Fig.1: Non-stationary temperature distribution in a circular plate.

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Virginia Nichols
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1 LECTURE Problem : Fi.1: Non-stationary temperature distribution in a circular plate. The non-stationary, radial symmetric temperature distribution ur, t in a circular plate of radius c is determined as solution to the boundary and initial value problem, see 13.- u k r + 1 u = u r r t uc, t =, t, ur, = fr, r, c, r, t, c, 1 A solution to the partial differential equation is searched for on the separated form ur, θ = RrT t k R r + 1 r R r T t = RrT t R r + 1 r R r Rr = T t kt t = λ r R r + rr r + λ r Rr = T t + λ k T t = 3 Assume that the separation constant is real and positive, i.e. λ >, in which case represents the parametric Bessel equation of order ν =. Then, the eneral solution to and 3 reads, cf

2 Rr = c 1 J λr + c Y λr 4 T t = c 3 e λ kt 5 The temperature field must be bounded at r =. Since, Y λr for r, a bounded temperature at the center of the plate requires c =. The boundary condition at r = c then requires that J λ c = 6 The solutions λ 1, λ,... of 6 represent the eienvalues of the problem. The several lowest eienvalues become, see Table 6.1 λ 1 c =.488 λ c = 5.51 λ 3 c = λ 4 c = Product solutions to the differential equation, which fulfill the boundary condition at r = c and the eometrical boundary condition ur, t < for r =, then has the form u i r, t = A i e λ i kt J λ i r 8 where A i = c 1 c 3. The superposition principle provides the followin solution for the temperature field ur, t = u i r, t = A i e λ i kt J λ i r 9 The coefficients A i are determined from the radial symmetrical initial condition ur, = fr = A i J λ i r 1 1 is a Fourier-Bessel series of the order n = of the function fr, cf The expansion coefficients c i = A i are evaluated from Case I, p. 456 for b = c and n =. The solution becomes, cf

3 A i = c r fr J λ i r dr c r J λ i r dr = c J 1 λ i c Then, the final solution follows from 9 and 11 c r fr J λ i r dr 11 ur, t = c c r fr J λ i r dr J1 λ e λ i kt J λ i r 1 i c Problem : a b Fi. 1: Oscillatin rope. Fi. 1 shows a vertical rope of the lenth L, which is performin plane vibrations in the horizontal direction. The rope is assumed to be perfectly flexible in bendin and infinite stiff aainst axial deformations. The mass per unit lenth is µ, and the acceleration of ravity is. A x, y- coordinate system is placed as shown on the fiure, so the free end and the support point of the rope have the abscissas x = and x = L, respectively. Only small vibrations ux, t of the rope in the y-direction are considered, i.e. ux, t L. Then, the vibrations may be analyzed by linear vibration theory. At first the partial differential equation of motion for the rope is determined. At a point of the rope with the abscissa x the axial force T x in the static referential state is equal to the weiht of the rope below this point, i.e. T x = µx 1 A differential rope element of lenth dx, placed at the abscissa x is cut free, and the axial rope forces are applied tanentially at the ends of the free rope element. Inorin possible axial load increments from centrifual accelerations the vertical components of the indicated axial rope forces 3

4 are unchaned equal to T x and T x + dx durin the vibration. Then, the horizontal component of the rope force at the lower end, actin in the neative y-direction becomes exactly T x ux, t, where ux, t specifies the slope of the displacement curve. The correspondin horizontal force at the upper end, actin in the positive y-direction, becomes T x + dx ux + dx, t. Application of Newton s second law of motion for the rope element in the y-direction then provides µdx ux, t ux + dx, t t = T x + dx ux, t T x + µ ux, t t = T x T x ux, t ux, t ux, t T x ux, t dx T x A first order Taylor expansion has been used for the first term on the riht hand side in the first line. Insertion of 1 in provides the followin partial differential equation for undamped eienvibrations of the rope ux, t t = x ux, t 3 3 is solved with the followin boundary and initial conditions ul, t =, t, ux, = fx, ux, t =, x, L 4 Physically, the initial conditions specifies undamped eienvibrations startin at rest with the initial deflection fx. The indicated boundary value problem will be solved by means of the separation method. Here, it should be noticed that the separation method is completely identical to modal analysis in structural dynamics. To emphasize this identity the product solutions to the partial differential equation are written on the form ux, t = qtφx 5 The time function qt represents a modal coordinate, and the function Φx is the correspondin undamped mode shape function. Insertion into 3 provides Φx qt = Φ x + x Φ x qt Φ x + x Φ x Φx = qt qt = ω 6 4

