Math Spring Review Material - Exam 3

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1 Math 85 - Spring 01 - Review Material - Exam 3 Section 9. - Fourier Series and Convergence State the definition of a Piecewise Continuous function. Answer: f is Piecewise Continuous if the following to conditions are satisfied: 1) It is continuous except possibly at some isolated points. ) The left and right limits f (x + ) and f (x ) exist (finite) at the points of discontinuity. For example, Square-Wave functions are piecewise continuous. But f (x) = sin 1 x for 0 < x < 1, extended to R periodically, is not piecewise continuous, since the right limit at zero DNE. Also f (x) = tanx for 0 < x < π and for π < x < π, extended to R periodically, is not piecewise continuous, since the right and left limits at π DNE (infinite). Define Fourier Series for functions for a Piecewise Continuous periodic function with period L. Answer: where for n 0 and for n 1 f (x) a 0 + a n cosn π L x + b n sinn π L x a n = 1 L b n = 1 L L L L L f (x)cosn π L x dx f (x)sinn π L x dx State the definition of a Piecewise Smooth function. Answer: f (x) is Piecewise Smooth if both f (x) and f (x) are piecewise continuous. For example, Square-Wave functions are piecewise smooth. State the Convergence Theorem for Fourier Series. Answer: If f (x) is periodic and piecewise smooth, then its Fourier Series converges to 1) f (x) at each point x where f is continuous. ) 1 ( f (x+ ) + f (x )) at each point where f is NOT continuous. Compute the Fourier Series of Square-Wave Function with period L: { +1 0 < x < L f (x) = 1 L < x < L For which values of f (x) at the points of discontinuity the Fourier series is convergent for all x? 1

2 Answer: a n = 0 for all n 0, and b n = nπ (1 cosnπ) for all n 1. Thus f (x) = nπ sinnπ L x Note that f (x) is discontinuous at x = integer multiples of L at which the average of left and right limit of f (x) is zero. Thus by Convergence Theorem f (x) must be 0 at those points. Letting x = L in the Fourier Series representation of the Square-Wave Function, obtain the following relation: ( 1) k k=0 k + 1 = = π Answer: Note that if we let x = L, then the Fourier series is convergent to f (x) = 1, thus we have ( ) L 1 = f = sin(k + 1) π π = ( 1) k=0 k + 1 π k k=0 k + 1 Hence ( 1) k k=0 k + 1 = = π

3 3 Section Fourier Sine and Cosine Series Recall f (x) is odd if f ( x) = f (x), i.e. its graph is symmetric w.r.t y-axis. Example: x n+1,sinnx for all integers n 0, the square-wave functions { 1 L < x < 0 f (x) = 1 0 < x < L Recall f (x) is even if f ( x) = f (x), i.e. its graph is symmetric w.r.t origin. Example: x n,cosnx for all integers n. Remarks: 1) If f (x) is odd, then L L f (x)dx = 0 for any L ) If f (x) is even, then L L f (x)dx = L 0 f (x)dx for any L 3) If f and g are odd, then f g is even. ) If f and g are even, then f g is even. 3) If f is odd and g is even, then f g is odd. Fourier series of odd functions with period L: a 0 = L 1 L L f (x)dx = 0, a n = L 1 L L f (x)cosn L πxdx = 0 since f (x)cosnπ Lx is odd. b n = L 1 L L f (x)sinn L πxdx = L L0 f (x)sinn L πxdx since f (x)sinnπ Lx is even. In this case, if f is piecewise smooth, f (x) = b n sinn L π x only involves sine. Example: f (x) = x on ( π,π), then f (x) = sin nx ( 1) n+1 n Fourier series of even functions with period L: a n = L 1 L L f (x)cosn L πxdx = L L0 f (x)cosn L πxdx since f (x)cosnπ Lx is even. b n = L 1 L L f (x)sinn L πxdx = 0 since f (x)sinnπ Lx is odd. In this case, if f is piecewise smooth, f (x) = a 0 + a n cosn L π x only involves cosine. Example: Try f (x) = x on ( π,π). Even and odd extensions of a function: Suppose f is a piecewise continuous function defined on interval (0,L). Even extension of f to the interval ( L,0) is { f (x) 0 < x < L f E (x) = f ( x) L < x < 0 Example: f (x) = x + x + 1 on (0,L) Then { x + x < x < L f E (x) = x x + 1 L < x < 0 Odd extension of f to the interval ( L,0) is { f (x) 0 < x < L f O (x) = f ( x) L < x < 0

