Lecture 1, 9/8/2010 The Laplace operator in Rn is 2 2 @ ··· @ ∆ = 2 + + 2 (1) @x1 @xn It was introduced by P.-S.Laplace in a 1784 paper in connection with gravita- tional potentials. 3 Recall that if we have masses M1;:::;Ml at points a1; : : : ; an of R , then the gravitational potential of the configuration is given by M M u(x) = −{ 1 · · · − { l ; (2) jx − a1j jx − alj where { is the gravitational constant. The Newton law of gravity can be reformulated in terms of the potential u: The force on a point particle of mass m located at x is force = −mru(x):1 (3) The value of the potential u(x) at a point x 2 R3 can be interpreted as the potential energy due to gravity of a particle of unit mass located at x: if we would like to move this particle to the spatial infinity, we need to act against the gravitational force, and the total work we do while moving the particle to the spatial infinity is exactly −u(x).2 Assume now that instead of point masses we are dealing with a continuous distribution of mass descrivedR by a density ρ(x): the mass an an arbitrary O ⊂ R3 domain is given by O ρ(x) dx. Often we think of ρ being non-zero only in some domain Ω (representing a planet or a star, for example). The gravitational potential due to the mass described by the density ρ is Z ρ(y) u(x) = −{ dy (4) R3 jx − yj If we wish to emphasize that ρ is only non-zero in a domain Ω, we can write Z −{ ρ(y) u(x) = j − j dy : (5) Ω x y Now Laplace's observation is that at a point x in a neighborhood of which we have no masses (i. e. ρ vanishes close to x) the potential u satisfies ∆u = 0 : (6) This can be easily seen from the following two facts: 1 1. ∆ ( jx−aj ) = 0 away from x = a, and 1Excercise: Check by direct calculation that (3) indeed gives Newton's law. 2Excercise: Verify this statement 1 2. equation (6) is linear, i. e. a linear combination of solutions is again a solu- tion.3 One could perhaps think that since we have the explicit representation (5), we can answer any questions about the gravitational field by calculating the corresponding integrals. So what is the point of bringing in equation (6)? Let us illustrate the usefulness of equation (6) on a very simple example. Con- sider the situation when the total mass M is uniformly distributed over a solid 3 sphere BR of radius R > 0. So we have ρ = M=(4πR ) in BR and ρ = 0 outside BR. What is the gravitational potential (5) outside Ω = BR? First, try to calculate the potential by direct integration. It may look difficult, but with some effort one can still evaluate the integral explicitly4. If we use equation (6), we can avoid the explicit calculation of the integrals, and still get the exact u. We first note that the potential u is rotationally symmetric: u(x) depends only on r = jxj. 5 With some abuse of notation we will write u(x) = u(r). For such symmetric functions u in R3 we have 2u0 ∆u = u00 + ; (7) r where u0 denotes the derivative of u with respect to r .6 As a simple exercise in solving ordinary differential equations, we can find the general solution of 2u0 u00 + = 0 : (8) r It is given by A u(r) = + B: (9) r We now determine the value of the constants A and B by looking at the potential for large x. We have Z ρ(y) u(x) = −{ dy jx − yj ZBR ρ(y) = −{ dy jxj BR Z − { 1 1 + ρ(y)(j − j + j j) dy BR x y x M 1 = −{ + O( ) as x ! 1 ; jxj jxj2 3Excercise: Verify all these statements 4Excercise: Do this calculation 5Excercise: Verify that this is indeed the case 6Excercise: Verify the formula 2 where we have used the usual \O-notation", which has the following meaning: O(1=jxj2) denotes any function whose absolute value for large x is below C=jxj2, for some (fixed) C > 0. Combining the last equation with (9), we see that we must have A = −{M; B = 0 : We see that M u(x) = −{ for jxj > R . jxj We have shown the classical result (probably going back to Newton) that the gravitational potential outside the sphere is exactly the same as the potential of a point change at the origin, with the same mass as the total mass of the sphere. Let us replace in the above example the solid sphere be a spherical shell fR1 < jxj < R2g, with the total mass M distributed uniformly over the shell. The same −{ M argument as above shows that outside the shell we again have u(x) = jxj . As an exercise, you can show that inside the shell (i. e. for jxj < R1) the potential u is constant, i. e. u(x) = B for some constant B. This means that the gravitational force inside the shell vanishes. All this can be done again by calculating directly the integral (5), but the above argument is computationally much simpler. 