First-Order Substitution Methods. Rescaling and Bernoulli Equations

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Recall. Algebraic and differential equations.

An algebraic equation is a mathematical statement asserting that two algebraic expressions are equal.. These expressions are constructed using:

1. Dependent and independent variables. Variables are symbols representing unknown quantities. Variables can be:
   Independent Variables: Variables or input that can be chosen (or change) freely.
   Dependent Variables: Variables that depend on the independent variables.
2. Constants. Fixed numerical values that do not change, e.g., 2, π, 7.
3. Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, powers, and root.

An algebraic equation typically has the following general form: Expression₁ = Expression₂, where the two expressions are set equal to each other.

Examples: y = 2x + 4 (simple linear equation, represents a straight line with a slope of 2 and a y-intercept of 4), ax² + bx + c = 0 (quadratic equation, it represents a parabola), $ln(xy + t) = z·sin(\sqrt{x})$.

Definition. A differential equation is an equation that involves one or more functions and their derivatives. It relates the function itself (the dependent variable), its derivatives with respect to one or more independent variables (rates of change), and the independent variable(s) themselves, e.g., $\frac{dy}{dx} = 3x +5y, y' + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

If an algebraic equation is a “snapshot” of a relationship, a differential equation is the “instruction manual” for how that relationship moves and changes.

Key Components (e.g., $\frac{dy}{dx} = 3x +5y$):

• Dependent variables: The function y (or y(x)) we are trying to find or solve for. Variable(s) (y) that depend on one or more other variables.
• Independent variables: Variable(s) (x) upon which the dependent variables depend and we differentiate with respect to (often time or space).
• Derivatives: Expressions like $\frac{dy}{dx}$ or y’ that represent the rate at which the dependent variable (y) change with respect to the independent variable (x).

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It answers two fundamental questions: Is there a solution? and Is it the only one? It states that if:

• The function $f(x, y)$ (the right-hand side of the ODE when written as $y' = f(x, y)$) is continuous in some neighborhood around a specific point $(x_0, y_0)$ and
• Its partial derivative with respect to y, denoted as $\frac{\partial f}{\partial y}$, is also continuous near $(x_0, y_0)$.

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it perfectly satisfies the initial condition $y(x_0) = y_0$.

This theorem is a mathematician’s safety net: it ensures that under these well-behaved conditions, a solution definitively exists and is absolutely unique near $x = x_0$. The physical system modeled by the ODE will only follow one distinct path.

Rescaling in a Temperature Model

Consider a physical system where the internal temperature T of an object changes over time due to heat exchange with its environment. The environment has a constant external temperature M. The rate at which the internal temperature changes is given by the differential equation: $\frac{dT}{dt} = k(M^4-T^4)$ where:

• $\frac{dT}{dt}$ is the rate of change of the internal temperature over time (dependent on the object’s surface area and emissivity).
• k is a positive constant that governs the rate of heat exchange.
• M is the constant external temperature (often measured in Kelvin).
• T is the internal temperature of the object.

This model is relevant in situations involving large temperature differences, such as radiative heat transfer governed by the Stefan-Boltzmann law. In these extreme environments (like spacecraft in orbit or industrial furnaces), heat exchange is proportional not to the simple temperature difference, but to the difference of the fourth powers of the temperatures.

Our goal is to simplify the given differential equation by rescaling the temperature variable to make it dimensionless, thereby removing the constant M from the equation and simplifying the mathematical analysis.

Step 1: Define a Dimensionless Temperature Variable. We aim to eliminate the units of temperature from the equation by introducing a dimensionless variable. Let’s define: $T_1 = \frac{T}{M}$.

This rescaling means that $T_1$ now represents the ratio of the internal temperature to the external temperature. Since both T and M have units of temperature, $T_1$ is entirely dimensionless.

When T = M, $T_1 = 1$, meaning the internal and external temperatures are at equilibrium. $T_1 > 1$ indicates the object is hotter than the environment $T_1 < 1$ indicates the object is colder than the environment.

Step 2. Express T in terms of T₁.

By simply rearranging our definition, we get: $T = M T_1$.

Step 3. Substitute into the original differential equation.

Now, we plug $T = M T_1$ into our initial ODE. $\frac{d(M T_1)}{dt} = k(M^4 - (M T_1)^4)$. Remember that $M$ is a constant, so it factors out of the derivative: $M \frac{dT_1}{dt} = k(M^4 - M^4 T_1^4)$

Factor out $M^4$ on the right side: $M \frac{dT_1}{dt} = k M^4(1 - T_1^4)$

Divide both sides by $M$: $\frac{dT_1}{dt} = k M^3(1 - T_1^4)$

Step 4. Introduce a new constant to simplify further: $k_1 = kM^3$, which simplifies the equation: $\frac{dT_1}{dt} = k_1(1-T_1^4)$ where:

• $T_1$ is dimensionless, representing the normalized temperature.
• $k_1 = kM^3$ is the new effective heat exchange rate.

