Comprehensive University Study Notes: Advanced Multivariable Calculus, Vector Analysis, and Ordinary Differential Equations
Dimensional Analysis, Similitude, and Physical Modelling
Foundations of Mathematical Modelling:
Mathematical modelling is the process of creating, analysing, and validating mathematical representations of physical, biological, chemical, economic, and engineering phenomena.
The modelling cycle proceeds from identifying key variables and interactions in the natural world to formulating equations, solving and analysing those equations, and validating and interpreting predictions against experimental data.
Units and Base Physical Dimensions:
Postulate 1: There exist base quantities providing units in terms of which the units of all other physical quantities can be expressed.
In classical mechanics, the three primary base dimensions are:
Mass:
Length:
Time:
Additional fundamental base quantities in the International System of Units (SI) include temperature (, ), electric current (, ), amount of substance (, ), and luminous intensity (, ).
The dimensions of any derived physical variable are denoted , expressed as a product of integer powers of the base dimensions: where .
For composite quantities, the dimension operator satisfies the product property: .
Fundamental mechanical examples:
Velocity:
Acceleration:
Force:
A quantity is defined as dimensionless if its dimensional representation is .
Dimensional Homogeneity and Complete Sets:
Postulate 2: Equations representing physical laws are dimensionally homogeneous.
An equation is dimensionally homogeneous if and only if the dimensions of every additive term on both the left-hand side and right-hand side are identical: .
In transcendental functions (such as , , , ), the argument must be strictly dimensionless () for dimensional compatibility, as established by their Maclaurin series expansions (e.g., ).
A complete set of dimensionless products is a maximal set of independent dimensionless combinations formed from the variables of a model.
The Buckingham -Theorem:
An equation is dimensionally homogeneous if and only if it can be expressed in the form: where is a function and constitutes a complete set of independent dimensionless products corresponding to the mathematical model.
The set of exponents yielding dimensionless products is obtained by determining a basis for the null space of the dimensional matrix representing the system of linear equations for the exponents.
Worked Case Studies in Dimensional Analysis:
A Falling Mass:
An object of mass falls from rest in a vacuum near Earth's surface over distance in time .
In metric units (Australia), experiments yield , whereas in imperial units (USA), experiments yield .
Dimensional consistency requires , where , revealing that .
The Simple Pendulum:
Period depends on pendulum length , mass , initial release angle , and gravitational acceleration : .
Dimensions: , , , , ().
Product: .
Dimensional balance: .
Linear system for exponents:
Exponent is free.
Independent basis choices and generate the complete set:
Applying the Buckingham -theorem yields .
Archimedes' Buoyancy Law:
A box of mass displacing fluid volume with fluid density () under gravity .
Testing a proposed equation :
Since , the proposed law is dimensionally invalid.
Testing an alternative equation :
The equation is dimensionally homogeneous and physically viable.
By the Buckingham -theorem, any valid relationship between total force , , , , and must satisfy: Choosing directly recovers .
Stokes' Drag Law:
Drag force () on a sphere of radius () moving at velocity () in fluid of dynamic viscosity ().
Dimensionless combination .
Setting isolates the dependent variable: , , .
The solitary dimensionless group is , implying .
Stokes' exact theoretical solution yields , giving .
Fluid and Solute Mixing Dynamics:
A tank initially contains litres of water with grams of dissolved salt. Brine containing enters at , and well-stirred solution drains at .
Let be the mass of salt at time . The volume is , and solute density is .
Rate of mass inflow: . Rate of mass outflow: .
Governing Initial Value Problem:
Dimensional consistency check:
Scaling and Similitude:
Laboratory scale models preserve dimensionless numbers to match the physical behaviour of full-scale prototypes.
Fundamental dimensionless numbers in fluid mechanics:
Reynolds number:
Froude number:
Mach number: where is fluid density, is characteristic length, is velocity, is dynamic viscosity, is gravitational acceleration, and is speed of sound.
Ship Scaling Example:
Full-scale vessel: length , cruising velocity .
Model scale: length (scale ratio ).
Maintaining dynamic similitude via the Froude number:
Multivariable Functions, Limits, and Continuity
Functions of Multiple Variables:
A scalar-valued function of several variables is a mapping , where the domain is a subset of .
The graph of a two-variable function represents a surface in .
Multivariate Limits and Definitions:
Formal - Definition: Let be defined on an open domain containing points arbitrarily close to . The limit of as is , written: if for every , there exists a corresponding such that:
Here, denotes the absolute distance in , while denotes the Euclidean distance in .
Generalisation to : for , if for every , there exists such that .
The Path-Dependent Non-Existence Criterion:
For to exist, must approach the exact same value along every path in terminating at .
If along path and along path with , the limit does not exist.
Case Evaluations:
Function at :
Along (): .
Along (): .
Since , the limit does not exist.
Function at :
Along : .
Along : .
Since , the limit does not exist.
Function at :
Along coordinate axes ( or ): .
Along (): .
Since , the limit does not exist.
Techniques for Establishing Multivariate Limits:
Polar Coordinate Substitution:
Evaluate .
Set , , with :
Since , we have:
For any , select . Then .
Hence, the limit exists and equals .
Algebraic Factorisation:
Evaluate .
Factorise numerator: .
For , cancel common factors:
Multivariate Continuity and Properties:
A function is continuous at if:
All polynomial functions of variables are continuous across all of .
