Advanced Engineering Mathematics

Dennis G. Zill, Michael R. Cullen

ISBN #9780763740955

3rd Edition

4,310 Questions

17,647 Students Helped

Homework Questions

Summary

Learning Objectives

Key Concepts

Example Problems

Explanations

Common Mistakes

Summary

The reviewed material emphasizes that separation of variables transforms complex PDEs into simpler ODEs based on assumed product solutions. Carefully applying boundary conditions leads to eigenvalue problems whose eigenfunctions form an orthogonal basis. Both homogeneous and nonhomogeneous boundary-value problems can be solved via series expansions (Fourier or orthogonal series), and these techniques extend from rectangular to polar coordinates. Mastery of these methods is essential in modeling physical phenomena like heat conduction, wave propagation, and potential flows.

Learning Objectives

Explain the method of separation of variables for solving partial differential equations (PDEs) in various coordinate systems.

Identify and derive eigenvalue problems and their eigenfunctions from boundary?value conditions.

Construct Fourier and orthogonal series representations to express solutions in both rectangular and polar coordinates.

Apply substitution techniques to convert nonhomogeneous boundary?value problems into homogeneous ones and solve them.

Critically analyze advanced example problems (heat equation, Laplace’s equation) and recognize the role of separation constants.

Key Concepts

CONCEPT

DEFINITION

Separation of Variables

A method that assumes a solution in product form (e.g. u(x,t)=X(x)T(t)) to reduce a PDE into a set of ordinary differential equations.

Eigenvalue Problem

A problem of the form L(X)=−λX, where nontrivial solutions exist only for specific values of λ (eigenvalues), and the corresponding functions are eigenfunctions.

Boundary Value Problem (BVP)

A differential equation together with a set of constraints (boundary conditions) defined at the boundary of the domain.

Fourier Series

A representation of periodic functions as infinite sums of sines and cosines, used to expand solutions of PDEs.

Orthogonal Series Expansion

An expansion of a function in terms of mutually orthogonal eigenfunctions derived from a Sturm–Liouville problem.

Sturm–Liouville Problem

A type of eigenvalue problem involving a second‐order linear differential operator with associated orthogonal eigenfunctions.

Example Problems

Example 1

Substituting $u(x, y)=X(x) Y(y)$ into the partial differential equation yields $X^{\prime} Y=X Y^{\prime} .$ Separating variables and using the separation constant $-\lambda,$ where $\lambda \neq 0,$ we obtain $$\frac{X^{\prime}}{X}=\frac{Y^{\prime}}{Y}=-\lambda.$$ When $\lambda \neq 0$ $$X^{\prime}+\lambda X=0 \quad \text { and } \quad Y^{\prime}+\lambda Y=0$$ so that $$X=c_{1} e^{-\lambda x} \quad \text { and } \quad Y=c_{2} e^{-\lambda y}.$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-\lambda(x+y)}, \quad \lambda \neq 0.$$ When $\lambda=0$ the differential equations become $X^{\prime}=0$ and $Y^{\prime}=0,$ so in this case $X=c_{4}, Y=c_{5},$ and $u=X Y=c_{6}$.

Example 2

Substituting $u(x, y)=X(x) Y(y)$ into the partial differential equation yields $X^{\prime} Y+3 X Y^{\prime}=0 .$ Separating variables and using the separation constant $-\lambda$ we obtain $$\frac{X^{\prime}}{-3 X}=\frac{Y^{\prime}}{Y}=-\lambda.$$ When $\lambda \neq 0$ $$X^{\prime}-3 \lambda X=0 \quad \text { and } \quad Y^{\prime}+\lambda Y=0$$ so that $$X=c_{1} e^{3 \lambda x} \quad \text { and } \quad Y=c_{2} e^{-\lambda y}.$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{\lambda(3 x-y)}.$$ When $\lambda=0$ the differential equations become $X^{\prime}=0$ and $Y^{\prime}=0,$ so in this case $X=c_{4}, Y=c_{5},$ and $u=X Y=c_{6}$.

Example 3

Substituting $u(x, y)=X(x) Y(y)$ into the partial differential equation yields $X^{\prime} Y+X Y^{\prime}=X Y .$ Separating variables and using the separation constant $-\lambda$ we obtain $$\frac{X^{\prime}}{X}=\frac{Y-Y^{\prime}}{Y}=-\lambda.$$ Then $$X^{\prime}+\lambda X=0 \quad \text { and } \quad Y^{\prime}-(1+\lambda) Y=0$$ so that $$\begin{aligned} &X=c_{1} e^{-\lambda x} \quad \text { and }\\ &Y=c_{2} e^{(1+\lambda) y} \end{aligned}$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{y+\lambda(y-x)}.$$

