The Laplacian - University of Plymouth

1 BasicMathematicsTheLaplacianR Horan & M LavelleTheaim of this package is to provide a short selfassessment programme for students who want toapply the Laplacian , Revision Date: April 4, 2005 Version of (Grad, Div, Curl) Laplacian of a Product of Laplacian and Vector QuizSolutions to ExercisesSolutions to QuizzesThe full range of these packages and some instructions,should they be required, can be obtained from our webpageMathematics Support 1: Introduction (Grad, Div, Curl)31. Introduction (Grad, Div, Curl)Thevector differential operator , called del or nabla , is definedin three dimensions to be: = xi+ yj+ zkThe result of applying this vector operator to a scalar field is calledthegradient of the scalar field:gradf(x,y,z) = f(x,y,z) = f xi+ f yj+ f zk.

2 (See the package onGradients and Directional Derivatives.)Thescalar productof this vector operator with a vector fieldF(x,y,z)is called thedivergence of the vector field:divF(x,y,z) = F= F1 x+ F2 y+ F3 1: Introduction (Grad, Div, Curl)4 Thevector productof the vector with a vector fieldF(x,y,z)is thecurl of the vector field. It is written ascurlF(x,y,z) = F F(x,y,z) = ( F3 y F2 z)i ( F3 x F1 z)j+ ( F2 x F1 y)k= ijk x y zF1F2F3 ,where the last line is a formal representation of the line above. (Seealso the package onDivergence and Curl.)Here are some revision zandF=xi xyj+z2kcalculate thefollowing (click on thegreenletters for the solutions).(a) f(b) F(c) F(d) f FSection 2: The Laplacian52. The LaplacianTheLaplacianoperator is defined as: 2= 2 x2+ 2 y2+ 2 ascalar operator.

3 If it is applied to ascalar field, itgenerates ascalar 1 TheLaplacianof the scalar fieldf(x,y,z) =xy2+z3is: 2f(x,y,z)= 2f x2+ 2f y2+ 2f z2= 2 x2(xy2+z3) + 2 y2(xy2+z3) + 2 z2(xy2+z3)= x(y2+ 0) + y(2xy+ 0) + z(0 + 3z2)=0 + 2x+ 6z= 2x+ 6zSection 2: The Laplacian6 Exercise theLaplacianof the following scalar fields:(click on thegreenletters for the solutions).(a)f(x,y,z) = 3x3y2z3(b)f(x,y,z) = xz+y(c)f(x,y,z) = x2+y2+z2(d)f(x,y,z) =1 x2+y2+z2 QuizChoose theLaplacianoff(r) =1rnwherer= x2+y2+z2.(a) 1rn+2(b)nrn+2(c)n(n 1)rn+2(d)n(n+ 5)rn+2 The equation 2f= 0is calledLaplace s is an impor-tant equation in science. From the above exercises and quiz we seethatf=1ris a solution of Laplace s equation except atr= 2: The Laplacian7 The Laplacian of a scalar field can also be written as follows: 2f= , as thedivergence of the gradient off.

4 To see this consider f= ( f xi+ f yj+ f zk)= x( f x) + y( f y) + z( f z)= 2f x2+ 2f y2+ 2f z2= 2fExercise theLaplacianof the scalar fieldsfwhose gradients fare given below (click on thegreenletters for the solutions).(a) f= 2xzi+x2k(b) f=1yzi xy2zj xyz2k(c) f= ezi+yj+xezk(d) f=1xi+1yj+1zkSection 3: The Laplacian of a Product of Fields83. The Laplacian of a Product of FieldsIf a field may be written as a product of two functions, then: 2(uv) = ( 2u)v+u 2v+ 2( u) ( v)A proof of this is given at the end of this 2 TheLaplacianoff(x,y,z) = (x+y+z)(x 2z)may bedirectly calculated from the above rule 2f(x,y,z) = ( 2(x+y+z))(x 2z) + (x+y+z) 2(x 2z)+2 (x+y+z) (x 2z)Now 2(x+y+z) = 0and 2(x 2z) = 0so the first line on theright hand side calculate the second line we note that (x+y+z) = (x+y+z) xi+ (x+y+z) yj+ (x+y+z) zk=i+j+kSection 3.

