THE LAPLACE EQUATION. The Laplace (or potential) equation is the equation. u = 0. = 2 x 2. x y 2 in R 2
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1 THE LAPLACE EQUATION The Laplace (o potential) equation is the equation whee is the Laplace opeato = 2 x 2 u = 0. in R = 2 x y 2 in R 2 = 2 x y z 2 in R 3 The solutions u of the Laplace equation ae called hamonic functions and play an impotant ole in many aeas of mathematics. The Laplace opeato is one of the most impotant opeatos in mathematical physics. It is associated with the gavitational and electical fields. Fo instance, we know fom Newton s law of univesal gavitation that two points A and B with masses M A and M B attact each othe with foces F A and F B as in figue, each foce with magnitude F = GM AM B AB 2 whee G is the univesal gavitational constant and AB is the distance fom A to B F A A F B Figue 1. Mass at A exet a foce F A on the mass at B B. As vectos these foces ae opposite and we have BA F A = F AB AB and F B = F AB Suppose that A is located at the oigin of the (x, y, z)-space and that B has a unit mass (M B = 1) and it is located at the point (x, y, z), then the foce F (x, y, z) with which A will attact B is ( x F = C x2 + y 2 + z 23 i + y x2 + y 2 + z 23 j + ) z x2 + y 2 + z 23 k Date: Januay 19,
2 2 THE LAPLACE EQUATION whee the constant C is C = GM A. This vecto-valued function F is the gavitational field geneated by the point mass A. The function Φ (which is eal valued) and defined by satisfies Hence, Φ(x, y, z) = C x2 + y 2 + z 2 = C(x2 + y 2 + z 2 ) 1/2 Φ x = Cx(x2 + y 2 + z 2 ) 3/2 = Φ y = Cy(x2 + y 2 + z 2 ) 3/2 = Φ z = Cz(x2 + y 2 + z 2 ) 3/2 = F (x, y, z) = gadφ(x, y, z). Cx x2 + y 2 + z 23 Cy x2 + y 2 + z 23 Cz x2 + y 2 + z 23 The function Φ is called a potential of the vecto field F. Now, we compute the second patial deivatives of Φ. We use the notation = x 2 + y 2 + z 2. We have 2 Φ x 2 = C( 3 3x 2 5 ) 2 Φ y 2 = C( 3 3y 2 5 ) 2 Φ z 2 = C( 3 3z 2 5 ) When we add the thee patial deivatives, we obtain Φ = C(3 3 3(x 2 + y 2 + z 2 ) 5 ) = 0 since 2 = x 2 + y 2 + z 2. This means that the potential Φ satisfies the Laplace equation. This is the eason why the Laplace equation is also efeed to as the potential equation. If we have N point masses A 1 A N, each geneates a gavitational field F i, (i = 1, N) and each field has a potential Φ i, the esulting field of all the mass points is the sum of the fields and the potential is the sum of the potentials 1 : N N Fi = gadφ i. i=1 i=1 1. Some BVP Fo The Laplace Equation The following ae typical poblems associated with the Laplace opeato The Diichlet Poblem. The poblem is to find a hamonic function u inside a domain D so that the values of u ae pescibed on the bounday D of D (u = f is given on the bounday D). 1 Suppose that a body occupies a egion R in the (x, y, z)-space and has mass density ρ(x, y, z) at the point (x, y, z). Then the body geneates a gavitational field F whose potential function Φ is given by the following integal Φ(x, y, z) = R ρ(ξ, η, ζ)dξdηdζ (x ξ) 2 + (y η) 2 + (z ζ) 2
3 THE LAPLACE EQUATION 3 Find a function u so that u=f on the bounday D u=0 D inside the domain D Figue 2. The Diichlet Poblem 1.2. The Neumann Poblem. The poblem is to find a hamonic function u inside the domain D so that the nomal deivatives of u, (i.e. ) ae pescibed on the bounday ( = g on D.) Recall that the nomal deivative at a point Find a function u so that Exteio unit nomal vectos u/=g on the bounday D u=0 D inside the domain D (x, y) on the bounday D is Figue 3. The Neumann Poblem (x, y) = gadu(x, y) n (x, y) whee n (x, y) is the exteio unit nomal at the point (x, y) The Poblem with mixed bounday conditions. The poblem is to find a hamonic function u inside the domain D so that on the bounday D it satisfies au + b = h, whee a, b, and h ae given.
