A partial differential equation (PDE) is a type of mathematical equation that involves a function of severable variables and its partial derivatives with respect to those variables, e.g., $\frac{∂f}{∂t} = \frac{∂^2f}{∂x^2}, \frac{∂w}{∂t}-\frac{∂^2w}{∂x^2} = 0, etc.$ Partial differential equations are incredibly important in many fields of science and engineering.

The Existence and Uniqueness Theorem provides critical insight into the behavior of the solutions to first-order differential equations. It states that:

If f(x, y) (the right-hand side of the ODE) is continuous in a neighborhood around a point (x_{0}, y_{0}) and its partial derivative with respect to y, $\frac{∂f}{∂y}$, 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 (x_{0}, y_{0}).

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

Consider a second-order linear differential equation with constant coefficients: y’’ + Ay’ + By = f(x)

This equation is inhomogeneous because of the non-zero forcing term f(x). Our goal is to find a particular solution y_{p} to this equation. The general solution to the ODE will then be the sum of the particular solution and the complementary solution y_{c}, which solves the associated homogeneous equation y’’ + Ay’ + By = f(x). Hence, the general solution y = y_{p} + c_{1}y_{1} + c_{2}u_{2} where c_{1}y_{1} + c_{2}u_{2} is the complementary or homogeneous solution.

We are particularly interested in finding y_{p} when f(x) takes certain common form such as:

- e
^{ax}, where a is a constant. - sin(wx) and cos(wx), where w is a constant.
- More generally, e
^{(a+iw)x}, which is a complex exponential. For simplicity, we’ll denote this as e^{αx}, where α is a complex number.

To make the process clearer, we rewrite the original ODE using differential operator notation. Let D represent the derivative with respect to x. The original ODE: y’’ + Ay’ + By = f(x) becomes (D^{2} + AD + B)y = f(x).

We can then introduce the operator p(D) defined as: p(D) = D^{2} + AD + B.

p(D) can be understood as a linear operator on functions and also a formal polynomial in D.

This allows us to rewrite the ODE as: p(D)y = f(x)

**The substitution Rule**. A key result for solving such equations is the Substitution Rule for exponentials. This rule states that: p(D)e^{αx} = p(α)e^{αx}.

Proof:

Apply the differential operator to the exponential function e^{αx}:

p(D)e^{αx} = (D^{2} + AD + B)e^{αx} =[By linearity, calculate each of them] D^{2}e^{αx} + ADe^{αx} + Be^{αx} = α^{2}e^{αx} + Aαe^{αx} + Be^{αx} =[Factor out e^{ax}] (α^{2} +Aα + B)e^{αx} = p(α)e^{αx} ⇒ p(D)e^{αx} = p(α)e^{αx}∎

**Exponential input theorem**. It states that if f(x) = e^{ax}, the inhomogeneous ODE: y'' + Ay' + By = e^{αx} has a particular solution of the form $y_p = \frac{e^{αx}}{p(α)}$ assuming that p(α) ≠ 0.

Proof.

Start with the equation p(D)y_{p} = e^{αx} and assume $y_p = \frac{e^{αx}}{p(α)}$

Substituting this into the equation:

$p(D)y_p = p(D)\frac{e^{αx}}{p(α)} =[\text{Use the substitution Rule}] \frac{p(α)e^{αx}}{p(α)} = e^{αx}$ ∎

Exercise. Find the particular solution for the equation y’’ -y’ +2y = 10e^{-x}sin(x)

**Step 1: Convert to Complex Form**
To handle the trigonometric function sin(x), express it as the imaginary part of the complex exponential e^{(-1+i)x}. Thus, we solve the complex ODE: $(D^2-D+2)\tilde{y} = 10e^{(-1+i)x}$

**Step 2: Apply the Exponential Input Theorem**

The characteristic polynomial is p(α) = α^{2} - α + 2. Substituting α = −1+i: $p(-1+i) = (-1+i)^2-(-1+i)+2 = 3 -3i$

Thus, the complex particular solution is: $\tilde{y_p} = \frac{e^{αx}}{p(α)} = \frac{10^{(-1+i)x}}{3 -3i} = \frac{10e^{(-1+i)x}}{3(1 -i)} = \frac{10}{3}\frac{1+i}{(1-i)(1+i)}e^{(-1+i)x} = \frac{10}{3}\frac{1+i}{2}e^{-x}(cos(x)+isin(x)) = \frac{5}{3}(1+i)e^{-x}(cos(x)+isin(x))$

Extract the imaginary part:

$y_p = Im(\tilde{y_p}) = \frac{5}{3}e^{-x}(cos(x)+sin(x)) =[\text{Alternatively, we can express this as:}] \frac{5}{3}e^{-x}\sqrt{2}cos(x-\frac{π}{4})$

Recall: Acos(x) + Bsin(x) = Rcos(x -ϕ) where A = B = 1, R is the amplitude, R = $\sqrt{1+1}=\sqrt{2}$ and ϕ is the phase shift, cos(ϕ) = sin(ϕ) = ^{1}⁄_{√2}. Since both the sine and cosine of ϕ are equal, we recognize that: ϕ = ^{π}⁄_{4}

When p(α)=0, the simple approach using the Exponential Input Theorem fails. This occurs when α is a root of the characteristic equation. In this case, we must modify the form of the particular solution.

