MAE 105 Introduction to Mathematical Physics HOMEWORK 1. Due on Thursday October 1st 2015

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1 MAE 5 Introduction to Mathematical Physics HOMEWORK Due on Thursday October st 25 PROBEM : Evaluate the following integrals (where n =, 2, 3,... is an integer) and show all your steps: (a) x nπx We use integration by parts. Note that d nπx x = x nπx nπ x nπx = nπx x nπ = ( )n+ 2. nπ nπ + nπ nπx, nπx, Notice that because n is an integer, the second integral is zero, ce nπ =, and that nπ = ( ) n. (b) e x nπx For this we use Euler s identity, which asserts that This implies that nπx/ can be written e x nπx e iθ = θ + i θ. nπx = imag e inπx/, = imag Exponentials aren t hard to integrate. We get exp {x + inπ/} = = { } exp x( + inπ/). { } + inπ/ exp x( + inπ/) + inπ/ e nπ, = inπ/ + (nπ/) 2 ( ) n e.,

2 In the last step, we multiplied by ( inπ/)/( inπ/), allowing us to clearly separate the real and imaginary parts of the expression. We also used the fact that nπ = ( ) n. We conclude at last that e x nπx { } = imag exp x( + inπ/)., nπ/ = + (nπ/) 2 ( ) n e. (c) First, we note the product-to-sum trigonometric identity θ φ = 2 (φ + θ) + 2 (φ θ), which can be derived by converting both θ and φ to complex exponential form ug Euler s identity, doing some algebra, and converting back to s. This identity implies that (d) 2 nπx which implies that Putting this into the integral yields PROBEM 2: = 6πx 2, = 2π =. 6πx, This integral can be tackled by noting another product-to-sum identity, θ φ = 2 (θ φ) 2 (θ + φ), 2 nπx = 2 2 2nπx. 2 nπx = x 2 2nπx, 4nπ = 2. Solve the following ordinary differential equations and show all your work: (a) dh dt = λkh with h() = h. The most efficient way to solve this differnetial equation is to realize that this is a constant coefficient equation with a gle solution of the form h = Ae rt, where r is a constant determined by the equation, and A is a constant determined by the initial condition. Our guess is thus that h = Ae rt, and 2 dh dt = raert.

3 Plugging these into the differential equation yields rae rt = λkae rt. A little squinting (or alternatively, rearranging terms in the equation) reveals that we must have r = λk for our guess to be a valid solution. Thus h(t) = Ae λkt. The initial condition then requires h = h() = A, which means that our solution is h(t) = h e λkt. A small amount of extra effort verifying that this h does indeed satisfy the differential equation and initial condition confirms its correctness. (b) d2 φ dφ = 9 φ with φ() = and () =. 2 This is another constant coefficient equation with solutions of the form φ = Ae rx. Because this is a second-order differential equation, we should find two independent solutions. Differentiating our guess for φ twice and plugging it into the equation yields r 2 Ae rx = 9Ae rx, which implies we must have r ± 3. The general solution is therefore The first initial condition implies that φ = Ae 3x + Be 3x. φ() = A + B =, or that A = B. The second initial condition implies that dφ () = 3A 3B =. Ug A = B this implies that 6A = or that A = /6 and B = /6. The solution is therefore (c) d2 f 2 + λ2 f = x with λ >, f() = f, parts, φ = 6 e3x 6 e 3x. df () =. This is an inhomogeneous differential equation. To tackle it we split our solution into two where f h is the homogeneous solution satisfying f = f h + f p, d 2 f h 2 + λ2 f h =, and f p is the particular solution for which d 2 f p / 2 + λ 2 f p equals x. It is good practice to find the homogeneous solution first (the reason for this will become apparent in (d)). Again the solution consists of exponentials of the form f h = Ae rx. Plugging in we find Ae rx r 2 + λ 2 =. 3

