Differential Equations

Transcription

1 Differential Equations Problem Sheet 1 3 rd November 2011 First-Order Ordinary Differential Equations 1. Find the general solutions of the following separable differential equations. Which equations are non-linear? (a) (b) (c) 1 / 2cot 2. Consider the differential equation /, where the right-hand side is an arbitrary function of /. Show that the substitution (or equivalently, /) in general renders this differential equation separable. Hence find the general solution of. 3. Use integrating factors to find the general solutions of the following differential equations: (a) 3cos 2 (b) Use integrating factors to solve the following initial value problems: (a) (b), 0 4, 1 0

2 Differential Equations Problem Sheet 2 14 th November 2011 Second-Order Ordinary Differential Equations 1. Find the general solutions of the following two equations: (a) (b) In each case, show that the two functions in the general solution are linearly independent by calculating the Wronskian determinant. 2. Solve both of the equations in Q1 for the initial conditions 0 1 and Using the results of Q1, obtain the general solutions of the following differential equations: (a) (b) (c) Hint: special case 4. Find the general solution of the following third-order homogeneous equation: (You will need to factorise the characteristic equation.) Compute the Wronskian determinant of the three functions. 5. The second equation in Q1 has equal roots. In the lectures we used the method of variation of parameters (closely related to the method of reduction of order) to derive the general solution. Another approach looks at the limit as the double root is approached. Starting with the equation: 0 the characteristic equation has two solutions. Instead of the basis,, choose the linear combination, / (are these independent?). Then consider the behaviour of the second function as tends to zero.

3 Differential Equations Problem Sheet 3 24 th November 2011 Power Series and Generalised Power Series (Frobenius) Solutions 1. Evaluate the radius of convergence for the three series below. Recall that is defined as the distance from the centre (about which the expansion is made) out to which the series converges. 2, 1, Hermite s equation is: 2 20, where is a constant. Determine the solution of the equation by expanding as a power series about the origin..., and, in this way, establish the following recurrence relation between successive terms: Express your solution as two series, each with an arbitrary constant. Examine your solution for different integer values of 0,1,2,3, and in each case identify which of the two series terminates after a few terms. Compare these polynomials to the Hermite polynomials from quantum mechanics. 3. The first four egendre polynomials are 1,, 3 1, and 5 3. The coefficients l of the egendre series expansion of a function in the range 1 1 are given by: l 2l 1 2 l. a) Evaluate the integrals to find the first two non-zero terms in the egendre series expansion of. Verify that the sum of these two terms is exactly. b) Demonstrate that each of the polynomials,, is orthogonal to the other two, on the interval Using the Frobenius method, show that the differential equation

4 has the generalised power series solution 4 2! 4, 2 1! where and are arbitrary constants. 5. Using the Frobenius method, show that the aguerre differential equation has one series solution of the form: 1 0, !, where is an arbitrary constant. Show that this series terminates whenever is a non-negative integer. (The aguerre polynomials defined in this way are closely related to the energy eigenfunctions of the Hydrogen atom.)

5 Differential Equations Problem Sheet 4 5 th December 2011 Partial Differential Equations 1. Determine the Fourier sine series that represents a string plucked at its mid-point, i.e. represent the function 2/ 2 / 0 /2 /2 by its Fourier sine series. [Hint: To evaluate the Fourier coefficients you integrate the product of the function with each sine term. Sketch the relevant functions and identify ways of reducing the number of terms you need to integrate]. 2. (From 2007 exam) A taut wire is fixed at points 0,. The displacement, is governed by the wave equation where is the speed of propagation. 1, (i) By assuming a separable solution of the form,, derive ordinary differential equations to be solved for each of the functions,. [6 marks] (ii) Show that the boundary conditions impose a general solution of the form sin sin cos [6 marks] (iii) At 0 the displacement is everywhere zero, when the wire is hit at the mid point, such that, /2, /2 Determine expressions for the coefficients, of the solution. [8 marks] 3. A cylindrical rod of length is insulated over its curved surface. The end of the rod at 0 is in contact with a heat bath at temperature Θ and the end of the rod at is in contact with a heat bath at temperature Θ. After some time, a steady state is reached. The heat equation that describes the temperature profile of the rod is 1, where is a constant and Θ, is the temperature at position and time.