Example 4

Substituting $u(x, y)=X(x) Y(y)$ into the partial differential equation yields $X^{\prime} Y=X Y^{\prime}+X Y .$ Separating variables and using the separation constant $-\lambda$ we obtain $$\frac{X^{\prime}}{X}=\frac{Y+Y^{\prime}}{Y}=-\lambda.$$ Then $$X^{\prime}+\lambda X=0 \quad \text { and } \quad y^{\prime}+(1+\lambda) Y=0$$ so that $$\begin{aligned} &X=c_{1} e^{-\lambda x} \quad \text { and }\\ &Y=c_{2} e^{-(1+\lambda) y}=0. \end{aligned}$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-y-\lambda(x+y)}.$$

Example 5

Substituting $u(x, y)=X(x) Y(y)$ into the partial differential equation yields $x X^{\prime} Y=y X Y^{\prime} .$ Separating variables and using the separation constant $-\lambda$ we obtain $$\frac{x X^{\prime}}{X}=\frac{y Y^{\prime}}{Y}=-\lambda.$$ When $\lambda \neq 0$ $$x X^{\prime}+\lambda X=0 \quad \text { and } \quad y Y^{\prime}+\lambda Y=0$$ so that $$X=c_{1} x^{-\lambda} \quad \text { and } \quad Y=c_{2} y^{-\lambda}.$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3}(x y)^{-\lambda}.$$ When $\lambda=0$ the differential equations become $X^{\prime}=0$ and $Y^{\prime}=0,$ so in this case $X=c_{4}, Y=c_{5},$ and $u=X Y=c_{6}.$

Step-by-Step Explanations

QUESTION

How do you solve a homogeneous heat equation using separation of variables?

STEP-BY-STEP ANSWER:

Step 1: Assume a product solution u(x,t)=X(x)T(t). Substitute into the PDE to obtain X(x)T'(t)=kX''(x)T(t).
Step 2: Separate the variables by dividing both sides by kX(x)T(t), obtaining T'(t)/(kT(t)) = X''(x)/X(x) = -λ.
Step 3: Solve the ODE for X(x): X'' + λX = 0 subject to spatial boundary conditions. Depending on conditions (Dirichlet, Neumann, mixed) determine eigenvalues and eigenfunctions.
Step 4: Solve the ODE for T(t): T' + kλT = 0 to obtain T(t)=Ce^{-kλt}.
Step 5: Combine separated solutions (apply superposition principle) to express u(x,t) as a Fourier series with coefficients determined by the initial condition.
Final Answer:

Separable PDE in Rectangular Coordinates

QUESTION

How is the method of splitting u(x,t)=v(x,t)+ψ(x) used for nonhomogeneous boundary conditions?

STEP-BY-STEP ANSWER:

Step 1: Choose ψ(x) to satisfy the nonhomogeneous boundary conditions. For example, if u(0,t)=u0 and u(L,t)=u1, set ψ(x)=u0+(x/L)(u1−u0).
Step 2: Define v(x,t)=u(x,t)−ψ(x) so that v(x,t) obeys homogeneous boundary conditions.
Step 3: Substitute u(x,t)=v(x,t)+ψ(x) into the original PDE. The steady-state function ψ(x) is chosen so that the extra terms cancel the nonhomogeneous part.
Step 4: Solve the resulting homogeneous PDE for v(x,t) using separation of variables and expand the initial condition in terms of the eigenfunctions.
Step 5: Express the final solution u(x,t)=v(x,t)+ψ(x) as the sum of the steady state and the transient response.
Final Answer:

Substitution for Nonhomogeneous Boundary Conditions

QUESTION

How do eigenfunction expansions arise when solving Laplace’s equation in polar coordinates?

STEP-BY-STEP ANSWER:

Step 1: Write Laplace’s equation in polar coordinates (r,θ) and assume a separable solution u(r,θ)=R(r)Θ(θ).
Step 2: Substitute into Laplace’s equation and separate variables to obtain equations of the form r^2R''+rR'-λR=0 and Θ''+λΘ=0.
Step 3: Solve the angular equation Θ''+λΘ=0 subject to periodicity (e.g. Θ(θ+2π)=Θ(θ)), leading to expansions in terms of sines and cosines.
Step 4: Solve the radial equation, often resulting in Bessel’s equations for problems in circular domains.
Step 5: Combine the solutions and determine series coefficients by applying the boundary conditions on the domain.
Final Answer:

Separation in Polar Coordinates

Common Mistakes

Not correctly separating variables – failing to fully divide by the product solution might lead to mixing of terms dependent on different variables.
Misapplying boundary conditions, such as using an incorrect eigenvalue sign or forgetting to enforce homogeneous conditions via substitution.
Errors in computing Fourier coefficients (e.g., sign errors, missing normalization factors) when representing the initial condition.
Overlooking the importance of orthogonality in eigenfunction expansions, leading to improper series solutions.
Confusing the roles of the separation constant and parameters in the differential equations, especially when substituting nonhomogeneous terms.