4 4 THE LAPLACE EQUATION 2. Steady-State Tempeatue Poblems The above poblems fo the Laplace equation ae illustated by the steady-state solutions of the 2-D and 3-D heat equation. By a steady-state function u, we mean a function that is independent on time t. Thus, u t 0. In paticula if u satisfies the heat equation u t = u and u is steady-state, then it satisfies u = Example 1. Wite the BVP fo the steady-state tempeatue u(x, y) in a 1 2 ectangula plate if the bottom hoizontal side is kept at 0 0, the top hoizontal side at 100 0, the left vetical side at 10 0 and the ight vetical side at u=100 u= 10 u =0 u=200 u=0 Figue 4. A Diichlet poblem fo the steady-state tempeatue This is an example of a Diichlet poblem. We can wite it as u(x, y) = 0, 0 < x < 1, 0 < y < 2; u(x, 0) = 0, u(x, 2) = 100, 0 < x < 1; u(0, y) = 10, u(1, y) = 200, 0 < y < Example 2. This time we have steady-state tempeatue in a 1 2 ectangula plate. Assume that the bounday conditions ae as follows: the bottom and ight sides ae insulated and left and top sides ae kept at constant tempeatues of 0 and 100 degees, espectively. The BVP can be witten as u(x, y) = 0, 0 < x < 1, 0 < y < 2; (x, 0) = 0, u(x, 2) = 100, 0 < x < 1; y u(0, y) = 0, (1, y) = 200, 0 < y < 2. x 2.3. Example 3. Conside a plate in the shape of a quate of a cicle with adius 1. Suppose that the tempeatue is steady-state, the cicula side is insulated, one adial side is kept at 100 degees and the othe at 50 degees.
5 THE LAPLACE EQUATION 5 u=100 u=0 u =0 Figue 5. A mixed poblem fo the steady-state tempeatue u=50 u=100 Figue 6. A BVP fo the steady-state tempeatue in a cicula domain The BVP can be witten as u(x, y) = 0, 0 < x 2 + y 2 < 1; x > 0, y > 0 u(x, 0) = 50, 0 < x < 1; u(0, y) = 100, 0 < y < 1; (x, y) = 0, x2 + y 2 = 1, x > 0, y > 0. We expess the nomal deivative in tems of x = gadu n, and y by using whee n is the unit nomal vecto to the unit cicle. Fo a cicle x 2 + y 2 = R 2 the unit nomal at point (x 0, y 0 ) is n(x 0, y 0 ) = x 0 R i + y 0 R j. In ou case R = 1, so that at each point (x, y) on the bounday (x 2 + y 2 = 1) the unit nomal vecto is just. n(x, y) = x i + y j. Hence, (x, y) = x x (x, y) + y (x, y) y
6 6 THE LAPLACE EQUATION. The BVP can then be ewitten as u(x, y) = 0, 0 < x 2 + y 2 < 1; x > 0, y > 0 u(x, 0) = 50, 0 < x < 1; u(0, y) = 100, 0 < y < 1; x (x, y) + y x x (x, y) = 0, x2 + y 2 = 1, x > 0, y > 0. Remak Since the domain is cicula, this poblem is in fact bette suited fo pola coodinates. We will evisit these poblems. 3. The Laplacian In Pola, Cylindical, And Spheical Coodinates Fo BVP that deal with non ectangula shaped domains, it is useful to use coodinates systems othe than the ectangula coodinates. In paticula, fo cylindically o spheically shaped domains, the appopiate coodinates ae the cylindical and spheical coodinates. To use these coodinates, it is necessay to expess the Laplace opeato in these coodinates The 2D-Laplacian in pola coodinates. Fist ecall that a point p R 2 can be expessed in ectangula coodinates as (x, y) o in pola coodinates as (, θ) y P θ x Figue 7. Rectangula and pola coodinates The elations between these coodinates is given by and x = cos θ and y = sin θ 2 = x 2 + y 2 and tan θ = y x o cot θ = x y Let u be a function defined in the the plane R 2. Then u can be expessed in tems of the ectangula coodinates as u(x, y) o in tems of the pola coodinates u(, θ). Its Laplacian u is also a function in R 2. We know how to expess it in ectangula coodinates: u(x, y) = u xx (x, y) + u yy (x, y). We would like to expess u in pola coodinates only (so that x and y will not appea at all but only and θ ae involved). Fo this we need to use the chain ule to elate the fist and second patial deivatives of u given in tems of x, y to thei countepats in tems of and θ.
