Functions 4 PDF

# Chapter 4: Partial Differentiation ## (4.1) Definitions Consider a function u = u(x,y) of 2 independent variables x,y. We can think of u as being the height of a surface above the (x,y) plane. - It is often helpful to visualize the surface using *contour lines*. - w(x,y) = c (constant) for di...

# Chapter 4: Partial Differentiation ## (4.1) Definitions Consider a function u = u(x,y) of 2 independent variables x,y. We can think of u as being the height of a surface above the (x,y) plane. - It is often helpful to visualize the surface using *contour lines*. - w(x,y) = c (constant) for different values of c - Example: u = x^2 + y^2 - 5 has circular contours - Physically, u could represent a geometrical object or temperature or pressure, etc. - We now look at (spatial) rates of change. - First, start at P(x,y) and move a small distance δx = h in the x direction to Q (x+h,y) i.e. keeping y fixed. - Then we define (if this limit exists) - - du/dx = lim (h->0) [u(x+h,y)-u(x,y)]/h. - This is the rate of change of u with respect to x at P (keeping y fixed) - Notations: du/dx, d^2u/dx^2, etc. - Similarly, P(x,y) -> R(x,y+k): - - du/dy = lim (k->0) [u(x,y+k)-u(x,y)]/k. - This is the rate of change of u with respect to y at P (keeping x fixed). - Notations: du/dy, d^2u/dy^2, etc. - Example: u = x^2 sin(y) + y^3 - du/dx = 2x sin(y) - du/dy = x^2 cos(y) + 3y^2 - We can, of course, consider *higher derivatives*. - d^2u/dx^2 = d/dx(du/dx) = 2 sin(y) - d^2u/dydx = d/dx(du/dy) = 2x cos(y) - Note the order - For the example above, we have: - d^2u/dx^2 = 2 sin(y) - d^2u/dy^2 = -x^2 sin(y) + 6y - d^2u/dxdy = 2x cos(y) - d^2u/dydx = 2x cos(y) - We note that d^2u/dxdy = d^2u/dydx in this case. This is a general result (requiring only continuity of LHS and RHS - usually the case!). - Similarly, we generally have: - u_xxyy = u_yyxx = etc. - Example: u(x,t) = a sin(x-ct) - a, c constants, - du/dx = a cos(x-ct) - du/dt = -a sin(x-ct) - d^2u/dx^2 = -a cos(x-ct) - d^2u/dt^2 = -a^2 sin(x-ct) - u(x,t) satisfies d^2u/dt^2 = c^2 d^2u/dx^2. This is known as the (ideal) *wave equation*. - In fact, any reasonable function f(x-ct) will satisfy this equation! This represents a *wave form* moving (with c>0 here) to the right. - Example: u = tan^(-1)(y/x) - u satisfies d^2u/dx^2 + d^2u/dy^2 = 0. - This is the (2-dim) *Laplace's equation*. ## (4.2) The Total Differential - When we have a function of a single variable f(x) and we make a small change x->x + δx, so that f-> f + δf, then δf ~ df/dx δx. - In the limit ('small δx->0'), df = df/dx dx. - These are known as *differentials*. - Now for a function of two variables, small changes x->x+δx, y->y+δy lead to u(x,y)->u+δu with δu ~ (du/dx) δx + (du/dy) δy. - In the limit ('small δx->0'), we get du = (du/dx) dx + (du/dy) dy. - These are known as *differentials*. - This is called the *total differential* of u(x,y) - Note: conventions! d, δ, δ... are important! - Example: u = x^2 sin(y) + y^3 - δu = (2xsin(y)) δx + (x^2cos(y) + 3y^2) δy - du = (2xsin(y)) dx + (x^2cos(y) + 3y^2) dy ## (4.3) Function of a Function - The Chain Rule - When we have had previously u = f(x) and x = g(t) (say), we have as a consequence du/dt = (du/dx)(dx/dt) = f'(g(t))g'(t). - Now consider u = u(x,y) where x(t) and y(t). - In section (4.2) we saw that δu ~ (du/dx) δx + (du/dy) δy. - Dividing by δt and taking the limit δt->0, we get du/dt = (du/dx)(dx/dt) + (du/dy)(dy/dt). - This is the chain rule. - If x(t) and y(t), then x and y are functions only of t. - Example: dx/dt = (du/dx)(dx/dt) + (du/dy)(dy/dt) = etc. ## (4.4) Cartesians -> Polars - u(x,y) = u(r,θ) - Cartesians: x = r cos(θ) and y = r sin(θ) - Plane Polars: r = (x^2 + y^2)^1/2 and θ = tan^(-1)(y/x) - Both are orthogonal systems. - We need to be careful! - Example: ∂x/∂r = cos(θ) keeping θ constant and ∂x/∂θ = - r sin(θ) keeping r constant. - Hence we should not be tempted! - Note also the overbar on u above. - The Cartesian and polar versions of our function have the same function values, but are described differently! - Example: u(x,y) = x^2 + y^2 = r^2 = u(r,θ), Not (r^2 + θ^2) - Chain Rule - du/∂r = (∂u/∂x)(∂x/∂r) + (∂u/∂y)(∂y/∂r) - du/∂r = (cos(θ))(∂u/∂x) + (sin(θ))(∂u/∂y) - And du/∂θ = (∂u/∂x)(∂x/∂θ) + (∂u/∂y)(∂y/∂θ) - du/∂θ = (sin(θ))(∂u/∂x) + (cos(θ))(∂u/∂y) - We have therefore **partial differential operators**: - ∂/∂x = (cos(θ)) ∂/∂r - (sin(θ)/r ) ∂/∂θ - ∂/∂y = (sin(θ)) ∂/∂r + (cos(θ)/r ) ∂/∂θ - Which relate rates of change in the two different coordinate systems. - Example: u(x,y) = x^2 y^2 = r^4 (cos^2(θ) sin(2(θ)) = u(r,θ) - - ∂u/∂x = 2xy^2 = 2r^3 (cos(θ) sin^2(θ)) using the above results - ∂u/∂y = 2x^2y = 2r^3 (sin(θ)cos^2(θ) ) - We can express e.g. (∂u/∂x)^2 + (∂u/∂y)^2 in plane polars as [(cos(θ)) ∂/∂r - (sin(θ)/r) ∂/∂θ]^2 + [(sin(θ) ∂/∂r + (cos(θ)/r) ∂/∂θ]^2 - This is equal to (∂u/∂r)^2 + (1/r^2)(∂u/∂θ)^2. ## (4.5) Implicit Functions - If we have a function defined implicitly F(x,y) = 0, then F does not change as x and y do so. - (x,y) constrained to a curve - This implies the total derivative is dF = (∂F/∂x) dx + (∂F/∂y) dy = 0. - So the derivative of y with respect to x is given by dy/dx = - (∂F/∂x) / (∂F/∂y). - Example: F(x,y) = x^2 sin(y) + xy - 1 = 0 - dy/dx = -(2x sin(y) + y)/(x^2 cos(y) + x) as before. - If we have an implicit function of 3 variables F(x,y,z)=0, this constrains our point (x,y,z) to be on a particular surface. We can certainly regard, if we wish, x=x(y,z) or y=y(x,z) or z = z(x,y). - Now, no change in F on the surface dF = (∂F/∂x) dx + (∂F/∂y) dy + (∂F/∂z) dz = 0. - Then, at constant y: (∂z/∂x)y = - (∂F/∂x)/(∂F/∂z); at constant x: (∂z/∂y)x = - (∂F/∂y)/(∂F/∂z); at constant z: (∂y/∂x)z = - (∂F/∂x)/(∂F/∂y). - Note: e.g. (∂z/∂x)y = 1 / (∂x/∂z)y - Here, because of course the variable y is being kept constant on both sides: we are looking at variation on a constant y slice of the F=0 surface! - Example: In thermodynamics, the equation of state of a gas/liquid is written F(P,V,T)=0 where P=pressure, V=volume, T=absolute temperature. - This implies an implicit definition of P=P(V,T). - Only in simple cases can we express this relation explicitly e.g. Ideal gas, P = RT/V. - In any case, from the general relation above, we can show that (∂P/∂V)(∂V/∂T)P(∂T/∂P)V = -1. This is an example of an exact thermodynamic identity. ## (4.6) Exact Differentials - We know that for a function of two variables w(x,y), the total differential (derivative) is du = (∂u/∂x) dx + (∂u/∂y) dy. - And of course ∂u/∂x and ∂u/∂y will both, in general, be functions of x and y. - Now consider the converse