All rational functions are continuous everywhere across their defined domains.
Differential Calculus of Several Variables: Tangents, Gradients, and Chain Rules
Partial Derivatives:
The first partial derivatives of at represent the instantaneous rates of change along cross-sections where the other coordinate is held constant:
Higher-Order Partial Derivatives and Clairaut's Theorem:
Second partial derivatives are denoted , , , .
Clairaut's Theorem: If the mixed second partial derivatives and are continuous on an open disc containing , then .
Example: For , we find:
Pathology of Partial Derivatives:
The existence of partial derivatives and does not imply continuity at .
For () and : Both partial derivatives exist at the origin, but the function itself is discontinuous at .
The Tangent Plane and Differentiability:
Equation of the Tangent Plane: For a smooth surface at , the tangent plane has equation:
Vector parametric representation:
Definition of Multivariable Differentiability: Let . A function is differentiable at if there exists a vector such that:
Theorems on Differentiability:
If is differentiable at a point, it is continuous at that point.
If all first-order partial derivatives of exist and are continuous in a neighbourhood of , then is differentiable at .
Linear Approximations, Total Differentials, and Error Estimation:
The linear (first-order) approximation near is:
The total differential of is:
If independent variables have maximum absolute errors and , the estimated worst-case error in is:
Example (Cylindrical Barrel): Volume . Nominal dimensions , . Nominal volume . Partial derivatives: ; .
Given a error in radius () and height (): Relative percentage error: .
Directional Derivatives and the Gradient Vector:
Directional Derivative: For a unit vector (), the directional derivative of at in direction is:
When is differentiable:
For an arbitrary non-zero direction vector :
Properties of the Gradient Vector :
.
By , the directional derivative is maximized when ( parallel to ). Hence, points in the direction of maximal increase of , with maximum rate of change equal to .
The directional derivative is minimized when , pointing in the direction of steepest descent () with rate .
When , . Thus, is orthogonal to the level curve (or level surface) of passing through .
Multivariable Chain Rules and Implicit Differentiation:
Chain Rule: Let where each is differentiable. Then:
If each depends on multiple variables , then:
Implicit Differentiation: Let a plane curve be defined implicitly by . Differentiating with respect to via the chain rule:
Example (Folium of Descartes): Curve . Define . Then , . Thus:
Implicit Function Theorem: Let be an open set and be continuously differentiable. Let satisfy and . Then there exists an interval and a unique continuously differentiable function such that and for all .
Multivariable Optimisation, Critical Points, and Constrained Extrema
Quadratic Approximations:
The second-order Taylor approximation of about is:
Classification of Critical Points and the Second Derivative Test:
A point is a critical point of if or if is undefined.
The Second Derivative Test: Let have continuous second partial derivatives in a neighbourhood of a critical point where . Define the Hessian determinant (discriminant):
If and , then has a local minimum at .
If and , then has a local maximum at .
If , then has a saddle point at .
If , the test is inconclusive.
Global Extrema on Closed and Bounded Sets:
By the Extreme Value Theorem, a continuous function on a closed and bounded (compact) set attains an absolute maximum and an absolute minimum.
Procedure: (1) Find values of at all critical points inside ; (2) Find extreme values of on the boundary of using single-variable calculus or Lagrange multipliers; (3) The largest and smallest values from these steps are the absolute maximum and minimum.
Constrained Optimisation via Lagrange Multipliers:
Single Constraint: To maximize or minimize subject to , where , solve: where is the Lagrange multiplier. Geometrically, extrema occur where level sets of are tangent to the constraint manifold .
Multiple Constraints: To optimize subject to constraints ( with ): This is equivalent to rendering stationary the Lagrangian function:
For inequality constraints , the formulation generalises to the Karush-Kuhn-Tucker (KKT) conditions.
Envelopes and Economic Applications:
Envelope of a Family of Curves: Given a family of curves parameterized by , the envelope is a curve tangent to each member of the family.
The envelope satisfies the dual conditions: , giving the envelope as .
Economic Profit Maximisation (Hotelling's Lemma):
Let be a production function (, , , ) yielding output for input .
Selling price is , unit input cost is . Profit is .
Optimal input satisfies .
The maximized profit function represents the envelope of the family of profit functions with respect to , possessing the envelope property:
Example: For , . Maximized profit: . For , , with .
Line Integrals, Vector Fields, and Conservative Systems
Arc Length and Scalar Line Integrals:
The arc length of a smooth curve parameterized by for is:
In polar coordinates for :
The scalar line integral of along represents the area of a vertical ribbon of height above :
Vector Line Integrals and Work:
The work done by a continuous vector force field moving a particle along curve is:
Reversing path orientation inverts the sign: .
Conservative Vector Fields and Potential Theory:
A vector field is conservative if is independent of path, depending solely on the endpoints and .
Fundamental Theorem for Line Integrals: A vector field is conservative if and only if it is a gradient field, meaning there exists a scalar potential function such that . In this case:
Component Criterion: On a simply connected domain in , is conservative if and only if:
Conservation of Energy and Central Forces in 3D Space:
Under a conservative force , Newton's second law reads .
Taking the inner product with , the multivariable chain rule yields:
Hence, total mechanical energy is conserved: .
Central Force Dynamics:
A force is central if , where and .
Every continuous central force field is conservative, with potential .
Angular Momentum: .
Differentiating with respect to time yields torque: .
For central forces, , proving that angular momentum is strictly conserved.