4 Example: f (x) = x + x + 1 on (0,L) Then { x + x < x < L f O (x) = x + x 1 L < x < 0 Remark: 1) f E (x) is an even function with Fourier Series of the form a 0 + a n cosn L πx Hence f (x) = a 0 + a n cosn L π x for x in (0,L). This is called the Fourier cosine series of f ) f O (x) is an odd function with Fourier Series of the form b n sinn L πx Hence f (x) = b n sinn L π x for x in (0,L). This is called the Fourier sine series of f Remark: Note that for x in (0,L), f (x) = f E (x) = f O (x). In many cases we are not concerned about f (x) on ( L,0), so the choice between (1) and () depends on our need for representing f by sine or cosine. Example: f (t) = 1 on (0,π). Compute the Fourier sine and cosine series and graph the two extensions. 1) The Even extension is f E (t) = 1 on ( π,π), period π. Then a 0 =,a n = 0,b n = 0, so the cosine series is just f (t) = a 0 = 1 ) The Odd extension is f O (t) = 1 on (0,π) and 1 on ( π,0), period π. Then a 0 = a n = 0,b n = nπ (1 ( 1)n ), so the cosine series is f O (t) = nπ sinnx Example: f (t) = 1 t on (0, 1). Compute the Fourier sine and cosine series and graph to the to extensions. 1) The Even extension is f E (t) = 1 t on (0,1) and 1 +t on ( 1,0), period L, L = 1. Then a 0 = 1,a n = 1 cosnπ,b n π n = 0, so the cosine series is f (t) = 1 + n π cosnπx ) The Odd extension is f O (t) = 1 t on (0,1) and 1 t on ( 1,0), period L, L = 1. Then a n = 0 for all n 0 and b n = nπ, so the cosine series is f (t) = nπ sinnπx Termwise differentiation of a Fourier series Theorem: Suppose f (x) is Continuous for all x, Periodic with period L, and f is

5 5 Piecewise Smooth for all t. If then f (x) = a 0 + a n cosn π L x + b n sinn π L x f (x) = nπ ( a n L )sinnπ L x + (b n nπ L )cosnπ L x Remarks: 1) RHS it the Fourier series of f (x). ) It is obtained by termwise differentiation of the RHS for f (x). 3) Note that the constant term in the FS of f (x) is zero as L f (x)dx = f (L) f ( L) = 0 L ) The Theorem fails if f is not continuous! For example: consider f (x) = x on ( L, L) Then f (x) = x = L nπ ( 1)n+1 sinn π L x if we differentiate 1 π ( 1)n+1 cosn π L x For example equality fails at t = 0 or L! Example: Verify the above Theorem for f (x) = x on (0, L) and f (x) = x on ( L, 0). Answer: b n = 0 as f is even, a 0 = L, a n = L n π (( 1) n 1). Thus f (x) = 1 L + L n π cosnπ L x Then term wise derivative gives f (x) = nπ sinnπ L x On the other hand, directly computing the derivative of f (x) we have f (x) = 1 on (0,L) and f (x) = 1 on ( L,0). Thus a n = 0 for all n and b n = nπ (1 cosnπ) which gives the same Fourier Series as in above. Applications to BVP s Consider the BVP of the form Example: ax + bx + cx = f (t), x(0) = x(l) = 0 or x (0) = x (L) = 0 x + x = 1, x(0) = x(π) = 0 Here f (t) = 1, restrict to the interval (0,π) as in Boundary Values. The idea is to find a formal Fourier Series solution of the equation. Since we want x(0) = x(π) = 0, we prefer to consider the Fourier sine series of x(t) on the interval (0,π), x(t) = b n sinnt