7 In more complicated situations, e. g. in electrostatics of conducting surfaces of general shape, the direct calculation of the integrals in many cases is no longer feasible, whereas arguments using the equation ∆u = 0 still work very well8 although we have to study the properties of general solutions (not necessarily symmetric) in more detail. This will be our program in the next few lectures. 7Excercise: Do the direct calculation of the integral in this case 8In fact, during the 19th century there has been an important shift in thinking about the basic laws of physics. One can probably say that in the times of Newton, the law of force exemplified by explicit representation formulae, such as (2) and (3), were considered as fundamental. However, the importance of partial differential equations satisfied by the “fields of force" gradually grew, and eventually the equations themselves became viewed as fundamental. The theory of the electromagnetic field by Faraday and Maxwell (completed in 1865) played an important role in this transformation. 3 Lecture 2, 9/10/2010 Last time we looked at the gravitational potential Z ρ(y) u(x) = −{ dy (10) R3 jx − yj in the regions where ρ vanishes, and we saw that in these regions we have ∆u = 0. What happens in the regions where ρ does not vanish? Around 1812, S. D. Poisson discovered the following fundamental fact: If ρ is sufficiently regular9, then ∆u = 4π{ρ . This is why the equation ∆u = f is usually called the Poisson equation. Before going the the proof, we will adjust our notation and recall some definitions. We set 1 G(x) = − (11) 4πjxj and for a sufficiently regular function f (say, continuous, compactly supported) we set Z Gf(x) = G(x − y)f(y) dy ; (12) R3 which is one of the standard forms of the gravitational potential used in PDE textbooks10. We can think of G as a linear operator taking functions to func- tions. Recalling that the convolution of two functions f; g on Rn is defined11 by Z Z f ∗ g(x) = f(x − y)g(y) dy = g(x − y)f(y) dy ; (13) Rn Rn we can also write Gf = G ∗ f : (14) The above notation is somewhat \heavy". You will soon see that it makes sense to use G for both the kernel (11) and the linear operator G. If we write Gf = G∗f instead of G = G∗f, it may look quite ambiguous at first, as Gf can in principle mean both the pointwise product of the kernel G with the function f and the result of applying the operator G to the function f. However, in practice such confusion does not arise, as the intended meaning is usually clear from the context. With the above notation, we can now write the Poisson's result as follows: ∆(Gf) = f : (15) We will see that the proof also gives12 G(∆u) = u : (16) 9We will specify what this exactly means later. 10The other standard form is the one one gets by changing the sign of the kernel. 11under some assumptions on f; g, e. g. f bounded measurable and compactly supported and g locally integrable 12under appropriate assumptions 4 Therefore we can write, somewhat loosely, G = (∆)−1 and ∆ = (G)−1 , (17) i. e. the integral operator G is the inverse of the differential operator ∆, and vice versa. This illustrates an important general point that, roughly speaking, the inverses of differential operators are closely related to integral operators.13 The oldest example of this is of course the Fundamental Theorem of Calculus: the inverse of taking derivatives is integration.14 We recall the following rule for differentiating convolutions: @ @ @ (f ∗ g) = ( f) ∗ g = f ∗ ( g) : (18) @xi @xi @xi Of course, one needs some assumptions. For example, if f is locally integrable and g is continuously differentiable and compactly supported, you can check as an exercise that with the \classical interpretation" of all the expressions the part @ @ (f ∗ g) = f ∗ ( g) ; (19) @xi @xi is fine, while the expression ( @ f) ∗ g needs some interpretation, as the term @xi @ f may not be well-defined by the classical point-wise definitions. We will @xi deal with issues such as this later, for now we will only use the expressions which are defined classically.