4 This implies that and that r = ±iλ, f h = Ae iλx + Be iλx. Now, notice that Euler s identity (given in the solution to (b)) implies that we can also write f h as f h = C λx + D λx, where C = A + B and D = i (B A). Since A, B, C, and D are arbitrary constants at this point, we can choose whichever representation we like. And because the solution is real (not imaginary), es and coes make the most sense to me right now. The particular solution can be found in a number of ways. By far the fastest way is to recognize that when the inhomogeneous forcing (the term x on the right hand side) is a polynomial, the particular solution is also a polynomial. It is clear in this case that if we guess the particular polynomial f p = Ex, where E is a constant, we find that d 2 f p / 2 =, so that the differential equation becomes d 2 f p 2 + λ2 f p = λ 2 Ex. For this to equal x, we choose E = /λ 2. Bringing it all together, the general solution is f = C λx + D λx + x λ 2. The initial condition f() = f implies that C = f. The initial condition df()/ = implies or that D = λ 3. Our final answer is therefore = Dλ + λ 2, f = f λx λ 3 λx + x λ 2. (d) d2 y dt 2 + dy t dt 4 t 2 y = This differential equation is solved by first recognizing that it is an equidimensional equation. Equidimensional equations have the form t n y (n) + t n y (n ) + + ty + y = f(t), where y (n) denotes the n th derivative of y(t) and y denotes the first derivative of y. Our differential equation can be brought into this form by multiplying by t 2. Equidimensional equations are special they become constant coefficient equations after making the substitution x = ln t, and can be solved by a similar easy method. The solution to equidimensional equation takes the form y = At r. Note that in terms of x = ln t, this becomes y = Ae rx, which is the solution form for constant coefficient equations. This simply confirms that the universe makes sense and we have (so far) retained our sanity. To solve our inhomogeneous, equidimensional differential equation, we split the problem into a homogeneous and inhomogeneous part, such that y = y h + y p, 4

5 where y h contains the two solutions to the homogeneous problem, d 2 y h dt 2 + t dy h dt 4 t 2 y h =, and y p, again, is the particular solution that satisfies the differential equation with the on the right hand side. With the guess y h = At r we have dy h dt = ratr, Plugging these into the differential equation yields and At r 2 r(r ) + r 4 d 2 y h dt 2 = r(r )Atr 2. =. This implies that r = ±2, and that the homogeneous solution is y h = At 2 + Bt 2. ike the constant coefficient problem in 2(c), the particular solution can be found by a number of methods. The fastest method is to recognize that the forcing term is a polynomial as good as any other, and that the particular solution to an equidimensional equation forced by a polynomial is, indeed, itself a polynomial. We can see by inspection (for example, multiple the differential equation by t 2 ) that the correct polynomial is simply proportional to t 2, and thus our first guess is y p = Ct 2. Unfortunately, this won t work. The reason is that Ct 2 is one of the homoegeneous solutions, which means that no matter what C is, this solution will always yield = when it is plugged into the differential equation. Since, Ct 2 cannot possibly be the correct particular solution. The problem is that the forcing term is resonant with the differential equation on the left side it is going to produce an interestingly large response in y. If the differential equation were constant coefficient (which it becomes when cast in terms of x = ln t), then the solution is to propose a new particular solution of the form y p2 (x) = x y p (x), where y p is our first guess for the particular solution. Note that if y p2 is also one of the homogeneous solutions, we must try again and multiply by x again. For the equidimensional problem, the analagous approach is to multiply by ln t; in other words, our new guess is y p2 (t) = (ln t) y p (t). In this particular case, where our first guess is y p = Ct 2, we propose y p = C (ln t) t 2. We thus have dy p dt = C (t + 2t ln t), and d 2 y p = C (3 + 2 ln t). dt2 Putting these into the differential equation yields C ln t ln t 4 ln t =, which implies that C = /4. Our general solution is therefore y(t) = At 2 + Bt t2 ln t. 5