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Review of 2 nd -order PDEs. 1. 1 st -order PDEs 2. 2 nd -order PDEs 3. An example: acoustic waves (optional) 4. Another example: surface waves in liquids (optional) 5. Separation of variables. Week 1. 1. 1 st -order PDEs.
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Review of 2nd-order PDEs 1. 1st-order PDEs 2. 2nd-order PDEs 3. An example: acoustic waves (optional) 4. Another example: surface waves in liquids (optional) 5. Separation of variables Week 1
1. 1st-order PDEs Consider an unknown function u depending on free variables x and y, and let u satisfy (1) where B, C, and R are given functions. Eq. (1) is called a quasi-linear 2D PDE of the 1st order. Example 1: (2) It can be verified by inspection that (2) is satisfied by (3) where f(x) is an arbitrary function.
Solution (3) implies that u doesn’t change on the straight lines (4) which are called the characteristics of Eq. (2). For the general equation (1), the characteristics cannot be found explicitly, but are determined by the equations (5) Example 1 (revisited): For Eq. (2), the characteristics equation (5) yields the correct expression (4).
Theorem 1 (Method of Characteristics for 1st-order quasi-linear 2D PDEs): The general solution of Eq. (1) satisfies the following set of ODEs: If R = 0, the above ODEs should be, strictly speaking, replaced with
Example 1 (revisited and expanded): (a) For Eq. (2), the equation for the characteristics yields the correct solution (4). (b) Find the solution of Eq. (2) with a boundary condition: Why cannot the solution be found in case (b)? Example 2: Solve a ‘new’ term
The characteristics equations yield Let these equalities be valid on a characteristic ‘emanating’ from a point (x0, 0), with the value of u = (x0)2. Hence, hence, hence, eliminating x0 (by expressing it from the first equation and substituting it in the second),
Example 3 (obtaining solution in parametric form): Solve the following boundary-value problem: Solution: The standard approach yields which can be followed through to the end. We shall, however, try a different approach (which is more convenient in some cases).
Introduce a parameter τ such that the ‘old’ characteristics equations τ = 0 at the boundary Re-arrange the characteristics equations: And solve them, (6)
Parameterise the boundary curve, Don’t confuse the parameters τ and s: the former varies along a characteristic, whereas the latter varies along the boundary. Now, fix c1,2,3 by substituting (6) into the boundary conditions, x, y, andu at the boundary Thus, (6) yields a (doubly parameterised) solution,
Example 4 (‘overturning’ of solutions of quasi-linear PDEs):
2. 2nd-order PDEs Consider a function u depending on free variables x and y, and let u satisfy (7) where the functions A, B, C, and R depend on x, y, and u.
If A<0 for all x, y, and u, then Eq. (7) is called hyperbolic, • If A>0 for all x, y, and u, then Eq. (7) is called elliptic, • IfA = 0 and C ≠ 0for all x, y, and u, then Eq. (7) is called parabolic, • IfA = C = 0for all x, y, and u, then Eq. (7) is an ODE, not a PDE (no y-derivatives), • If, depending on x, y, and u, A changes its sign, we say that Eq. (7) changes its type.
Example 5: the wave equation: wherec is a coefficient (the wave speed). This equation is hyperbolic. Example 6: the Helmholtz equation: whereλ is a coefficient (the wavelength). This equation is elliptic.
Example 7: The linear diffusion equation: whereα is a coefficient (the diffusivity). This equation is parabolic. This equation also describes heat transfer, in which case α is the heat conductivity.
Comments: • The general form of a 2nd-order, 2D PDE is The extra term makes classification more convoluted, but the essence of the matter remains the same. If, for example, A, B, C, and D are constants, the extra term can be easily eliminated by a change of variables. If A, B, C, and D depend on x and/or y and/or u, the extra term can still be eliminated, though not easily. • 2nd-order PDEs can also be classified by their characteristics.
A similar classification can be applied to equations with 3 or more free variables, e.g., the following equations, are hyperbolic, elliptic, and parabolic, respectively. Sometimes, when describing the dimension of a PDE, only those variables are counted that represent spatial coordinates (not the time variable) – which might be confusing. The best way to refer to an equation with, say, 3 coordinates and the time is “a (3+1)-dimensional PDE”.
3. An example: acoustic waves (optional) An ideal gas (no viscosity, no heat conductivity) is described by where p and ρ are the pressure and density of the gas; u, v, and w are the x-, y-, and z-components of the velocity; the coefficient γ is called the adiabatic index (depends on the number of atoms per molecule, for diatomic gases γ = 7/5).
The same equations can be written in vector notation, (8) where u = [u, v, w]. We’ll separate the mean density and pressure from the deviations from these: (9) (observe that there’s no mean velocity).
Let the mean values be uniform in space and time, i.e. not depend on x,y,z, and t. Then, for example, Assume also that the deviations are smaller than the mean values, so let’s linearise Eqs. (8). To do so, substitute (9) into (8) and omit all nonlinear terms (those where the ‘primed’ variables occur more than once), which yields (10) (11) (12)
To eliminate the velocity, differentiate Eqs. (11) and (12) w.r.t. t, (13) and re-write Eq. (10) in the form (14) Substitute (14) into (13), (15) (16)
(16) is a (3+1)-dimensional linear wave equation for p’. Once p’ is found, it can be substituted into (14) and (15) which yield u’ and ρ’. Comment: According to Eq. (16) the wave speed is (17) Given that, at sea level and 20°C, formula (17) yields c ≈ 343 m/s.