6 (i) Find the general steady-state (time-independent) solution of the heat equation and hence show that the steady-state temperature distribution is: Θ Θ. (ii) (iii) At time 0, the rod is disconnected from the heat baths. Assuming that no heat subsequently leaves or enters the rod, write down the boundary/initial conditions: (a) at 0; (b) at ; and (c) at 0 for 0. (Hint: recall Fourier s law of heat flow, where is the conductivity.) Writing Θ, use the method of separation of variables to derive ordinary differential equations satisfied by the functions and. Note that the separation constant may be zero, positive ( ) or negative ( ). Using the homogeneous boundary conditions investigate each of these possibilities. In this way show that the distribution of temperature along the rod may be expressed in the form Θ, cos, which is a linear combination of all possible solutions. Here and ( 1,2, ) are constants. (iv) The initial condition may be satisfied by expanding Θ, 0, 0, in a half-range Fourier cosine series. Make a suitable extension of Θ, 0 into the range x 0 and find the required Fourier expansion coefficients. (Note that you will need to include the constant term.) Hence show that: Θ, cos, 0. (v) What happens as? 4. (From 2005 exam) The partial differential equation: 1, 0, with the boundary conditions: 0, 0,, 0,, 0, may be solved using the method of separation of variables. (i) Show that the partial differential equation has separable solutions of the form,, where and 1. You may assume that the separation constant is negative. [5 marks] (ii) Solve the ordinary differential equation for subject to the homogeneous boundary

7 conditions to find the allowed values of and the corresponding spatial eigenfunctions. Hence show that the separable solutions are:, / 1, 1,2,3,, where is an arbitrary constant. [7 marks] (iii) Given that 1 sin, write down an expression for,. [4 marks] (iv) As, every term in the summation that gives, tends to zero. Which term tends to zero the most slowly? Write down an approximation for, valid when is very large but finite. Sketch, in this limit. [4 marks]

8 Differential Equations Problem Sheet 5 13 th December 2011 PDEs in polar and spherical coordinates and Sturm-iouville problems 1. In solving aplace s equation in spherical coordinates, separation of variables,, leads to the following equation for : 1 sin 1 0. Using the substitution cos show that transforming the above differential equation for the function into a differential equation for the function, where cos, leads to the associated egendre equation: Hint: show that 1 / (Exam question) The displacement,, of a circular drum head, radius, satisfies the wave equation: 1 1 1, in cylindrical coordinates, where is the wave speed. a) If the displacement is independent of, show that the wave equation has separable solutions of the form,, where,, and is a separation constant. [7 marks] b) The homogeneous boundary condition, 0 fixes the eigenvalues and eigenfunctions of the equation for. The eigenvalues, 1,2,3,, are negative and hence may be written as, where is a positive constant. For the rest of this question, you may assume that the eigenvalues and eigenfunctions are known. Your answers may be expressed in terms of these quantities. (i) Find the function that appears in the separable solution,. [4 marks] (ii) Write down the general solution of the wave equation subject to the homogeneous boundary condition at. [4 marks] (iii) Write down the specific solution that also satisfies the initial conditions, 0 3,, 0 0. [5 marks] 1

9 3. Consider a second-order differential equation of the form:. Show that by multiplying this equation by an appropriate function, an equation of the Sturm-iouville form may be obtained. Find the multiplicative factors required to bring the following equations into Sturm-iouville form: Hermite s equation 2 2, aguerre s equation 1, (Q4, 5 are for an extra challenge, and are not essential.) It may be shown that the function 1 Φ, 1, 1 2 which is known as the generating function for the egendre polynomials, satisfies the following useful equation: Φ,. Recalling the binomial expansion ! 1 2, 3! and setting 2 and, use the generating function to show that 1,, The electrostatic potential at the point arising from a charge at the point is 4. Given that the squared length of any vector is equal to, show that 4 4 2, where is the angle between the vectors and. (You have just derived the cosine rule.) Using the generating function from Q4, and considering the case when, show that 4 4 cos,. This is called the multipole expansion of the Coulomb interaction. It converges rapidly when and is often used to work out the potential far away from finite charge distributions such as atoms or electronic components. 2

10 Differential Equations Answers to Problem Sheet 1 First-Order Ordinary Differential Equations 1. They are all non-linear because none can be expressed in the form dy + f(x)y = g(x). dx (a) dy dx = x 2 y(1 + x 3 ) y dy = x x 3 dx 1 2 y2 = 1 3 ln(1 + x3 ) + c (b) where c is an arbitrary constant. x dy ( dx = 1 y 2) 1/2 dy dx (1 y 2 ) 1/2 = x cos θdθ (1 sin 2 θ) 1/2 = ln(x) + c (substitution y = sin θ) θ = ln(x) + c arcsin y = ln(x) + c y = sin(ln(x) + c) (c) dr = 2r cot θ dθ 1 dr = cot θdθ 2 r 1 ln r 2 = ln(sin θ) + c ln r 1/2 = ln(c sin θ) r = c 2 sin 2 θ where c, c are arbitrary constants. 2. Given the differential equation dy dx = f(y/x), where f is an arbitrary function, the substitution y = xu, 1 dy dx = u + xdu dx,