problem! Given P(x,y) dx + Q(x,y) dy (i.e. given P and Q), when is it the case that this is the total differential of some (as yet unknown) function u(x,y)? - If it is such, then P(x,y) = ∂u/∂x and Q(x,y) = ∂u/∂y for that function u(x,y). - This *implies* and is *implied by* the condition of integrability ∂P/∂y = ∂Q/∂x. - Example: - (y^2) dx + (x^2 + 2y) dy - P = y^2, Q = x^2 + 2y - ∂P/∂y = 2y != ∂Q/∂x = 2x. - This is a test fail, so it's not exact. - (2xy - cos(x)cos(y)) dx + (x^2 - sin(x)sin(y)) dy - P = 2xy - cos(x)cos(y) - Q = x^2 - sin(x)sin(y) - ∂P/∂y = 2x - cos(x)sin(y) = ∂Q/∂x - This is a test pass. - So ∂u/∂x = 2xy + cos(x)cos(y). - This implies that u(x,y) = x^2 y + sin(x)cos(y) + f(y). - Then either (∂u/∂y) = -sin(x)sin(y) + df/dy = x^2 - sin(x)sin(y) - This implies that df = 0 and f(y) = K, a constant with respect to x and y. - Or (∂u/∂y) = x^2 - sin(x)sin(y) + dg/dy = x^2 - sin(x)sin(y), a constant of integration with respect to y. - This implies that u(x,y) = x^2 y + sin(x)cos(y) + K. - Note: P dx + Q dy = 0 => du/dx = -P/Q, so solution u(x,y) = constant. ## (4.7) Stationary Points - In section (2.3) we looked at stationary points for functions of 1 independent variable. What happens for 2 independent variables? - There are 3 types of stationary points, all with a locally horizontal tangent plane: - (Local) maximum, w(x,y) decreasing away from the point. - (Local) minimum, w(x,y) increasing away from the point. - Saddle point, w(x,y) increasing/decreasing depending on direction. - For functions of two variables w(x,y), stationary points are located at simultaneous solutions of the two equations ∂u/∂x = 0 and ∂u/∂y = 0. - This makes the tangent plane horizontal. - Each (x_0, y_0) has character determined by E = (∂^2u/∂x^2)(∂^2u/∂y^2) - (∂^2u/∂x∂y)^2 = B^2 - AC via: - E > 0 => Sadddle - E < 0 => A < 0 => (Local) maximum => A > 0 => (Local) minimum - E = 0 => Higher order derivatives determine the issue, not considered here. - Proof of the above criteria in chapters.... - Example: w(x,y) = x^3 + xy^2 - x - yx^2 - y^3 + ty = (x - y)(x^2 + y^2 - 1) - ∂u/∂x = 3x^2 + y^2 - 1 - 2xy = 0 - ∂u/∂x = 2xy -x^2 - 3y^2 + 1 = 0 - 2x^2 - 2y^2 = 0 => y = x - 2x^2 + x^2 - 1 - 2x^2 = 0 => x^2 = 1 => x = ±1 - There are 4 stationary points: (1,1), (-1,-1), (1,-1), (-1,1). - P | A = (∂^2u/∂x^2) | B = (∂^2u/∂y∂x) | C = (∂^2u/∂y^2) | B^2 - AC = E | u_0 | Type ----|---|---|---|---|---|---| - P_1 | (6x - 2y) | (2y - 2x) | (2x - 6y) | 8 | 0 | Saddle - P_2 | 4 | 0| - 4 | 8 | 0 | Saddle - P_3 | 8/16 | -4/16 | 8/16 | -8 | -313/312 | Minimum - P_4 | -8/16 | 4/16 | -8/16 | -8 | 313/312 | Maximum - Armed with this information, we can sketch the contours. - The zero contour is y=x together with x^2 + y^2 = 1. - Warning! When we are faced with a function of several variables and we need to find stationary points (and potential local max and min), we need to make sure that our independent variables are indeed independent. - Example: Maximise volume V of a rectangular box given the surface area A (fixed). - max V = xyz - given that A = 2xy + 2yz + 2xz: fixed - x, y, z are not independent. - Simple method: Write z = A-2xy => V = xy(A-2xy) - Now, x, y are independent. - x = A/(2(x+y)) = y => (x+y) = A/2 => x=y=A/4, box is cubical! - And V_max = 1/6 (A/2)^3.