d Wave Equation. Rectangular membrane.

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1 1 ecture d Wave Equation. Rectangular membrane. The first problem is for the wave equation on a rectangular domain. You can interpret this as a problem for determining the displacement of a flexible elastic membrane stretched over a rectangular frame. The problem reads: 2 ( u 2 t 2 k u x u y 2 = u(xy = v(xy (xy = w(xy t u(xt = u(xht = for x [] and y [H]. The separation of variables leads to a series: (1.1 u(xyt = + + m=1 with coefficients given by the formulas B nm = A nm = 4 H u(yt = u(yt = ( A nm cos( λ mn kt + B nm sin( λ nm kt 4 λnm kh H H v(xysin mπy nπx sin H dxdy w(xysin mπy H nπx sin dxdy. 1.2 Radial vibrations of a circular membrane sin mπy H nπx sin Now the wave equation is considered on a circular domain x 2 + y 2 a 2 which represents an elastic membrane stretched on a cirle of radius a. We use polar coordinates (rθ to represent a circle as a rectangle The wave equation becomes: (1.2 2 u t 2 k r a π θ π. ( 1 r r ( r u r r 2 θ 2 =. First we consider a simpler case when vibrations are radial: u = u(rt that is u is independent of angle θ. This happens when the initial data are independent of θ : v = v(r w = w(r. 1

2 The wave equation of in polar coordinates with u independent of θ : 2 u t 2 k ( r =. r r r Separation of variable u = h(tφ(r leads to the equation for φ : and equation for h : rφ + φ + λrφ = φ(a = h + kλh =. First we solve for φ and λ. It has a general solution of φ equation: φ(r = AJ ( λr + BY ( λr where J (zy (z are two linearly independent solutions of Bessel s equation of zero order: z f + f + z f =. By J (z we denote a solutions which is bounded at z = and by Y (z solution that behaves like lnz near z =. We are not using Y (z because it is singular at z =. In this way φ(r = J ( λr. We also know that J (z has infinitely many roots that we label: z = z n n = They have the property that lim n z n = +. Now to solve for eigenvalues we have to use the boundary condition and write Solving this for λ we get The general solution is expressed as u(rt = φ(a = J ( λa =. λ n = ( zn 2. a + ( A n cos( λ n kt + B n sin( λ n kt J ( λ n r. The coefficients are determined from the orthogonality conditions. Note that they differ from the conditions we used so far. J ( λ n rj ( λ m rrdr = n m. 2

3 There is a factor r in front of dr. This leads to the formulas: A n = v(rj ( λ n rrdr J2 ( λ n rrdr B n = w(rj ( λ n rrdr λn k J2 ( λ n rrdr. 1.3 Circular membrane. General case. Now we work with equation (1.2. To separate variable we use function to find that h + kλh = Now we separate r and θ : We find that and s + µs = Solving for s and µ we get u(rθt = h(tφ(rθ r r ( r Φ + 2 u r θ 2 + r2 λφ =. Φ(rθ = φ(rs(θ. s( π = s(π s ( π = s (π. r(rφ + r 2 λφ µφ = φ(a =. s m = cos(mθ sin(mθ µ = m 2 m = Then we use µ = m 2 in the equation for φ : r(rφ + r 2 λφ m 2 φ = φ(a =. The solution of this problem is exprssed in terms of the Bessel s function of order m (see below. φ(r = J m ( λr. Using the boundary condition φ(a = and the roots of J m (z we find that λ mn = ( zmn 2 a n = The corresponding eigenfunctions: φ mn = J m ( λ mn r. 3

4 Finally we can solve for h. It is given by exactly the same expression as before: h mn = cos( λ mn kt sin( λ mn kt. To construct a general solution we take all possible products of The gernal solution: u(rθt = m= + m=1 + m= h nm (tφ nm (rs m (θ. A mn J m ( λ mn rcos( λ mn ktcos(mθ B mn J m ( λ mn rcos( λ mn ktsin(mθ C mn J m ( λ mn rsin( λ mn ktcos(mθ + m=1 D mn J m ( λ mn rsin( λ mn ktsin(mθ. To simplify things a little bit lets find the formula for coefficients for the intial data: u(rθ = v(rθ (rθ =. t The last condition means that C mn D mn = for all choices of m and n. This leaves us an equation v(rθ = A mn J m ( m=( λ mn r cos(mθ + m=1( If we separate m = terms from the first series we get v(rθ = ( A n J ( λ n r + m=1( + A mn J m ( λ mn r m=1( B mn J m ( λ mn r cos(mθ B mn J m ( λ mn r sin(mθ. sin(mθ. Now think of r being fixed and the left-hand side being function of θ. Then the righthand side is a full Fourier Series of that function with Fourier coefficients appearing inside the parentheses. This gives: A n J ( λ n r = 1 π v(rθdθ 2π π 4