Planar Motion: Since , the position vector must remain orthogonal to the constant vector , confining the motion strictly to a two-dimensional plane.
Kepler's First Law: In an inverse-square gravitational field , the planar trajectory is a conic section: where is the eccentricity and is the semi-latus rectum. Bound orbits (planets) have (ellipses/circles with maximum distance apoapsis), whereas unbound trajectories have (parabolas/hyperbolas).
First-Order Ordinary Differential Equations: Theory, Geometry, and Numerics
Classification and Fundamentals:
An Ordinary Differential Equation (ODE) involves derivatives of an unknown function with respect to a single independent variable.
A Partial Differential Equation (PDE) involves partial derivatives with respect to two or more independent variables.
The order of an ODE is defined as the order of the highest derivative appearing in the equation.
A function is a solution to an ODE if direct substitution of and its derivatives satisfies the equation identically over an interval .
The general solution contains all possible solutions, incorporating arbitrary constants of integration ( constants for an -th order equation).
An Initial Value Problem (IVP) pairs an ODE with specified initial conditions (e.g., ), uniquely fixing arbitrary constants.
Geometric Interpretation: Slope Fields and Solution Curves:
For a first-order ODE , the right-hand side prescribes the slope of the solution curve passing through any point .
A slope field (or direction field) is constructed by drawing short line segments with slope at grid points in the -plane, providing a qualitative portrait of solution trajectories without integration.
Equilibrium Solutions and Stability Analysis:
An equilibrium solution is a constant solution satisfying for all . It corresponds to roots of .
In autonomous systems , equilibrium points partition the phase space.
Stability Classification:
Stable: An equilibrium is stable if any solution with initial condition sufficiently close to satisfies: Solution trajectories converge toward .
Unstable: Solution trajectories diverge away from as .
Semi-stable: Trajectories approach from one side but diverge on the opposite side.
Example Analysis:
ODE : Equilibrium at . For , (increasing away); for , (decreasing away). Hence, is unstable.

ODE : Equilibrium at . For , (decreasing toward 1); for , (increasing toward 1). Hence, is stable.

ODE : Direct integration yields . Since cannot hold for all , there are no equilibrium solutions.

ODE : Equilibria at and . For , ; for , ; for , . Hence, is stable, and is unstable.

ODE : Setting for . Infinite equilibrium solutions:
For (even integers): are unstable.
For (odd integers): are stable.

* ODE : Equilibrium at . For , (diverges); for , (approaches 0). Hence, is **semi-stable** (and classified overall as unstable).
Picard-Lindelöf Existence and Uniqueness Theorem:
Let and be continuous functions on a closed rectangle containing . Then the IVP: has a unique solution defined on an open interval containing .
Crucial Consequence: Solution trajectories cannot cross or intersect one another. If two distinct solution curves intersected, the ODE would have multiple distinct slopes at the same point, violating uniqueness. Therefore, equilibrium solutions act as impermeable barriers partitioning the solution space.
Numerical Methods for Initial Value Problems:
Euler's Method:
Approximates the solution curve by stepping along tangent lines. At step , the tangent line at gives:
Worked Manual Example: , , . Compute .
Exact analytical solution: .
Step 0: , .
Step 1: , .
Step 2: , .
Step 3: , .
Numerical error at : .
Implementing with a smaller step size ( steps) yields .
Python Implementation of Euler's Method:
import matplotlib.pyplot as plt
import numpy as np
t = np.arange(0, 0.6 + 0.05, 0.05)
y = np.zeros(len(t))
y[0] = 0
for i in range(1, len(t)):
y[i] = y[i - 1] + 2 * t[i - 1] * 0.05
print(y[-1]) # outputs 0.33
plt.plot(t, y, "-")
plt.show()
Heun's Method (Improved Euler):
Derived by integrating the ODE across a single time step using the Second Fundamental Theorem of Calculus:
While Euler's method approximates the integral as a left-endpoint rectangle of area , Heun's method uses a trapezoid of area .
To resolve the implicit dependence on , the standard Euler step acts as a predictor:
Separable First-Order ODEs:
A first-order ODE is separable if it can be written as:
Solution procedure:
Separate variables: , valid for .
Integrate both sides: .
Evaluate integrals and solve explicitly for when feasible.
Singular/Equilibrium Solutions: Points where yield constant equilibrium solutions . These solutions may not emerge from the general integral and must be identified separately.
Worked Example 1: , .
.
Applying yields .
Explicit solution: .
Worked Example 2: , .
.
Applying yields .
Selecting the negative branch matching the initial condition: .
Linear First-Order ODEs and Integrating Factors:
A first-order ODE is linear if it can be arranged in standard form:
Linear Operator Property: The differential operator is linear: .
Integrating Factor Derivation:
Multiply by an integrating factor : .
Demand that the LHS matches the derivative of a product: .
Equating terms yields .
Any constant multiple of divides out, so the integration constant can be taken as zero.
The general solution is obtained by direct integration:
Worked Example: Solve \frac{dy}{dt} + \left(\frac{2t}{t^2 + 1}\right)y = t^2 + 1$.\n * Here p(t) = \frac{2t}{t^2 + 1}q(t) = t^2 + 1.\n * Integrating factor: I(t) = \exp\left(\int \frac{2t}{t^2 + 1}\,dt\right) = \exp(\ln(t^2 + 1)) = t^2 + 1.\n * Multiplying through: \frac{d}{dt}[(t^2 + 1)y] = (t^2 + 1)^2 = t^4 + 2t^2 + 1.\n * Integrating: (t^2 + 1)y = \frac{t^5}{5} + \frac{2t^3}{3} + t + C.\n * General solution: y(t) = \frac{\frac{t^5}{5} + \frac{2t^3}{3} + t + C}{t^2 + 1}.\n\n# Applications of First-Order Ordinary Differential Equations\n\n* **Newton's Law of Cooling:**\n * The rate of change of temperature T(t)T_m:\n \frac{dT}{dt} = -k(T - T_m), \quad k > 0\n * Standard linear form: \frac{dT}{dt} + kT = kT_m.\n * Integrating factor I(t) = e^{kt}T(t) = (T_0 - T_m)e^{-kt} + T_mT(0) = T_0.\n * **Forensic Post-Mortem Analysis (Carl W. Case):**\n * Body discovered at 11:00 am on 19 April 2010 in a prison cell at T_m = 21.1^\circ\text{C}.\n * Measurements: at 11:00 am (t = 0T(0) = 34.8^\circ\text{C}t = 1.5\,\text{hours}T(1.5) = 34.1^\circ\text{C}.\n * Normal living body temperature at death is assumed to be 37.0^\circ\text{C}.\n * Determine cooling constant k:\n T(t) = (34.8 - 21.1)e^{-kt} + 21.1 = 13.7e^{-kt} + 21.1\n T(1.5) = 13.7e^{-1.5k} + 21.1 = 34.1 \implies 13.7e^{-1.5k} = 13.0 \implies e^{-1.5k} = \frac{13.0}{13.7}\n k = -\frac{1}{1.5}\ln\left(\frac{13.0}{13.7}\right) \approx 0.03496\,\text{hr}^{-1}\n * Determine time of death t_d:\n 37.0 = 13.7e^{-k t_d} + 21.1 \implies 15.9 = 13.7e^{-k t_d} \implies e^{-k t_d} = \frac{15.9}{13.7}\n t_d = -\frac{1}{k}\ln\left(\frac{15.9}{13.7}\right) = -\frac{\ln(15.9 / 13.7)}{0.03496} \approx -4.26\,\text{hours}\n * Calculating time of death: 11:00\,\text{am} - 4.26\,\text{hours} = 11:00\,\text{am} - 4\,\text{hours } 16\,\text{minutes} = 6:44\,\text{am} \approx 6:45\,\text{am}.\n\n* **Pharmacokinetics: Drug Absorption and Elimination:**\n * Drug amount y(t)\text{mg}a(t)k yk > 0):\n \frac{dy}{dt} + ky = a(t)\n * For a single oral dose Dk_ak_a e ka(t) = k_a D e^{-k_a t}.\n * Integrating factor I(t) = e^{kt}:\n \frac{d}{dt}[e^{kt}y] = k_a D e^{(k - k_a)t} \implies e^{kt}y = \frac{k_a D}{k - k_a} e^{(k - k_a)t} + B\n * Initial condition y(0) = 0 \implies B = \frac{k_a D}{k_a - k}. The exact solution is:\n y(t) = \frac{k_a D}{k_a - k}\left(e^{-kt} - e^{-k_a t}\right)\n * *Numerical Case:* Dose D = 100\,\text{mg}k_a = 2\,\text{hr}^{-1}k = 1\,\text{hr}^{-1}.\n y(t) = \frac{2(100)}{2 - 1}\left(e^{-t} - e^{-2t}\right) = 200\left(e^{-t} - e^{-2t}\right)\n * After 1 hour: y(1) = 200(e^{-1} - e^{-2}) \approx 46.5\,\text{mg}.\n * Peak time: Setting y'(t) = 200(-e^{-t} + 2e^{-2t}) = 0 \implies e^t = 2 \implies t_{\text{peak}} = \ln(2) \approx 0.693\,\text{hours}.\n * Peak amount: y(\ln(2)) = 200(1/2 - 1/4) = 50\,\text{mg}.\n * Continuous IV infusion: If a(t) = Ry(t) = \frac{R}{k}(1 - e^{-kt})y_{\infty} = \frac{R}{k}.\n\n* **Single-Species Population Dynamics:**\n * **Malthusian Growth Model:** Unlimited growth rate proportional to population P(t)\frac{dP}{dt} = rP \implies P(t) = P_0 e^{rt}.\n * *Projection of Land Exhaustion:* Current population growth rate r = 0.0125\,\text{yr}^{-1}P(2000) = 6\,\text{billion}150 \times 10^6\,\text{km}^2 = 1.5 \times 10^{14}\,\text{m}^2.\n * Carrying capacity constraint: 1 person per square metre yields maximum capacity P_{\text{crit}} = 1.5 \times 10^{14}\,\text{people} = 150000\,\text{billion}.\n * Setting 6 e^{0.0125(t - 2000)} = 150000 \implies e^{0.0125(t - 2000)} = 25000.