7 THE LAPLACE EQUATION 7 Then By diffeentiating 2 with espect to x and y, we obtain 2 x = 2x x = x = cos θ 2 y = 2y y = y = sin θ xx = x x x x 2 = 2 = 2 x 2 3 = y2 3 = sin2 θ yy = y y y y 2 = 2 = 2 y 2 3 Now we diffeentiate tan θ with espect to x and y = x2 3 = cos2 θ (tan θ) x = (sec 2 θ) θ x = y x 2 θ x = y cos2 θ x 2 (tan θ) y = (sec 2 θ) θ y = 1 θ y = cos2 θ x x Then fo the second patial deivatives, we get θ xx = ( y ) 2 ( x ) θ yy = 2 y x = 2x y 4 = 2y 2y x x 4 = 3 2x y = 3 = 2xy 4 = 2xy 4 3 = y 2 = sin θ = x 2 = cos θ 2 sin θ cos θ = 2 = 2 sin θ cos θ 2 Now we go back to a function u defined in the plane and elate its deivatives fom one system of coodinates to the othe by using the chain ule. We have Fo the second deivatives, we have Hence, u xx Similaly, we have u x = u x + u θ θ x and u y = u y + u θ θ y. = (u x ) x = (u x + u θ θ x ) x = (u ) x x + (u θ ) x θ x + u xx + u θ θ xx = [(u ) x + (u ) θ θ x ] x + [(u θ ) x + (u θ ) θ θ x ]θ x + u xx + u θ θ xx u xx = u ( x ) 2 + 2u θ x θ x + u θθ (θ x ) 2 + u xx + u θ θ xx u yy = u ( y ) 2 + 2u θ y θ y + u θθ (θ y ) 2 + u yy + u θ θ yy. By adding these last elations, we obtain u = u xx + u yy = u (( x ) 2 + ( y ) 2 ) + 2u θ ( x θ x + y θ y ) +u θθ ((θ x ) 2 + (θ y ) 2 ) + u ( xx + yy ) + u θ (θ xx + θ yy ) By using the fomulas given above fo the deivatives of and θ, we get ( x ) 2 + ( y ) 2 = 1, x θ x + y θ y = 0, xx + yy = 1, (θ x ) 2 + (θ y ) 2 = 1 2, and θ xx + θ yy = 0.
8 8 THE LAPLACE EQUATION With these, the expession fo u becomes u = u xx + u yy = u + 1 u u θθ The ight expession contains only the vaiables and θ. We have established the following Poposition 3.1. The Laplace opeato in pola coodinates is: = θ 2. Example Conside a plate in the shape of a secto of a ing with inne adius 1 and oute adius 2. Suppose that the steady state tempeatue in the plate satisfies the bounday conditions as shown in the figue. To wite the BVP fo the steady u=100 u=5θ u =0 n Figue 8. Steady-state tempeatue in a 45 0 secto state tempeatue, we need to wite the PDE inside the secto (Laplace equation) u (, θ) + 1 u (, θ) u θθ(, θ) = 0, 1 < < 2, 0 < θ < π/4. wite the specified tempeatue on the slanted edge u(, π/4) = < < 2. wite the specified tempeatue on the oute cicula side u(2, θ) = 5θ, 0 < θ < π/4. wite the insulation condition on the hoizontal edge wite the insulation condition on the inne cicula side The last two condition need an explanation. Recall that insulating a suface means that the nomal deivative of the tempeatue u is 0. Now fo the hoizontal side, it means that u y (x, 0) = 0. But, we need to wite this in pola coodinates. At each point (x, 0), we have then 0 = u y = u y + u θ θ y Fom the pevious calculation we have y = y/ and so y = 0 when y = 0 (and of couse > 0). Also, θ y = x/ 2 0 since x > 1. Hence, the insulation condition on the hoizontal side is simply u θ (, 0) = 0 fo 1 < < 2.
