Laplace ’s Equation 1. Equilibrium Phenomena Consider a general conservation statement for a region U in Rn containing a material which ⃗ is being transported through U by a flux field, F = F⃗x,t. Let u = u⃗x,t denote the scalar concentration field for the material (u equals the concentration at ⃗x,t). Note that u is a scalar valued function while F⃗x,t is a vector valued function whose value at each ⃗x,t is a vector whose direction is the direction of the material flow at ⃗x,t and whose magnitude is proportional to the speed of the flow at ⃗x,t. In addition, suppose there is a scalar source density field denoted by s⃗x,t. This value of this scalar at ⃗x,t indicates the rate at which material is being created or destroyed at ⃗x,t. If B denotes an arbitrary ball inside U, then for any time interval t1,t2 conservation of material requires that ⃗ ⃗ t2 ⃗ ⃗ ⃗ t2 ⃗ ∫ ux,t2 dx = ∫ ux,t1 dx − ∫ ∫ Fx,t ⋅ nxdS xdt + ∫ ∫ sx,tdxdt B B t1 ∂B t1 B Now ⃗ ⃗ t2 ⃗ ∫ ux,t2 dx − ∫ ux,t1 dx = ∫ ∫ ∂tux,tdxdt B B t1 B and t t ∫ 2 ∫ F⃗x,t ⋅ ⃗n⃗xdS xdt = ∫ 2 ∫ divF⃗x,t dxdt t1 ∂B t1 B hence t 2 ⃗ + ⃗ ⃗ = ∫ ∫ ∂tux,t divFx,t − sx,tdxdt 0 for all B ⊂ U, and all t1,t2 . 1.1 t1 B Since the integrand here is assumed to be continuous, it follows that ⃗ ⃗ ⃗ ⃗ ∂tux,t + divFx,t − sx,t = 0, for all x ∈ U, and all t. 1.2 Equation (1.1) is the integral form of the conservation statement, while (1.2) is the differential form of the same statement. This conservation statement describes a large number of physical processes. We consider now a few special cases, a) Transport u = u⃗x,t ⃗ ⃗ F⃗x,t = ux,t V where V = constant, s⃗x,t = 0. ⃗ ⃗ ⃗ In this case, the equation becomes ∂tux,t + V ⋅ gradux,t = 0 b) Steady Diffusion u = u⃗x F⃗x,t = −K∇ux where K = constant > 0, s = s⃗x. In this case, the equation becomes − K divgradu⃗x = s⃗x or − K ∇2u⃗x = s⃗x. This is the equation that governs steady state diffusion of the contaminant through the region U. The equation is called Poisson ’s equation if s⃗x ≠ 0, 1 and Laplace ’s equation when s⃗x = 0. These are the equations we will study in this section. Another situation which leads to Laplace’s equation involves a steady state vector field ⃗ ⃗ ⃗ ⃗ V = V⃗x having the property that div Vx = 0. When V denotes the velocity field for an ⃗ ⃗ incompressible fluid, the vanishing divergence expresses that V conserves mass. When V denotes the magnetic force field in a magnetostatic field, the vanishing divergence asserts ⃗ that there are no magnetic sources. In the case that V represents the vector field of electric force, the equation is the statement that U contains no electric charges. In addition to the ⃗ ⃗ ⃗ equation div Vx = 0, it may happen that V satisfies the equation, curl Vx = 0. This ⃗ condition asserts that the field V is conservative (energy conserving). Moreover, it is a ⃗ ⃗ standard result in vector calculus that curl Vx = 0 implies that V = −grad u⃗x, for some scalar field, u = u⃗x. Then the pair of equations, ⃗ ⃗ div Vx = 0 and curl Vx = 0, taken together, imply that ⃗ ∇2u⃗x = 0 and V = −grad u⃗x. ⃗ We say that the conservative field V is ”derivable from the potential, u = u⃗x”. To say that u is a potential is to say that it satisfies Laplace’s equation. The unifying feature of all of these physical models that lead to Laplace’s equation is the fact that they are all in a state of equilibrium. Whatever forces are acting in each model, they have come to a state of equilibrium so that the state of the system remains constant in time. If the balance of the system is disturbed then it will have to go through another transient process until the forces once again all balance each other and the system is in a new equilibrium state. 