By making this substitution, we’ve simplified the original equation by removing M, leaving a more manageable differential equation to solve.

Solving the Simplified Differential Equation

For visual simplicity during the integration steps, let’s temporarily drop the subscripts and denote $T_1$ as $T$ and $k_1$ as $k$. The equation is: $\frac{dT}{dt} = k(1 - T^4)$.

Our task is to solve this differential equation for $T(t)$.

1. Separate variables. Move all $T$ terms to the left and $t$ terms to the right: $\frac{dT}{1-T^4} = kdt$.

2. Integrate both sides: $\int \frac{dT}{1-T^4} = \int kdt$

3. Solve the integral using Partial Fractions. The integral on the left can be challenging due to the quartic term T⁴. However, it can be simplified using partial fractions.

   First, factor the denominator completely using the difference of squares: $1-T^4 = (1-T^2)(1+T^2) = (1-T)(1+T)(1+T^2).$

   Now, set up the partial fraction decomposition: $\frac{1}{(1 - T)(1 + T)(1 + T^2)} = \frac{A}{1 - T} + \frac{B}{1 + T} + \frac{CT + D}{1 + T^2}$

   Multiply both sides by the common denominator $(1-T)(1+T)(1+T^2)$ to clear the fractions: $1 = A(1+T)(1+T^2) + B(1-T)(1+T^2)+(CT+D)(1-T)(1+T)$

   Expand and Simplify Each Term: $A(1+T)(1+T^2) = A[1·(1+T^2)+T·(1+T^2)] = A(1+T^2+T+T^3) = A(1+T+T^2+T^3)$.
   $B(1-T)(1+T^2) = B[1·(1+T^2)-T(1+T^2)] = B(1 + T^2 -T - T^3) = B(1-T+T^2-T^3)$.
   $(CT+D)(1-T)(1+T) = (CT+D)(1-T^2) = (CT)(1-T^2) + D(1-T^2) = (CT-CT^3)+(D-DT^2)$

   Now, group the coefficients by powers of $T$ (constant, $T$, $T^2$, $T^3$) and equate them to the left side (which is just $1$, or $1 + 0T + 0T^2 + 0T^3$): $1 = A(1+T+T^2+T^3) + B(1-T+T^2-T^3) + (CT-CT^3)+(D-DT^2)$

   Group Like Terms: (constant, T, T², T³) $\begin{cases} A + B + D = 1 (i) \\\ A -B + C = 0 (ii)\\\ A +B -D = 0 (iii) \\\ A -B -C = 0 (iv)\end{cases}$

   Let’s solve this system:
   Subtract equation (iv) from equation (ii): A -B + C -(A -B -C) = A - B + C -A + B + C = 0 ↭ 2C = 0 ↭ C = 0.
   Substitute C = 0 into equation (2): A -B + C = 0 ⇒ A = B.
   Substitute A = B into (i) and (iii): 2A + D = 1, 2A -D = 0 ⇒[Solve for A and D] $4A = 1 ↭ A = B = \frac{1}{4}$ ⇒[Now calculate D using 2A -D = 0] $D = \frac{1}{2}$

   Plug these back into the partial fraction setup: $\frac{1}{(1-T)(1+T)(1+T^2)} = \frac{A}{1-T}+ \frac{B}{1+T} + \frac{CT+D}{1+T^2} = \frac{1}{4}\frac{1}{1-T}+\frac{1}{4}\frac{1}{1+T}+\frac{\frac{1}{2}}{1+T^2}$

   Now, integrate each term individually: $\int (\frac{1}{4}\frac{1}{1-T}+\frac{1}{4}\frac{1}{1+T}+\frac{\frac{1}{2}}{1+T^2})dT = \int kdt ↭ -\frac{1}{4}ln|T-1|+\frac{1}{4}ln|1+T| + \frac{1}{2}arctan(T) = kt + C$.

   Note: $\int \frac{1}{4}\frac{1}{1-T}dT =[u = 1-T, du = -dt] \frac{-1}{4}\int \frac{1}{u}du = \frac{-1}{4}ln|u| = \frac{-1}{4}ln|1-T|$

   Using logarithm properties, we can condense the log terms: $\frac{1}{4}\ln\left|\frac{1 + T}{1 - T}\right| + \frac{1}{2}\arctan(T) = kt + C$

4. Final Solution. The implicit solution to the differential equation is: $\boxed{\frac{1}{4}\ln\left|\frac{1 + T}{1 - T}\right| + \frac{1}{2}\arctan(T) = kt + C}$ where C is the constant of integration determined by the initial temperature.