5 A mn J m ( λ mn r = 1 π π π π v(rθcos(mθdθ mn J m ( B λ mn r = 1 v(rθsin(mθdθ. π π We use the orthogonality of Bessel s functions to find the coefficients. A mn = B mn = π π A n = v(rθj ( λ n rrdrdθ 2π J2 ( n = λ n rrdr π π v(rθj m( λ mn rcos(mθrdrdθ π J2 m( λ mn rrdr nm = π π v(rθj m( λ mn rsin(mθrdrdθ π J2 m( λ mn rrdr nm = Bessel s equation of order m. To be able to solve the problem for vibrations of a membrane when u depends on both r and θ we need to use Bessel s functions of order m. We d like to solve the eigenvalue problem: r 2 φ + rφ + (r 2 λ m 2 φ = f (a = where m = We can eliminate λ from the equation if we use a change of variables: φ(r = f ( λr. If we call λr = z then f = f (z is the solution of Bessel s equation of order m: z 2 f + z f + (z 2 m 2 f =. There are two linearly independent solutions: one is bounded at z = and the other is singular at z =. We call them J m (z and Y m (z. The functions can be expressed as power series in which the coefficients are determined by recursive relations. It can also be shown that function oscillate between positive and negative values and there are countable many roots of the equation which we list as J m (z = = z m < z m1 < z m2 < z m3 <... The roots have property that lim n + z mn = +. We will not be using function Y m (z because it is singular at z =. We also not be using the zero root because the corresponding eigenfunction is trivial. The orthogonality conditions hold: J m ( λ mn rj m ( λ mk rrdr = n k. 5

6 1.5 Summary of eigenvalue problems and Formulas for coefficients Problem φ + λφ = φ( = φ( = Formulas Problem Fourier Sine Series: ( nπ 2 λ n = φn = sin n = f (x = B n = 2 B n sin f (xsin dx. φ + λφ = φ ( = φ ( = Formulas Problem ( nπ 2 λ n = φn = cos n = Fourier Cosine Series: A = 1 f (x = A + f (xdx A n = 2 A n cos f (xcos dx. φ + λφ = φ( = φ( φ ( = φ ( = Formulas ( nπ 2 λ n = φ n = cos φ n = sin n = n =

7 Problem Fourier Series: f (x = A + A = 1 f (xdx 2 B n = 1 A n cos + B n sin A n = 1 f (xsin rφ + φ + rλφ = φ(a = φ(r is bounded f (xsin dx. dx. Formulas Bessel Series λ n = ( zn 2 a φ n (r = J ( λ n r n = f (r = B n J ( λ n r B n = f (rj ( λ n rrdr J2 (. λ n rrdr Problem Formulas for m = Bessel Series r 2 φ + φ + (r 2 λ m 2 φ = λ mn = φ(a = φ(r is bounded ( zmn 2 a φ mn (r = J m ( λ mn r n = f (r = B mn J ( λ mn r B mn = f (rj m( λ mn rrdr J2 m(. λ mn rrdr 7

8 11 ecture Non-homogeneous problems In this section we consider methods to solve PDEs in which either the equation is nonhomogeneous: t k 2 u x 2 = Q(xt or boundary conditions are non-homogeneous for example u(t = T 1 (t u(t = T 2 (t. We start with a simple problem t k 2 u = x ( t > x2 u(x = u (x x [] u(t = T 1 u(t = T 2 t where T 1 T 2 are two fixed temperatures at the ends of the rod. Solving for an equilibrium solution we find that u e (x = T 2 T 1 x + T 1. Now we introduce new function v(xt by the formula v(xt = u(xt u e (x. You can verify that v is the solution of the problem (11.1 v t k 2 v = x ( t > x2 v(x = u (x u e (x x [] v(t = v(t = t. This problem is solved by the method of separation of variables and we find v(xt = B n e kλnt sin( λ n x with coefficients Then B n = 2 l (u (x u e (xsin( λ n xdx. u(xt = v(xt + u e (x. 8