2. Harmonic Functions A function u = ux is said to be harmonic in U ⊂ Rn if: i) u ∈ C2U; i.e., u, together with all its derivatives of order ≤ 2, is continuous in U ii) ∇2u⃗x = 0 at each point in U Note that in Cartesian coordinates, ∂u/∂x1 ⃗ = = 2 2 + + 2 2 div ∇ux ∂/∂x1, ...,∂/∂xn ⋅ ⋮ ∂ u/∂x1 ... ∂ u/∂xn ∂u/∂xn = ∂∂u⃗x = ∇2u⃗x It is clear from this that all linear functions are harmonic. = 2 + + 2 A function depending on x only through the radial variable, r x1 ... xn is said to be a radial function . If u is a radial function then 2 1 = ′ = 1 2 + + 2 − 2 = ∂u/∂xi u r∂r/∂xi and ∂r/∂xi 2 x1 ... xn 2xi xi/r 2 2 2 ” 2 ′ r − xi xi/r ” 2 ′ 1 xi ∂ u/∂x = u r∂r/∂xi + u r = u r∂r/∂xi + u r − i r2 r r3 and 2 2 ⃗ = n 2 2 = ” n 2 + ′ n 1 xi ∇ ux ∑ ∂ u/∂xi u r ∑ xi/r u r ∑ − i=1 i=1 i=1 r r3 = ” + ′ n 1 = ” + n − 1 ′ u r u r r − r u r r u r We see from this computation that the radial function u = unr is harmonic for various n if: n = 1 u1”r = 0; i.e., u1r = Ar + B = + 1 ′ = 1 d ′ = = n 2 : u2”r r u r r dr ru2r 0; i.e., u2r C ln r > ” + n−1 ′ = 1−n d n−1 ′ = = 2−n n 2 : unr r unr r dr r unr 0; unr Cr 2 2 Note also that since ∇ ∂u/∂xi = ∂/∂xi∇ u, for any i, it follows that every derivative of a harmonic function is itself, harmonic. Of course this presupposes that the derivative exists but it will be shown that every harmonic function is automatically infinitely differentiable so every derivative exists and is therefore harmonic. It is interesting to note that if u and u2 are both harmonic, then u must be constant. To see this, write ∇2u2 = div gradu2 = div 2u∇u = 2∇u ⋅ ∇u + 2u∇2u = 2|∇u|2 Then ∇2u2 = 0 implies |∇u|2 = 0 which is to say, u is constant. Evidently, then, the product of harmonic functions need not be harmonic. It is easy to see that any linear combination of harmonic functions is harmonic so the harmonic functions form a linear space. It is also easy to see that if u = ux is harmonic on Rn then for any z ∈ Rn, the translate, vx = ux − z is harmonic as is the scaled function, w = wλx for all scalars λ. Finally, ∇2 is invariant under orthogonal transformations. To see this suppose coordinates x and y are related by Q11x1 + ... + Q1nxn Qx⃗ = ⃗y = ⋮ Qn1x1 + ... + Qnn xn Then ∇x = ∂/∂x1, ...,∂/∂xn and = + + = + + ∂xi ∂y1/∂xi∂y1 ... ∂yn/∂xi∂yn Q1i ∂y1 ... Qni ∂yn = (i-th row of Q)⋅∇y i.e., ∇x = Q ∇y and ∇x = Q ∇y = ∇y Q 3 2 Then ∇x = ∇x ⋅ ∇x = ∇y QQ ∇y = ∇y ∇y, for QQ = I. A transformation Q with this property, QQ = I, is said to be an orthogonal transformation. Such transformations include rotations and reflections. Problem 6 Suppose u and v are both harmonic on R3. Show that, in general, the product of u times v is not harmonic. Give one or more examples of a special case where the product does turn out to be harmonic. 3. Integral Identities Let U denote a bounded, open, connected set in Rn having a smooth boundary, ∂U. This is ⃗ sufficient in order for the divergence theorem to be valid on U. That is, if F⃗x denotes a smooth vector field over U, (i.e., F ∈ CŪ ∩ C1U and if ⃗nx denotes the outward unit normal to ∂U at x ∈ ∂U, then the divergence theorem asserts that ⃗ ⃗ ∫ divF dx = ∫ F ⋅ ⃗ndS x 3.1 U ∂U ⃗ Consider the integral identity (3.1) in the special case that Fx = ∇ux for u ∈ C1Ū ∩ C2U. Then ⃗ divFx = div ∇ux = ∇2ux ⃗ ⃗ ⃗ and F ⋅ n = ∇u ⋅ n = ∂N ux the normal derivative of u) Then (3.1) becomes 2 = ∫ ∇ uxdx ∫ ∂N uxdS x 3.2 U ∂U The identity (3.2) is known as Green ’s first identity . If functions u and v both belong to ⃗ C1Ū ∩ C2U and if Fx = vx∇ux, then ⃗ divFx = div vx∇ux = vx∇2ux + ∇u ⋅ ∇v ⃗ ⃗ ⃗ and F ⋅ n = vx∇u ⋅ n = v∂N ux ⃗ and, with this choice for F, (3.1) becomes Green ’s second identity , 2 + = ∫ vx∇ ux ∇u ⋅ ∇v dx ∫ vx∂N uxdS x 3.3 U ∂U Finally, writing (3.3) with u and v reversed, and subtracting the result from (3.3), we obtain Green ’s symmetric identity , 2 2 = ∫ vx∇ ux − ux∇ vx dx ∫ vx∂N ux − ux∂N vx dS x 3.4 U ∂U Problem 7 Let u = ux,y,z be a smooth function on R3 and let A denote a 3 by 3 matrix ⃗ whose entries are all smooth functions on R3 Let F = A∇u. If U denotes a bounded open set in R3 having smooth boundary ∂U, then find a surface integral over the boundary whose ⃗ value equals the integral of the divergence of F over U.