\n * Solving for t:\n t = 2000 + \frac{\ln(25000)}{0.0125} = 2000 + 80 \ln(25000) \approx 2000 + 810.13 \approx \text{Year } 2810\n * **Verhulst Logistic Growth Model:** Incorporates environmental carrying capacity \theta:\n \frac{dP}{dt} = rP\left(1 - \frac{P}{\theta}\right)\n * Nondimensionalisation: Define y = \frac{P}{\theta}\tau = rt. The equation simplifies to:\n \frac{dy}{d\tau} = y(1 - y)\n * Separating variables via partial fractions:\n \int \left(\frac{1}{y} + \frac{1}{1 - y}\right)\,dy = \int d\tau \implies \ln\left|\frac{y}{1 - y}\right| = \tau + C \implies \frac{y}{1 - y} = A e^\tau\n * Solving for y(\tau)y(0) = y_0:\n y(\tau) = \frac{y_0 e^\tau}{1 - y_0 + y_0 e^\tau}\n * In terms of original dimensional variables with P(0) = P_0:\n P(t) = \frac{\theta P_0 e^{rt}}{\theta - P_0 + P_0 e^{rt}}\n * **Logistic Model with Harvesting (Pacific Halibut Fishery):**\n * Proportional harvesting rate E P modifies the logistic equation:\n \frac{dP}{dt} = rP\left(1 - \frac{P}{\theta}\right) - EP = (r - E)P\left(1 - \frac{P}{\theta(1 - E/r)}\right)\n * Equilibria occur at P = 0P^* = \theta\left(1 - \frac{E}{r}\right).\n * To prevent population collapse and extinction, the harvesting intensity must satisfy E < r.\n\n* **Coupled Systems of First-Order ODEs:**\n * **The Lotka-Volterra Predator-Prey System:**\n * Let prey population be R(t)W(t) (wolves):\n \frac{dR}{dt} = \alpha R - \gamma RW = R(\alpha - \gamma W)\n \frac{dW}{dt} = -\beta W + \delta RW = W(-\beta + \delta R)\n where \alpha, \beta, \gamma, \delta > 0.\n * Equilibria occur when \dot{R} = 0\dot{W} = 0:\n (R, W) = (0, 0) \quad \text{and} \quad (R^, W^) = \left(\frac{\beta}{\delta}, \frac{\alpha}{\gamma}\right)\n * Phase plane differential equation:\n \frac{dW}{dR} = \frac{W(-\beta + \delta R)}{R(\alpha - \gamma W)}\n * *Numerical Parameter Study:* \alpha = 0.08\beta = 0.02\gamma = 0.001\delta = 0.00002.\n * Non-zero equilibrium: R^* = \frac{0.02}{0.00002} = 1000W^* = \frac{0.08}{0.001} = 80.\n * Initial condition: R(0) = 1000W(0) = 60.\n\n\n\n\n\n* Extended integration (t = 400) demonstrates closed periodic orbits in phase space and boom-bust population cycles.\n\n\n\n\n\n\n\n* **Competing Species Model:**\n * Two species P_1P_2 competing for shared resources:\n \frac{dP_1}{dt} = P_1(\alpha_1 - \beta_1 P_1 - \gamma_{12} P_2)\n \frac{dP_2}{dt} = P_2(\alpha_2 - \beta_2 P_2 - \gamma_{21} P_1)\n * *Worked Parameter System:* \alpha_1 = \beta_1 = \beta_2 = \gamma_{12} = 1\alpha_2 = 0.75\gamma_{21} = 0.5.\n \frac{dP_1}{dt} = P_1(1 - P_1 - P_2)\n \frac{dP_2}{dt} = P_2(0.75 - P_2 - 0.5P_1)\n * Determining all four equilibrium points (\dot{P}1 = 0\dot{P}_2 = 0):\n * If P_1 = 0 \implies P_2(0.75 - P_2) = 0 \implies P_2 = 0P_2 = 0.75.\n * If 1 - P_1 - P_2 = 0 \implies P_1 = 1 - P_2\dot{P}_2 = 0:\n P_2(0.75 - P_2 - 0.5(1 - P_2)) = 0 \implies P_2(0.25 - 0.5P_2) = 0\n Yielding P_2 = 0 \implies P_1 = 1P_2 = 0.5 \implies P_1 = 0.5.\n * The four equilibria are: (0, 0)(0, 0.75)(1, 0)(0.5, 0.5).\n\n\n\n\n\n* Evolution from P_1(0) = 0.6P_2(0) = 0.4(0.5, 0.5).\n\n\n\n# Second-Order Linear Differential Equations\n\n* **Definitions and Structural Foundations:**\n * General second-order linear differential equation:\n \frac{d^2y}{dt^2} + p(t)\frac{dy}{dt} + q(t)y = r(t)\n * Functions p(t)q(t)r(t) = 0, the equation is **homogeneous**; otherwise it is **inhomogeneous**.\n * **Linear Differential Operator:** Defining L[y] = y'' + p(t)y' + q(t)y, linearity requires:\n L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]\n * **Superposition Principle:** If y_1(t)y_2(t)L[y] = 0y(t) = c_1 y_1(t) + c_2 y_2(t) is also a solution.\n * **Linear Independence and General Solutions:** Two solutions y_1(t)y_2(t)y_2(t) e C y_1(t). The general solution of a homogeneous second-order linear ODE is spanned by exactly two linearly independent solutions:\n y(t) = c_1 y_1(t) + c_2 y_2(t)\n\n* **Reduction of Order:**\n * If one non-zero solution y_1(x)y'' + p(x)y' + q(x)y = 0y_2(x) = u(x)y_1(x).\n * Derivatives: y_2' = u'y_1 + uy_1'y_2'' = u''y_1 + 2u'y_1' + uy_1''.