Example: quiz answers

GREEN’S FUNCTION FOR LAPLACIAN - University of Michigan

GREEN S FUNCTION FOR LAPLACIANThe Green s FUNCTION is a tool to solve non-homogeneous linear equations. We will illus-trate this idea for the LAPLACIAN .Suppose we want to find the solutionuof the Poisson equation in a domainD Rn: u(x) =f(x),x Dsubject to some homogeneous boundary condition. Imaginefis the heat source anduis thetemperature. The idea of Green s FUNCTION is that if we know the temperature respondingto an impulsive heat source at any pointx0 D, then we can just sum up the result withthe weight functionf(x0) (it specifies the strength of the heat at pointx0) to obtain thetemperature responding to the heat sourcef(x) inD.

and for x on the boundary of D, we have u(x) = 0 because G(x,x0) = 0 by the definition of G in (0.5). Verification of (0.3) for u and G satisfying Neumann or Robin conditions can be done similarly. Now let’s see how to find the Green’s function for some particular domains. 1

Tags:

  Conditions, Functions, Robin, Boundary, Robin conditions

Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

Other abuse

Transcription of GREEN’S FUNCTION FOR LAPLACIAN - University of Michigan

1 GREEN S FUNCTION FOR LAPLACIANThe Green s FUNCTION is a tool to solve non-homogeneous linear equations. We will illus-trate this idea for the LAPLACIAN .Suppose we want to find the solutionuof the Poisson equation in a domainD Rn: u(x) =f(x),x Dsubject to some homogeneous boundary condition. Imaginefis the heat source anduis thetemperature. The idea of Green s FUNCTION is that if we know the temperature respondingto an impulsive heat source at any pointx0 D, then we can just sum up the result withthe weight functionf(x0) (it specifies the strength of the heat at pointx0) to obtain thetemperature responding to the heat sourcef(x) inD.

2 Mathematically, one may expressthis idea by defining the Green s FUNCTION as the following:Letu=u(x),x= (x1, x2, .. xn) be the solution of the following problem:{ u(x) =f(x),x Dusatisfies some homogeneous boundary condition along the boundary D( )Define the Green s functionG=G(x,x0) to be the solution of{ G(x,x0) = (x x0),x DG(x,x0) satisfies the same homogeneous boundary condition as in ( )( )herex0 Dis a fixed point. Thenu(x) = DG(x,x0)f(x0)dx0( )should be the solution of the boundary value problem ( ). We can check this statementfor the Dirichlet boundary value problem:{ u(x) =f(x),x Du(x) = 0x D( )We define the Green s functionGwith Dirichlet BC by{ G(x,x0) = (x x0),x DG(x,x0) = 0x D( )Applying = ni=1 2xito both sides of ( ), we get u(x) = D G(x,x0)f(x0)dx0= D (x x0)f(x0)dx0=f(x)and forxon the boundary ofD, we haveu(x) = 0 becauseG(x,x0) = 0 by the definitionofGin ( ).}}}}

3 Verification of ( ) foruandGsatisfying Neumann or robin conditions canbe done let s see how to find the Green s FUNCTION for some particular S FUNCTION FOR LAPLACIANTo simplify the discussion, we will be focusing onD R2, the same idea extends todomainsD Rnfor anyn 1, and to other linear equations. In what follows we letx= (x, y) fundamental first look for the FUNCTION (x) in the whole spaceR2so that (x) = (x)forx R2 Since (x) is the responding temperature to the point heat source at the origin, it must beradially symmetric, that is (x) = (r),r=|x|= x2+y2We know the LAPLACIAN in polar coordinates gives (r) = (r) +1r (r)we want (x) = (x), therefore (r) +1r (r) = 0forr >0( )Solving ( ), we get (r) =c1lnr+c2for some constantsc1andc2.

4 We takec2= 0. We now want to determinec1so that (x) = (x). We use Divergence divergence theorem states that for any vector fieldF(x) = (F1(x, y), F2(x, y))defined on a domainD R2, we have D F(x)dx= Dn F(x)dS( )here divF(x) = F(x) = xF1(x, y) + yF2(x, y),nis the unit outward normal to theboundary of the domainD, D.. dSis the line integral along the boundary curve applying the divergence theorem toF(x) = u(x) = ( xu(x, y), yu(x, y)). Since u= u, we have D u(x)dx= Dn u(x)dSWe want (x) = (x). Therefore, we want for any diskB(0, R) centered at the originwith radiusR,1 = (x)dx= (x)dx= B(0,R) (x)dx= B(0,R)n (x)dSNotice that the normal derivative of along the circle B(0, R) is the same as the derivativeof in the radial direction evaluated atR:n (x) = (R)for|x|=Rand (R) is a constant along the circle B(0, R), therefore B(0,R)n (x)dS= B(0,R) (R)dS= 2 R (R)GREEN S FUNCTION FOR LAPLACIAN3finally we arrive at1 = 2 R (R)this gives that (R) =12 R, therefore (R) =12 lnR.

