Equations of Mathematical Physics and Separation of Variables Notes

Course Overview and Introduction to Mathematical Physics Equations

  • Content Summary: The course covers four primary methods, three fundamental equations, and two special functions used in mathematical physics.     - Four Methods: Separation variable method, Traveling wave method, integral-transform method, and Green’s function method.     - Three Equations: Wave equation, Heat equation, and Laplace equation.     - Two Special Functions: Bessel Function and Legendre Function.
  • Definition of Mathematical Equations: Using mathematical equations to describe physical phenomena.
  • Chapter 2 Focus: Separation Variable Method, applied specifically to:     - A. The free vibration of a finite string.     - B. Heat conduction on a finite rod.     - C. Laplace’s Equation.     - D. Solution of non-homogeneous Partial Differential Equations (PDE).     - E. Treatment of non-homogeneous boundary conditions.

The Separation of Variables Method

  • Main Concept: The core idea is to assume that the solution to a PDE, such as u(α,β,γ)u(\alpha, \beta, \gamma), can be decomposed into a product of three independent functions: u(α,β,γ)=f(α)g(β)h(γ)u(\alpha, \beta, \gamma) = f(\alpha)g(\beta)h(\gamma).
  • Procedure:     - 1. Determine three ordinary differential equations (ODE) that each function (ff, gg, hh) must satisfy.     - 2. Solve each ODE independently (these are usually familiar, elementary equations).     - 3. "Paste" the solutions together to reconstruct the full PDE solution.
  • Formal Steps:     - Step 1: Find particular solutions (satisfying boundary or initial conditions) in variable separation form.     - Step 2: Use the superposition principle to create a linear combination of these solutions.     - Step 3: Use the remaining initial or boundary conditions to determine the superposition coefficients.
  • Scope of Application: Wave equations, heat conduction problems, stable field (steady-state) problems, etc.
  • Characteristics:     - Physical Guarantee: The superposition principle ensures physical validity.     - Mathematical Guarantee: The uniqueness of the solution ensures correctness.     - Simplification: Transforms complex PDEs into manageable ODEs.

Free Vibration of a Finite String

  • Problem Definition: Vibration behavior of a string with two ends fixed.
  • General PDE Formulation:     - Equation: 2ut2=a22ux2\frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2} where 0<x<l0 < x < l and t>0t > 0.     - Boundary Conditions (BCs): u(0,t)=0u(0, t) = 0 and u(l,t)=0u(l, t) = 0 (ends are fixed).     - Initial Conditions (ICs): u(x,0)=ϕ(x)u(x, 0) = \phi(x) and utt=0=ψ(x)\frac{\partial u}{\partial t} |_{t=0} = \psi(x).
  • Derivation via Separation of Variables:     - Let u(x,t)=X(x)T(t)u(x, t) = X(x)T(t).     - Substitute into the initial equation: X(x)T(t)=a2X(x)T(t)X(x)T''(t) = a^2 X''(x)T(t).     - Rearrange to separate variables: X(x)X(x)=T(t)a2T(t)=λ\frac{X''(x)}{X(x)} = \frac{T''(t)}{a^2 T(t)} = -\lambda.     - Resulting ODEs:         - X(x)+λX(x)=0X''(x) + \lambda X(x) = 0         - T(t)+λa2T(t)=0T''(t) + \lambda a^2 T(t) = 0     - Boundary conditions for X(x)X(x): X(0)=0X(0) = 0 and X(l)=0X(l) = 0.

The Eigenvalue Problem

  • Definition: The set of nonzero values λ\lambda is called the Eigenvalue, and the corresponding non-zero functions X(x)X(x) are the Eigenfunctions.
  • Case Analysis for $\lambda$:     - Case 1: λ<0\lambda < 0:         - Let λ=β2\lambda = -\beta^2. The general solution is X(x)=Aeβx+BeβxX(x) = Ae^{\beta x} + Be^{-\beta x}.         - Applying X(0)=0X(0) = 0: A+B=0B=AA + B = 0 \rightarrow B = -A.         - Applying X(l)=0X(l) = 0: A(eβleβl)=0A(e^{\beta l} - e^{-\beta l}) = 0. Since βl0\beta l \neq 0, then A=0A = 0.         - Trivial solution: X(x)=0X(x) = 0.     - Case 2: λ=0\lambda = 0:         - The general solution is X(x)=Ax+BX(x) = Ax + B.         - Applying X(0)=0B=0X(0) = 0 \rightarrow B = 0.         - Applying X(l)=0Al=0A=0X(l) = 0 \rightarrow Al = 0 \rightarrow A = 0.         - Trivial solution: X(x)=0X(x) = 0.     - Case 3: λ>0\lambda > 0:         - Let λ=β2\lambda = \beta^2. The general solution is X(x) = A \cos(\beta x) + B \sin(̢\beta x).         - Applying X(0)=0A=0X(0) = 0 \rightarrow A = 0.         - Applying X(l)=0Bsin(βl)=0X(l) = 0 \rightarrow B \sin(\beta l) = 0. For non-zero solution, sin(βl)=0\sin(\beta l) = 0.         - Eigenvalues: β=nπl\beta = \frac{n\pi}{l}, so λn=(nπl)2\lambda_n = ( \frac{n\pi}{l} )^2 for n=1,2,3,n = 1, 2, 3, \dots.         - Eigenfunctions: Xn(x)=Bnsin(nπxl)X_n(x) = B_n \sin( \frac{n\pi x}{l} ).