\n * Substitution into the ODE gives:\n u''y_1 + u'(2y_1' + py_1) + u(y_1'' + py_1' + qy_1) = 0\n * Since L[y_1] = 0u vanishes identically, leaving:\n u'' + \left(p + \frac{2y_1'}{y_1}\right)u' = 0\n * Setting v = u'vI(x) = \exp\left(\int \left(p + \frac{2y_1'}{y_1}\right)dx\right) = y_1^2 \exp\left(\int p\,dx\right).\n * *Worked Example:* Solve x^2 y'' + 2x y' - 6y = 0y_1(x) = x^2.\n * Set y_2 = u(x)x^2y_2' = u'x^2 + 2uxy_2'' = u''x^2 + 4u'x + 2u.\n * Substitute: x^2(u''x^2 + 4u'x + 2u) + 2x(u'x^2 + 2ux) - 6ux^2 = 0.\n * Simplification: x^4 u'' + 6x^3 u' = 0 \implies u'' + \frac{6}{x}u' = 0.\n * Set v = u'v' + \frac{6}{x}v = 0 \implies \frac{dv}{v} = -\frac{6}{x}\,dx \implies v(x) = C_1 x^{-6}.\n * Integrate u' = C_1 x^{-6} \implies u(x) = -\frac{C_1}{5}x^{-5} + C_2 = A x^{-5} + B.\n * Second solution: y_2(x) = x^2(x^{-5}) = x^{-3}.\n * General solution: y(x) = c_1 x^2 + c_2 x^{-3}.\n\n* **Homogeneous Linear Equations with Constant Coefficients:**\n * General form: a y'' + b y' + c y = 0a e 0.\n * Characteristic equation ansatz: y = e^{\lambda t} \implies a\lambda^2 + b\lambda + c = 0.\n * Roots via quadratic formula: \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\n * **Classification of Solution Regimes:**\n * **Case 1: Two Distinct Real Roots (b^2 - 4ac > 0):**\n y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}\n * *Example:* y'' - 3y' + 2y = 0 \implies \lambda^2 - 3\lambda + 2 = 0 \implies (\lambda - 1)(\lambda - 2) = 0.\n * Roots: \lambda_1 = 1, \lambda_2 = 2y(t) = c_1 e^t + c_2 e^{2t}.\n * **Case 2: Repeated Real Root (b^2 - 4ac = 0):**\n \lambda_1 = \lambda_2 = \mu = -\frac{b}{2a}\n By reduction of order, the two linearly independent solutions are e^{\mu t}t e^{\mu t}, giving:\n y(t) = (c_1 + c_2 t)e^{\mu t}\n * *Example (IVP):* y'' + 6y' + 9y = 0y(0) = 2y'(0) = 0.\n * Characteristic equation: \lambda^2 + 6\lambda + 9 = (\lambda + 3)^2 = 0 \implies \lambda = -3.\n * General solution: y(t) = (c_1 + c_2 t)e^{-3t}.\n * Initial conditions: y(0) = c_1 = 2.\n * Derivative: y'(t) = c_2 e^{-3t} - 3(c_1 + c_2 t)e^{-3t} \implies y'(0) = c_2 - 3c_1 = 0 \implies c_2 = 6.\n * Unique solution: y(t) = (2 + 6t)e^{-3t}.\n * **Case 3: Complex Conjugate Roots (b^2 - 4ac < 0):**\n \lambda = \alpha \pm i\beta, \quad \alpha = -\frac{b}{2a}, \quad \beta = \frac{\sqrt{4ac - b^2}}{2a}\n Using Euler's formula e^{i\theta} = \cos(\theta) + i\sin(\theta), the real general solution is:\n y(t) = e^{\alpha t}(b_1 \cos(\beta t) + b_2 \sin(\beta t))\n * *Example (IVP):* y'' - 4y' + 5y = 0y(0) = 1y'(0) = 0.\n * Characteristic roots: \lambda = \frac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i \implies \alpha = 2, \beta = 1.\n * General solution: y(t) = e^{2t}(b_1 \cos(t) + b_2 \sin(t)).\n * y(0) = b_1 = 1.\n * Derivative: y'(t) = 2e^{2t}(b_1 \cos(t) + b_2 \sin(t)) + e^{2t}(-b_1 \sin(t) + b_2 \cos(t)).\n * y'(0) = 2b_1 + b_2 = 0 \implies b_2 = -2.\n * Unique solution: y(t) = e^{2t}(\cos(t) - 2\sin(t)).\n\n* **Inhomogeneous Linear Second-Order Equations:**\n * General equation: a y'' + b y' + c y = r(x).\n * Structure of General Solution: y(x) = y_H(x) + y_P(x)y_H(x) = c_1 y_1(x) + c_2 y_2(x)y_P(x) is any particular solution to the inhomogeneous equation.\n * **Method of Undetermined Coefficients:**\n * *Case 1: Polynomial Forcing:* For r(x)ny_P(x) = A_n x^n + \dots + A_1 x + A_0.\n * Example: y'' + y' - 2y = x^2 - 2x + 3.\n * Homogeneous roots: \lambda^2 + \lambda - 2 = 0 \implies \lambda = 1, -2y_H = c_1 e^x + c_2 e^{-2x}.\n * Particular trial: y_P = A x^2 + B x + Cy_P' = 2Ax + By_P'' = 2A.\n * Substitute: 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2 - 2x + 3.\n * Match coefficients:\n * x^2: -2A = 1 \implies A = -1/2\n * x^1: 2A - 2B = -2 \implies -1 - 2B = -2 \implies B = 1/2\n * x^0: 2A + B - 2C = 3 \implies -1 + 1/2 - 2C = 3 \implies -2C = 7/2 \implies C = -7/4\n * Particular solution: y_P(x) = -\frac{1}{2}x^2 + \frac{1}{2}x - \frac{7}{4} = \frac{1}{4}(-2x^2 + 2x - 7).