5 The Laplacian of a Product of Fields9and (x 2z) = (x 2z) xi+ (x 2z) yj+ (x 2z) zk=i 2kand taking their scalar product we obtain 2f(x,y,z) = 0 + 2 (x+y+z) (x 2z)= 2(i+j+k) (i 2k) = 2(1 + 0 2)= example may be checked by expanding(x+y+z)(x 2z)anddirectly calculating the this rule to calculate theLaplacianof the scalarfields given below (click on thegreenletters for the solutions).(a)(2x 5y+z)(x 3y+z)(b)(x2 y)(x+z)(c)(y z)(x2+y2+z2)(d)x x2+y2+z2 Section 3: The Laplacian of a Product of Fields10 Proofthat 2(uv) = ( 2u)v+u 2v+ 2( u) ( v).By definition 2(uv) = 2 x2(uv) + 2 y2(uv) + 2 z2(uv). Considertherefore: 2 x2(uv)= x( x(uv))= x( u xv+u v x)= 2u x2v+ u x v x+ u x v x+u 2v x2= 2u x2v+ 2 u x v x+u 2v x2where theproduct rulewas repeatedly symmetry we also have: 2 y2(uv) = 2u y2v+ 2 u y v y+u 2v y2and 2 z2(uv) = 2u z2v+ 2 u z v z+u 2v z2.

6 Adding these results yields thedesired 4: The Laplacian and Vector Fields114. The Laplacian and Vector FieldsIf the scalar Laplacian operator is applied to avector field, it acts oneach component in turn and generates avector 3 TheLaplacianofF(x,y,z) = 3z2i+xyzj+x2z2kis: 2F(x,y,z) = 2(3z2)i+ 2(xyz)j+ 2(x2z2)kCalculating the components in turn we find: 2(3z2)= 2 x2(3z2) + 2 y2(3z2) + 2 z2(3z2) = 0 + 0 + 6 = 6 2(xyz)= 2 x2(xyz) + 2 y2(xyz) + 2 z2(xyz) = 0 + 0 + 0 = 0 2(x2z2)= 2 x2(x2z2) + 2 y2(x2z2) + 2 z2(x2z2) = 2z2+ 0 + 2x2So the Laplacian ofFis: 2F= 6i+ 0j+ (2z2+ 2x2)k= 6i+ 2(x2+z2)kSection 4: The Laplacian and Vector Fields12 QuizSelect from the answers below the Laplacian of the vector fieldF=x3yi+ ln(z)j+ ln(xy)k.(a)6xyi(b)6xyi 1z2j x2+y2x2y2k(c)3xi+1z2j (1x2 1y2)k(d)y3i+x2j y2 x2z2kQuizChoose the Laplacian ofF= 3x2zi sin( y)j+ln(2x3)kat thepoint(1, 2,1)?

7 (a)0(b)6i 3k(c)3i 3k(d)6i 2j 6kQuizWhich of the following choices is the Laplacian of the vector fieldF= ln(y)i+z2j sin(2 x)kat(1,1, )?(a) i+ 2j(b)0(c)4j+ 4 2k(d) i+j+kSection 5: Final Quiz135. Final QuizBegin QuizChoose the solutions from the options the Laplacian off(r) = 5x3y4z2.(a)30xy4z2+60x3y2z2+10x3y4(b)30x + 20y2+ 10(c)30xy4z2+75x3y2z2+15x3y4(d)30xy4z2+1 2x3y2z2+ the Laplacian off(x,y,z) = ln(r)wherer= x2+y2+z2.(a)0(b)2r2(c)12r2(d) the Laplacian ofF=x3i+ 7yj 3 sin(2y)k.(a)6xi(b)6xi+ 12 sin(2y)k(c)6xi+ 3 sin(2y)k(d)6xi 12 sin(2y)kEnd QuizSolutions to Exercises14 Solutions to ExercisesExercise 1(a)To find thegradientof the scalar fieldf=x2y z, we need thepartial derivatives: f x= x(x2y z) = 2x2 1 y= 2xy, f y= y(x2y z) =x2 y1 1=x2, f z= z(x2y z) = 0 z1 1= thegradientoff=x2y zis f(x,y,z) = f xi+ f yj+ f zk= 2xyi+x2j on thegreensquare to return Solutions to Exercises15 Exercise 1(b)To find thedivergenceof the vector fieldF=xi xyj+z2k, werecognise that its components areF1=x,F2= xy,F3=z2,So the divergence is thescalarexpression F= F1 x+ F2 y+ F3 z= x(x) + y( xy) + z(z2)=x1 1 x y1 1+ 2 z2 1= 1 x+ on thegreensquare to return Solutions to Exercises16 Exercise 1(c)Thecurlof the vector fieldFwhose components areF1=x,F2= xy,F3=z2,is given by thevectorexpression.