Consider a resonator, i.e. a region enclosed by a curved wall. The no-flow boundary condition at the wall is (18) where B is the boundary (wall) and n is a normal to B. Note that the length of n is unimportant, and it doesn’t matter whether it’s internal or external normal. Observe that (18) involves u’, while we need a BC for p’. To eliminate u’, differentiate (18) w.r.t. t and take into account Eq. (14), which yields (19) Given an initial condition, Eq. (16) and BC (19) fully determine p(x, y, z, t).
Let the dependence of the solution on t be harmonic, i.e. (20) where ω is the frequency and x = [x, y, z]. Then, Eq. (16) yields hence, (21) where λ = c/ω is, physically, the wavelength. (21) is a 3D Helmholtz equation. To obtain a BC for it, substitute (20) into (19), which yields (22)
4. Another example: surface waves in liquids (optional) Consider waves on the surface of a liquidlayer and let h(x, y, t) be the layer’s thickness and u(x, y, z, t), the liquid’s velocity (observe that the latter depends on all three spatial coordinates, and the former depends only on the horizontal ones). This problem is characterised by 3 non-dimensional parameters: where ρ and μ are the liquid’s density and dynamic viscosity. Physically, a is the waves’ amplitude, s is the typical slope of the surface, and Re is the Reynolds number. We’ll assume that
We’ll consider two opposite limits: Re << 1 and Re >> 1. In the former case, an asymptotic (approximate) equation can be derived for h’, (23) where (23) is a linear diffusion equation. If Re >> 1, h’ is governed by a linear wave equation, (24) where
It can be shown that, for both Eqs. (23) and (24), the BC are (25) where B is the boundary and n is a normal to B. Comments: • Under certain approximations, the linear wave equation also describes electromagnetic waves (visible light, x-rays, radio waves), waves in plasma, star matter, etc. • The linear diffusion equation comes up in various problems of physics and biology (spreading of species, etc.).
5. Separation of variables Example 8: Oscillations of a string with fixed ends Oscillations of a string are described by (26) where his the string’s displacement and c2 = T/ρ (T and ρ are thestring’s tension and density). We shall assume that the ends of the string are fixed: (27) where L is the string’s length.
BVP (26)-(27) is separable, i.e. admits a substitution of the form for which (26) yields hence, The l.-h.s. of this equation appears to depend on t, whereas the r.-h.s. appears to depend on x – which actually means that neither depends on anything, i.e. equals (the same) constant, say, −ω2...
hence, hence, where A = A1 A2. BCs (27) require which, generally, implies...
where m and n are arbitrary integers – hence, hence, where l = m – n. W.l.o.g. we can put n = 0. Indeed, an even n doesn’t change h at all. An odd n, in turn, changes the sign of h, i.e. nπ can be omitted with a simultaneous change A → −A, then the extra minus can be ‘incorporated’ into A (which is arbitrary anyway). Thus, omitting the subscript from θ1, we obtain a ‘standing wave’ solution...
Comments: • W.l.o.g. we can assume that A ≥ 0 and 0 ≤ θ ≤ 2π. The latter is obvious, while the former is true because we can always change the sign of the solution by changing θ → θ±π). • LetL = π, c = 1. • It is instructive to sketch the ‘frame-by-frame’evolution of a standing wave with, say, A = 1, θ = 0, l = 1. The times can be, say, t = 0, π/2, π, 3π/2. • Draw also a frame-by-frame sketch for l = 2 (the rest of the parameters remain the same). What frame times would be convenient in this case? • Can you predict (without sketching) what happens for l = 3?
Comments (continued): • The standing-wave solution includes 3parameters: A, l, and θ. • A determines the wave’s amplitude (solutions of linear equations always include an amplitude factor). • l is the wavenumber. It represents the number of ‘half-oscillations’ per string’s length, and also determines the time period of the oscillations (see below). • θ is called the phase of the wave (it characterises the initial state, t = 0). • The period of a standing wave is T = 2L/lc, and its frequency is ν = 1/T (the angular frequency is ω = 2π/T). • The wavelength of a standing wave is λ = 2L/l.
Even though the standing wave is just a particular solution, the set of all standing waves with various A, l, and θ is complete, i.e. any solution of (26)-(27) can be represented by Physically, this is a superposition of standing waves with various wavenumbers, amplitudes, and phases. • Comment: • In linear problems, solutions with separated variables often form a complete set, which is why they are important.
Example 9: waves in a rectangular pond Omit the primes in Eqs. (24)-(25) for surface waves: (28) (29)
Consider (28)-(29) in a rectangle 0 < x < Lx , 0 < y < Ly : (0, Ly) (Lx , Ly) (0, 0) (Lx , 0)
Seek a solution of (28)-(29) in the form (30) where A, kx , ky , ω, θx , θy , and θt are undetermined coefficients. Then Eq. (28) yields hence, (31)
Substitute (30) into BCs (29), Left BC Right BC Lower BC Upper BC
The left BC holds for all y and t if and only if hence, The former ‘option’ eliminates the x-dependence from solution (30)– hence, should be discarded. The latter ‘option’ can be reduced w.l.o.g. to θx = 0 (similarly to how we treated θ1 in Example 8).
Going, in a similar fashion, through all of the BCs, we obtain (32) where lx and ly are integers. Observe that θt remains arbitrary. Replacing θt→ θ, let’s summarise solution (30)-(32): where
Comments: • Using the identity we can rearrange the standing-wave solution into Using the identity again and rearranging the resulting expression in such a way that ω always appears with a minus, we obtain...
Those who have done “Mathematics of natural phenomena” should be able to see that the four terms in the above expression describe four ‘inter-scattering’ waves with wavevectors k2 k3 k1 k4
Comments (continued): • The standing-wave solution includes 4 arbitrary parameters: the amplitude A, the phase θ, and the x- and y-wavenumbers lx andly.