6 6 substitute in the equation and compare the coefficients with that of the sine Fourier Series of f (t) = 1 which is Note that by term wise differentiation, f (t) = nπ sinnt x (t) + x(t) = ( n )b n sinnt Thus b n = 0 for n even and b n = for n odd. Hence a formal power series of x(t) πn( n ) is x(t) = πn( n ) sinnt Example: x + x = t, x (0) = x (π) = 0 Here f (t) = t, restrict to the interval (0,π) as in Boundary Values. Since we want x (0) = x (π) = 0, we prefer to consider the Fourier Cosine Series of x(t) on (0,π), x(t) = a 0 + a n cosnt substitute in the equation and compare the coefficients with that of the Cosine FS of f (t) = t which is f (t) = π Note that by termwise differentiation, πn cosnt x (t) + x(t) = a 0 + ( n )a n cosnt Thus a 0 = π, a n = 0 for n even and a n = πn ( n ) series of x(t) is x(t) = π πn ( n ) cosnt for n odd. Hence a formal power Termwise Integration of the Fourier Series. Theorem: Suppose f (t) is Piecewise Continuous (not necessarily piecewise smooth) periodic with period L with FS representation f (t) a 0 + a n cosn π L t + b n sinn π L t Then we can integrate term by term as t0 f (x)dx = t a a t0 n cosn L πxdx + b t0 n sinn L πxdx = a 0 t + a n nπ L sinnπ L x b n nπ L (cosnπ L x 1) Note that the RHS is not the Fourier series of LHS unless a 0 = 0. Example: Consider f (x) = 1 on (0,π) and f (x) = 1 on ( π,0), then f (t) = πn sinnt

7 Then F(t) = t 0 f (x)dx = t for t (0,1) and t for t ( π,0), whose Fourier Series is F(t) = 1 π π 1 n cosnt Term by term integration gives the same thing! π 1 n (1 cosnt) = π ( 1 n = π 8 ) π 1 n cosnt 7

8 8 Section 9. - Applications of Fourier Series Finding general solutions of nd order linear DE s with constant coefficients: Example: x + 5x = F(t), where F(t) = 3 on (0,π) and 3 on ( π,0), is odd with period π. We obtain a particular solution in the following way: Since L = π and the FS of F(t) is n odd 1 nπ sinnt, we may assume x(t) is odd and we consider its Fourier sine series F(t) = b n sinnt. Substitute in the equation: x + 5x = (5 n 1 )b n sinnt = F(t) = n odd nπ sinnt Hence, comparing the coefficients of sinnt on both sides we get b n = 1 n(5 n )π if n is odd and zero otherwise. Thus a particular solution is x(t) = n odd 1 n(5 n )π sinnt Definition: We call this a steady periodic solution, denoted by x sp (t). Thus, if x 1 (t),x (t) are the solutions of the associated homogeneous equation, then the general solution is x(t) = c 1 x 1 (t) + c x (t) + x sp (t) = c 1 cos( 5t) + c sin( 5t) + n odd 1 n(5 n )π sinnt Remark: Consider the equation x + 9x = F(t). Then when trying to find a particular solution we get x + 9x = (9 n 1 )b n sinnt = F(t) = n odd nπ sinnt We can not find b 3 as 9 n = 0 for n = 3. In this case we need to use the method of undetermined coefficients to find a function y(t) such that y + 9y = 1 3π sin3t Take y = At sin3t + Bt cos3t, and substitute to find A = 0 and B = 3π. Therefore, the general solution is x(t) = c 1 cos(3t) + c sin(3t) + n odd,n 3 Definition: We say in this case a pure resonance occurs. 1 n(9 n )π sinnt 3π t cos3t Remark: To determine the occurrence of pure resonance, just check if for some n, sinn L π t is a solution of the associated homogeneous equation.

9 Application: Forced Mass-Spring Systems Let m be the mass, c be the damping constant, and k the constant of spring. Then mx + cx + kx = F(t) Consider the case that the external force F(t) is odd or even periodic function. Remark: If F(t) is periodic for t 0, it can be arranged to be odd or even by passing to odd or even extension for values of time t. Case 1) Undamped Forced Mass-Spring Systems: c = 0 mx + kx = F(t) k Let ω 0 = m be the natural frequency of the system, then we can write x + ω 0x = 1 m F(t) Assume F(t) is periodic odd function with period L. Then F(t) = F n sinn π L t Consider the odd extension of x(t), so that x(t) = b n sinn π L t substituting in the equation we get ( x + ω 0x = ω 0 ( nπ L )) b n sinn π L t = 1 m F(t) = 1 m F n sinn π L t Thus ( ω 0 ( nπ L )) b n = 1 m F n If for all n 1, ω 0 ( nπ L ) k 0, or equivalently m πl is not a positive integer, then we can solve for b n as 1 m b n = F n ω 0 ( nπ L ) Hence we have a steady periodic solution x sp (t) = Example: If m = 1,k = 5,L = π, 1 m F n ω 0 ( nπ L ) sinnπ L t x + 5x = F(t) then ω 0 ( nπ L ) = 5 n 0 for all positive integers n 1. k On the other hand, if n 0 = m πl is a positive integer, then a particular solution is x(t) =,n n 0 1 m F n ω 0 ( nπ L ) sinnπ L t F n 0 mω 0 t cosω 0 t 9

