2. THE HEAT EQUATION (Joseph FOURIER ( ) in 1807; Théorie analytique de la chaleur, 1822).

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1 mpc2w4.tex Week THE HEAT EQUATION (Joseph FOURIER ( ) in 1807; Théorie nlytique de l chleur, 1822). One dimension. Consider uniform br (of some mteril, sy metl, tht conducts het), of cross-sectionl re S, with sides insulted so tht het flows only prllel to the x-xis. The rte of flow of het cross surfce is K u/ x, where: K is constnt, the therml diffusivity of the mteril, u = u(x, t) is the temperture, n is the outwrd norml to the surfce (minus s het flows form hotter to colder). Consider the slb x b. At the right-hnd end x = b, the outwrd norml is in the direction of incresing x, so u/ n = + u/ x, while t the left-hnd end x =, u/ n = u/ x. So the rte of het flow into the s lb minus the rte of het flow out is This cn be written s But the het content is dq/dt = K[ u(b, t) u(, t)]s. x x K Q = 2 u(x, t)dx.s. x2 cρu(x, t)dx.s, where c is the specific het, ρ is the density. So dq/dt = cρ u(x, t)dx.s t (ssuming tht we cn tke the t-differentition inside the x-integrtion differentiting under the integrl sign ; this is justified under sufficient smoothness conditions, which we ssume here). Equting, (dq/dt =) K 2 x u(x, t)dx.s = 2 1 cρ u(x, t)dx.s. t

2 Cncel S. This holds for ll, b. So the integrnds must be equl: K 2 x u(x, t) = cρ u(x, t). 2 t Write k := K/cρ for the therml diffusivity. Then the het eqution. Subscript nottion: Higher dimensions: 2 x u(x, t) = cρ u(x, t)/k, 2 t u xx = u t /k. u xx + u yy = u t /k (2D); u xx + u yy + u zz = u t /k (3D). Seprtion of vribles. For u(x, t) = X(x)T (t), u xx = u t /k: X T = X T /k, sy: X /X = T /T k = const = C 2, T /T = kc 2, T = const.e kc2 t (so temperture decreses with time exponentil increse with time is unphysicl, hence the minus sign in C 2 ). As for the wve eqution, X /X = C 2 gives X = cos Cx, sin Cx. Suppose we impose the BCs u(0, t) = 0, u(l, t) = 0 (br of length l, with temperture fixed t both ends t 0 o C); similrly for other fixed tempertures (which we cn choose s 0 by ltering the origin of temperture), nd IC u(x, 0) = f(x), the initil temperture distribution (given by the initil het distribution). Then X = A cos Cx + B sin Cx nd X(0) = 0 gives A = 0. Then X = B sin CX nd X(l) = 0 gives Cl = nπ, C = nπ/l. So B n sin nπ/l e kn2 π 2 t/l 2

3 is solution. By linerity (= superposition), u = B n sin nπ/l e kn2 π 2 t/l is solution. ICs: B n sin nπ/l = f(x). This is Fourier series for f, from which we cn determine the constnts B n see Ch. V. Stedy stte. As t, the time-dependence dies wy. The het eqution then simplifies to u xx = 0, giving u(x) = A + Bx. If the BCs re (l = 1 sy) u(0) = u 0, u(1) = u 1, this gives A = u 0, B = u 1 u 0. So u = u P I = u 0 + (u 1 u 0 )x is prticulr integrl (PI). But it does not stisfy the IC u(x, 0) = f(x). To solve the full het eqution u xx = u t /k, with BCs u(0, t) = u 0, u(1, t) = u 1 nd IC u(x, 0) = f(x), tke the complementry function nd dd it to u P I : u = u CF = B n sin nπ/l e kn2 π 2 t/l u = u P I + u CF = u 0 + (u 1 u 0 )x + B n sin nπ/l e kn2 π 2t/l, with B n the Fourier coefficients of f (Ch. V). 3. LAPLACE S EQUATION (P. S. LAPLACE ( ), Mécnique céleste ( , Vols 1-5); S. D. POISSON ( ) in 1813). The stedy-stte solution of the het eqution stisfies u xx = 0, u xx + u yy = 0, u xx + u yy + u zz = 0 in one, two nd three dimensions. For resons which will emerge lter (Ch. VI), this eqution rises in other 3