11 yields u + x du dx = f(u), which is separable. Applying this method to or equivalently xdu dx = f(u) u, gives the equation dy dx = x2 + xy + y 2 x 2 = 1 + (y/x) + (y/x) 2, x du dx = (1 + u + u2 ) u = 1 + u 2. Separating variables and integrating: du 1 + u 2 = dx x sec 2 θdθ 1 + tan 2 θ = ln(x) + c (substitution u = tan θ, du = sec 2 θdθ) dθ = ln(x) + c (since 1 + tan 2 θ = sec 2 θ) θ = ln(x) + c arctan(y/x) = ln(x) + c y = x tan(ln(x) + c). 3. (a) The integrating factor for the differential equation dy dt + 1 ( ) 1 t y = 3 cos(2t) is exp t dt = exp (ln t) = t. After multiplying by this factor, the ODE becomes Integrating: t dy + y dt = 3t cos(2t) or d (ty) dt = 3t cos(2t). ty = 3t cos(2t)dt = 3t 1 2 sin(2t) 3 1 sin(2t)dt (integration by parts) 2 = 3 2 t sin(2t) cos(2t) + c, where c is an arbitrary constant. So the general solution is y = 3 2 sin(2t) + 3 4t cos(2t) + c t 2

12 (b) The differential equation (1 + t 2 ) dy dt + 4ty = 1 (1 + t 2 ) 2 rearranges to dy dt + 4t 1 + t 2 y = 1 (1 + t 2 ) 3. So the integrating factor is ( ) 4t exp 1 + t 2 dt = exp(2 ln(1 + t 2 )) = exp(ln[(1 + t 2 ) 2 ]) = (1 + t 2 ) 2, where the integral was evaluated by noting that 2t (half the numerator) is the derivative of (1 + t 2 ) (the denominator). Multiplying by the integrating factor gives or Integrating: (1 + t 2 ) 2 y = = = = θ + c (1 + t 2 ) 2 dy dt + 4t(1 + t2 )y = t 2 d ( ) (1 + t 2 ) 2 y = 1 dt 1 + t t 2 dt tan 2 θ sec2 θdθ (substitution t = tan θ, dt = sec 2 θdθ) dθ (since 1 + tan 2 θ = sec 2 θ) = arctan(t) + c, where c is an arbitrary constant. Hence, the general solution is y = arctan(t) + c (1 + t 2 ) (a) By inspection, the integrating factor for the differential equation dy dt + 2 t y = cos t t 2 is t 2. Multiplying by this factor, the equation becomes d ( ) t 2 y = cos t. dt Integrating: t 2 y = sin t + c, where c is an arbitrary constant. So the general solution is: y = sin t t 2 + c t 2 The initial condition, y(π) = 0, implies that c = 0 and so the specific solution we seek is 3 y = sin t t 2

13 (b) By inspection, the integrating factor for the differential equation t 3 dy dt + 4t2 y = e t is t. Multiplying by this factor, the equation becomes Integrating: d ( ) t 4 y = te t. dt t 4 y = te t dt = te t + e t dt = (1 + t)e t + c, where c is an arbitrary constant. So the general solution is y = 1 + t t 4 e t + c t 4 (integration by parts) The initial condition, y( 1) = 0, implies that c = 0. So the specific solution we seek is y = 1 + t t 4 e t 4

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17 Differential Equations Answers to Problem Sheet 3 Power Series and Generalised Power Series (Frobenius) Solutions 1. For each of the series, we apply the ratio test for absolute convergence, i.e. compute the ratio r which is the limit m of the ratio between successive terms: r m = t m+1 t. m The series converges over the region where r < 1. For a) we have so and r < 1 out to x = 2, so R = 2. For b) we have so R = 1. 2 m x m+1 r m = 2 m+1 x m, r = x 2, ( 1) m+1 (m + 1)x m+2 r m = ( 1) m (m + 2)x m+1 x For c), the series is expanded about the point x = a. We have (x a) 2(m+1) 2 m r m = (x a) 2m 2 (m+1). So the condition for convergence is (x a) 2 r = 2 < 1. To obtain R we ask how large may (x a) become such that the above condition is true, which obtains when the above expression is an equality. So the answer is R = We seek the coefficients of the power series that satisfies Hermite s equation Differentiating the series successively we have: y = C 0 + C 1 x + C 2 x 2 + C 3 x d 2 y dy 2x dx2 dx + 2py = 0. 1