10 10 Example: m = 1,k = 9,L = π, x + 9x = F(t) then ω 0 ( nπ L ) = 9 n = 0 for n = 3. There will be a pure resonance. Example: m = 1,k = 9,L = 1, x + 9x = F(t) then ω 0 ( nπ L ) = 9 (nπ) 0 for all positive integers n. We have a steady periodic solution. Case ) Damped Forced Mass-Spring Systems: c 0 mx + cx + kx = F(t) Assume F(t) is periodic odd function with period L. Then F(t) = F n sinn π L t For each n 1, we seek a function x n (t) such that mx n + cx n + kx n = F n sinn π L t = F n sinω n t where ω n = n L π. Note that there will never be a duplicate solution as c 0. Hence using the method of undetermined coefficients, we can show F n x n (t) = (k mω n ) + (cω n ) sin(ω nt α n ) where ( ) α n = tan 1 cωn k mω n Therefore, we have a steady periodic solution x sp (t) = x n (t) = 0 α π F n (k mω n ) + (cω n ) sin(ω nt α n ) Example: m = 3,c = 1,k = 30, F(t) = t t for 0 t 1 is odd and periodic with L = 1. Compute the first few terms of the steady periodic solution. So, we have 3x + x + 30x = F(t) = x sp (t) = F n sinn π L t = 8 n 3 π 3 sinnπt F n (k mω n ) +(cω n ) sin(ω nt α n ) 8 n = 3 π 3 n= odd sin(nπt α (30 3n n) π +n π ( ) α n = tan 1 nπ 30 n π 0 α π The fist two terms are sin(πt 1.69) sin(3π ) +

11 11 Section Heat Conduction and Separation of Variables Until now, we studied ODE s - which involved single variable functions. In this section will consider some special PDE s - Differential Equations of several variable functions involving their Partial Derivatives - and we will apply Fourier Series method to solve them. Heat Equations: Let u(x,t) denote the temperature at pint x and time t in an ideal heated rod that extends along x-axis. then u satisfies the following equation: u t = ku xx where k is a constant - thermal diffiusivity of the material - that depends on the material of the rod. Boundary Conditions: Suppose the rod has a finite length L, then 0 x L. 1) Assume the temperature of the rod at time t = 0 at every point x is given. Then we are given a function f (x) such that u(x,0) = f (x) for all 0 x L. ) Assume the temperature at the two ends of the rod is fixed (zero) all the time - say by putting two ice cubes! Then u(0,t) = u(l,t) = 0 for all t 0. Thus we obtain a BVP, u t = ku xx u(x,0) = f (x) for all 0 x L u(0,t) = u(l,t) = 0 for all t 0 Remark: Other possible boundary conditions are, insulating the endpoints of the rod, so that there is no heat flow. This means u x (0,t) = u x (L,t) = 0 for all t 0 Remark: Geometric Interpretation of the BVP. We would like to find a function u(x,t) such that on the boundary of the infinite strip t 0 and 0 x L satisfies the conditions u = 0 and u = f (x). Remark: If f (x) is piecewise smooth, then the solution of the BVP is unique. Important observations: 1) Superposition of solutions: If u 1,u,... satisfy the equation u t = ku xx, then so does any linear combination of the u i s. In other words, the equation u t = ku xx is linear! ) Same is true about the boundary condition u(0,t) = u(l,t) = 0 for all t 0. We say this is a linear or homogeneous condition. 3) The condition u(x,0) = f (x) for all 0 x L is not homogenous, or not linear!