4 physicl contexts. It is clled Lplce s eqution. It rises in: Electromgnetism (EM) nd grvittion (celestil mechnics, strophysics). The potentil (potentil function) stisfies Lplce s eqution if there re no sources of mss or chrge present, Poisson s eqution u xx + u yy + u zz = 4πρ if mss or chrge density ρ is present. Seprtion of vribles. In two dimensions, u xx = u yy. Tke u(x, y) = X(x)Y (y): X Y = XY, X /X = Y /Y = k 2, sy, the seprtion constnt (w.l.o.g., k > 0). X(x) = e ±kx, Y (y) = sin ky, cos ky, = e ±kx cos / sin ky. As before, we cn superpose solutions, nd use BCs or ICs to determine constnts. Agin s before, for unbounded regions, exponentilly growing potentils re unphysicl: the potentil generted by chrge or mss decreses to zero t infinity. Lplcin. 2, or, := 2 x y z 2 = D xx + D yy + D zz or D 11 + D 22 + D 33 is clled the Lplcin opertor, del ( ) or nbl squred ( 2 ); the Lplcin of u is 2 u, or u, := 2 u x u y u z 2 = u xx + u yy + u zz or u 11 + u 22 + u 33. Other coordinte systems. As well s Crtesin coordintes (x, y, z), other possible coordinte systems include: Plne polrs (r, θ); cylindricl polrs (r, θ, z) plne polrs + z in the third dimension; Sphericl polrs: (r, θ, ϕ): r = distnce from the origin; θ = longitude; ϕ = coltitude (think of θ, ϕ s ngle vribles on the erth s surfce nd r s the rdius of the erth). Select the coordinte system to reflect ny SYMMETRY present 4

5 in the problem. Clssifiction. Recll tht the generl lgebric eqution of the second degree in two vribles is x 2 + bxy + cy 2 + dx + ey + f = 0. By completing the squre nd chnging vribles, one cn reduce to one of x 2 + by 2 = c (two second-order terms), x 2 = by (one second-order term). In the first, we cn tke > 0 w.l.o.g.; then the sign of the other second=order term b is crucil. There re three stndrd forms: x 2 / 2 + y 2 /b 2 = 1 (ellipse both coefficients > 0); x 2 / 2 y 2 /b 2 = 1 (hyperbol one coefficient > 0, one < 0); y 2 = 4x (prbol only one second-order terms). These curves re (including limiting cses line-pir, line, point) the conic sections or conics intersections of (doubly infinite) cone with plne. By nlogy, we clssify liner 2nd-order PDEs similrly: 2 u xx + b 2 u yy + 1st-order liner differentil opertor = 0 elliptic, prototype Lplce s eqution; 2 u xx b 2 u yy + 1st-order liner differentil opertor = 0 hyperbolic, prototype the wve eqution; 2 u xx + 1st-order differentil opertor = 0 prbolic, prototype the het eqution. Lplce s eqution in (cylindricl) polrs. We quote: u = u rr + 1 r u r + 1 r 2 u θθ.(+u zz ) = 0. So with symmetry (xil symmetry if z is present): u rr + 1 r u r = 0. 5

6 Write v := u r : v r = dv/dr v/r; dv/v = dr/r; log v = log r + const; vr = c; du/dr = c/r; du = cdr/r; u = c log r + d. Lplce s eqution in sphericl polrs. We quote: u = 1 u r 2 (r2 r r ) + 1 r 2 sin θ θ (sin θ θ ) u r 2 sin 2 θ θ. 2 Sphericl symmetry: / θ = 0, / ϕ = 0. So u = 0 is 1 u r 2 (r2 r r ) = 0; u u (r2 ) = 0; r2 r r r = c; u/ r = c/r 2 ; u = c/r +. If u 0 s r (s is needed for the potentil to be physicl), = 0: u = u(r) = c/r. The force (electromgnetic or grvittionl) is the derivtive of the potentil (more generlly, the grdient of the potentil, in the lnguge of Vector Clculus see Ch. VI), i.e. Force = c/r 2. This expresses two fundmentl lws: 1. Newton s Lw of Grvity (Sir Isc NEWTON ( ); Principi, 1687): The force due to grvity between two msses m 1, m 2 distnce r prt is F = Cm 1 m 2 /r 2, where C is the grvittionl constnt (n bsolute constnt). 2. Coulomb s Inverse Squre Lw (C. A. COULOMB ( ), in 1785). Similrly for the electrosttic force. Note. Tht two of the four fundmentl forces of Nture electromgnetism, wek nucler force (governing rdioctivity), strong nucler force (holding the nucleus together or protons would repel ech other by electrosttic repulsion!) nd grvity re governed by the sme Inverse Squre Lw is quite remrkble. As you my know, the first three forces hve been unified, but not the first three with grvity. 6

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