General Solution and Coefficients

  • Time-dependent Solution: For λn\lambda_n, the time equation is Tn(t)+(nπal)2Tn(t)=0T''_n(t) + ( \frac{n\pi a}{l} )^2 T_n(t) = 0.     - Solution: Tn(t)=Cncos(nπatl)+Dnsin(nπatl)T_n(t) = C'_n \cos( \frac{n\pi a t}{l} ) + D'_n \sin( \frac{n\pi a t}{l} ).
  • Superposition: The general solution is the sum of particular solutions:     - u(x,t)=n=1[Cncos(nπatl)+Dnsin(nπatl)]sin(nπxl)u(x, t) = \sum_{n=1}^{\infty} [ C_n \cos( \frac{n\pi a t}{l} ) + D_n \sin( \frac{n\pi a t}{l} ) ] \sin( \frac{n\pi x}{l} ).
  • Determining Coefficients (Cn,DnC_n, D_n):     - Use orthogonality of trigonometric functions: 0lsin(nπxl)sin(mπxl)dx={0amp;nm l2amp;n=m\int_0^{l} \sin(\frac{n\pi x}{l}) \sin(\frac{m\pi x}{l}) \, dx = \begin{cases} 0 &amp; n \neq m \ \frac{l}{2} &amp; n = m \end{cases}.     - From Initial Condition u(x,0)=ϕ(x)u(x, 0) = \phi(x): Cn=2l0lϕ(x)sin(nπxl)dxC_n = \frac{2}{l} \int_0^l \phi(x) \sin( \frac{n\pi x}{l} ) \, dx.     - From Initial Condition ut(x,0)=ψ(x)u_t(x, 0) = \psi(x): Dn=2nπa0lψ(x)sin(nπxl)dxD_n = \frac{2}{n\pi a} \int_0^l \psi(x) \sin( \frac{n\pi x}{l} ) \, dx.

Characteristics of Vibration Solutions

  • Standing Wave Method: A particular solution can be expressed as un(x,t)=Ansin(nπxl)cos(ωntθn)u_n(x, t) = A_n \sin( \frac{n\pi x}{l} ) \cos( \omega_n t - \theta_n ).     - Amplitude: An=Cn2+Dn2A_n = \sqrt{C_n^2 + D_n^2}.     - Frequency: ωn=nπal\omega_n = \frac{n\pi a}{l}.     - Initial Phase: θn=arctan(DnCn)\theta_n = \arctan( \frac{D_n}{C_n} ).
  • Nodes: Points where the displacement is always zero (sin(nπxl)=0\sin( \frac{n\pi x}{l} ) = 0).
  • Antinodes: Points where the displacement amplitude is maximal.
  • Harmonics:     - N=1N=1: Fundamental mode (1st harmonic).     - N=2,3,N=2, 3, \dots: Overtones (2nd, 3rd harmonics).
  • Visual Concepts: The material includes diagrams of the 1st through 5th harmonics and mentions application to musical instruments.

Detailed Example 1: Finite String Vibration

  • Parameters: Length l=10l = 10, fixed ends. Initial speed ψ(x)=0\psi(x) = 0. Initial displacement ϕ(x)=x(10x)1000\phi(x) = \frac{x(10-x)}{1000}. a=100a = 100.
  • Equation: 2ut2=10022ux2\frac{\partial^2 u}{\partial t^2} = 100^2 \frac{\partial^2 u}{\partial x^2}.
  • Separation Result:     - Eigenvalues: λn=(nπ10)2\lambda_n = ( \frac{n\pi}{10} )^2.     - Eigenfunctions: Xn(x)=sin(nπx10)X_n(x) = \sin( \frac{n\pi x}{10} ).     - Time solution: Tn(t)=Cncos(10nπt)+Dnsin(10nπt)T_n(t) = C'_n \cos( 10n\pi t ) + D'_n \sin( 10n\pi t ).
  • Coefficient Solution:     - Since velocity is 0, Dn=0D_n = 0.     - Cn=210010[x(10x)1000]sin(nπx10)dxC_n = \frac{2}{10} \int_0^{10} [ \frac{x(10-x)}{1000} ] \sin( \frac{n\pi x}{10} ) \, dx.     - Calculation result: Cn=15000010(10xx2)sin(nπx10)dx=45(nπ)3(1cos(nπ))C_n = \frac{1}{5000} \int_0^{10} (10x - x^2) \sin( \frac{n\pi x}{10} ) \, dx = \frac{4}{5(n\pi)^3} (1 - \cos(n\pi)).     - If nn is even, Cn=0C_n = 0. If nn is odd, Cn=85n3π3C_n = \frac{8}{5n^3\pi^3}.
  • Final Series Solution: u(x,t)=n=1,3,585n3π3cos(10nπt)sin(nπx10)u(x, t) = \sum_{n=1, 3, 5 \dots} \frac{8}{5n^3\pi^3} \cos( 10n\pi t ) \sin( \frac{n\pi x}{10} ).

