomes, 0( y) @2 tu
(y; t) = @2 yu
(y; t) (8.3) This equation is known as theWave Equation. It states that the force acting on an in nitesimal segment of our string, represented on the left by the density multiplied by an acceleration, is equal to an elastic force, represented on the right by the constantmultiplied by a second-derivative with respect to space of the vertical displacement. Given that this equation is nothing but Newton's law written in a local form, we could consider the terms on the left as aDensity of Vertical Force. Rewriting the known terms through the use of a new variable, c(x) =r  0( x)(8.4) which we call theSpeed of Propagationwe obtain the nal form of the wave equation, @2 tu
(y; t)c2 (x) @2 xu
(y; t) = 0 (8.5) 8.1 Conservation of Energy for the Wave Equa-tion TheKinetic Energyof this an interval t( I) at timetis given by, EI K( t) =Z I12 ( @ tu )2 (y; t) 0( y) dy(8.6) In order to nd thePotential Energyof the forces acting on the same interval t( I) we must rst compute the in nitesimal change in length of the segment of string as it vibrates,
I=Z t( I)1 d tZ I1 d y=Z x+
x xq1 + j(@ yu )(y; t)j2 dyZ x+
x x1 d y=Z x+
x x q1 + j(@ yu )(y; t)j2 1 dy For small values of
xwe can approximate the terms in the integral as, q1 + j(@ yu )(y; t)j2 11 +j (@ yu )(y; t)j22 1 =j (@ yu )(y; t)j22 8.1. CONSERVATION OF ENERGY FOR THE WAVE EQUATION 85 so that, taking the limit as
x!0,
I12 Z x+
x xj @ yu (y; t)j2 dy12 j @ yu (y; t)j2
x(8.7) ThePotential Energycontained in an intervalIof the string at timet thus becomes, EI p( t) =Z I2 j @ yu (y; t)j2 dy(8.8) TheTotal Energyof our intervalIwill be given by the sum of kinetic and potential energy, EI T( t) =EI K( t) +EI P( t) In order to nd a unique solution to our problem we must impose some boundary conditions, such as the D.B.C., on our interval of interestI= [a; b], ((@ tu )(a; t) = 0 (@ tu )(b; t) = 0 These conditions demand that the extreme points of the string remain xed in their initial position on thex-axis as the string vibrates. Choosing N.B.C. of the type, ((@ yu )(a; t) = 0 (@ yu )(b; t) = 0 instead would have meant demanding that though the string might travel freely in space, the derivative of the displacement at its extremes must always be zero, meaning that these point travel parallel to thex-axis. Subsequently one can prove the following theorem, Theorem 8.1(Conservation of total energy of the solutions of the wave equation under D.B.C. or N.B.C.). ET( t) =E T(0) 8t2R(8.9) Proof.Let's take the total time derivative of the energy at timet, dd tE T( t) =dd t 12 Z b a 0( @ tu )2 (x; t) +(@ tu ) (x; t)2 dx 86 CHAPTER 8. THE WAVE EQUATION Assuming that the solution is suciently smooth we can interchange the integration and derivation order, bringing the derivative with respect tot inside the integral, dd tE T( t) =12 Z b a@@ t  0( @ tu )2 (x; t) +(@ tu ) (x; t)2 dx =12 Z b a 2 0( @ tu ) @2 tu
(x; t) + 2(@ tu )@ t( @ xu ) dx =Z b a 0( @ tu ) @2 tu
(x; t) +(@ tu )@ x( @ tu ) (x; t) dx Integrating the second term in the integral by parts, =Z b a 0( @ tu ) @2 tu
(x; t)(@ xu )2 (@ tu ) (x; t) dx+ [(@ xu ) (@ tu )]b a The last term of the equation is zero due to the B.C. and we see that,since u(x; t) is a solution to the wave equationn =Z b a( @ tu ) 0 @2 tu
(x; t)(@ xu )2  dx= 0 We have just shown that the time derivative of the total energy of the system is zero, meaning that the total energy must remain constant in time (it can be shown that this holds under any Euclidean domain).8.2 D'Alembert's Formula D'Alemebert's formulaallows us to nd the solution to the wave equation on the whole real lineR(not just in an interval) in terms of the initial conditions. This formula is obtained by exploiting the relationship between the wave equation and the transport equation seen in previous chapters. By using regular derivation rules we can see that, @2 t c2 @2 x |{z} Wave Operator= ( @ t c@ x) ( @ t+ c@ x) |{z} Transport Operators(8.10) the wave operator can be written as the combination of two transport operators with the same velocity termctraveling in opposite direction. 