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Introduction to Homogeneous Linear PDEs
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Today we are going to explore some examples related to Homogeneous Linear Partial Differential Equations or PDEs. Who can remind us what makes a PDE 'homogeneous'?
I think it means it doesnβt have any free terms?
Right! A homogeneous PDE contains no terms that are independent of the dependent variable or its derivatives. Let's recall that we typically use the operator method for these. Can anyone tell me what this method includes?
It involves transforming the PDE into operator form by replacing derivatives with operators.
Exactly! Weβll take our first example which involves the equation \( \frac{\partial^2 z}{\partial x^2} - 2 \frac{\partial^2 z}{\partial x \partial y} + \frac{\partial^2 z}{\partial y^2} = 0\). Now, who can convert this into operator form?
That would be \((D^2 - 2DD' + D'^2) z = 0\).
Excellent! Now, what's the next step?
We form the auxiliary equation by substituting \(D\) with \(m\) and \(D'\) with 1, right?
Precisely, let's move on to that!
Solving Example 1
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So, now we have \(m^2 - 2m + 1 = 0\). Who can solve for m here?
That gives us a repeated root, \(m = 1\).
Great! And what does the presence of a repeated root imply for our solution?
It means our solution will take the form \(z = f(y - x) + x f(y - x)\).
Correct! Now remember that this function f could represent various functions based on the context of the problem. Letβs look at Example 2, shifting gears to involve complex roots. Does anyone remember how to handle such roots?
Yeah, we use both sine and cosine functions in our general solution.
Spot on! Let's dive deeper into Example 2!
Exploring Example 2
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Now our second example is \( \frac{\partial^2 z}{\partial x^2} + 4 \frac{\partial^2 z}{\partial y^2} + 5 \frac{\partial^2 z}{\partial x \partial y} = 0\). How do we start here?
Again, we convert to operator form, so it becomes \((D^2 + 4D'D' + 5D'D)z = 0\).
Exactly! Whatβs our auxiliary equation when we substitute the operators with m?
It is \(m^2 + 4m + 5 = 0\).
Perfect! What are the roots?
Those roots are complex: \(m = -2 \pm i\).
Great recovery! And how does this affect our general solution?
It results in \(z = f(y + 2x)\cos(x) + f(y + 2x)\sin(x)\).
Excellent work! So, can anyone summarize the key points we learned today?
We learned to form operator equations and solve for the roots to construct the general solution forms based on whether they are repeated or complex!
Spot on! Letβs continue practicing these techniques.
Introduction & Overview
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Quick Overview
Standard
In this section, we discuss two example problems that illustrate the steps for solving homogeneous linear PDEs with constant coefficients. These examples utilize the operator method and the auxiliary equation to derive the general solutions, showcasing how the nature of the roots influences the solutionβs form.
Detailed
Detailed Summary
This section addresses the practical application of solving Homogeneous Linear Partial Differential Equations (PDEs) with constant coefficients through concrete examples. The key steps highlighted include:
- Operator Form Conversion: Each PDE is converted into operator form, which simplifies further manipulation.
- Example 1: The equation is transformed into $$ (D^2 - 2DD' + D'^2)z = 0 $$
- Auxiliary Equation Formation: The converted form leads to the formation of an auxiliary equation through substitution of operators with variables (e.g., \( D \rightarrow m \)).
- Example 1: Results in \( m^2 - 2m + 1 = 0 \), leading to repeated roots.
- Finding Roots: Determining the nature of the roots (distinct, repeated, complex) unveils the form of the general solution.
- Example 1 illustrates that repeated roots (1) yield a solution of the form \( z = f(y - x) + x f(y - x) \).
- Constructing the General Solution: Based on the auxiliary equationβs roots, the complementary function is built, applicable to a variety of problem types.
- Example 2 shows how complex roots influence the solution to include cosine and sine functions.
Through these examples, the process is elaborated comprehensively, reinforcing the method's significance in solving homogeneous PDEs effectively.