18 dy dx = C 1 + 2C 2 x + 3C 3 x 2 + 4C 4 x 3... d 2 y dx 2 = 2C 2 + 6C 3 x + 12C 4 x C 5 x So writing out each element of the equation, and lining up the terms, we have: 2py = 2pC 0 +2pC 1 x +2pC 2 x 2 +2pC 3 x x dy dx = 2C 1x 4C 2 x 2 6C 3 x 3... d 2 y dx2 = 2C 2 +6C 3 x +12C 4 x 2 +20C 5 x 3... The first column in which all elements have a term is x 1, so we start the series there. 2py = 2pC 0 + n=1 x n 2pC n 2x dy dx = n=1 x n 2nC n d 2 y dx2 = 2C 2 + n=1 x n (n + 1)(n + 2)C n+2 So summing the elements and equating to zero yields: 0 = 2pC 0 + 2C 2 + x n [2pC n 2nC n + (n + 1)(n + 2)C n+2 ]. n=1 We now set the coefficients of each power to zero. For x 0, the orphan terms give C 2 = pc 0. ooking at the sum term, we find the recursion relation C n+2 = 2(n p) (n+1)(n+2) C n. Because this relates C n+2 to C n we can anticipate that this will lead to two series. Note also that the recursion relation works even for n = 0, i.e. gives C 2 in terms of C 0. Nevertheless, this does not always apply, and as a general rule the orphan terms should be treated separately. Inserting n = 1, we get C 3 = 2(1 p) 6 C 1, and then inserting successively n = 3, 5, 7, we can get an odd series in terms of C 1, and inserting successively n = 2, 4, 6, we can get an even series in terms of C 0. The general solution of the differential equation may be written: where y 1 is the even series: and y 2 is the odd series: y 2 = x + y 1 = 1 px 2 2(1 p) x 3 + 3! y = C 0 y 1 + C 1 y 2, 4(2 p)p x 4 4! 4(3 p)(1 p) x 5 + 5! 8(4 p)(2 p)p x ! 8(5 p)(3 p)(1 p) x ! Then, if p is a non-negative integer, we find that one or other of these two series terminates. We may write down the terminating series, successively, for p = 0 y = C 0 (1), for p = 1 y = C 1 (x), 2

19 for p = 2 for p = 3 for p = 4 for p = 5 y = C 0 (1 2x 2 ), y = C 1 (x 2 3 x3 ), y = C 0 (1 4x x4 ), y = C 1 (x 4 3 x x5 ). The terms in brackets, when suitably normalised (i.e. multiplied by constants), are the Hermite polynomials. 3. (a) If f(x) = x 3, the first few coefficients are: c 0 = x3 dx = 0 (odd integrand) c 1 = 3 [ ] x x3 dx = 3 x = c 2 = (3x2 1) x 3 dx = 0 (odd integrand) c 3 = 7 [ (5x3 3x) x 3 dx = x x5] 1 = Hence, the first two non-zero terms in the egendre series are 3 5 P 1(x) P 3(x) = 3 5 x (5x3 3x). The sum of these two terms gives x 3 exactly. This is an example of a general result: any n th order polynomial may be expanded exactly in terms of the egendre polynomials up to and including P n (x). The integrals for the expansion coefficients c l with l > n are all equal to zero. (b) We need to show that (i) P 1 and P 2 are orthogonal, (ii) P 1 and P 3 are orthogonal, and (iii) P 2 and P 3 are orthgonal, which is to say that, in each case, the integral of the product of the two functions over the interval 1 < x < 1 is zero. For (i): I = 1 1 P 1 P 2 dx = x 1 2 (3x2 1)dx = [ 1 1 3x 3 xdx = 2 (x4 x /2)] 2 1. Evaluated this is I = 1 [(1 1/2) (1 1/2)] = 0, 2 as required. Clearly if the integrand is an odd function, the result will always be zero. For (ii): 1 1 P 1 P 3 dx = x 1 2 (5x3 3x)dx = [ 1 1 5x 4 3x 2 dx = 2 (x5 x )] 3 1. Evaluated this is I = 1 [(1 1) ( 1 + 1)] = 0, 2 3

20 as required. For (iii): I = 1 1 P 2 P 3 dx = (3x2 1) 1 2 (5x3 3x)dx = x 5 14x 3 + 3xdx 4 1 Since the integrand is n odd function, from (i) we know that the integral is zero, as required. 4. We insert the generalised power series solution: y = C 0 x r + C 1 x r+1 + C 2 x r+2 + C 3 x r+3... into the equation. Differentiating once, and then again, we obtain: dy dx = rc 0x r 1 + (r + 1)C 1 x r + (r + 2)C 2 x r+1 + (r + 3)C 3 x r+2... d 2 y dx 2 = r(r 1)C 0x r 2 + (r + 1)rC 1 x r 1 + (r + 2)(r + 1)C 2 x r + (r + 3)(r + 2)C 3 x r+1... Using these, we now assemble the three elements of the equation, with the powers of x aligned: 2y = 2C 0 x r +2C 1 x r+1 +2C 2 x r+2... dy dx = rc 0x r 1 +(r + 1)C 1 x r +(r + 2)C 2 x r+1 +(r + 3)C 3 x r x d2 y = dx 2 2r(r 1)C 0 x r 1 +2(r + 1)rC 1 x r +2(r + 2)(r + 1)C 2 x r+1 +2(r + 3)(r + 2)C 3 x r+2... We have orphan terms in x r 1 and can start the series at x r. 2y = + n=0 x r+n 2C n dy dx = rc 0x r 1 + n=0 x r+n C n+1 (r + n + 1) 2x d2 y = 2r(r 1)C dx 2 0 x r 1 + n=0 x r+n C n+1 2(r + n + 1)(r + n) Summing the three elements of the equation and equating to zero, we have: 0 = x r 1 C 0 [r + 2r(r 1)] + x r+n [2C n + C n+1 (r + n (r + n + 1)(r + n))]. n=0 We equate the coefficient of each power of x to zero, and since we disallow C 0 = 0, the x r 1 term gives us the indicial equation: r + 2r(r 1) = 0, while the coefficient in the sum term provides a recursion relation between C n+1 and C n C n+1 = 2C n (r + n + 1)(2r + 2n + 1), from which we can get C 1, C 2, C 3,... in terms of C 0. The indicial equation gives 2r 2 r = 0, with solutions r = 0 and r = 1/2. Since these do not differ by an integer, we can obtain both series solutions. In the case when r = 0, the recursion relation may be written: C n+1 = 2C n (n + 1)(2n + 1) = 4C n (2n + 2)(2n + 1). 4

