Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 6 pps

Solution 37.19 uxx + uyy = f (x, y), < x < a, < y < b, u(0, y) = u(a, y) = 0, uy (x, 0) = uy (x, b) = We will solve this problem with an eigenfunction expansion in x To determine a suitable set of eigenfunctions, we substitute the separation of variables u(x, y) = X(x)Y (y) into the homogeneous partial differential equation uxx + uyy = (XY )xx + (XY )yy = X Y =− = −λ2 X Y With the boundary conditions at x = 0, a, we have the regular Sturm-Liouville problem, X = −λ2 X, X(0) = X(a) = 0, which has the solutions, nπ nπx , Xn = sin , a a We expand u(x, y) in a series of the eigenfunctions λn = ∞ u(x, y) = un (y) sin n=1 n ∈ Z+ nπx a We substitute this series into the partial differential equation and boundary conditions at y = 0, b ∞ − n=1 nπ a un (y) sin ∞ un (0) sin n=1 nπx nπx + un (y) sin a a nπx = a ∞ un (b) sin n=1 1774 = f (x) nπx =0 a We expand f (x, y) in a Fourier sine series ∞ fn (y) sin nπx a f (x, y) sin nπx a f (x, y) = n=1 fn (y) = a a dx We obtain the ordinary differential equations for the coefficients in the expansion nπ un (y) = fn (y), un (0) = un (b) = 0, a We will solve these ordinary differential equations with Green functions Consider the Green function problem, un (y) − gn (y; η) − nπ a n ∈ Z+ gn (y; η) = δ(y − η), gn (0; η) = gn (b; η) = The homogeneous solutions nπy nπ(y − b) and cosh a a satisfy the left and right boundary conditions, respectively We compute the Wronskian of these two solutions cosh cosh(nπy/a) cosh(nπ(y − b)/a) nπ sinh(nπy/a) a sinh(nπ(y − b)/a) nπy nπy nπ nπ(y − b) = cosh − sinh sinh cosh a a a a nπ nπb =− sinh a a W (y) = nπ a The Green function is gn (y; η) = − a cosh(nπy< /a) cosh(nπ(y> − b)/a) nπ sinh(nπb/a) 1775 nπ(y − b) a The solutions for the coefficients in the expansion are b un (y) = gn (y; η)fn (η) dη Solution 37.20 utt + a2 uxxxx = 0, < x < L, t > 0, u(x, 0) = f (x), ut (x, 0) = g(x), u(0, t) = uxx (0, t) = 0, u(L, t) = uxx (L, t) = 0, We will solve this problem by expanding the solution in a series of eigen-solutions that satisfy the partial differential equation and the homogeneous boundary conditions We will use the initial conditions to determine the coefficients in the expansion We substitute the separation of variables, u(x, t) = X(x)T (t) into the partial differential equation (XT )tt + a2 (XT )xxxx = X T =− = −λ4 2T a X Here we make the assumption that ≤ arg(λ) < π/2, i.e., λ lies in the first quadrant of the complex plane Note that λ4 covers the entire complex plane We have the ordinary differential equation, T = −a2 λ4 T, and with the boundary conditions at x = 0, L, the eigenvalue problem, X = λ4 X, X(0) = X (0) = X(L) = X (L) = For λ = 0, the general solution of the differential equation is X = c1 + c2 x + c3 x2 + c4 x3 1776 Only the trivial solution satisfies the boundary conditions λ = is not an eigenvalue For λ = 0, a set of linearly independent solutions is {eλx , eıλx , e−λx , e−ıλx } Another linearly independent set, (which will be more useful for this problem), is {cos(λx), sin(λx), cosh(λx), sinh(λx)} Both sin(λx) and sinh(λx) satisfy the left boundary conditions Consider the linear combination c1 cos(λx)+c2 cosh(λx) The left boundary conditions impose the two constraints c1 + c2 = 0, c1 − c2 = Only the trivial linear combination of cos(λx) and cosh(λx) can satisfy the left boundary condition Thus the solution has the form, X = c1 sin(λx) + c2 sinh(λx) The right boundary conditions impose the constraints, c1 sin(λL) + c2 sinh(λL) = 0, −c1 λ2 sin(λL) + c2 λ2 sinh(λL) = c1 sin(λL) + c2 sinh(λL) = 0, −c1 sin(λL) + c2 sinh(λL) = This set of equations has a nontrivial solution if and only if the determinant is zero, sin(λL) sinh(λL) = sin(λL) sinh(λL) = − sin(λL) sinh(λL) Since sinh(z) is nonzero in ≤ arg(z) < π/2, z = 0, and sin(z) has the zeros z = nπ, n ∈ N in this domain, the eigenvalues and eigenfunctions are, λn = nπ , L Xn = sin 1777 nπx , L n ∈ N The differential equation for T becomes, T = −a2 nπ L T, which has the solutions, nπ nπ t , sin a L L The eigen-solutions of the partial differential equation are, cos a u(1) = sin n nπx nπ cos a L L t nπx nπ sin a L L u(2) = sin n t , 2 t , n ∈ N We expand the solution of the partial differential equation in a series of the eigen-solutions ∞ sin u(x, t) = n=1 nπx L cn cos a nπ L t + dn sin a nπ L t The initial condition for u(x, t) and ut (x, t) allow us to determine the coefficients in the expansion ∞ u(x, 0) = nπx = f (x) L cn sin n=1 ∞ ut (x, 0) = nπ L dn a n=1 sin nπx = g(x) L cn and dn are coefficients in Fourier sine series cn = dn = L L f (x) sin 2L aπ n2 nπx L L g(x) sin 1778 nπx L dx dx Solution 37.21 ut = κuxx + I αu, < x < L, t > 0, u(0, t) = u(L, t) = 0, u(x, 0) = g(x) We will solve this problem with an expansion in eigen-solutions of the partial differential equation We substitute the separation of variables u(x, t) = X(x)T (t) into the partial differential equation (XT )t = κ(XT )xx + I αXT I 2α X T − = = −λ2 κT κ X Now we have an ordinary differential equation for T and a Sturm-Liouville eigenvalue problem for X (Note that we have followed the rule of thumb that the problem will be easier if we move all the parameters out of the eigenvalue problem.) T = − κλ2 − I α T X = −λ2 X, X(0) = X(L) = The eigenvalues and eigenfunctions for X are λn = nπ , L nπx , L Xn = sin n ∈ N The differential equation for T becomes, Tn = − κ nπ L − I α Tn , which has the solution, Tn = c exp − κ nπ L 1779 − I 2α t From this solution, we see that the critical current is κπ αL ICR = If I is greater that this, then the eigen-solution for n = will be exponentially growing This would make the whole solution exponentially growing For I < ICR , each of the Tn is exponentially decaying The eigen-solutions of the partial differential equation are, un = exp − κ nπ L − I α t sin nπx , L n ∈ N We expand u(x, t) in its eigen-solutions, un ∞ an exp − κ u(x, t) = n=1 nπ L − I α t sin nπx L We determine the coefficients an from the initial condition ∞ u(x, 0) = an sin n=1 an = L L g(x) sin nπx = g(x) L nπx L dx If α < 0, then the solution is exponentially decaying regardless of current Thus there is no critical current Solution 37.22 1780 a) The problem is ut (x, y, z, t) = κ∆u(x, y, z, t), −∞ < x < ∞, −∞ < y < ∞, < z < a, u(x, y, z, 0) = T, u(x, y, 0, t) = u(x, y, a, t) = t > 0, Because of symmetry, the partial differential equation in four variables is reduced to a problem in two variables, ut (z, t) = κuzz (z, t), < z < a, t > 0, u(z, 0) = T, u(0, t) = u(a, t) = We will solve this problem