5. Interpretation.

The solution describes how the dimensionless temperature $T$ evolves over time. Due to the transcendental nature of combining logarithmic and inverse trigonometric functions, solving for $T$ explicitly as a function of $t$ is not mathematically feasible using elementary functions.

However, we can still analyze the physics! As $t \to \infty$, the right side goes to infinity. For the left side to go to infinity, the denominator of the log term, $(1 - T)$, must approach 0. Therefore, $T \to 1$. Returning to our original variables ($T_1 \to 1 \implies T/M \to 1$), we see that the object’s temperature successfully approaches the environment’s temperature $M$ over time.

Solving a Bernoulli Differential Equation Using Direct Substitution

Bernoulli differential equations are a special class of nonlinear first-order differential equations of the form: $y' = p(x)y + q(x)y^n$ where:

• y = y(x) is the unknown function of x,
• p(x) and q(x) are continuous functions on an interval I,
• n is a real number not equal to 0 or 1.

These equations are nonlinear due to the term $y^n$ and can be challenging to solve directly. However, they can be transformed into linear differential equations using an appropriate substitution, making them solvable using standard methods.

General Method for Solving Bernoulli Equations

Given the Bernoulli equation: $y' = p(x)y + q(x)y^n$, our goal is to eliminate the nonlinear term $y^n$ through substitution.

1. Rewrite the Equation if Necessary. Ensure the equation is in the standard Bernoulli form. If not, rearrange terms to achieve this form.
2. Divide both sides by $y^n$: $\frac{y'}{y^n} = p(x)\frac{1}{y^{n-1}}+q(x)$
3. To simplify the nonlinear term, we introduce a new variable v defined as: $v = \frac{1}{y^{n-1}} = y^{1-n}$. Compute v’ (the derivative of v with respect to x) using the chain rule: $v' = (1-n)y^{-n}y' = (1-n)\frac{1}{y^n}y'$
4. Substitute Back into the Equation. Our original equation becomes: $\boxed{\frac{v'}{1-n} = p(x)v+q(x)}$. This is now a linear differential equation in v, which we can solve using standard techniques like integrating factors.

Example. Solve $y' = \frac{y}{x}-y^2$

Let’s apply the Bernoulli substitution method to solve the differential equation: $y' = \frac{y}{x}-y^2$.

Step 1: Rewrite the equation in standard form. $y'-\frac{y}{x} = -y^2$. We identify that this is a Bernoulli equation y’ + p(x)y = q(x)yⁿ with $p(x) = \frac{-1}{x}, q(x)=-1$, and n = 2. Since n = 2 ≠ 0, 1, we can proceed with the Bernoulli substitution method.

Step 2: Divide both sides by y² (n = 2, the method requires $y \ne 0$): $\frac{y'}{y^2} -\frac{1}{xy} = -1$

Step 3: Introduce the Substitution. We define a new variable $v = \frac{1}{y}$ (since v = y^1-n = y^1-2 = y^-1). Compute the derivative $v' = \frac{d}{dx}(\frac{1}{y}) = \frac{-1}{y^2}y'$. Solving for y: $y' = -y^2v' $

Step 4: Substitute Back into the Original Equation: $-v' - \frac{v}{x} = -1$, and we rewrite it in standard form (multiply both sides by -1) $v' + \frac{v}{x} = 1$

Step 5: Solve the linear equation:

Now, this is a linear first-order differential equation in v. The standard method for solving this is to find an integrating factor. The integrating factor is $μ(x)=e^{\int p(x)dx} = e^{\int \frac{1}{x}dx} = e^{ln(x)} = x$ (assuming x > 0 to avoid absolute values).

Step 6: Multiply through by the integrating factor: v’x + v = x ↭[Recognize the left-hand side as a product derivative] So the equation becomes (xv)’ = x

Step 7: Integrate both sides of the equation with respect to x: $xv = \int xdx = \frac{x^2}{2}+C⇒[\text{Solve for v}] v = \frac{1}{2}x+ \frac{C}{x} = \frac{x^2+2C}{2x}$.

Step 8: Back-substitute to y: $v = \frac{1}{y},\text{ so } y = \frac{2x}{x^2+c_1}$ where c₁ = 2c (for simplicity), and this is a constant of integration determined by initial conditions, if provided.

Thus, the general solution is $\boxed{y = \frac{2x}{x^2+c_1}}, x \ne 0$.

Trivial/Singular Solution: In Step 2, we divided by $y^2$. We must check if $y = 0$ is a solution. Substituting $y=0$ into the original equation $y' = \frac{y}{x} - y^2$ gives $0 = 0 - 0$, which is true. Therefore, $y = 0$ is also a solution.

Final Result:

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