21 Hence, or, generally, n=0 term: C 1 = C 0 = ( 4) 2! C 0 n=1 term: C 2 = C 1 = ( 4)2 4! C 0 n=2 term: C 3 = C 2 = ( 4)3 6! C 0 C n = ( 4)n (2n)! C 0. In the case when s = 1/2, the recursion relation becomes: C n+1 = 2C n (n )(2n + 2) = 4C n (2n + 3)(2n + 2). Hence, or, generally, n=0 term: C 1 = C 0 = ( 4) 3! C 0 n=1 term: C 2 = C 1 = ( 4)2 5! C 0 n=2 term: C 3 = C 2 = ( 4)3 7! C 0 C n = ( 4)n (2n + 1)! C 0. The two series are two independent functions y 1, y 2. Hence, the generalised series solution of the ODE is ( 4x) n y = c 1 + c 2 x 1/2 ( 4x) n (2n)! (2n + 1)!, n=0 where c 1 and c 2 are arbitrary constants. 5. Following the same procedure as in 5., above, we substitute the generalised power series solution, and assemble the four elements of the equation, lining up the powers of x: py = pc 0 x r +pc 1 x r+1 +pc 2 x r+2... x dy dx = rc 0x r (r + 1)C 1 x r+1 (r + 2)C 2 x r+2... dy dx = rc 0x r 1 +(r + 1)C 1 x r +(r + 2)C 2 x r+1 +(r + 3)C 3 x r+2... x d2 y = r(r 1)C dx 2 0 x r 1 +(r + 1)rC 1 x r +(r + 2)(r + 1)C 2 x r+1 +(r + 3)(r + 2)C 3 x r+2... We have orphan terms in x r 1 and can start the series at x r. py = + n=0 x r+n pc n x dy dx = n=0 x r+n (r + n)c n dy dx = rc 0x r 1 + n=0 x r+n (r + n + 1)C n+1 x d2 y = r(r 1)C dx 2 0 x r 1 + n=0 x r+n (r + n + 1)(r + n)c n+1 Summing the four elements of the equation and equating to zero, we have: 0 = x r 1 C 0 (r+r 2 r)+ x r+n [pc n (r+n)c n +(r+n+1)c n+1 +(r+n+1)(r+n)c n+1 ]. n=0 We equate the coefficient of each power of x to zero, and since we disallow C 0 = 0, the x r 1 term gives us the indicial equation: r 2 = 0, i.e. a single repeated solution r = 0, so we can only get a single series solution by the standard method. n=0 5

22 Setting r = 0, the coefficient in the sum term provides the recursion relation: C n+1 = C n(p n) (n + 1) 2, from which we can get C 1, C 2, C 3,... in terms of C 0, as follows: or, generally, n=0 term: C 1 = p (0+1) 2 C 0 = p 1 2 C 0 n=1 term: C 2 = p 1 (1+1) 2 C 1 = (p 1)p C 0 n=2 term: C 3 = p 2 (2+1) 2 C 2 = (p 2)(p 1)p C 0 C n = ( 1)n p(p 1)(p 2)... (p (n 1)) (n!) 2 C 0. A series solution of the aguerre equation is thus y = C 0 n=0 ( 1) n p(p 1)(p 2)... (p (n 1)) x n (n!) 2. If p=n is an integer, the numerator of every term with n > N contains a factor (N ((N +1) 1)) = 0. The series therefore terminates after the N th term, yielding the N th order aguerre polynomial N (x). 6