8.2. D'ALEMBERT'S FORMULA 87 We recall that the solution to a transport problem of the form,((@ tu ) (x; t) +c(@ xu ) (x; t) =f(x; t) u(x;0) =g(x) has solution, u(x; t) =g(xct) +Z t 0f (xc(ts); s)ds Theorem 8.2(D'Alembert's Formula).Given,g(x; t)2C1 (R)representing the initial state of the solution andk(x; t)2C(R)representing the initial velocity, the solution to the transport problem, 8 > < > :( @2 tu )(x; t)c2 (@2 xu )(x; t) = 0x2R; t2R u(x;0) =g(x) (@ tu ) (x;0) =h(x) is given by, u(x; t) =g (x+ct) +g(xct)2 + 12 cZ x+ct xcth (y)dy What this theorem tells us is that the solution to the wave equation is the sum of the arithmetic mean of two traveling waves, both starting in the origin att= 0 and traveling in opposite directions and an integral term that describes the in uence of the initial velocity of the waves. Proof.Letu(x; t) be the solution to the wave equation. Let's create the function, v(x; t) = (@ t+ c@ x) u(x; t) It is easy to see that the original wave equation becomes, @2 t c2 @2 x
u(x; t) = (@ t c@ x) ( @ t+ c@ x) u(x; t) = (@ t c@ x) v(x; t) = 0 Moreover, the value of this function at timet= 0 is given by, v(x;0) = (@ tu ) (x;0) +c(@ xu ) (x;0) =h(x) +cg0 (x) = (x) Therefore, this functionv(x; t) solves the following transport problem, 88 CHAPTER 8. THE WAVE EQUATION (@tv (x; t)c@ xv (x; t) = 0 v(x;0) = (x) and must be equal to, v(x; t) = (x+ct) =h(x+ct) +cg0 (x+ct) Utilizing the de nition ofv(x; t) we can create a second transport problem for the functionu(x; t), (v(x; t) = (@ tu ) (x; t) +c(@ xu ) (x; t) = (x+ct) u(x;0) =g(x) whose solution must again be given by, u(x; t) =g(xct) +Z t 0 (xct+ 2cs) ds by substituting, y=xct+ 2cs dy= 2cds we get, u(x; t) =g(xct) +12 cZ x+ct xct (y)dy=g(xct) +12 cZ x+ct xcth (y) +cg0 (y)dy =g(xct) +12 Z x+ct xctg 0 (y)dy+12 cZ x+ct xcth (y)dy =g(xct) +12 [ g(x+ct)g(xct)] +12 cZ x+ct xcth (y)dy obtaining D'Alembert's formula, u(x; t) =g (xct) +g(x+ct)2 + 12 cZ x+ct xcth (y)dy 8.2. D'ALEMBERT'S FORMULA 89 RemarksUnlike in the heat equation, the wave equation provides no reg- ularizing eects on the initial distribution. Ifg(x)= 2C2 (RR) the solution u(x; t)= 2C2 (RR). The solution simply takes the original pro le of the string and translates it to the right and to the left as the wave(s) propagates, without smoothing out any eventual singularities. We can more explicitly write the solution as the sum of apro-gressive and are-gressive wave by creating the functions, F(z) =12 g (z) +12 cZ z 0h (y)dy G(z) =12 g (z) +12 cZ 0 zh (y)dy The solution then can be written as, u(x; t) =F(x+ct) Regressive wave+ G(xct) Progressive Wave These functions remain constant along the lines in Space-Time whose equations are given by, xct=const Looking at the image we have that, F(A) =F(C)G(A) =G(B) F(B) =F(D)G(C) =G(D) So that, u(A) +u(D) =F(A) +G(A) +F(D) +G(D) =F(C) +G(B) +F(B) +G(C) = F(C) +G(C) +F(B) +G(B) =u(B) +u(C) The values of the solutionu(x; t) at a generic point in Space-Timex 0; t 0 can be found by following these lines, 8.2.1 Method of Separation of Variables for the Heatand Wave Equation Let 2Rn be a bounded domain. We look for solutions to the heat equation, @tu (x; t) r2 u(x; t) = 0x2 ; t >0 90 CHAPTER 8. THE WAVE EQUATION of the form,u(x; t) =v(x)w(t) The equation then becomes, v(x)w0 (t) r2 v(x)w(t) = 0 as before, we can divide byu(x; t) and notice that the equation can then be split into two independent terms that must be equal to a constant, r2 vv ( x) =w 0w ( t) =2R alternatively,(r2 v(x) =v(x)x2 w0 (t) =w(t)t >0 By imposing D.B.C. on the solutionu(x; t) which translate into D.B.C on the functionv(x), u(x; t) = 08x2@ ;8t >0 v(x) = 08x2@ Multiplying the rst of our two problems in the variablexbyv(x) and then integrating over we get, Z v (x)r