with an expansion in eigen-solutions of the partial differential equation that satisfy the homogeneous boundary conditions We substitute the separation of variables u(z, t) = Z(z)T (t) into the partial differential equation ZT = κZ T T Z = = −λ2 κT Z With the boundary conditions at z = 0, a we have the Sturm-Liouville eigenvalue problem, Z = −λ2 Z, which has the solutions, λn = nπ , a Z(0) = Z(a) = 0, Zn = sin nπz , a n ∈ N The problem for T becomes, Tn = −κ nπ a Tn , with the solution, Tn = exp −κ 1781 nπ a t The eigen-solutions are un (z, t) = sin nπz nπ exp −κ a a t The solution for u is a linear combination of the eigen-solutions The slowest decaying eigen-solution is u1 (z, t) = sin πz π exp −κ a a t Thus the e-folding time is ∆e = a2 κπ b) The problem is ut (r, θ, z, t) = κ∆u(r, θ, z, t), < r < a, < θ < 2π, −∞ < z < ∞, u(r, θ, z, 0) = T, u(0, θ, z, t) is bounded, u(a, θ, z, t) = t > 0, The Laplacian in cylindrical coordinates is 1 ∆u = urr + ur + uθθ + uzz r r Because of symmetry, the solution does not depend on θ or z ut (r, t) = κ urr (r, t) + ur (r, t) , < r < a, t > 0, r u(r, 0) = T, u(0, t) is bounded, u(a, t) = We will solve this problem with an expansion in eigen-solutions of the partial differential equation that satisfy the homogeneous boundary conditions at r = and r = a We substitute the separation of variables u(r, t) = 1782 For < t < δ we have L = L L nπx dx L L π(x − ξ) πt nπx v cos dx sin sin 2d δ L 8dLv nπd nπξ πt = cos sin sin − 4d2 n2 ) π(L L L δ sn (t) = s(x, t) sin For t > δ, sn (t) = Substituting this into the partial differential equation yields, nπc un + L un = 8dLv π(L2 −4d2 n2 ) cos nπd L sin nπξ L sin πt δ , for t < δ, for t > δ Since the initial position and velocity of the string is zero, we have un (0) = un (0) = First we solve the differential equation on the range < t < δ The homogeneous solutions are cos nπct L , nπct L sin Since the right side of the ordinary differential equation is a constant times sin(πt/δ), which is an eigenfunction of the differential operator, we can guess the form of a particular solution, pn (t) pn (t) = d sin πt δ We substitute this into the ordinary differential equation to determine the multiplicative constant d pn (t) = − 8dδ L3 v cos π (L2 − c2 δ n2 )(L2 − 4d2 n2 ) 1798 nπd L sin nπξ L sin πt δ The general solution for un (t) is un (t) = a cos nπct L + b sin 8dδ L3 v cos − π (L − c2 δ n2 )(L2 − 4d2 n2 ) nπct L nπd L sin nπξ L sin πt δ We use the initial conditions to determine the constants a and b The solution for < t < δ is un (t) = 8dδ L3 v cos π (L2 − c2 δ n2 )(L2 − 4d2 n2 ) nπd L sin nπξ L L sin δcn nπct L − sin πt δ The solution for t > δ, the solution is a linear combination of the homogeneous solutions This linear combination is determined by the position and velocity at t = δ We use the above solution to determine these quantities nπd nπξ nπcδ 8dδ L4 v cos sin sin δcn(L2 − c2 δ n2 )(L2 − 4d2 n2 ) π L L L 8dδ L v nπd nπξ nπcδ un (δ) = cos sin + cos π δ(L2 − c2 δ n2 )(L2 − 4d2 n2 ) L L L un (δ) = The fundamental set of solutions at t = δ is cos nπc(t − δ) L , L sin nπc nπc(t − δ) L From the initial conditions at t = δ, we see that the solution for t > δ is 8dδ L3 v nπd nπξ cos sin (L2 − c2 δ n2 )(L2 − 4d2 n2 ) π L L L nπcδ nπc(t − δ) π nπcδ sin cos + + cos sin δcn L L δ L un (t) = 1799 nπc(t − δ) L Width of the Hammer The nth harmonic has the width dependent factor, d cos L2 − 4d2 n2 nπd L Differentiating this expression and trying to find zeros to determine extrema would give us an equation with both algebraic and transcendental terms Thus we don’t attempt to find the maxima exactly We know that d < L The cosine factor is large when nπd ≈ mπ, m = 1, 2, , n − 1, L mL d≈ , m = 1, 2, , n − n Substituting d = mL/n into the width dependent factor gives us L2 (1 d (−1)m − 4m2 ) Thus we see that the amplitude of the nth harmonic and hence its kinetic energy will be maximized for d≈ L n The cosine term in the width dependent factor vanishes when d= (2m − 1)L , 2n m = 1, 2, , n The kinetic energy of the nth harmonic is minimized for these widths L For the lower harmonics, n , the kinetic energy is proportional to d2 ; for the higher harmonics, n 2d kinetic energy is proportional to 1/d2 1800 L , 2d the Duration of the Blow The nth harmonic has the duration dependent factor, δ2 L2 − n c δ L sin ncδ nπcδ L cos nπc(t − δ) L + π δ + cos nπcδ L sin nπc(t − δ) L If we assume that δ is small, then L sin ncδ and π δ nπcδ L ≈ π + cos nπcδ L ≈ 2π δ Thus the duration dependent factor is about, L2 δ sin − n2 c2 δ nπc(t − δ) L L Thus for the lower harmonics, (those satisfying n ), the amplitude is proportional to δ, which means that the cδ L kinetic energy is proportional to δ For the higher harmonics, (those with n ), the amplitude is proportional to cδ 1/δ, which means that the kinetic energy is proportional to 1/δ Solution 37.29 Substituting u(x, y, z, t) = v(x, y, z) eıωt into the wave equation will give us a Helmholtz equation −ω v eıωt −c2 (vxx + vyy + vzz ) eıωt = vxx + vyy + vzz + k v = We find the propagating modes with separation of variables We substitute v = X(x)Y (y)Z(z) into the Helmholtz equation X Y Z + XY Z + XY Z + k XY Z = Y Z X = + + k2 = ν − X Y Z 1801 The eigenvalue problem in x is X = −ν X, which has the solutions, νn = nπ , L X(0) = X(L) = 0, Xn = sin nπx L We continue with the separation of variables − Z Y nπ = + k2 − Y Z L = µ2 The eigenvalue problem in y is Y = −µ2 Y, Y (0) = Y (L) = 0, which has the solutions, mπ , L Now we have an ordinary differential equation for Z, µn = Z + k2 − Ym = sin π L mπy L n2 + m2 Z = We define the eigenvalues, π L λ2 = k − n,m If k − π L n2 + m2 (n2 + m2 ) < 0, then the solutions for Z are, exp ± π L (n2 + m2 ) − k z We discard this case, as the solutions are not bounded as z → ∞ 1802 If k − π L (n2 + m2 ) = 0, then the solutions for Z are, {1, z} The solution Z = satisfies the boundedness and nonzero condition at infinity This corresponds to a standing wave π If k − L (n2 + m2 ) > 0, then the solutions for Z are, e±ıλn,m z These satisfy the boundedness and nonzero conditions at infinity For values of n, m satisfying k − there are the propagating modes, un,m = sin π L (n2 + m2 ) ≥ 0, mπy ı(ωt±λn,m z) nπx e sin L L Solution 37.30 utt = c2 ∆u, < x < a, < y < b, u(0, y) = u(a, y) = u(x, 0) = u(x, b) = We substitute the separation of variables u(x, y, t) = X(x)Y (y)T (t) into Equation 37.12 T X Y = + = −ν 2T c X Y X Y =− − ν = −µ X Y This gives us differential equations for X(x), Y (y) and T (t) X = −µX, X(0) = X(a) = Y = −(ν − µ)Y, Y (0) = Y (b) = T = −c2 νT 1803 (37.12) First we solve the problem for X µm = mπ a , Xm = sin mπx a Then we solve the problem for Y νm,n = mπ a + nπ b nπy b , Ym,n = sin m a Finally we determine T Tm,n = cos cπ sin + n b t The modes of oscillation are um,n = sin nπy cos mπx sin cπ sin a b m a + n b t The frequencies are m n + a b Figure 37.4 shows a few of the modes of oscillation in surface and density plots ωm,n = cπ Solution 37.31 We substitute the separation of variables φ = X(x)Y (y)T (t) into the differential equation φt = a2 (φxx + φyy ) XY T = a2 (X Y T + XY T ) T X Y = + = −ν 2T a X Y T X Y = −ν, = −ν − = −µ 2T a X Y 1804 (37.13) First we solve the eigenvalue problem for X X + µX = 0, µm = X(0) = X(lx ) = mπ lx , Xm (x) = sin mπx lx m ∈ Z+ , Then we solve the eigenvalue problem for Y Y + (ν − µm )Y = 0, nπ ly νmn = µm + Y (0) = Y (ly ) = , nπy ly Ymn (y) = cos , n ∈ Z0+ Next we solve the differential equation for T , (up to a multiplicative constant) T = −a2 νmn T T (t) = exp −a2 νmn t The eigensolutions of Equation 37.13 are sin(µm x) cos(νmn y) exp −a2 νmn t , m ∈ Z+ , n ∈ Z0+ We choose the eigensolutions φmn to be orthonormal on the xy domain at t = φm0 (x, y, t) = φmn (x, y, t) = sin(µm x) exp −a2 νmn t , lx ly sin(µm x) cos(νmn y) exp −a2 νmn t , lx ly The solution of Equation 37.13 is a linear combination of the eigensolutions ∞ φ(x, y, t) = cmn φmn (x, y, t) m=1 n=0 1805 m ∈ Z+ m ∈ Z+ , n ∈ Z+ We determine the coefficients from the initial condition φ(x, y, 0) = ∞ cmn φmn (x, y, 0) = m=1 n=0 ly lx φmn (x, y, 0) dy dx cmn = ly lx lx ly cm0 = sin(µm x) dy dx 0 − (−1)m 2lx ly , mπ cm0 = cmn = lx ly lx m ∈ Z+ ly sin(µm x) cos(νmn y) dy dx 0 cmn = 0, m ∈ Z+ , n ∈ Z+ ∞ φ(x, y, t) = cm0 φm0 (x, y, t) m=1 ∞ φ(x, y, t) = m=1 odd m 2lx ly sin(µm x) exp −a2 νmn t mπ Addendum Note that an equivalent problem to the one specified is φt = a2 (φxx + φyy ) , < x < lx , −∞ < y < ∞, φ(x, y, 0) = 1, φ(0, y, t) = φ(ly , y, t) = Here we have done an even periodic continuation of the problem in the y variable Thus the boundary conditions φy (x, 0, t) = φy (x, ly , t) = 1806 are automatically satisfied Note that this problem does not depend on y Thus we only had to solve φt = a2 φxx , < x < lx φ(x, 0) = 1, φ(0, t) = φ(ly , t) = Solution 37.32 Since the initial and boundary conditions not depend on θ, neither does φ We apply the separation of variables φ = u(r)T (t) φt = a2 ∆φ φt = a2 (rφr )r r T = (ru ) = −λ a2 T r (37.14) (37.15) (37.16) We solve the eigenvalue problem for u(r) (ru ) + λu = 0, u(0) bounded, u(R) = First we write the general solution √ √ u(r) = c1 J0 λr + c2 Y0 λr The Bessel function of the second kind, Y0 , is not bounded at r = 0, so c2 = We use the boundary condition at r = R to determine the eigenvalues λn = j0,n R , un (r) = cJ0 1807 j0,n r R We choose the constant c so that the eigenfunctions are orthonormal with respect to the weighting function r j0,n r R J0 un (r) = R rJ0 j0,n r R √ = J0 RJ1 (j0,n ) j0,n r R Now we solve the differential equation for T T = −a2 λn T Tn = exp − aj0,n R2 t The eigensolutions of Equation 37.14 are √ φn (r, t) = J0 RJ1 (j0,n ) j0,n r R aj0,n R2 exp − exp − aj0,n R2 t The solution is a linear combination of the eigensolutions ∞ √ φ= cn J0 RJ1 (j0,n ) n=1 j0,n r R 1808 t We determine