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The 2D wave equation Separation of variables Superposition Examples The two dimensional wave equation Ryan C. Daileda Trinity University Partial Diﬀerential Equations March 1, 2012 Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesPhysical motivation Consider a thin elastic membrane stretched tightly over a rectangular frame. Suppose the dimensions of the frame are a × b and that we keep the edges of the membrane ﬁxed to the frame. Perturbing the membrane from equilibrium results in some sort of vibration of the surface. Our goal is to mathematically model the vibrations of the membrane surface. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples We let deﬂection of membrane from equilibrium at u(x, y , t) = position (x, y ) and time t. For a ﬁxed t, the surface z = u(x, y , t) gives the shape of the membrane at time t. Under ideal assumptions (e.g. uniform membrane density, uniform tension, no resistance to motion, small deﬂection, etc.) one can show that u satisﬁes the two dimensional wave equation utt = c 2 ∇2 u = c 2 (uxx + uyy ) (1) for 0 < x < a, 0 < y < b. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Remarks: For the derivation of the wave equation from Newton’s second law, see exercise 3.2.8. As in the one dimensional situation, the constant c has the units of velocity. It is given by τ c2 = , ρ where τ is the tension per unit length, and ρ is mass density. The operator ∂2 ∂2 ∇2 = 2 + 2 ∂x ∂y is called the Laplacian. It will appear in many of our subsequent investigations. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples The fact that we are keeping the edges of the membrane ﬁxed is expressed by the boundary conditions u(0, y , t) = u(a, y , t) = 0, 0 ≤ y ≤ b, t ≥ 0, u(x, 0, t) = u(x, b, t) = 0, 0 ≤ x ≤ a, t ≥ 0. (2) We must also specify how the membrane is initially deformed and set into motion. This is done via the initial conditions u(x, y , 0) = f (x, y ), (x, y ) ∈ R, ut (x, y , 0) = g (x, y ), (x, y ) ∈ R, (3) where R = [0, a] × [0, b]. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesSolving the 2D wave equation Goal: Write down a solution to the wave equation (1) subject to the boundary conditions (2) and initial conditions (3). We will follow the (hopefully!) familiar process of using separation of variables to produce simple solutions to (1) and (2), and then the principle of superposition to build up a solution that satisﬁes (3) as well. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesSeparation of variables We seek nontrivial solutions of the form u(x, y , t) = X (x)Y (y )T (t). Plugging this into the wave equation (1) we get XYT ′′ = c 2 X ′′ YT + XY ′′ T . If we divide both sides by c 2 XYT this becomes T ′′ X ′′ Y ′′ = + . c 2T X Y Because the two sides are functions of diﬀerent independent variables, they must be constant: T ′′ X ′′ Y ′′ =A= + . c 2T X Y Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples The ﬁrst equality becomes T ′′ − c 2 AT = 0. The second can be rewritten as X ′′ Y ′′ =− + A. X Y Once again, the two sides involve unrelated variables, so both are constant: X ′′ Y ′′ =B =− + A. X Y If we let C = A − B these equations can be rewritten as X ′′ − BX = 0, Y ′′ − CY = 0. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples The ﬁrst boundary condition is 0 = u(0, y , t) = X (0)Y (y )T (t), 0 ≤ y ≤ b, t ≥ 0. Since we want nontrivial solutions only, we can cancel Y and T , yielding X (0) = 0. When we perform similar computations with the other three boundary conditions we also get X (a) = 0, Y (0) = Y (b) = 0. There are no boundary conditions on T . Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Fortunately, we have already solved the two boundary value problems for X and Y . The nontrivial solutions are mπ Xm (x) = sin µm x, µm = , m = 1, 2, 3, . . . a nπ Yn (y ) = sin νn y , νn = , n = 1, 2, 3, . . . b with separation constants B = −µ2 and C = −νn . m 2 Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Recall that T must satisfy T ′′ − c 2 AT = 0 with A = B + C = − µ2 + νn < 0. It follows that for any choice m 2 of m and n the general solution for T is ∗ Tmn (t) = Bmn cos λmn t + Bmn sin λmn t, where m2 n2 λmn = c µ2 + νn = cπ m 2 + 2. a2 b These are the characteristic frequencies of the membrane. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Assembling our results, we ﬁnd that for any pair m, n ≥ 1 we have the normal mode umn (x, y , t) = Xm (x)Yn (y )Tmn (t) ∗ = sin µm x sin νn y (Bmn cos λmn t + Bmn sin λmn t) where mπ nπ µm = , νn = , λmn = c µ 2 + νn . m 2 a b Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Remarks: Note that the normal modes: oscillate spatially with frequency µm in the x-direction, oscillate spatially with frequency νn in the y -direction, oscillate in time with frequency λmn . While µm and νn are simply multiples of π/a and π/b, respectively, λmn is not a multiple of any basic frequency. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesSuperposition According to the principle of superposition, because the normal modes umn satisfy the homogeneous conditions (1) and (2), we may add them to obtain the general solution ∞ ∞ ∗ u(x, y , t) = sin µm x sin νn y (Bmn cos λmn t + Bmn sin λmn t). n=1 m=1 We must use a double series since the indices m and n vary independently throughout the set N = {1, 2, 3, . . .}. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesInitial conditions Finally, we must determine the values of the coeﬃcients Bmn and ∗ Bmn that are required so that our solution also satisﬁes the initial conditions (3). The ﬁrst of these is ∞ ∞ mπ nπ f (x, y ) = u(x, y , 0) = Bmn sin x sin y a b n=1 m=1 and the second is ∞ ∞ ∗ mπ nπ g (x, y ) = ut (x, y , 0) = λmn Bmn sin x sin y. a b n=1 m=1 These are examples of double Fourier series. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesOrthogonality (again!) To compute the coeﬃcients in a double Fourier series we can appeal to the following result. Theorem The functions mπ nπ Zmn (x, y ) = sin x sin y , m, n ∈ N a b are pairwise orthogonal relative to the inner product a b f ,g = f (x, y )g (x, y ) dy dx. 0 0 This is easily veriﬁed using the orthogonality of the functions sin(nπx/a) on the interval [0, a]. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Using the usual argument, it follows that assuming we can write ∞ ∞ mπ nπ f (x, y ) = Bmn sin x sin y, a b n=1 m=1 then a b mπ nπ f (x, y ) sin x sin y dy dx f , Zmn 0 0 a b Bmn = = a b Zmn , Zmn mπ nπ sin2 x sin2 y dy dx 0 0 a b a b 4 mπ nπ = f (x, y ) sin x sin y dy dx (4) ab 0 0 a b Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesRepresentability The question of whether or not a given function is equal to a double Fourier series is partially answered by the following result. Theorem If f (x, y ) is a C 2 function on the rectangle [0, a] × [0, b], then ∞ ∞ mπ nπ f (x, y ) = Bmn sin x sin y, a b n=1 m=1 where Bmn is given by (4). To say that f (x, y ) is a C 2 function means that f as well as its ﬁrst and second order partial derivatives are all continuous. While not as general as the Fourier representation theorem, this result is suﬃcient for our applications. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesConclusion Theorem Suppose that f (x, y ) and g (x, y ) are C 2 functions on the rectangle [0, a] × [0, b]. The solution to the wave equation (1) with boundary conditions (2) and initial conditions (3) is given by ∞ ∞ ∗ u(x, y , t) = sin µm x sin νn y (Bmn cos λmn t + Bmn sin λmn t) n=1 m=1 where mπ nπ µm = , νn = , λmn = c µ 2 + νn , m 2 a b Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Theorem (continued) ∗ and the coeﬃcients Bmn and Bmn are given by a b 4 mπ nπ Bmn = f (x, y ) sin x sin y dy dx ab 0 0 a b and a b ∗ 4 mπ nπ Bmn = g (x, y ) sin x sin y dy dx. abλmn 0 0 a b We have not actually veriﬁed that this solution is unique, i.e. that this is the only solution to the wave equation with the given boundary and initial conditions. Uniqueness can be proven using an argument involving conservation of energy in the vibrating membrane. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesExample 1 Example A 2 × 3 rectangular membrane has c = 6. If we deform it to have shape given by f (x, y ) = xy (2 − x)(3 − y ), keep its edges ﬁxed, and release it at t = 0, ﬁnd an expression that gives the shape of the membrane for t > 0. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples ∗ We must compute the coeﬃcients Bmn and Bmn . Since g (x, y ) = 0 we immediately have ∗ Bmn = 0. We also have 2 3 4 mπ nπ Bmn = xy (2 − x)(3 − y ) sin x sin y dy dx 2·3 0 0 2 3 2 2 mπ 3 nπ = x(2 − x) sin x dx y (3 − y ) sin y dy 3 0 2 0 3 2 16(1 + (−1)m+1 ) 54(1 + (−1)n+1 ) = 3 π 3 m3 π 3 n3 576 (1 + (−1)m+1 )(1 + (−1)n+1 ) = 6 . π m3 n3 Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples The coeﬃcients λmn are given by m2 n2 λmn = 6π + =π 9m2 + 4n2 . 4 9 Assembling all of these pieces yields ∞ ∞ 576 (1 + (−1)m+1 )(1 + (−1)n+1 ) mπ u(x, y , t) = 6 sin x π m3 n3 2 n=1 m=1 nπ × sin y cos π 9m2 + 4n2 t . 3 Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition ExamplesExample 2 Example Suppose in the previous example we also impose an initial velocity given by g (x, y ) = sin 2πx. Find an expression that gives the shape of the membrane for t > 0. Because Bmn depends only on the initial shape, which hasn’t changed, we don’t need to recompute them. ∗ We only need to ﬁnd Bmn and add the appropriate terms to the previous solution. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Using the values of λmn computed above, we have 2 3 ∗ 2 mπ nπ Bmn = √ sin 2πx sinx sin y dy dx 3π 9m2 + 4n2 0 0 2 3 2 3 2 mπ nπ = √ sin2πx sin x dx sin y dy . 3π 9m2 + 4n2 0 2 0 3 The ﬁrst integral is zero unless m = 4, in which case it evaluates to 1. Evaluating the second integral, we have ∗ 1 3(1 + (−1)n+1 ) 1 + (−1)n+1 B4n = √ = √ 3π 36 + n2 nπ π 2 n 36 + n2 ∗ and Bmn = 0 for m = 4. Daileda The 2D wave equation
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The 2D wave equation Separation of variables Superposition Examples Let u1 (x, y , t) denote the solution obtained in the previous example. If we let ∞ ∞ ∗ mπ nπ u2 (x, y , t) = Bmn sin x sin y sin λmn t 2 3 m=1 n=1 ∞ sin 2πx 1 + (−1)n+1 nπ = √ sin y sin 2π 36 + n2 t π2 n 36 + n2 3 n=1 then the solution to the present problem is u(x, y , t) = u1 (x, y , t) + u2 (x, y , t). Daileda The 2D wave equation
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