8 F=( F3 y F2 z)i ( F3 x F1 z)j+( F2 x F1 y)k=( y(z2) z( xy))i ( x(z2) z(x))j+( x( xy) y(x))k= (0 + 0)i (0 0)j+ ( y 0)k= on thegreensquare to return Solutions to Exercises17 Exercise 1(d)To subtract thecurlof the vectorF=xi xyj+z2kfrom the gradientof the scalar fieldf=x2y z f use the results ofExercise 1aandExercise 1c, where it was foundthat f= 2xyi+x2j kand F= the difference of these two vectors is f F= 2xyi+x2j k ( y)k= 2xyi+x2j (1 y) on thegreensquare to return Solutions to Exercises18 Exercise 2(a)TheLaplacianof the scalar fieldf= 3x3y2z3is: 2f= 2f x2+ 2f y2+ 2f z2= 2 x2(3x3y2z3) + 2 y2(3x3y2z3) + 2 z2(3x3y2z3)= x(9x2y2z3) + y(6x3yz3) + z(9x3y2z2)= 18xy2z3+ 6x3z3+ common factors, the scalar 2fcan also be written as 2f= 6xz(3y2z2+x2z2+ 3x2y2)= 6xz(3y2(z2+x2) +x2z2).

9 Click on thegreensquare to return Solutions to Exercises19 Exercise 2(b)TheLaplacianof the scalar fieldf= xz+y=x1/2z1/2+yis: 2f= 2 x2(x12z12+y)+ 2 y2(x12z12+y)+ 2 z2(x12z12+y)= x(12x(12 1)z12)+ y(1) + z(12x12z(12 1))= x(12x 12z12)+ 0 + z(12x12z 12)=12( 12)x( 12 1)z12+12( 12)x12z( 12 1)= 14x 32z12 14x12z result may be rewritten as 2f= 14x12z12(x 2+z 2)= 14 xz(1x2+1z2).Click on thegreensquare to return Solutions to Exercises20 Exercise 2(c)To calculate 2 x2+y2+z2, defineu=x2+y2+z2,sof=u1/2. From the chain rule we have f x= u12 u u x=12u(12 1) 2x=xu the second derivative is (from the product and chain rules): 2f x2= x(x) u 12+x u 12 u u x=u 12 x2u symmetric inx,yandzwe have 2f y2=u 12 y2u 32and 2f z2=u 12 z2u these results and usingx2+y2+z2=uyields 2f=(u 12 x2u 32)+(u 12 y2u 32)+(u 12 z2u 32)= 3u 12 (x2+y2+z2)u 32= 2u 12=2 x2+y2+ on thegreensquare to return Solutions to Exercises21 Exercise 2(d)To find theLaplacianoff= (x2+y2+z2) 12, weagain defineu=x2+y2+z2.

10 From the chain rule f x= u 12 u u x= 12u( 12 1) 2x= xu the second order derivative is: 2f x2= x(x) u 32 x u 32 u u x= u 32+ 3x2u to the symmetry under interchange ofx,yandz: 2f y2= u 32+ 3y2u 52, 2f z2= u 32+ 3z2u we find that theLaplacianoffvanishes: 2f=( u 32+ 3x2u 52)+( u 32+y2u 52)+( u 32+ 3z2u 52)= 3u 32+ 3(x2+y2+z2)u 52= 3u 32+ 3u 32= on thegreensquare to return Solutions to Exercises22 Exercise 3(a)If thegradientof the scalar functionfis f= 2xzi+x2k, then theLaplacianoffis given by thedivergenceof this vector 2f= div ( f).Therefore theLaplacianoffis 2f= div(2xzi+x2k)= x(2xz) + y(0) + z(x2)= 2z+ 0 + 0= on thegreensquare to return Solutions to Exercises23 Exercise 3(b)If a scalar fieldfhasgradient f=1yzi xy2zj xyz2k, itsLaplacianis given by thedivergence 2f= div ( f).