12 1 General Strategy: Find solutions that satisfy the linear conditions and then take a suitable linear condition that satisfies the non-linear conditions. Example: Verify that u n (x,t) = e nt sinnx is a solution of u t = u xx (here k = 1) for any positive integer n. For example, u 1 (x,t) = e t sinx and u (x,t) = e t sinx. Example: Use the above example to construct a solution of the following BVP. u t = u xx u(0,t) = u(π,t) = 0 for all t 0 u(x,0) = sinx + 3sinx for all 0 x π Answer: Here L = π. Note that u n (x,t) = e nt sinnx also satisfy the linear condition u(0,t) = u(π,t) = 0. Thus it is enough to take u(x,t) to be a linear combination of u n s that satisfies the non homogenous condition u(x,0) = sinx + 3sinx. Since u n (x,0) = sinnx, we take so that u(x,t) = u 1 + 3u = e t sinx + 3e t sinx u(x,0) = e 0 sinx + 3e 0 sinx = sinx + 3sinx Remark: The above method in the example for u t = u xx works whenever f (x) is a finite linear combination of sin x, sin x,... Example: Use the above example to construct a solution of the following BVP. u t = u xx u(0,t) = u(π,t) = 0 for all t 0 u(x,0) = sinxcosx for all 0 x L Solution: Again we have L = π. Note that f (x) = sinxcosx = 1 sin(x + x) + 1 sin(x x) = 1 sin5x + 1 sin3x Thus we take u(x,t) = 1 u u 5 = 1 e 9t sin3x + 1 e 5t sin5x Remark: When f (x) is a not a finite linear combination of the sine functions, then represent it as an infinite sum using Fourier sine series. Example: Construct a solution of the following BVP. u t = u xx u(0,t) = u(π,t) = 0 for all t 0 u(x,0) = 1 for all 0 x π Answer: Note that f (x) = 1 and L = π, so represent f (x) as a Fourier sine series with period L = π f (x) = nπ sinnx n= odd

13 13 Thus we take u(x,t) = n= odd nπ u n(x,t) = n= odd nπ ent sinnx This is a formal series solution of the BVP, one needs to check the convergence,... We only take finitely many terms for many applications. In general, to solve the following BVP (k is anything, not necessarily 1, and L is anything, not just π) u t = ku xx u(0,t) = u(π,t) = 0 for all t 0 u(x,0) = f (x) for all 0 x L we observe that u n (x,t) = e k( nπ L )t sinn π L x satisfies the equation u t = ku xx and the homogenous boundary condition u(0,t) = u(l,t) = 0 for all t 0. Thus, represent f (x) as a Fourier sine series with period L, f (x) = b n sinn π L x Then u(x,t) := b n u n (x,t) = b n e k( nπ L )t sinn π L x satisfies the non homogenous condition u(x,0) = f (x) for all 0 x L. Case of a rod with insulated endpoints: Consider the BVP corresponding to a heated rod with insulated endpoint, u t = ku xx u x (0,t) = u x (π,t) = 0 for all t 0 u(x,0) = f (x) for all 0 x L we observe that for n 0, u n (x,t) = e k( nπ L )t cosn π L x satisfies the equation u t = ku xx and the homogenous boundary condition u x (0,t) = u x (L,t) = 0 for all t 0. Remark: for n = 0, we get u 0 = 1 which satisfies the linear conditions! Thus, represent f (x) as a Fourier cosine series with period L, Then f (x) = a 0 + a n cosn π L x u(x,t) := a 0 + a n u n (x,t) = a 0 + a n e k( nπ L )t cosn π L x satisfies the non homogenous condition u(x,0) = f (x) for all 0 x L.

14 1 Remarks: 1) In the BVP for heated rod with zero temperature in the endpoints, we have lim t u(x,t) = lim t b n e k( nπ L )t sinn π L x = 0 in other words, heat goes away with no insulation, thus temperature is zero at the end! ) In the BVP for heated rod with insulated endpoints, we have a 0 lim u(x,t) = lim t t + b n e k( nπ L )t cosn π L x = a 0 which means, with insulation heat distributes evenly throughout the rod, which is the average of the initial temperature as a 0 = 1 L L 0 f (x)dx