23 Differential Equations Answers to Problem Sheet 4 Partial Differential Equations 1. The Fourier sin series for f(x) = { 2hx/ 0 x /2 2h( x)/ /2 x is the Fourier series for the periodic antisymmetric extension of f(x). Since the extended function is antisymmetric, its Fourier series only has sin terms, and since its repeat length is 2, all those sin terms must also have repeat length 2. The Fourier representation of the antisymmetric extended function therefore takes the form ( ) 2nπx f extended (x) = c n sin 2 n=1 where c n = 2 ( ) 2nπx f extended (x) sin dx. 2 2 This is the standard formula for the Fourier sin coefficients of a periodic function with repeat length 2. Because f extended is an odd function, the integrand appearing in the integral that gives c n is even and we obtain c n = 2 ( ) 2nπx f extended (x) sin dx = 2 ( ) nπx f(x) sin dx This is the standard formula for the coefficients of a half-range Fourier sin series. To find the Fourier sin series, we have to evaluate the integral c n = 2 /2 0 2hx ( ) nπx sin dx + 2 ( 2h 1 x ) sin /2, ( nπx ) dx. Now start to make your life easy by noting that for n even the second integral will always be the negative of the first integral (sketch the functions), so the sum will be zero for n even. For n odd, observe that the second integral will be equal to the first integral, so we only have to evaluate the integral for n odd. Integrating by parts we obtain c n = 8h /2 2 x sin 0 ( nπx ) dx, c n = 8h [ ( ) ( ) ] x nπx 2 nπ cos + 2 /2 nπx n 2 π 2 sin = 8h ( nπ n 2 π 2 sin 2 0 ). 1

24 So the required half-range Fourier sin series is f(x) = 8h ( 1 nπ π 2 n 2 sin 2 n=1,3,5,... ) sin ( ) nπx. 2. (a) Inserting the trial solution gives T d2 X dx 2 = X v 2 d 2 T dt 2. Rearranging this becomes 1 d 2 X X dx 2 = 1 d 2 T v 2 T dt 2. Since the HS depends only on X and the RHS depends only on T each term must be equal to a constant which we write as S = ±k 2. Therefore there are two ordinary differential equations: d 2 X dx 2 = ±k2 X, and d 2 T dt 2 = ±k2 v 2 T. [6 marks] (b) Considering the first equation, for S = +k 2 it is impossible to satisfy the boundary conditions, whereas for S = 0 there is only a trivial solution X = 0. Therefore S = k 2, and the solution is X = C 1 sin(kx) + C 2 cos(kx). To satisfy the boundary conditions, C 2 = 0 and the eigenvalues are k = nπ/, n = 1,. Therefore the second equation becomes d 2 T dt 2 = π 2 v 2 n2 2 T, with solution ( ) ( ) nπvt nπvt T = C 3 sin + C 4 cos. Therefore the solution for particular n is given by ( ) [ ( ) ( )] nπx nπvt nπvt u = sin A n sin + B n cos, where A n, B n are arbitrary constants. The general solution is a linear combination of all possible solutions, and therefore equates to the expression in the question. [6 marks] (c) Since u(x, 0) = 0, B n = 0 for all n. Then, differentiating yields u t (x, 0) = n=1 A n nπv sin( nπx ) = f(x), where f(x) is the triangular function given in the question. In the exam you would evaluate the Fourier series, but here we can use the answer to question 2 to obtain the coefficients A n, by simply scaling the result from 2 by the factor a 0 /(nπvh), i.e. A n = 8a ( ) 0 nπ n 3 π 3 v sin 2 2 [8 marks]

25 3. (a) Any function Θ(x) that satisfies 2 Θ x 2 = 0 is a steady-state solution of the heat equation. The general steady-state solution is Θ(x, t) = a + bx, where a and b are arbitrary constants. The boundary conditions Θ(x = 0) = a = Θ 0 and Θ(x = ) = a + b = Θ imply that a = Θ 0 and b = Θ Θ 0 Hence, the initial steady-state temperature distribution is. (b) Θ(x) = Θ 0 + Θ Θ 0 i. The boundary condition at x=0 is that no heat is flowing in or out of the end of the rod. This implies that the temperature gradient at x=0 is zero: Θ(x, t) x = 0. x=0 ii. The boundary condition at x= is that no heat is flowing in or out of the end of the rod. This implies that the temperature gradient at x= is zero: Θ(x, t) x = 0. x= iii. The initial condition at t=0 for 0 x is that the initial temperature distribution is the steady-state temperature distribution: x. Θ(x, t = 0) = Θ 0 + Θ Θ 0 x, 0 x. (c) Writing Θ(x, t) = X(x)T (t) and substituting into the heat equation gives and hence X(x) T (t) D t 1 T DT (t) t = T (t) 2 X(x) x 2, = 1 2 X X(x) x 2. Since the HS is a function of t only and the RHS is a function of x only, both sides must be equal to the same constant, which we write as ±k 2 : dt = ±Dk 2 T (t), dt d 2 X dx 2 = ±k 2 X(x), where the equations are written as ordinary differential equations, since in each equation there is only one independent variable. [In the following, subscript notation is used, e.g. Θ x implies Θ/ x.] Consider the equation for X(x) first, treating the three cases (separation constant zero, separation constant k 2, and separation constant k 2 ) separately. Note that the homogeneous boundary conditions Θ x (0, t) = Θ x (, t) = 0 imply that X x (0) = X x () = 0. 3