9 Problem Solve the problem t k 2 u = x ( t > x2 u(x = x x [] u(t = u(t = 1 t Answer u(xt = x + n+1 2 ( 1 nπ e k( nπ 2t sin. The method works whenever the original problem has an equilibrium solution. Consider for example t k 2 u x 2 = kx2 x ( t > u(x = u (x x [] u(t = T (t = t x In this problem both the equation and the boundary conditions are non-homogeneous. Solving for the equilibrium solution we get Introducing u e (x = x x 3 + T. v(xt = u(xt u e (x we find that v solves a homogeneous problem: v t k 2 v = x ( t > x2 (11.2 v(x = u (x u e (x x [] v(t = (t = t. v x This problem is solved by the method of separation of variables and we find with coefficients Then v(xt = B n e kλnt sin( λ n x λ n = B n = 2 l (u (x u e (xsin( λ n xdx. u(xt = v(xt + u e (x. ( (n 1/2π 2 9

10 11.2 Method of eigenfunction expansion Now we allow the right-hand side to be an arbitrary function Q(xt but assume that the boundary conditions are homogeneous. t k 2 u = Q(xt x ( t > x2 u(x = u (x x [] u(t = u(t = t Using the eigenfunctions we can represent Q(xt = q n (tsin( λ n x and u (x = u n sin( λ n x. We can even do this for the unknown solution u(xt : u(xt = b n (tsin( λ n x where b n (t are yet undetermined coefficients. Substituting series into the equation and rearranging terms (assuming we allowed to do this we get ( b n + kλ n b n q n (t sin( λ n x = which implies that for any n = we must have Since we have initial conditions b n + kλ n b n = q n (t. b n ( = u n. In this way we have to solve first order linear ODEs for coefficients b n. Solving them we get t b n = u n e kλnt + e kλ nt q n (te kλnt dt. Problem Solve t k 2 u = xt x ( t > x2 u(x = 1 x [] u(t = u(t = t Solution Use the Fourier Series from the last problem n+1 2 q n (t = ( 1 nπ t u n = 2 nπ and the formula for b n. (1 cos(nπ 1

11 11.3 Non-homogeneous problems with time-dependent boundary conditions This type of problems can be reduced to the case of non-homogenous equation and homogeneous boundary conditions which is solved then by the method of eigenfunction expansion. Consider t k 2 u = x ( t > x2 u(x = u (x x [] u(t = T 1 (t u(t = T 2 (t t Introduce a function that matches the boundary conditions for example w(xt = T 2(x T 1 (t x + T 1 (t. This is not an equilibrium solution even though it looks like it. Set v(xt = u(xt w(xt. Then you verify that v solves v t k 2 v x 2 = T 2 (t T 1 (t x T 1 (t x ( t > u(x = u (x T 2( T 1 ( x T 1 ( x [] u(t = u(t = t Method of eigenfunction expansion in several dimensions Consider heat equation on a disk x 2 + y 2 a 2. ( 2 t k u x u y 2 = Q(rt r (a t > u(xyt = x 2 + y 2 = a 2 u(xy =. Changing to polar coordinates (rθ and solving to radial solutions u = u(rt we obtain the problem ( ( 1 t k r = Q(rt r (a t > r r r u(at = x 2 + y 2 = a 2 u(r =. Using eigenfunctions J ( λ n r λ n = 11 ( zn 2 a

12 we can write where Q(rt = q n (tj ( λ n r q n (t = Q(rtJ ( λ n rrdr J2 ( λ n rrdr and Each eigenfunction verifies equation u(rt = 1 r r b n (tj ( λ n r. φ n (r = J ( λ n r ( r φ n = kλ n φ n. r Using this from the heat equation we obtain ODEs for b n (t : b n(t + kλ n b n = q n (t b n ( =. Notice that the equation is exactly the same as before except that formulas for λ n are different. Thus t b n (t = e kλ nt q n (te kλnt dt. 12 Wave equation. D Alambert formula 12.1 Wave equation as a model for propagation of sound waves The motion of a gas is described by the equations where p(ρ is the pressure function: assuming that the temperature is constant. t ρ + u x ρ + ρ x u = ρ t u + ρu x u + x p(ρ = p = RρT 12