9 THE LAPLACE EQUATION 9 The last condition is = 0 on the inne cicula side. Hee, is the nomal deivative. The oute unit nomal to the inne cicle is simply n = (1/)(x i + y j). Hence, = gadu n = 1 (xu x + yu y ) = 1 (x(u x + u θ θ x ) + y(u y + u θ θ y )) = x x + y y u xθ x + yθ y u θ Fom the above calculation, we have x x + y y 2 = 1, and xθ x + yθ y = 0. All of this simply means that in pola coodinates = u. Theefoe, the insulation of the inne cicle eads u (1, θ) = 0, Now, we can wite the BVP as 0 < θ < π/4. u (, θ) + 1 u (, θ) + 1 u 2 θθ (, θ) = 0, 1 < < 2, 0 < θ < π/4; u(, π/4) = 100, 1 < < 2; u(2, θ) = 5θ, 0 < θ < π/4; u θ (, 0) = 0, 1 < < 2; u (1, θ) = 0, 0 < θ < π/ The 3-D Laplacian in cylindical coodinates. Recall that if a point p in R 3 has catesian coodinates (x, y, z), then its cylindical coodinates ae (, θ, z) with and θ as above: x = cos θ, y = sin θ and z = z. Fom the pevious calculations, we get the following P z x θ y Figue 9. Rectangula and cylindical coodinates Poposition 3.2. The expession fo the thee dimensional Laplacian in cylindical coodinates is = θ z 2
10 10 THE LAPLACE EQUATION 3.3. The 3-D Laplacian in spheical coodinates. Recall that if a point p in R 3 has catesian coodinates (x, y, z), then its spheical coodinates ae (ρ, θ, ϕ) (see figue) with z φ P x θ ρ y Figue 10. Rectangula and spheical coodinates x = ρ cos θ sin ϕ, y = ρ sin θ sin ϕ, z = ρ cos ϕ. We would like to expess the Laplacian in tems of only ρ, θ, and ϕ. This can be achieved by using the chain ule and so we would need to compute the fist and second patial deivatives ρ, θ, ϕ with espect to x, y, z. This is not difficult to do, but thee is a moe economical (just a bit moe economical) way to each the same esult by using the tansition fom cylindical (, θ, z) to spheical coodinates (ρ, θ, ϕ) instead of going fom ectangula to spheical. The tansition fom these systems of coodinates is given by and = ρ sin ϕ, θ = θ, z = ρ cos ϕ ρ 2 = 2 + z 2, θ = θ, tan ϕ = z. Notice that the coodinate θ is the same in both systems and so θ = θ ϕ = 0. This is why it is a little bit easie to use this tansition. Fom the pevious section, we know the expession of in cylindical coodinates. The action of on a function u is: u = u + 1 u + u zz u θθ. We need theefoe to expess u, u, and u zz in tems of the spheical vaiables. By the chain ule, we have the following u = u ρ ρ + u ϕ ϕ (no θ involved!); u z = u ρ ρ z + u ϕ ϕ z ; u = (u ρ ) ρ + (u ϕ ) ϕ + u ρ ρ + u ϕ ϕ = (u ρρ ρ + u ρϕ ϕ ) ρ + (u ϕρ ρ + u ϕϕ ϕ ) ϕ + u ρ ρ + u ϕ ϕ so u = u ρρ (ρ ) 2 + 2u ρϕ ρ ϕ + u ϕϕ (ϕ ) 2 + u ρ ρ + u ϕ ϕ u zz = u ρρ (ρ z ) 2 + 2u ρϕ ρ z ϕ z + u ϕϕ (ϕ z ) 2 + u ρ ρ zz + u ϕ ϕ zz Fom these, we get then u + 1 u ( + u zz = u 2 ρρ ρ + ρ ) ( 2 z + 2u ρϕ (ρ ϕ + ρ z ϕ z ) + u 2 ϕϕ ϕ + ϕ ) 2 ( z +u ρ ρ + ρ zz + 1 ρ ( ) + uϕ ϕ + ϕ zz + 1 ϕ )
11 THE LAPLACE EQUATION 11 Now we expess the tems between paentheses in tems of spheical coodinates. We have, ρ = sin ϕ, ρ z = cos ϕ, ρ = cos2 ϕ, ρ zz = sin2 ϕ, ρ ρ ϕ = cos ϕ ρ, ϕ sin ϕ cos ϕ sin ϕ cos ϕ z = sin ϕρ, ϕ = 2 ρ 2, ϕ zz = 2 ρ 2 We obtain the paentheses tems as ρ 2 + ρ z 2 = 1, ρ ϕ + ρ z ϕ z = 0, ϕ 2 + ϕ z 2 = 1 ρ 2, ρ + ρ zz + 1 ρ = 2 ρ, and ϕ + ϕ zz + 1 ϕ = cot ϕ ρ 2 This gives then u + 1 u + u zz = u ρρ + 2 ρ u ρ + 1 ρ 2 u ϕϕ + cot ϕ ρ 2 u ϕ. To obtain u, need to add the tem 1 2 u θθ. This gives the following Poposition 3.3. The expession of the Laplacian in spheical coodinates is = 2 ρ ρ ρ ρ 2 ϕ 2 + cot ϕ ρ 2 ϕ + 1 ρ 2 sin 2 ϕ θ 2 Example Conside the steady-state tempeatue in the potion of a spheical shell contained in the fist octant. Assume that that the inne adius is 1 and the oute adius is 2 and that the bounday conditions ae as indicated in the figue. The 2 z Inne suface kept at 0 degees yz Side kept at 0 degees xz Side insulated Oute suface insulated y x xy Side kept at 200 degees Figue 11. Potion of a spheical shell steady-state tempeatue u(ρ, θ, ϕ) satisfies The Laplace equation inside the solid u ρρ + 2 ρ u ρ + 1 ρ 2 u ϕϕ + cot ϕ ρ 2 u ϕ + 1 ρ 2 sin ϕ u θθ = 0 The xy-face: u = 200 The yz-face: u = 0
12 12 THE LAPLACE EQUATION The xz-face: u θ = 0 The inne spheical bounday u = 0 The oute spheical bounday u ρ = 0 The BVP is theefoe u ρρ + 2 ρ u ρ + 1 ρ u 2 ϕϕ + cot ϕ ρ u 2 ϕ + 1 u ρ 2 sin ϕ θθ = 0 u(ρ, θ, π/2) = 200 u(ρ, π/2, ϕ) = 0 u θ (ρ, 0, ϕ) = 0 u(1, θ, ϕ) = 0 u ρ (2, θ, ϕ) = 0 1 < ρ < 2, 0 < θ < π/2, 0 < ϕ < π/2; 1 < ρ < 2, 0 < θ < π/2; 1 < ρ < 2, 0 < ϕ < π/2; 1 < ρ < 2, 0 < ϕ < π/2; 0 < θ < π/2, 0 < ϕ < π/2; 0 < θ < π/2, 0 < ϕ < π/2. 4. Execises All of the following execises deal with the steady-state distibution of the tempeatue in eithe 2-dimensional plates o 3-dimensional egions. Fo each execise, wite the coesponding BVP. Execise 1. A ectangula plate. The vetical sides ae kept at constant tempeatues. The left at 10 degees and the ight at 50 degees. The hoizontal sides ae insulated. (Can you guess the solution fo this poblem?) Execise 2. A ectangula plate with bounday conditions as in figue. At the lowe side whee thee is poo insulation the nomal deivative of the tempeatue is equal to 0.5 times the tempeatue. Kept at 20 degees Kept at 10 degees Poo insulation Figue 12. Rectangula plate Execise 3. A solid ectangula box. with bounday conditions as indicated in the figue. Execise 4. A plate in the shape of a quate of a disk of adius 10 and with bounday conditions as indicated in the figue.
13 THE LAPLACE EQUATION 13 Top and back faces insulated Left face u= 10 Right face u=10 Font face insulated Bottom face u=0 Figue 13. Rectangula box u=2θ u=0 Figue 14. A quate of a disk shaped plate Execise 5. A plate in the shape of half of a ing with inne adius 1 and oute adius b, (with b > 1), whee the bounday conditions ae as indicated in the figue. Execise 6. A plate in the shape of a secto of a ing with adii 1 and b, with b > 1, and whee the bounday conditions ae as indicated Execise 7. A solid cylinde of adius 10 and height 20. The top suface is insulated, the bottom suface is kept at tempeatue 20 degees and the lateal suface is kept at tempeatue 100 degees. Execise 8. A solid hollow cylinde (cylindical shell) with adii 10 and 15 and with height 20. Assume that the inne lateal suface is insulated, the oute lateal suface is kept at tempeatue 100 degees, the bottom suface is insulated and the top suface is kept at 50 degees. Execise 9. A solid sphee with adius 10. The top hemisphee is kept at tempeatue 100 degees and the lowe hemisphee is kept at tempeatue 0 degees.
14 14 THE LAPLACE EQUATION u=100θ /pi u=100 u=0 Figue 15. A half-ing shaped plate u=100 θ u=100 Figue 16. A 45 0 secto of a ing shaped plate Execise 10. A hollow solid sphee (spheical shell) with adii 1 and 5. The inne and oute sufaces ae kept at tempeatues 100 and 50 degees, espectively. (Can you guess the tempeatue distibution?)
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