26 Zero separation constant: The general solution of X xx = 0 is X(x) = α + βx, where α and β are arbitrary constants. The boundary conditions X x (0)=0 and X x ()=0 imply that β=0, leading to the spatial eigenfunction X(x) = α. The corresponding function of time, T (t), satisfies the equation T t =0, the general solution of which is T (t)=γ, where γ is another arbitrary constant. The full separable solution for zero separation constant is thus X(x)T (t) = c 0, where c 0 = αγ is also an arbitrary constant. Positive separation constant: The general solution of X xx =k 2 X is X(x) = αe kx + βe kx, where α and β are arbitrary constants. The boundary conditions X x (0)=0 and X x ()=0 imply that αk βk = 0 and αke k βke k = 0. Multiplying the first equation by e k and then subtracting it from the second equation gives ( αk e k e k) = 0. Since 0 and k is positive by assumption, it follows that α=0 and hence (substituting α=0 back into either of the simultaneous equations obtained from the boundary conditions) that β=0 also. This shows that the only solution with positive separation constant is the trivial solution X(x)=0. Negative separation constant: The general solution of X xx = k 2 X is X(x) = α cos(kx) + β sin(kx), where α and β are arbitrary constants. The boundary conditions X x (0)=0 and X x ()=0 imply that kβ = 0 and kα sin(k) + kβ cos(k) = 0. The first of these equations implies that β=0. For almost all values of k, the second equation then implies that α=0, leading only to the trivial solution X(x)=0. If k is an integer multiple of π/, however, sin(k)=0 and the value of α is undetermined. These special values of k yield the spatial eigenfunctions ( ) nπx X(x) = α cos, n = 1, 2,.... The function T (t) that corresponds to X(x) satisfies the equation T t = Dk 2 T = n2 π 2 D 2 T, ) t, where γ is an arbitrary constant. ( the solution of which is T (t) = γ exp n 2 π 2 D 2 The full separable solution for the negative separation constant k 2 = n 2 π 2 / 2 is thus ( nπx X(x)T (t) = a n cos where a n = αγ is an arbitrary constant. 4 ) e n2 π 2 D 2 t,

27 Since the separable solutions obtained for zero and negative separation constants all satisfy the heat equation and the homogeneous boundary conditions, so does the arbitrary linear combination ( ) nπx Θ(x, t) = c 0 + a n cos e n2 π 2 D 2 t. n=1 To obtain the temperature distribution in the rod for all times t>0, we have to find the constants c 0, a 1, a 2, a 3,... for which this linear combination satisfies the initial condition Θ(x, 0) = Θ 0 + Θ Θ 0 (d) We seek coefficients c 0 (which we shall call a 0 /2), a 1, a 2, a 3,... such that Θ(x, 0) = a n=1 ( ) nπx a n cos x. = Θ 0 + Θ Θ 0 To save typing, the constant (Θ Θ 0 )/ will be denoted by Λ from now on. The initial condition may then be written as Θ 0 + Λx = a n=1 a n cos ( ) 2nπx 2 in which form it is clear that the RHS is the Fourier expansion of a symmetric function of period 2. (We know that the function is symmetric because there are no sin terms, and we know that the repeat length is 2 because ( ) ( ) 2nπ(x + 2) 2nπx cos = cos nπ, = cos x. ( ) 2nπx 2 for any integer n.) This implies that the constants a 0, a 1, a 2,... must be the Fourier coefficients of the periodic symmetric extension, f symm (x) = Θ 0 + Λ x x f symm (x + 2n) = f symm (x) n any integer of the function Θ 0 + Λx. The Fourier coefficients of f symm (x) may be obtained using the standard procedure, which is to multiply both sides of the Fourier expansion f symm (x) = a n=1 a n cos ( ) 2nπx 2 ( ) by cos 2mπx 2 and integrate over one complete period (e.g., from to ): ( ) 2mπx f symm (x) cos dx = a 0 cos a n n=1 ( 2mπx 2 cos ) dx ( 2nπx 2 ) cos ( 2mπx 2 ) dx. Since the integral of any sin or cos function over a whole number of periods is zero, the first integral on the RHS vanishes unless m=0, in which case it is equal to 2. Using the 5

