(09/14) Lecture Comprehensive Study Guide on Integrating Factors, Euler's Method, and Existence Theorems

Integrating Factors for Non-Exact Differential Equations

• Exactness condition for a first-order differential equation $M(x, y) + N(x, y) y' = 0$ requires that:   $\frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$   where $\mu$ is an integrating factor.

• Expanding both sides using the product rule for partial derivatives yields:   $M \frac{\partial \mu}{\partial y} + \mu \frac{\partial M}{\partial y} = N \frac{\partial \mu}{\partial x} + \mu \frac{\partial N}{\partial x}$

• Rearranging all terms to one side gives the partial differential equation for $\mu$:   $-N \frac{\partial \mu}{\partial x} + M \frac{\partial \mu}{\partial y} + \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) \mu = 0$

• Because this partial differential equation contains partial derivatives of $\mu$ with respect to both $x$ and $y$, solving for $\mu$ generally requires simplification.

• Simplifying Assumption: Assume $\mu$ is a function of $x$ only, $\mu = \mu(x)$.

  • Under this assumption, $\frac{\partial \mu}{\partial y} = 0$ and $\frac{\partial \mu}{\partial x} = \frac{d\mu}{dx}$.

  • The partial differential equation reduces to an ordinary differential equation:     $-N \frac{d\mu}{dx} + \left( M_y - N_x \right) \mu = 0$

  • Rearranging terms:     $N \frac{d\mu}{dx} = \left( M_y - N_x \right) \mu$

  • Isolating $\frac{d\mu}{dx}$ gives:     $\frac{d\mu}{dx} = \frac{M_y - N_x}{N} \mu$

• Separability Condition:

  • Both $M$ and $N$ are original functions of two variables $x$ and $y$.

  • For the ordinary differential equation $\frac{d\mu}{dx} = \frac{M_y - N_x}{N} \mu$ to be linear or separable in $x$, the coefficient quantity $\frac{M_y - N_x}{N}$ must be a function of $x$ only (denoted as $p(x)$).

  • If $\frac{M_y - N_x}{N}$ depends on both $x$ and $y$, the equation cannot be separated to find $\mu(x)$.

  • If $\frac{M_y - N_x}{N} = Q(y)$ (a function of $y$ only), a corresponding integrated factor $\mu(y)$ can be found using an analogous formulation.

Questions & Discussion

• Question: For clarification, is it assumed that $y$ is not a function of $x$?

• Response: In the background of differential equations, $y$ is indeed treated as a function of $x$. However, when evaluating functions $M(x,y)$ and $N(x,y)$ and their partial derivatives, $x$ and $y$ are treated as independent input variables. Taking the assumption that $\mu$ depends strictly on $x$ does not contradict $y$ being a function of $x$ in the broader context of solving the initial differential equation.

Euler's Method for Numerical Approximations

• General Initial Value Problem Setup:   $y' = f(t, y), \quad y(t_0) = y_0$

  • The exact solution curve passes through the initial point $(t_0, y_0)$ with an initial slope of $y'(t_0) = f(t_0, y_0)$.

• Tangent Line Derivation:

  • Using the point-slope form, the equation of the tangent line at $(t_0, y_0)$ is:     $y = y_0 + f(t_0, y_0)(t - t_0)$

• Stepping Procedure:

  • Instead of following the true solution curve $y(t)$, Euler's method steps along tangent line segments to approximate values of $y$ at discrete time intervals $t_1, t_2, \dots, t_n$

  • First Step ($t_1$): Advance from $t_0$ to $t_1$. Evaluate the tangent line at $t_1$ to compute $y_1 \approx y(t_1)$:     $y_1 = y_0 + f(t_0, y_0)(t_1 - t_0)$

  • Second Step ($t_2$): Construct a new tangent line starting at $(t_1, y_1)$ with slope $f(t_1, y_1)$:     $y = y_1 + f(t_1, y_1)(t - t_1)$     Evaluate at $t_2$ to compute $y_2 \approx y(t_2)$:     $y_2 = y_1 + f(t_1, y_1)(t_2 - t_1)$

• General Euler's Formula:   $y_{n+1} = y_n + f(t_n, y_n) \times h$

  • $n = 0, 1, 2, \dots$

  • $f(t_n, y_n)$: The slope calculated at step $n$.

  • $h$: The uniform step size, where $h = t_{n+1} - t_n$ (e.g., $h = 0.2\,s$, $0.1\,s$, or $0.01\,s$).

• Numerical Performance and Accuracy Characteristics:

  • The initial point $y_0$ matches the exact solution $y(t_0)$ perfectly.

  • As $n$ increases, numerical drift occurs due to overshooting the linear tangent approximations relative to curved exact solution trajectories.

  • For example, an approximate value $y_1 = 1.5$ may correspond to an exact solution value $y(t_1) = 1.43$

  • Agreement between numerical approximation $y_n$ and exact solution $y(t_n)$ improves as step size $h$ decreases or total steps $n$ increase over a fixed interval.

Existence Theorems and Equivalence to Integral Equations

• Existence and Uniqueness Condition:

  • If $f(t, y)$ and $\frac{\partial f}{\partial y}$ are continuous on a defined rectangle $R$, then there exists an interval $|t| < h$ (bounded by upper bound $a$) such that a unique solution to the initial value problem exists.

• Coordinate Transformation Assumption:

  • Without loss of generality, proofs often assume $\ y(0) = 0$ (i.e., initial point at the origin $(0,0)$).

  • If an initial point $(t_0, y_0)$ is not at the origin, a coordinate transformation (change of variables) translates $(t_0, y_0)$ to $(0,0)$, making geometric and analytic arguments simpler without restricting generality.

• Fundamental Proof Principle:

  • It is analytically more convenient to demonstrate the existence of a solution to a related integral equation than directly to the differential equation.

• Derivation of the Related Integral Equation:

  1. Assume a differentiable function $y = \phi(t)$ solves the differential equation $y' = f(t, y)$, such that:      $\phi'(t) = f(t, \phi(t))$

  2. Introduce a dummy variable of integration $s$ and integrate both sides from $0$ to $t$:      $\int_0^t y'(s)\,ds = \int_0^t f(s, \phi(s))\,ds$

  3. Apply the Fundamental Theorem of Calculus to the left side:      $\int_0^t y'(s)\,ds = y(t) - y(0)$

  4. Substitute the initial condition $y(0) = 0$:      $y(t) - 0 = \int_0^t f(s, \phi(s))\,ds$

  5. Yields the related integral equation:      $y(t) = \int_0^t f(s, \phi(s))\,ds$

• Proof of Equivalence (Integral Solution Solves Initial Value Problem):

  • Verification of Initial Condition:     Evaluate the integral equation at $t = 0$:     $y(0) = \int_0^0 f(s, \phi(s))\,ds = 0$     An integral over an interval of length zero equals zero, satisfying $y(0) = 0$

  • Verification of Differential Equation:     Take the derivative with respect to $t$ on both sides of the integral equation:     $y'(t) = \frac{d}{dt} \left( \int_0^t f(s, \phi(s))\,ds \right)$     By the Fundamental Theorem of Calculus:     $y'(t) = f(t, \phi(t))$     This reproduces the original differential equation, proving that solving the integral equation is completely equivalent to solving the differential initial value problem.