Exponential-shift rule. If p(α) = 0, it states that: p(D)e^{ax}u(x) = e^{ax}p(D +a)u(x).

Proof:

Particular case p(D) = D

p(D)e^{ax}u(x) =[Particular case p(D) = D] De^{ax}u(x) = [Product rule] e^{ax}u’ + ae^{ax}u =[Factor e^{ax}] e^{ax}(u’ + au) = e^{ax}(D + a)u = e^{ax}p(D + a)u(x)

p(D) = D^{2}

D^{2}e^{ax}u = D(De^{ax}u) =[Previous result] D(e^{ax}(D + a)u) = [Previous result, but u is in this case (D + a)u] e^{ax}(D + a)(D + a)u = e^{ax}(D + a)^{2}u.

This result is derived from the product rule of differentiation and can be applied iteratively to higher-order terms.

Let y’’ + Ay’ + By = e^{ax} be an inhomogeneous ODE where a is a constant. Then, a particular solution is given by $y_p = \frac{xe^{ax}}{p’(a)}$ assuming that p(a) = 0 and a is a simple root of the characteristic equation p(D) = D^{2} + AD + B. If a is a double root, the particular solution is $y_p = \frac{x^2e^{ax}}{p’’(a)}$

Proof (Simple root case)

If a is a simple root of p(D), then p(D) = D^{2} +AD + B. If a is a root of this characteristic equation, we know that p(a) = a^{2} + Aa + B = 0. In the case of a simple root, we differentiate p(D) with respect to D, p’(D) = 2D + A. Substituting D = a, we get, p’(a) = 2a + A.

Establish the claim: p(D)y_{p} = $p(D)\frac{e^{ax}·x}{p’(a)} = e^{ax}$

$p(D)y_p = p(D)\frac{e^{ax}·x}{p’(a)}$

Now, use the Exponential Shift Rule, which states that: p(D)e^{ax}u(x) = e^{ax}p(D+a)u(x), and apply this rule to e^{ax}·x where u(x) = x, so

$p(D)y_p = p(D)\frac{e^{ax}·x}{p’(a)} = \frac{e^{ax}}{p’(α)}·p(D+a)x$

Next, calculate p(D + a) ↭[p(D) = D^{2} + AD + B] (D + a)^{2} + A(D + a) + B^{2} = D^{2} + 2aD + a^{2} + AD + Aa + B =[a is a root, a^{2} +Aa + B = 0] D^{2} +(2a + A)D

Now, applying this to x, we calculate: p(D + a)x = (D^{2} +(2a + A)D)x = D^{2}(x) + (2a + A)D(x) =[D(x) = 1, D’’(x) = 0] 0 + (2a + A)·1 = (2a + A) = p’(a)

The operator D is shorthand for “taking the derivative with respect to x”, D(x) is 1, D(1) = 0 because the derivative of a constant is zero, D^{2}(x) = 0.

$p(D)y_p = p(D)\frac{e^{ax}·x}{p’(a)} = \frac{e^{ax}}{p’(a)}p(D+a)x = \frac{e^{ax}}{p’(a)}p’(a) = e^{ax}$∎

- Find a particular solution to the inhomogeneous ODE y’’ -3y’ + 2y = e
^{x}.

To apply the theorem, we first need to determine the characteristic equation associated with the homogeneous part of the ODE: y’’ -3y’ + 2y = e^{x}.

The characteristic equation is obtained by replacing y′′ with D^{2}, y′ with D, and y with 1, where D is the differential operator: D^{2} -3D +2 = 0.

Solve the characteristic equation D^{2} -3D +2 =[Factoring the quadratic equation gives:] (D -1)(D -2), so the roots are D = 1 and D = 2. These roots corresponds to the solutions e^{x} and e^{2x} for the homogeneous ODE y’’ -3y’ + 2y = 0.

Since the right-hand side of the inhomogeneous ODE is e^{x}, we look at the root D = 1. 1 is a simple root of the characteristic equation because it only appears once in the factorization (D−1)(D−2).