15 15 Section Vibrating Strings and the One-Dimensional Wave Equation Consider a uniform flexible string of length L with fixed endpoints, stretched along x-axis in the xy-plane from x = 0 to x = L. Let y(x,t) denote the displacement of the points x on the string at time t (we assume the points move parallel to y-axis). Then y satisfies the One-dimensional wave equation: y tt = a y xx where a is a constant that depend on the material of the string and the tension! Boundary Conditions: 1) Since the endpoints are fixed, y(0,t) = y(l,t) = 0. ) The initial position of the string y(x,0) at each x is given as a function y(x,0) = f (x). 3) The solution also depends on the initial velocity y t (x,0) of the string at each x, given as a function y t (x,0) = g(x). Thus we obtain the following BVP y tt = a y xx y(0,t) = y(l,t) = 0 for all t 0 y(x,0) = f (x) for all 0 x L y t (x,0) = g(x) for all 0 x L Important Observations: 1) y tt = a y xx is a linear equation. Thus the superposition of solutions applies. ) Condition y(0,t) = y(l,t) = 0 is linear. 3) Conditions y(x,0) = f (x) and y t (x,0) = g(x) are not linear. General Strategy: Split the BVP into two problems, y tt = a y xx y tt = a y(0,t) = y(l,t) = 0 for all t 0 y xx y(0,t) = y(l,t) = 0 for all t 0 (A) (B) y(x,0) = f (x) for all 0 x L y(x,0) = 0 for all 0 x L y t (x,0) = 0 for all 0 x L y t (x,0) = g(x) for all 0 x L If y A (x,t) and y B (x,t) are the respective solutions, then satisfies the original BVP as and y(x,t) = y A (x,t) + y B (x,t) y(x,0) = y A (x,0) + y B (x,0) = f (x) + 0 = f (x) y t (x,0) = (y A ) t (x,0) + (y B ) t (x,0) = 0 + g(x) = g(x) Solving a BVP of type (A) y tt = a y xx y(0,t) = y(l,t) = 0 for all t 0 (A) y(x,0) = f (x) for all 0 x L y t (x,0) = 0 for all 0 x L

16 16 Verify directly that for all positive integers n, the function y n (x,t) = cosn π L at sinnπ L x satisfies the equation y tt = a y xx and the linear conditions y(0,t) = y(l,t) = 0 and y t (x,0) = 0. Thus, by superposition law, we need to find coefficients b n such that y(x,t) = satisfies y(x, 0) = f (x). Note that b n y n (x,t) = b n cosn π L at sinnπ L x f (x) = y(x,0) = b n y n (x,0) = b n sinn π L x Thus b n s are the Fourier sine coefficients of f (x). Example: Triangle initial position {(pulled from the midpoint) with zero initial velocity x if 0 < x < π Assume a = 1, L = π and f (x) = π x if π < x < π. Then the BVP is type (A) Fourier sine series of f (x) is y tt = y xx y(0,t) = y(π,t) = { 0 for all t 0 (A) x if 0 < x < π y(x,0) = f (x) = π x if π < x < π y t (x,0) = 0 for all 0 x π sin nπ πn sinnx = ( 1) n 1 πn sinnx Thus, since a = 1 and L = π, y(x,t) = b n cosn π L at sinnπ L x sin nπ πn = ( 1) n 1 = πn ( 1) n 1 = = π Remark: Using the identity we can write the solution as πn cost sinx 9π cosnt sinnx cosnt sinnx cosnt sinnx cos3t sin3x + 5π sinacosb = sin(a + B) + sin(a B) cos5t sin5x +

17 17 ( 1) n 1 y(x,t) = = 1 = 1 πn n 1 ( 1) πn n 1 ( 1) πn n 1 ( 1) cosnt sinnx cosnt sinnx (sin(nx + nt) + sin(nx nt)) = 1 sinn(x +t) + 1 πn ( 1) = 1 f O(x +t) + 1 f O(x t) πn n 1 sinn(x t) Solving a BVP of type (B) y tt = a y xx y(0,t) = y(l,t) = 0 for all t 0 (A) y(x,0) = 0 for all 0 x L y t (x,0) = g(x) for all 0 x L Verify directly that for all positive integers n, the function y n (x,t) = sinn π L at sinnπ L x satisfies the equation y tt = a y xx and the linear conditions y(0,t) = y(l,t) = 0 and y(x,0) = 0. Thus, by superposition law, we need to find coefficients c n such that y(x,t) = c n y n (x,t) = c n sinn π L at sinnπ L x satisfies y t (x,0) = g(x). Note that by termwise differentiation with respect to variable t we have y t (x,t) = c n (n π L a)cosnπ L at sinnπ L x Thus g(x) = y t (x,0) = c n (n π L a)sinnπ L x Then c n (n L π a) s are the Fourier sine coefficients of g(x). Hence c n = nla π n-th Fourier sine coefficients of g(x) Example: Assume a = 1, L = π and g(x) = 1. Then the following BVP is type (B) y tt = y xx y(0,t) = y(π,t) = 0 for all t 0 (B) y(x,0) = 0 y t (x,0) = g(x) = 1 for all 0 x π Fourier sine series of g(x) = 1 with L = π is πn sinnx