5 where the separation constant has been denoted ω. 6 implies that the mode shape and the modal coordinate must fulfill the followin differential equations x Φ x + Φ x + ω Φx = 7 qt + ω qt = 8 7 cannot at be solved analytically in its present form. Instead, the followin variable transformation is introduced x = r 4 r = x, dr dx = r 9 Next, Φx is implicitly considered a function of r via the substitution 9. Use of the chain rule provides dφx dx d Φx dx = dφr dr dr dx = r dφr dr = dr dφr r + dx dr r d Φr dr dr dx = 4 dφr r 3 dr + 4 r d Φr dr 1 Insertion of 9, 1 into 7 provides r 4 4 r 3 Φ r + 4 r Φ r + r Φ r + ω Φr = r Φ r + rφ r + ω r Φr = is reconized as the parametric Bessel equation of order ν = with α = ω, cf Then, the eneral solution becomes, cf ω ω Φr = c 1 J r + c Y r 1 The eneral solution of 8 reads qt = c 3 cosωt + c 4 sinωt 13 At the free end of the rope x = r = the displacement u, t must be bounded. Since, Y ω r for r, it is necessary to require that c =. Hence, the eienmodes have the form 5

6 ω x Φx = J r = J ω 14 The boundary condition ul, t ΦL = leads to the followin frequency condition, cf. Table 6.1, p. 64 J ω ω 1 ω ω 3 ω 4. L =.488 = L Let the vibration field be approximated by the SDOF model ux, t Φx qt Φx = 1 x L 16 The linearly varyin shape function implies that the rope is oscillatin as a physical pendulum. The mass moment of inertia around the support point becomes J = 1 3 µl3 and the restorin moment of the ravity force µl actin at the center of ravity becomes µl 1 Lθ. Hence, the undamped anular frequency becomes ω = µl 1 L 3 1 = 3 µl3 L 1.47 L 17 ω is an upper bound to ω 1 as a consequence of Rayleih s principle. The initial value ux, t = implies that q =. 13 then provides c 4 =. Product solutions, which fulfills the partial differential equation of motion, the eometrical boundary condition ux, t < for x, and the boundary condition ul, t have the form u i x, t = q i tφ i x = A i cosω i tj α i r, α i r = ω i x 18 where A i = c 3. Then, the superposition principle provides 6

7 ux, t = q i tφ i x = A i cosω i tj α i r 19 The coefficients A i are determined from the initial condition ux, = fx r fx = f = 4 A i J α i r is a Fourier-Bessel series of the order n = for the function fr /4, cf The expansion coefficients c i = A i are evaluated from Case I, p. 456 for x = L b = r = L and n =. The solution becomes, cf L r f r 4 J α i r dr A i = = L r f r 4 J α i r dr L r J α i r dr L J 1 α i = L L fx J ω x i dx 1 L J1 ω L i The solution follows from 19 and 1 ux, t = 1 L fu J ω u i du x J ω i J1 ω L i L cosω i t Problem : a b 7

8 Fi. 1: Steady-state temperature distribution in a hemisphere. The temperature distribution is independent of the azimuthal anle φ rotational symmetry around the z-axis. The stationary temperature distribution is determined from the boundary value problem u r + 1 u r θ + cot θ u, r θ =, r, θ, c π u = u r + r u r, π =, r, c uc, θ = fθ, θ, π 1 A solution to the partial differential equation is searched for on the separated form ur, θ = RrΘθ R r + r R r Θθ + 1 r Θ θ + cot θ Θ θ Rr = R r + r R r 1 r Rr = Θ θ + cot θ Θ θ Θθ = λ r R r + rr r λ Rr = Θ θ + cot θ Θ θ + λ Θθ = 3 3 cannot at first be solved in its present form. Instead, the followin variable transformation is introduced x = cos θ dx dθ = sin θ = 1 cos θ = 1 x cot θ = cos θ sin θ = x 1 x 4 Θθ is implicitly considered a function of x via the substitution 4. Use of the chain rule provides dθθ dθ = dθx dx dx dθ = 1 x dθx dx d Θθ dθ = x 1 x dx dθx dθ dx 1 x d Θx dx dx dθ = xdθx dx + 1 x d Θx dx 8