\n * *Case 2: Exponential-Polynomial Forcing:* For r(x) = P(x)e^{kx}y_P(x) = u(x)e^{kx}.\n * *Case 3: Trigonometric Forcing:* For r(x) = P(x)\cos(\omega x)P(x)\sin(\omega x)y_P(x) = A(x)\cos(\omega x) + B(x)\sin(\omega x).\n * *Case 4: Superposition of Forcing Terms:* If r(x) = r_1(x) + r_2(x)y{P1}, y_{P2}r_1, r_2y_P = y_{P1} + y_{P2}.\n\n# Physical Oscillations: Harmonic, Damped, Forced, and Resonant Systems\n\n* **The Undamped Mass-Spring Oscillator:**\n * A mass mk > 0sks = mg.\n * Measuring displacement x(t)F_s = -k(s + x).\n * Total force: F = -k(s + x) + mg = -kx.\n * By Newton's second law: m\ddot{x} + kx = 0.\n * Defining natural angular frequency \omega = \sqrt{\frac{k}{m}}, the equation of motion is:\n \ddot{x} + \omega^2 x = 0\n * Characteristic roots: \lambda = \pm i\omega.\n * General solution:\n x(t) = c_1 \cos(\omega t) + c_2 \sin(\omega t) = A \cos(\omega t - \phi)\n where amplitude A = \sqrt{c_1^2 + c_2^2}\phi = \tan^{-1}\left(\frac{c_2}{c_1}\right).\n\n* **The Damped Oscillator:**\n * Adding a viscous damping force proportional to velocity F_d = -\gamma \dot{x}\gamma > 0):\n m\ddot{x} + \gamma \dot{x} + kx = 0 \implies \ddot{x} + 2p\dot{x} + \omega^2 x = 0\n where \omega^2 = \frac{k}{m} > 02p = \frac{\gamma}{m} > 0.\n * Characteristic equation: \lambda^2 + 2p\lambda + \omega^2 = 0D = 4(p^2 - \omega^2).\n * **Classification of Damping Regimes:**\n * **Underdamping (p < \omega):**\n * Roots: \lambda = -p \pm i\omega_p\omega_p = \sqrt{\omega^2 - p^2} = \omega \sqrt{1 - (p/\omega)^2}.\n * General solution:\n x(t) = e^{-pt}\left(c_1 \cos(\omega_p t) + c_2 \sin(\omega_p t)\right) = A e^{-pt}\cos(\omega_p t - \phi)\n * Exhibits oscillatory decay under an exponentially decaying envelope A(t) = A e^{-pt}.\n * Because \omega_p < \omega, damping induces a frequency red-shift and increases the period.\n * **Critical Damping (p = \omega):**\n * Discriminant D = 0\lambda = -p.\n * General solution:\n x(t) = (c_1 + c_2 t)e^{-pt} = e^{-pt}[x(0) + (x'(0) + px(0))t]\n * Provides the most rapid return to equilibrium without sustained oscillations.\n * Solution crosses equilibrium at most once if x'(0) < -p x(0).\n * **Overdamping (p > \omega):**\n * Discriminant D > 0\lambda_1 = -p - \sqrt{p^2 - \omega^2}\lambda_2 = -p + \sqrt{p^2 - \omega^2}.\n * General solution:\n x(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}\n * Strictly non-oscillatory. Governed at large te^{\lambda_2 t}, making the return to equilibrium slower than in critical damping.\n\n* **Forced Oscillations and Mechanical Resonance:**\n * Applying periodic external forcing: \ddot{x} + 2p\dot{x} + \omega^2 x = \alpha \cos(\Omega t).\n * **Damped Forced Response (p < \omega):**\n * Solution: x(t) = x_H(t) + x_P(t) = e^{-pt}(c_1 \cos(\omega_p t) + c_2 \sin(\omega_p t)) + A \cos(\Omega t - \phi).\n * The homogeneous part x_H(t) \to 0t \to \infty (transient solution).\n * The long-time asymptotic response (steady-state solution) oscillates at the forcing frequency \Omega:\n x_{\text{steady}}(t) = A(\Omega)\cos(\Omega t - \phi)\n A(\Omega) = \frac{\alpha}{\sqrt{(\omega^2 - \Omega^2)^2 + 4p^2 \Omega^2}}, \quad \phi = \tan^{-1}\left(\frac{2p\Omega}{\omega^2 - \Omega^2}\right)\n * **Undamped Forced Oscillations and Pure Resonance (p = 0):**\n * Equation: \ddot{x} + \omega^2 x = \alpha \cos(\Omega t).