28 formula cos(a) cos(b) = 1 2 [cos(a + b) + cos(a b)], the second integral on the RHS may be written as 1 cos 2 ( 2(n + m)πx 2 ) + cos ( 2(n m)πx 2 ) dx, which evaluates to zero if m n or to if m=n. The formula for a m, which works for all m (including m=0), is therefore a m = 1 Setting m equal to 0 gives For non-zero m, we obtain a m = 2 = 2Λ ( ) 2mπx f symm (x) cos dx = 2 0 = 2Λ mπ a 0 = (Θ 0 + Λx) dx = 2Θ 0 + Λ. ( ) 2mπx (Θ 0 + Λx) cos dx = 2Λ 2 ( )] 2mπ 2mπ sin 2 x 2Λ 2 0 2mπ ( ) 2mπ sin 2 x dx = 2Λ 2 mπ 2mπ [ x 2 0 = 2Λ m 2 (cos(mπ) 1) = 2Λ π2 0 ( ) 2mπx f(x) cos dx [ cos m 2 π 2 (( 1)m 1). ( ) 2mπx x cos dx 2 ( ) 2mπ sin 2 x dx ( )] 2mπx Putting the expressions for a 0 and a m back into Fourier cos expansion of f symm (x) gives f symm (x) = Θ Λ 2Λ + π 2 n=1 ( 1) n 1 n 2 cos 2 0 ( ) 2nπx 2, or, replacing Λ by (Θ Θ 0 )/, f symm (x) = Θ 0 + Θ 2 + 2(Θ Θ 0 ) π 2 n=1 [( 1) n 1] n 2 ( ) nπx cos. This function is symmetric and periodic with period 2, but equals the initial condition Θ(x, 0) = Θ 0 + Θ Θ 0 x when 0 x. Finally, substituting the Fourier coefficients a 0 (= 2c 0 ), a 1, a 2,..., back into the general expression for the solution of the PDE derived in Q3, we obtain Θ(x, t) = Θ 0 + Θ 2 + 2(Θ Θ 0 ) π 2 n=1 [( 1) n 1] n 2 ( nπx cos ) e n2 π 2 D 2 t. (e) As t, the temperature at all points x tends to the average temperature Θ 0+Θ (i) Substituting the separable solution u(x, t) = X(x)T (t) into the PDE gives T X = (1 + t)xt 6

29 and hence, dividing through by XT, X X = (1 + t)t T Since the HS is a function of x only, while the RHS is a function of t only, the equality implies that both sides must be equal to a constant, k 2, which we are told to assume is negative. Hence X X = (1 + t)t k2 =. T This separates to give two ODEs, as required. X = k 2 X and T = k2 1 + t T,. [5 marks] (ii) The general solution of the ODE X = k 2 X is X(x) = c 1 sin(kx) + c 2 cos(kx). The homogeneous boundary conditions u(0, t) = u(, t) = 0 imply that X(0) = X() = 0 and hence c 2 = 0 and c 1 sin(k) = 0. The second of these implies that either c 1 is zero (yielding only the trivial solution X(x) = 0) or k = nπ, in which case c 1 is undetermined. The allowed values of k and the corresponding spatial eigenfunctions are thus k n = nπ, ( ) nπx X n(x) = a n sin, n = 1, 2, 3,..., where a n is an arbitrary constant. The function T n (t) that goes with X n (x) satisfies the ODE dt n dt = k2 n 1 + t T n = n2 π 2 2 (1 + t) T n. This is a separable equation: dtn = n2 π 2 T n 2 dt 1 + t ln(t n ) = n2 π 2 ln[c(1 + t)] 2 T n (t) = b n (1 + t) n2 π 2 2 where b n is an arbitrary constant. Combining the two parts, X n (x) and T n (t), of the separable solution u n (x, t) gives the required result: u n (x, t) = c n sin(nπx/) (1 + t) n2 π 2 / 2, n = 1, 2, 3,..., where c n = a n b n. [7 marks] 7

30 (iii) By taking an arbitrary linear combination of the separable solutions from part (ii), we can obtain a general solution of the PDE subject to the homogeneous boundary conditions: u(x, t) = sin(nπx/) c n. (1 + t) n=1 n2 π 2 / 2 Since we are told that u(x, 0) = n=1 ( ) 1 nπx n 2 sin it follows that c n = 1/n 2. The solution we require is thus u(x, t) = n=1 sin(nπx/) n 2 (1 + t) n2 π 2 / 2.. [4 marks] (iv) As t, the (1 + t) n2 π 2 / 2 factor appearing in the denominator of the n th term of the series tends to infinity. Every term in the series therefore tends to zero. The denominator that diverges the most slowly is the one with the smallest value of n. Hence, the term that tends to zero the most slowly is the n=1 term. When t is very large but finite, the n=1 term (small though it is) dominates the series and we can make the approximation u(x, t) sin(1πx/) 1 2 (1 + t) 12 π 2 / 2 sin(πx/) t π2 / 2, i.e. spatially a half sin wave, decaying as a power law in t. [4 marks] 8

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