the coefficients from the initial condition φ(r, θ, 0) = V √ j0,n r cn J0 =V RJ1 (j0,n ) R n=1 √ R j0,n r Vr cn = J0 dr RJ1 (j0,n ) R √ R J1 (j0,n ) cn = V RJ1 (j0,n ) j0,n /R √ VR cn = j0,n ∞ ∞ φ(r, θ, t) = 2V n=1 J0 j0,n r R j0,n J1 (j0,n ) aj0,n R2 exp − t Jν (r) ∼ πν π cos r − − , πr ν jν,n ∼ n + − π r → +∞ For large n, the terms in the series solution at t = are J0 j0,n r R j0,n J1 (j0,n ) 2R πj0,n r ∼ j0,n cos πj0,n j0,n r R − cos j0,n − π 3π cos (n−1/4)πr − π R R ∼ r(n − 1/4)π cos ((n − 1)π) 1809 The coefficients decay as 1/n Solution 37.33 We substitute the separation of variables Ψ = T (t)Θ(θ)Φ(φ) into Equation 37.7 a2 R2 1 ∂ (sin θ T Θ Φ) + T ΘΦ sin θ ∂θ sin2 θ Φ R2 T = (sin θ Θ ) + = −µ 2T a sin θ Θ sin2 θ Φ Φ sin θ (sin θ Θ ) + µ sin2 θ = − =ν Θ Φ T ΘΦ = We have differential equations for each of T , Θ and Φ T = −µ a2 T, R2 ν (sin θ Θ ) + µ − Θ = 0, sin θ sin2 θ Φ + νΦ = In order that the solution be continuously differentiable, we need the periodic boundary conditions Φ(0) = Φ(2π), Φ (0) = Φ (2π) The eigenvalues and eigenfunctions for Φ are νn = n2 , Φn = √ eınφ , 2π n ∈ Z Now we deal with the equation for Θ x = cos θ, Θ(θ) = P (x), sin2 θ = − x2 , d d = dx sin θ dθ 1 ν (sin2 θ Θ) + µ− Θ=0 sin θ sin θ sin2 θ n2 − x2 P + µ − P =0 − x2 P (x) should be bounded at the endpoints, x = −1 and x = 1810 √ If the solution does not depend on θ, then the only one of the Φn that will appear in the solution is Φ0 = 1/ 2π The equations for T and P become − x2 P + µP = 0, P (±1) bounded, a2 T = −µ T R The solutions for P are the Legendre polynomials µl = l(l + 1), Pl (cos θ), l ∈ Z0+ We solve the differential equation for T a2 T R2 a2 l(l + 1) Tl = exp − t R2 T = −l(l + 1) The eigensolutions of the partial differential equation are Ψl = Pl (cos θ) exp − a2 l(l + 1) t R2 The solution is a linear combination of the eigensolutions ∞ Al Pl (cos θ) exp − Ψ= l=0 1811 a2 l(l + 1) t R2 We determine the coefficients in the expansion from the initial condition Ψ(θ, 0) = cos2 θ − ∞ Al Pl (cos θ) = cos2 θ − l=0 cos2 θ − + · · · = cos2 θ − 2 A0 = − , A1 = 0, A2 = , A3 = A4 = · · · = 3 6a2 Ψ(θ, t) = − P0 (cos θ) + P2 (cos θ) exp − t 3 R A0 + A1 cos θ + A2 Ψ(θ, t) = − + cos2 θ − 3 exp − 6a2 t R2 Solution 37.34 Since we have homogeneous boundary conditions at x = and x = 1, we will expand the solution in a series of eigenfunctions in x We determine a suitable set of eigenfunctions with the separation of variables, φ = X(x)Y (y) φxx + φyy = X Y =− = −λ X Y (37.17) We have differential equations for X and Y X + λX = 0, X(0) = X(1) = Y − λY = 0, Y (0) = The eigenvalues and orthonormal eigenfunctions for X are λn = (nπ)2 , Xn (x) = √ sin(nπx), 1812 n ∈ Z+ ... solution for R(r) is Rn = j0 (λr) Applying the boundary condition at r = a, we see that the eigenvalues and eigenfunctions are λn = γn , a Rn = j0 γn r , a The problem for T becomes Tn = −κ 17 85 γn... dx for d = nπξ L for d = L , 2n L 2n The solution for u(x, t) is, 8dL2 v u(x, t) = π2c u(x, t) = v π2c ∞ n=1 ∞ n=1 n2 n(L2 cos − 4d2 n2 ) 2nπd + L sin 2nπd L sin nπξ L sin nπx sin L nπct L for. .. partial differential equation, we will expand the solution in a series of eigenfunctions in x for which the coefficients are functions of t The solution for u has the form, ∞ u(x, t) = un (t) sin n=1 nπx