Detailed Example 2: Modified Boundary Conditions

  • Problem: Solve the wave equation with u(0,t)=0u(0, t) = 0 and ux(l,t)=0u_x(l, t) = 0 (one end fixed, one end free).
  • Eigenvalue Problem:     - X(x)+λX(x)=0X''(x) + \lambda X(x) = 0 with X(0)=0X(0) = 0 and X(l)=0X'(l) = 0.     - For \lambda > 0, X(x)=Acos(βx)+Bsin(βx)X(x) = A \cos(\beta x) + B \sin(\beta x).     - X(0)=A=0X(0) = A = 0.     - X(l)=Bβcos(βl)=0cos(βl)=0X'(l) = B\beta \cos(\beta l) = 0 \rightarrow \cos(\beta l) = 0.     - Eigenvalues: βn=(2n1)π2l\beta_n = \frac{(2n-1)\pi}{2l}, so λn=[(2n1)π2l]2\lambda_n = [ \frac{(2n-1)\pi}{2l} ]^2.
  • General Solution: u(x,t)=n=1[Cncos((2n1)πa2lt)+Dnsin((2n1)πa2lt)]sin((2n1)πx2l)u(x, t) = \sum_{n=1}^{\infty} [ C_n \cos( \frac{(2n-1)\pi a}{2l} t ) + D_n \sin( \frac{(2n-1)\pi a}{2l} t ) ] \sin( \frac{(2n-1)\pi x}{2l} ).

Detailed Example 4: Robin Boundary Conditions

  • Problem: Lower end fixed (u(0,t)=0u(0, t) = 0), upper end satisfies a convective/spring-like condition (ux(l,t)+hu(l,t)=0u_x(l, t) + hu(l, t) = 0).
  • Separation ODEs:     - X(x)+λX(x)=0X''(x) + \lambda X(x) = 0     - BCs: X(0)=0X(0) = 0 and X(l)+hX(l)=0X'(l) + hX(l) = 0.
  • Eigenvalue Calculation:     - X(x)=Bsin(βx)X(x) = B \sin(\beta x).     - Substitute into second BC: Bβcos(βl)+hBsin(βl)=0B\beta \cos(\beta l) + hB \sin(\beta l) = 0.     - This yields the transcendental equation: tan(βl)=βh\tan(\beta l) = -\frac{\beta}{h}.
  • Analysis: The eigenvalues λn=βn2\lambda_n = \beta_n^2 are found at the intersections of y=tan(βl)y = \tan(\beta l) and the line y=βhy = -\frac{\beta}{h}.

Appendices of Supporting Mathematics

  • Appendix-I: Fourier Sine/Cosine Series:     - Non-periodic function f(x)f(x) on [0,l][0, l] can be represented as:     - Cosine Series: f(x)=a02+n=1ancos(nπxl)f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos( \frac{n\pi x}{l} ), where an=2l0lf(x)cos(nπxl)dxa_n = \frac{2}{l} \int_0^l f(x) \cos( \frac{n\pi x}{l} ) \, dx.     - Sine Series: f(x)=n=1bnsin(nπxl)f(x) = \sum_{n=1}^{\infty} b_n \sin( \frac{n\pi x}{l} ), where bn=2l0lf(x)sin(nπxl)dxb_n = \frac{2}{l} \int_0^l f(x) \sin( \frac{n\pi x}{l} ) \, dx.
  • Appendix-II: ODE Calculations:     - General form: y(x)+py(x)+qy(x)=0y''(x) + py'(x) + qy(x) = 0.     - Characteristic equation: k2+pk+q=0k^2 + pk + q = 0.     - Roots: k=p±p24q2k = \frac{-p \pm \sqrt{p^2 - 4q}}{2}.     - If p=0p=0 and q=λq=\lambda:         - λ<0y=Aeλx+Beλx\lambda < 0 \rightarrow y = Ae^{\sqrt{-\lambda}x} + Be^{-\sqrt{-\lambda}x}.         - λ=0y=Ax+B\lambda = 0 \rightarrow y = Ax + B.         - λ>0y=Acos(λx)+Bsin(λx)\lambda > 0 \rightarrow y = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x).