5 In other wards, an application ofdivergence theorem also gives us the same answer as above, with the constantc1=12 .Now we find that the FUNCTION (x) =12 lnr,wherer=|x|= x2+y2satisfies (x) = (x)We call this FUNCTION (x) the fundamental solution of the LAPLACIAN inR2. It is clear thatfor any pointx0 R2, (x x0) = (x x0). s FUNCTION for bounded can construct Green s FUNCTION for abounded domainD R2using the fundamental solution (x x0).For example if we want to find the Green s functionGwith Dirichlet BC onD R2:{ G= (x x0)x DG(x,x0) = 0forx D( )herex0 Dis a fixed point, we can constructGas the followsG(x,x0) = (x x0) +v(x,x0)( )In order forGto satisfy ( ),vmust satisfy{ v(x,x0) = 0x Dv(x,x0) = (x x0),forx D( )In other wards,vshould be a solution of the Laplace equation inDsatisfying a non-homogeneous boundary condition that nullifies the effect of on the boundary ofD.}}

6 Sim-ilarly we can construct the Green s FUNCTION with Neumann BCby settingG(x,x0) = (x x0) +v(x,x0) wherevis a solution of the Laplace equation with a Neumann bound-ary condition that nullifies the heat flow coming from .In what follows we construct the Green s functions for the upper half plane and for theunit point out that for a general domainD, it is usually difficult to solve explicitly forv,therefore difficult to find an explicit expression forG. Nevertheless, ( ) is an importantexpression from which we can draw qualitative informationsabout the Green s FUNCTION andtherefore the solution of the Poisson and Laplace call a solutionuof the Laplace equation u(x) = 0 forx Da harmonic S FUNCTION FOR s FUNCTION for the upper half plane{y >0}.

7 We first construct the Green sfunction in the upper half plane with the Dirichlet boundarycondition:{ G(x,x0) = (x x0)x= (x, y), y >0G(x,0,x0) = 0x R( )herex0= (x0, y0) is a fixed point in the upper half plane, > know (x x0) = (x x0, y y0) is the temperature response to the single heatsource (x x0) placed at the pointx0. However (x x0, y y0)6= 0 at the boundary {y= 0}. In order to construct the temperature responseGso thatGis zero on the boundary {y= 0}:G(x,0, x0, y0) = 0, we place another heat source (x x 0) of negative strengthat the image pointx 0= (x0, y0) and consider the temperature response to both of theheat sources atx0andx 0:G(x,x0) = (x x0) (x x 0)( )We claimGas defined in ( ) is the Green s FUNCTION satisfying ( ).}

8 First we know G(x,x0) = ( (x x0) (x x 0)) = (x x0) (x x 0)forx R2 Howeverx 0is a point in the lower half plane, and we know (x x 0) = 0 for allx6=x 0,therefore G(x,x0) = (x x0)forxin the upper half plane. And it is easy to check that on the boundary {y= 0}of theupper half plane,G(x,0,x0) = (x x0, y0) (x x0, y0) =12 (ln (x x0)2+ ( y0)2 ln (x x0)2+ (y0)2) = 0 Therefore we have constructed the Green s FUNCTION for the Dirichlet BC in the upper halfplane:G(x,x0) =12 (ln|x x0| ln|x x 0|) =14 ln(x x0)2+ (y y0)2(x x0)2+ (y+y0)2 Similarly we can construct the Green s FUNCTION for the upper half plane with the homo-geneous Neumann BC.

9 Notice that for the upper half plane, theoutward normal derivativeat the boundary is in the negativeydirection: G n= G ySo what we look for next is theGsatisfying{ G(x,x0) = (x x0)x= (x, y), y >0 G y(x,0,x0) = 0x R( )herex0= (x0, y0) is a fixed point in the upper half we want to make sure the heat sources are placed in such a way that the flux throughthe boundary {y= 0}is zero. This can be achieved by placing another heat source (x x 0)GREEN S FUNCTION FOR LAPLACIAN5with the same strength at the image pointx 0= (x0, y0). We claim the Green s functionfor ( ) isG(x,x0) = (x x0) + (x x 0) =14 ln{((x x0)2+ (y y0)2)((x x0)2+ (y+y0)2)}We leave the checking of this claim to the s representation that we have constructed the Green s func-tion for the upper half plane.}

10 Before we move on to construct the Green s FUNCTION for theunit disk, we want to see besides the homogeneous boundary value problem ( ), whatother problems can be solved by the Green s FUNCTION we derive the Green s identity from the divergence , vbe smooth functions defined on a domainD R2. LetF=u v v u, andwe apply the divergence theorem toF. Notice that F= u v+u v v u v u=u v v uwe get D(u(x) v(x) v(x) u(x))dx= D(un v vn u)dS( )wherenis the unit outward normal to D. ( ) is called the Green s the Green s FUNCTION defined in ( ) satisfying the Dirichlet BC.


Related search queries