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Example 1: Solving a Homogeneous PDE
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Example 1:
Solve β βΒ²π§ + 2βΒ²π§ + βΒ²π§ = 0
βπ₯Β² βπ₯βπ¦ βπ¦Β²
Solution: Operator form:
(π·Β² β 2π·π·β² + π·β²Β²)π§ = 0
Auxiliary equation:
πΒ² β 2π + 1 = 0 β (π β 1)Β² = 0
Repeated root: π = 1
So, the general solution is:
π§ = π(π¦ β π₯) + π₯π(π¦ β π₯)
Detailed Explanation
In this example, we start by observing the given PDE, which is homogeneous. First, we convert the PDE into operator form using differential operators. The auxiliary equation is then formed by substituting these operators with a variable (π). We solve the auxiliary equation to find the roots; in this case, we find a repeated root (π = 1). This indicates how the solution will behave. Given this repeated root, the general solution incorporates terms that include both a function of (π¦βπ₯) and a multiple of π₯ times that function, which can account for the repeated nature of the roots.
Examples & Analogies
Think of the process like finding a pattern in predicting weather. If you notice that two consecutive days are often similar (like needing to dress for two similarly warm days), youβd be looking to replicate that pattern in your planning. Similarly, in this problem, the repeated root tells us that thereβs a predictable trend in how the solution behaves based on given conditions.
Example 2: Another Homogeneous PDE Case
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Example 2:
Solve + βΒ²π§ + 4βΒ²π§ + 5 = 0
βπ₯Β² βπ₯βπ¦ βπ¦Β²
Solution: Operator form:
(π·Β² + 4π·π·β² + 5π·β²Β²)π§ = 0
Auxiliary equation:
πΒ² + 4π + 5 = 0 β π = -2 Β± π
Complex roots:
π§ = π(π¦ + 2π₯)cos(π₯) + π(π¦ + 2π₯)sin(π₯)
Detailed Explanation
In this second example, we once again start with a homogeneous PDE and convert it into operator form. The auxiliary equation derived from this form helps us find the characteristic roots. Here, we end up with complex roots, indicating that the solutions will include oscillatory behavior due to the sine and cosine terms; those describe the changes in solutions due to the underlying oscillations created by the imaginary part of the roots. The general solution thus comprises two parts: one related to the cosine function and the other to the sine function, reflecting the inherent oscillation in the problem.
Examples & Analogies
Think of the waves created when you throw a stone into a pond. The ripples spread out, creating waves that can be described mathematically by the sine and cosine functions. In our problem, these complex roots suggest the solutions will behave like these waves, fluctuating between positive and negative values as time or conditions change.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Homogeneous Linear PDEs: Equations with no free terms, of the form involving partial derivatives.
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Operator Form: Transitioning PDEs into a form using operators to simplify solution methods.
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Auxiliary Equation: The algebraic equation derived from the operator form that helps identify the nature of roots.
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Roots of the Auxiliary Equation: The roots determine the general form of the solution based on their distinctness, repetition, or complexity.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example 1 illustrates a repeated root scenario when solving the PDE \( \frac{\partial^2 z}{\partial x^2} - 2 \frac{\partial^2 z}{\partial x \partial y} + \frac{\partial^2 z}{\partial y^2} = 0 \).
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Example 2 demonstrates dealing with complex roots in \( \frac{\partial^2 z}{\partial x^2} + 4 \frac{\partial^2 z}{\partial y^2} + 5 \frac{\partial^2 z}{\partial x \partial y} = 0 \).
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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PDEs are fun, homogeneous and bright, No free terms in sight, just derivatives in their flight!
π Fascinating Stories
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Imagine a world where equations danced; each PDE in harmony, no lonely terms could prance. They paired up in lovely twirls, creating solutions that unfurl.
π§ Other Memory Gems
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To remember the roots: DRCC - Distinct, Repeated, Complex, and Consider.
π― Super Acronyms
Use the acronym PCO to remember the steps in the process
- P: - Formulate the PDE
- C: - Convert to operator form
- O: - Obtain the auxiliary equation.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
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Term: Partial Differential Equation (PDE)
Definition:
An equation that involves partial derivatives of a multivariable function.
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Term: Linear PDE
Definition:
A PDE where the dependent variable and all its derivatives are of the first power, without multiplication between them.
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Term: Homogeneous PDE
Definition:
A PDE where all terms contain the dependent variable or its derivatives, and there are no free terms.
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Term: Constant Coefficients
Definition:
Coefficients in the PDE that are constant, not functions of the independent variables.
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Term: Operator Method
Definition:
A systematic approach to solve a PDE by replacing derivatives with differential operators.
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Term: Auxiliary Equation
Definition:
An algebraic equation formed from the operator form of a PDE, used to find roots and construct the solution.
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Term: Complementary Function (CF)
Definition:
The general solution of a homogeneous equation derived from the roots of the auxiliary equation.