\n * For \Omega e \omegax(0) = x_0, \dot{x}(0) = v_0:\n x(t) = x_0 \cos(\omega t) + \frac{v_0}{\omega}\sin(\omega t) + \frac{\alpha}{\omega^2 - \Omega^2}(\cos(\Omega t) - \cos(\omega t))\n* Taking the resonant limit as \Omega \to \omega using L'Hôpital's rule on the forcing term:\n \lim_{\Omega \to \omega} \frac{\alpha(\cos(\Omega t) - \cos(\omega t))}{\omega^2 - \Omega^2} = \lim_{\Omega \to \omega} \frac{-\alpha t \sin(\Omega t)}{-2\Omega} = \frac{\alpha t}{2\omega}\sin(\omega t)\n* **Resonant Solution:**\n x(t) = x_0 \cos(\omega t) + \frac{v_0}{\omega}\sin(\omega t) + \frac{\alpha t}{2\omega}\sin(\omega t)\n* The amplitude of oscillation grows linearly without bound (\propto t), causing catastrophic structural failure in physical systems lacking adequate damping.\n\n# The Energy Principle in One-Dimensional Dynamical Systems\n\n* **Conservation of Mechanical Energy in 1D:**\n * Consider a mass myF(y)m\ddot{y} = F(y).\n * Potential energy V(y)F(y) = -V'(y) \implies V(y) = -\int F(y)\,dy.\n * Multiplying the equation of motion by \dot{y} and using the chain rule:\n m\ddot{y}\dot{y} = F(y)\dot{y} \implies \frac{d}{dt}\left(\frac{1}{2}m\dot{y}^2\right) = -\frac{dV}{dy}\frac{dy}{dt} = -\frac{dV}{dt}\n * Integrating with respect to time yields the total mechanical energy conservation law:\n E = \frac{1}{2}m\dot{y}^2 + V(y) = \text{constant}\n * Kinetic energy T = \frac{1}{2}m\dot{y}^2T \ge 0), which imposes the fundamental physical constraint:\n V(y) \le E\n\n* **Potential Energy Diagrams and Phase Space Topology:**\n * Plotting V(y)yE dictates system dynamics:\n * **Allowed Regions:** Intervals of yV(y) \le E. Motion is physically constrained within these regions.\n * **Forbidden Regions:** Intervals where V(y) > E\dot{y}^2 < 0), which is impossible in classical mechanics.\n * **Turning Points:** Points where V(y) = ET = 0\dot{y} = 0F = -V'(y).\n * **Equilibria and Stability:** Equilibrium points occur at critical points of the potential energy: F(y) = 0 \implies V'(y) = 0.\n * **Stable Equilibrium:** Local minimum of V(y)V''(y) > 0). Small perturbations produce restoring forces pushing the particle back toward the minimum.\n * **Unstable Equilibrium:** Local maximum of V(y)V''(y) < 0). Perturbations cause the particle to accelerate away from the crest.\n\n* **Worked Dynamical Applications:**\n * **Constant Gravitational Acceleration:**\n * Equation: m\ddot{y} = -mg.\n * Potential energy: V(y) = -\int (-mg)\,dy = mgyV(0) = 0).\n * Energy equation: \frac{1}{2}m\dot{y}^2 + mgy = E.\n * Maximum height y_{\max}\dot{y} = 0:\n mgy_{\max} = E = \frac{1}{2}m v_0^2 + mgy_0 \implies y_{\max} = y_0 + \frac{v_0^2}{2g}\n * For a projectile launched from ground level (y_0 = 0v_0 = 98\,m/sg = 9.8\,m/s^2:\n y_{\max} = \frac{98^2}{2(9.8)} = \frac{9604}{19.6} = 490\,m\n * **Harmonic Oscillator Energy:**\n * Restoring force: F(x) = -kx.\n * Elastic potential energy: V(x) = -\int (-kx)\,dx = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2.\n * Harmonic trajectory: x(t) = A\cos(\omega t - \phi)\dot{x}(t) = -A\omega\sin(\omega t - \phi).\n * Total energy:\n E = \frac{1}{2}m(-A\omega\sin(\omega t - \phi))^2 + \frac{1}{2}m\omega^2(A\cos(\omega t - \phi))^2 = \frac{1}{2}mA^2\omega^2(\sin^2(\dots) + \cos^2(\dots)) = \frac{1}{2}mA^2\omega^2\n * Total energy is strictly proportional to both the square of the amplitude and the square of the natural frequency.\n * *Example Calculation:* Mass m = 9\,kgk = 4\,N/mx(0) = 1\,m\dot{x}(0) = -0.5\,m/s.\n * Natural frequency: \omega = \sqrt{4/9} = \frac{2}{3}\,rad/s.\n * Amplitude: A = \sqrt{x(0)^2 + \left(\frac{\dot{x}(0)}{\omega}\right)^2} = \sqrt{1^2 + \left(\frac{-0.5}{2/3}\right)^2} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4} = 1.25\,m.\n * Total energy:\n E = \frac{1}{2}(9)\left(\frac{5}{4}\right)^2\left(\frac{2}{3}\right)^2 = \frac{1}{2}(9)\left(\frac{25}{16}\right)\left(\frac{4}{9}\right) = \frac{25}{8} = 3.125\,J\n * Speed at equilibrium (x = 0):\n \frac{1}{2}m v_{\text{eq}}^2 = E \implies \frac{9}{2}v_{\text{eq}}^2 = \frac{25}{8} \implies v_{\text{eq}}^2 = \frac{25}{36} \implies v_{\text{eq}} = \pm \frac{5}{6} \approx \pm 0.833\,m/s$$