9 5 Insertion of 4, 5 into 3 provides xθ x + 1 x Θ x + x 1 x Θ x + λ Θx = 1 x 1 x Θ x xθ x + λ Θx =, x [ 1, 1] 6 6 has for arbitrary values of λ the trivial solution Θθ. Nontrivial solutions exist for λ = λ n = nn + 1. The correspondin eienfunctions consist of the Leendre polynomials P n x, cf. 6.3-, The boundary condition u r, π on the x, y-plane implies that Θθ = Pn cos θ must fulfill the condition π π Θ = P n cos = P n = 7 7 is only fulfilled for the Leendre polynomials of odd order { P 1 x, P 3 x,... }, cf. Fi. 6.6, p. 67. Only these polynomials are included in the series expansion for the solution. is a differential equation of Cauchy-Euler s type, see p With λ = λ n = nn + 1 the solution is iven as, cf , 4.7- Rr = c 1 r n + c r n+1 8 The temperature field must be bounded for r. Hence, it is necessary to specify c = in the eneral solution 8. Then, product solutions fulfillin the partial differential equation, the eometrical boundary condition ur, θ < for r, and the boundary condition u r, π have the form u n r, θ = A n r n P n x, n = 1, 3, 5,... 9 Use of the superposition principle provides the followin series for the solution ur, θ = u n r, θ = A n r n P n x 1 n=1,3,5,... n=1,3,5,... The coefficients A n are determined from the boundary conditions 9

10 uc, θ = fθ = f arccosx = n=1,3,5,... A n c n P n x is only defined for θ [, π ] x [, 1]. In order to exploit the orthoonality conditions of the Leendre polynomials of the Leendre polynomials the definition interval of fθ is extended to the interval θ [, π]. The extension of the function is selected, so fθ becomes an odd function around θ = π, see Fi. 1b. The extended function is denoted x, and is defined as fθ = f arccosx [, θ, π [ x = fπ θ = f π arccosx ] π ] 1, θ, π x = x becomes an odd function of x, which follows from the identity arccos x = π arccosx verify this numerically!. Since x is defined on [ 1, 1] the followin Fourier- Leendre series is valid, cf , x = c n P n x, x [ 1, 1] 13 n= c n = n xp n xdx 1 n + 1 xp n xdx, n = 1, 3, 5,... =, n =,, 6, The last statement of 14 follows because x is an odd function of x, and P n x is an odd function for n = 1, 3, 5,..., and an even function for n =,, 4,.... This makes xp n x an even function for n odd, and an odd function for n even. Further, x = f arccosx = fθ for x [, 1]. Comparison of 11 and 13 then provides the followin solution for the expansion coefficients A n A n = n + 1 c n 1 f arccosx P n xdx = n + 1 c n π/ fϕp n cos ϕ sin ϕdϕ 15 where ϕ has been introduced as interation variable via the substitution x = cos ϕ dx = sin ϕdϕ. Finally, from 1 and 15 follows 1

11 ur, θ = n=1,3,5,... n + 1 π/ c n fϕp n cos ϕ sin ϕdϕ r n P n cos θ 16 Problem : Given the function ft = e t sin t 1 1 is of exponential order c = 1. Then, L{ft} = e st ftdt = e s+1t sin tdt = [ e s+1t ] s + 1 s + 1 sin t cos t + 1 = 1 s + s +, s > 1 Problem : Given the function ft = 4t 5 sin 3t 1 Due to the linearity property of the Laplace transform we have, cf L{4t 5 sin 3t} = 4 L{t } 5 L{sin 3t} = 4 8 s 3 15 s + 9! s s + 3 =, s > where Theorems 7.1b and 7.1d have been used. 11

Two special equations: Bessel s and Legendre s equations. p Fourier-Bessel and Fourier-Legendre series. p

LECTURE 1 Table of Contents Two special equations: Bessel s and Legendre s equations. p. 259-268. Fourier-Bessel and Fourier-Legendre series. p. 453-460. Boundary value problems in other coordinate system.