18 18 Thus, since a = 1 and L = π y(x,t) = c n sinn L πat sinnπ L x π = nπ(1) πn sinnt sinnx = sinnt sinnx πn

19 19 Section Steady-State Temperature and Laplace s Equation Consider the temperature in a -dimensional uniform thin plate in xy-plane bounded by a piecewise smooth curve C. Let u(x,y,t) denote the temperature of the point (x,y) at time t. Then the -dimensional eat equation states that u t = k(u xx + u yy ) where k is a constant that depends on the material of the plate. If we let u = u xx + u yy, which is called the laplacian of u, then we can write u t = k u Remark: The -dimensional wave equation is z tt = a (z xx + z yy ) = a z where z(x,y,t) is the position of the point (x,y) in a vibration elastic surface at time t. Case of the steady state temperature: i.e. we consider a -dimensional heat equation in which the temperature does not change in time (The assumption is after a while temperature becomes steady). Thus u t = 0 Therefore, u = u xx + u yy = 0. This equation is called the -dimensional Laplace equation. Boundary problems and Laplace equation: If we know the temperature on the boundary C of a plate, as a function f (x,y), can we determine the temp. at every point inside the plate? In other word, can we solve the BVP { u xx + u yy = 0 u(x,y) = f (x,y) on the boundary of the plate This BVP is called a Dirichlet Problem. Remark: If the Boundary C is Piecewise Smooth and the function f (x,y) is Nice!, then Dirichlet Problem has a unique solution. We will consider the cases that C is rectangular or circular! Case of Rectangular Plates: Suppose the plate is a rectangle positioned in xy-plane with vertices (0,0),(0,b),(a,b),(a,0). Assume we are given the temp. at each side of the rectangle. Then we have the following type BVP.

20 0 u xx + u yy = 0 u(x,0) = f 1 (x,y),0 < x < a u(x,b) = f (x,y),0 < x < a u(0,y) = g 1 (x,y),0 < y < b u(a,y) = g 1 (x,y),0 < y < b General Strategy: The main equation u xx + u yy = 0 is linear, but all of the boundary conditions are nonlinear, the idea is to split this BVP into simpler BVP denotes by A, B, C, D, in which only one of the boundary conditions in non-linear and apply Fourier series method there, and and the end, u(x,y) = u A (x,y) + u B (x,y) + u C (x,y) + u D (x,y) Solving a type BVP of type (A) For 0 < x < a and 0 < y < b, u xx + u yy = 0 u(x,0) = f 1 (x,y) u(x,b) = 0 u(0,y) = 0 u(a,y) = 0 In this case, one can show u n (x,y) = sin nπ a x sinh nπ a (b y) satisfies the linear conditions of the BVP. Recall that sinhx = 1 (ex e x ) and coshx = 1 (ex + e x ), so (sinhx) = coshx and (coshx) = sinhx Method of separation of variables: To actually find u n s Assume u(x, y) = X(x)Y (y). Then u xx + u yy = 0 implies X Y + XY = 0. Thus X X = Y Y. Since RHS only depends on y and LHS only depends on x, these fractions must be constant, say λ. Then X Y X = λ and Y = λ. Thus we obtain X and Y are nonzero solutions of X + λx = 0 and Y λy = 0, such that Y (b) = 0, X(0) = X(a) = 0 This is an endpoint problem on X, we seek those values of λ for which there are nonzero solutions X, such that X(0) = X(a) = 0. The eigen values are λ n = ( nπ a ) and the corresponding eigenfunctions are scalar multiples of X n (x) = sin nπ a x. Now substitute λ n = ( nπ a ) in Y λy = 0 with Y (b) = 0, and solve for Y (y). We obtain the solutions are a scalar multiple of Y n (y) = sinh nπ a (b y). Hence u n (x,y) = X n (x)y n (y) = sin nπ a x sinh nπ a (b y)