By the theorem, when the right-hand side of the ODE is of the form e^{ax} and a is a simple root of the characteristic equation, the particular solution is given by:

$y_p = \frac{xe^{ax}}{p’(a)} = \frac{xe^x}{p’(1)}$

The characteristic polynomial is: p(D) = D^{2} -3D +2. Differentiate this with respect to D:
p’(D) = 2D -3. Now, substitute D = 1: p’(1) = 2 -3 = -1.

y_{p} $= \frac{xe^x}{p’(1)} = \frac{xe^x}{-1} = -xe^x$

We are exploring the phenomenon of resonance through the following second-order linear inhomogeneous differential equation: y’’ + w_{0}^{2}y = cos(w_{1}t). This equation describes a system (such as a mass-spring system or a pendulum) that experiences external forcing, represented by cos(w_{1}t) where

- w
_{0}^{2}is the**natural frequency**of the system (how it naturally oscillates without external forces). - w
_{1}is the input frequency of the external force (the frequency of the external “push”, like pushing the pendulum at regular intervals).

We assume w_{1} ≠ w_{0}, meaning the input frequency is not exactly the same as the system’s natural frequency.

**Convert the Equation to Differential Operator Form:** First, we rewrite the differential equation using D, where D = d/dt represents the derivative with respect to t. So: (D^{2} + w_{0}^{2})y = cos(w_{1}t).

This is a second-order linear inhomogeneous differential equation.

**Switch to Complex Exponentials (Easier to Solve):** Since working with cosine functions can be tricky, we switch to using complex exponentials, as they simplify the process. Using Euler’s formula, we know that: $cos(w_1t)=Re(e^{iw_1t})$

Thus, solving the equation with $e^{iw_1t}$ will give us a complex solution, and we’ll take the real part at the end to get our solution for the cosine function.

The equation becomes: $(D^2+w_0^2)\tilde{y} = e^{iw_1t}$ where $\tilde{y}$ is the complex solution, and we are only interested in the real part of $\tilde{y}$ at the end.

**Find a Particular Solution:**

To solve for $\tilde{y_p}$, we use the fact that e^{iw1t} is a simple exponential, and we substitute into the equation: [**Exponential input theorem**. It states that if f(x) = e^{ax}, the inhomogeneous ODE: y'' + Ay' + By = e^{αx} has a particular solution of the form $y_p = \frac{e^{αx}}{p(α)}$ assuming that p(α) ≠ 0.]

p(D) is the characteristic polynomial of the differential operator D^{2} + w_{0}^{2}. p(iw_{1}) = $(iw_1)^2+w_0^2 = w_0^2-w_1^2$

Thus, the particular solution becomes:

$\tilde{y_p} = \frac{e^{iw_1t}}{(iw_1t)^2+w_0^2} =[\text{Simplifying the denominator:}] \frac{e^{iw_1t}}{w_0^2-w_1^2}$

Now, we take the real part of $y_p$, to find the solution for y_{p}:
$y_p = Re(\tilde{y_p}) = \frac{cos(w_1t)}{w_0^2-w_1^2}$. This is the particular solution when the input frequency is not equal to the system’s natural frequency, w_{1} ≠ w_{0}

**Resonance (When w _{1} ≈ w_{0}):**

Resonance occurs when the input frequency w_{1} is close to the natural frequency w_{0} of the system. In such cases, the response of the system becomes amplified (the denominator $w_0^2-w_1^2$ get closer to zero, which leads to the amplitude of the response growing significantly), as the forcing frequency aligns with the system’s natural oscillation (Refer to Figure A for a visual representation and aid in understanding it)

If w_{0} ≈ w_{1}, we have a resonance condition. In this case, the earlier approach doesn’t work because the denominator becomes zero, meaning the solution must be treated differently.

We start with (D^{2} + w_{0}^{2})y = cos(w_{1}t).

Again, we convert to complex exponentials:

$(D^2+w_0^2)\tilde{y} = e^{iw_0t}$ but iw_{0} is a root of the characteristic polynomial, $(D^2+w_0^2)(iw_0) = (iw_0)^2+w_0^2 = 0$

To find a particular solution [$y_p = \frac{xe^{ax}}{p’(a)}, p’(D) = 2D$], $p’(iw_0) = 2iw_0$ $\tilde{y_p} = \frac{te^{iw_0t}}{2iw_0}$

Finally, we take the real part of this to get the solution for the cosine case:

$y_p = Re(\tilde{y_p}) = \frac{tsin(w_0t)}{2w_0}$. When w_{1} ≈ w_{0}, the solution has a steadily increasing amplitude. In particular:

- The term tsin(w
_{0}t) shows that the amplitude grows linearly with time. - This is a characteristic of resonance: the system’s response builds over time, leading to larger and larger oscillations (Refer to Figure C for a visual representation and aid in understanding it).
If there’s no damping or limit to the oscillation, resonance can cause the amplitude to grow indefinitely, which is why resonance can be both powerful and dangerous (e.g., bridges collapsing due to resonance from wind or foot traffic)

We are examining resonance by working with the following second-order linear inhomogeneous differential equation: y’’ + w_{0}^{2}y = cos(w_{1}t)

This equation describes a system (such as a mass-spring system or a pendulum) that experiences external forcing, represented by cos(w_{1}t)

The general solution to this inhomogeneous differential equation has two parts:

- The complementary solution y
_{c}, which solves the associated homogeneous equation y’’ + w_{0}^{2}y = 0. - The particular solution y
_{p}, which accounts for the inhomogeneous part cos(w_{1}t)

The complementary solution corresponds to the general solution of the homogeneous equation: y’’ + w_{0}^{2}y = 0. The characteristic equation for this is r^{2} + w_{0}^{2}y = 0. Solving this gives complex roots r = ±iw_{0}. Therefore, the complementary solution is: y_{c} = c_{1}cos(w_{0}t) + c_{2}sin(w_{0}t) where c_{1} and c_{2} are arbitrary constants determined by initial conditions. This describes the natural oscillatory behavior of the system with frequency w_{0}.

We have previously found a particular solution:
$y_p = \frac{cos(w_1t)}{w_0^2-w_1^2}$. This is the particular solution when the input frequency is not equal to the system’s natural frequency, w_{0} ≠ w_{1}

When the forcing frequency w_{1} approaches the natural frequency w_{0}, the system experiences resonance. In this case, the particular solution $y_p = \frac{cos(w_1t)}{w_0^2-w_1^2}$ becomes problematic as the denominator approaches zero, leading to a singularity.

To deal with this, we find another particular solution that takes into account the resonance condition. We start with the complementary solution: $y_c = -\frac{cos(w_0t)}{w_0^2-w_1^2}$

So another particular solution is $\frac{cos(w_1t)}{w_0^2-w_1^2} -\frac{cos(w_0t)}{w_0^2-w_1^2}$

Taking the limit as w_{1} → w_{0}:

$\lim_{w_1 \to w_0} \frac{cos(w_1t)}{w_0^2-w_1^2} -\frac{cos(w_0t)}{w_0^2-w_1^2} = \lim_{w_1 \to w_0} \frac{cos(w_1t)-cos(w_0t)}{w_0^2-w_1^2} $[L’Hopital Rule, please note the variable is w_{1}] $\lim_{w_1 \to w_0} \frac{-sin(w_1t)·t}{-2·w_1} = \frac{sin(w_0t)}{2w_0}$ This result is the particular solution when w_{1} = w_{0}, which corresponds to the **resonance** case.

This solution grows linearly in time (t), reflecting the amplification of oscillations characteristic of resonance. The amplitude of the oscillation increases over time without bound, which can lead to destructive consequences if no damping is present.

We can also provide a geometric interpretation of the behavior as w_{1} approaches w_{0}.

Consider the difference:

$\frac{cos(w_1t)-cos(w_0t)}{w_0^2-w_1^2}= [\text{Using the identity trigonometry: }cosB -cosA = 2sin(\frac{A-B}{2})sin(\frac{A+B}{2})] = \frac{2sin(\frac{w_0-w_1}{2})t+sin(\frac{w_0+w_1}{2})t}{w_0^2-w_1^2}$ This equation consists of two parts: (Refer to Figure B for a visual representation and aid in understanding it)

- $\frac{2sin(\frac{w_0-w_1}{2})t}{w_0^2-w_1^2}$. This represents oscillations with very small frequency (since w
_{1}-w_{0}is small), resulting in**a slowly varying amplitude**. The period of this oscillation is large. Because, the term $\frac{2}{w_0^2-w_1^2}$, it increases at a steadily increasing amplitude as w_{1}approaches w_{0}. - $\frac{sin(\frac{w_0+w_1}{2})t}{w_0^2-w_1^2} ≈ \frac{sin(w_0)t}{w_0^2-w_1^2}$, this is
**a pure oscillatory term**with frequency close to w_{0}.

As w_{1} approaches w_{0}, the system exhibits a phenomenon known as beats, where the amplitude of the oscillations grows slowly over time. At exact resonance (w_{1} = w_{0}), the oscillation amplitude grows linearly over time, leading to increasingly large oscillations. This growth can continue indefinitely in undamped systems, resulting in increasingly large oscillations that may cause structural damage, failure or other dangerous effects.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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