PX275 Mathematical Methods for Physicists 2018 with Solutions PDF

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PX275 Mathematical Methods for Physicists 2018 with Solutions...

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Solutions for April 2018 PX275 exam (part 2) M S Turner January 15, 2018 Q3 a) Length2 /Time (by dimensional analysis of the heat equation or otherwise). [2] [Book] b) Substituting ∂T = 0 into the heat equation we require ∂t find T = Ax + B [2] [Book]

∂2T =0. ∂x2

Integrating this twice we

∂Q , reflecting the flow of heat from warmer regions c) The heat flux takes the form j = −κ ∂x to colder regions [2], with κ some material contstant. The continuity equation takes the ∂j which corresponds to conservation of (heat) energy. It arises by a balance form ∂Q = − ∂x ∂t of the heat flowing into a line segment of length dx in time dt with the heat change per unit length multiplied by the segment length dx. Thus [Q(x, t + dt) − Q(x, t)]dx = [j(x, t) − j(x + dx, t)]dt. Dividing both sides by dx dt and taking the limits dx, dt → 0 yields the continuity equation by the first-principles definition of differentiation [2]. Substitution of the form for the flux into the continuity equation gives, for constant κ,

∂Q ∂ 2Q =κ 2 ∂t ∂x Substituting for Q = CT and dividing both sides by C (a constant) one recovers the heat equation [2]. In the special case of a non-uniform thermal diffusivity κ(x) the equation instead takes the form ! ∂Q ∂ ∂ 2Q ∂κ(x) ∂Q ∂Q = + κ(x) 2 = κ(x) ∂x ∂x ∂t ∂x ∂x ∂x or, dividing by C ,

∂κ(x) ∂T ∂T ∂ 2T = + κ(x) 2 ∂x ∂x ∂t ∂x which is not the same as the heat equation [1]. [Book]

d) Using separation of variable according to T (x, t) = χ(x)τ (t) [1] we substitute this into the heat equation and divide both sides by κχ τ . We find 1 τ˙ /τ = χ′′/χ = λ κ where λ is a constant. Physical solutions follow from (i) λ = 0 and (ii) λ < 0 (solutions with λ > 0 grow unbounded in time and are therefore not physical). (i) Writing λ = 0 gives χ′′ = 0 and hence, integrating χ = A0 x + B0 [1]. 1

(ii) Writing λ = −k 2 [3] we obtain χ′′ = −k 2 χ with solutions that are sinusoidal χ = A sin kx + cos kx . We extend the domain of the solution to x ∈ [−L, L] and define it to be periodic, with period 2L. This periodicity leads to the requirement that kL = nπ with n an integer (positive integer without loss of generality), equivalently k = nπ [2]. L [An alternative approach in which the boundary condition of part (e) is anticipated in order to establish this restriction is also acceptable]. The solution for τ then follows by integrating the time equation. (i) For λ = 0 this yields τ = C where we can choose C = 1 without loss of generality. (ii) For λ = −k 2 we find τ = C exp(−κk 2 t) where, again, we can set C = 1 without loss of generality [2]. Hence, noting that we can combine these solutions, due to linearity, yields the general solution stated in the question [1]. e) We note that the boundary conditions that the temperature is zero on both ends for all t > 0 implies that Bn = 0 for all n ≥ 0 and that A0 = 0 [1]. Using the initial condition to determine the An by computing m an integer to show ∞ X

An

n=1

Hence

Z L 0

sin

RL 0

T (x, 0) sin mπx dx with L

Z L nπx mπx mπx sin To sin dx = dx L L 0 L

∞ X

L mπx L L ] = cos An δnm = To [− L 0 mπ 2 n=1

o (1 − (−1)m ) [3] This gives a solution for t > 0 according to the rubric. Hence Am = 2T mπ by substitution ∞ X nπx −κ( nπ )2 t 2To L T (x, t) = (1 − (−1)n ) sin e nπ L n=1 or X 4To nπx −κ( nπL )2 t e T (x, t) = sin L n,odd nπ

[3]R [Book/Unseen]. The heat density is Q(x, t) = CT (x, t) so the total heat Q(t) = R L L = [− nπ cos nπx C 0L T (x, t)dx. Using 0L sin nπx ]L (1 − (−1)n ) this evaluates to L L 0 = nπ Q(t) =

X

n,odd

8CLTo −κ( Lnπ)2 t e (nπ )2

[3] [Unseen]

Q4 a) Substituting into the definition given we find (i) F[a f (x)] = (ii) F[f (a x)] =

R∞

−∞

R∞

−∞

(iii) F[f (x + a)] =

a f (x)e−ikx dx = a f (a x)e−ikx dx =

R∞

−∞

f (x + a)e

−ikx

f (x)e−ikx dx = af˜(k) [2] [Book]. ′ f (x′ )e−ikx /adx′ /a = f˜(k/a)/a [2] [Book].

R∞

−∞

R∞

−∞

dx = 2

R∞

−∞

′ f (x′ )e−ik(x −a) dx′ = f˜(k)eika [2] [Book].

(iv) F[f (x)δ(x − a)] =

R∞

−∞

f (x)δ(x − a)e−ikx dx = f (a)e−ika [2] [Book]. 1

2

2

∞ ∞ e−(x/a) e−ikx dx = e−(ka) /4 −∞ e− a2 (x−ika /2) dx by completing the square. (v) F[f ] = −∞ By changing variables to x′ = x − ika2 /2 and using the result given in the rubric, √ 2 F[f ] = a π e−(ka) /4 [5] [Book]. 2

R

2

R

b) Using the expression given (i) Two pinholes at y = (0, l, 0) and y = (0, −l, 0) equates to an aperture a(y) = δ(y − (0, l, 0)) + δ(y − (0, −l, 0)) [up to an arbitrary constant]. Substituting and integrating, with k = kx/D, we find  e−iωt e−iωt  −ik·(0,l,0) klx2 e + e−ik·(0,−l,0) = 2 u(x, t) = cos D D D The intensity is [3] klx2 I = |u|2 ∝ cos2 D (with [3] for correct sketch, see below) (ii) A rectangular aperture, centred at the origin, with width 2w and height 2h with h > w equates to an aperture function that is zero outside this domain, hence

u(x, t) =

e−iωt Z w dy1 D −w

Z h

dy2 e−ik·y =

−h

Z h klx1 klx2 e−iωt Z w dy1 e−i D y1 dy2 e−i D y2 −h D −w

 e−iωt  D −i klx1 y1 w D −i klx2 y2 h D D [e [e = ]−w ]−h klx1 klx2 D

Hence

with sinc x =

klhx2 klwx1 D e−iω t D sin sin u(x, t) = 4 D D D klx2 klx1 sin x x

u(x, t) = 4wh

e−iωt klhx2 klwx1 sinc sinc D D D

The intensity is [3]

klhx2 klwx1 sinc2 D D (with [3] for correct sketch, see below) [Book/unseen]. I = |u|2 ∝ sinc2

3

!

4...