21 1 Back to solving a type BVP of type (A) We want u(x,y) = c n u n (x,y) such that u(x,0) = f 1 (x). Hence f 1 (x) = u(x,0) = c n u n (x,0) = c n sin nπ a x sinh nπ a (b) Therefore, c n sinh nπ a (b) is the n-th Fourier sine coefficient b n of f 1 (x) over interval (0, a). Hence c n = b n /sinh nπb a Example: Solve the BVP of type (A) if a = b = π and f 1 (x) = 1. Compute u(π/,π/), the temp at the center of the rectangle. Solution: Note that Thus Then u(x,y) = f (x) = u(x,y) = nπ sin nπ a x = nπ sinnx ( ) nπb /sinh sin nπ nπ a a x sinh nπ (b y) a ( nπsinhnπ ) sinnx sinhn(π y) Also note that after computing the first few terms and using sinht = sinht cosht, ( u(π/,π/) = (nπsinhnπ) ) sinnπ/ sinhnπ/ = sinnπ/.5 nπsinhnπ/ In fact, one can argue by symmetry that is is exactly.5. Case of a semi-infinite strip plate! Assume the plate is an infinite plate in the first quadrant whose vertices are (0,0) and (0,b), and u = 0 along the horizontal sides, u(0,y) = g(y) and u(x,y) < as x. Apply the separation of variable method to solve the corresponding Dirichlet problem. Answer: Let u(x,y) = X(x)Y (y). Then u xx + u yy = 0 implies X Y + XY = 0. Thus X X = Y Y. Since RHS only depends on y and LHS only depends on x, these fractions must be constant, say λ. Then X Y X = λ and Y = λ. Thus we obtain X and Y are nonzero solutions of X λx = 0 and Y + λy = 0, such that Y (0) = Y (b) = 0, and u(x,y) = X(x)Y (y) in bounded as x This is an endpoint problem on Y. We want those values of λ for which there are nonzero solutions Y, such that Y (0) = Y (b) = 0. The eigen values are λ n = ( nπ b )

22 and the corresponding eigenfunctions are scalar multiples of Y n (y) = sin nπ b y. Now substitute λ n = ( nπ b ) in X λx = 0 and solve for X(x). We obtain X n (x) = A n e nπ b x + B n e nπ b x Since u n = X n Y n and u and Y n = sin nπ b y are bounded, then X n must be bounded too. Hence A n = 0. Suppress b n. Thus u n (x,y) = X n (x)y n (y) = e nπ b x sin nπ b y Note that u n (x,0) = u n (x,b) = 0 and u n (x,y) < as x. Now, to solve the BVP: We want u(x,y) = c n u n (x,y) such that u(0,y) = g(y). Hence g(y) = u(0,y) = c n u n (0,y) = c n (1) sin nπ b y Therefore, c n is the n-th Fourier sine coefficient b n of g(y) over interval (0,b). Example: If b = 1 and g(y) = 1, then compute u(x, y). Solution: Thus g(y) = u(x,y) = nπ sin nπ b x = nπ sinnx nπ e nπx sinnπy Case of a circular disk: Using polar coordinates (r,θ) to represent the points in a disk, on can transform the Laplace equation into u rr + 1 r u r + 1 r u θθ = 0 Note that 0 < r < a = the radius of the disk, and 0 < θ < π. It turns out that the solution is of the form u(r,θ) = a 0 + (a n cosnθ + b n sinnθ)r n Boundary conditions: If we know the temp. on the boundary of the disk, want to determine the temp. inside. To give the temp. on the boundary means to give a function f (θ) such that u(a,θ) = f (θ) for 0 < θ < π Thus f (θ) = u(a,θ) = a 0 + (a n cosnθ + b n sinnθ)a n = a 0 + an a n cosnθ + a n b n sinnθ Therefore a n a n and a n b n are the n-th Fourier series coefficient of f (θ) over the interval (0,π).

23 Example: If f (θ) = 1 on (0,π) and 1 on (π,π), and radius r = 1, then compute u(r,θ). Answer: We compute the Fourier series of f (θ) = nπ sinnθ. Thus a n = 0 for all n. Hence u(r,θ) = a 0 + (a n cosnθ + b n sinnθ)r n = nπ rn sinnθ Remark: Intuitively, in the above example u(r,0) = 0, which is consistent with the solution: If θ = 0 or π, u(r,θ) = nπ rn sinnθ = 0 Remark: In the above example, if we change the boundary condition to f (θ) = 1 on (0,π), then intuitively, u(r,θ) = 1 everywhere by symmetry. This is consistent with the actual solution to BVP: In this case, the Fourier coefficients of f are a 0 =,a n = b n = 0 for all n 1. Thus u(r,θ) = 1 for all 0 < r < 1 and 0 < θ < π. 3