Questions: Separation of Variables for Partial Differential Equations

Question 1

After substituting u = X(x)T(t) into the heat equation u_t = κu_xx, you obtain X''/X = T'/(κT). Why must both sides equal the same constant?

• A. It is a mathematical convention adopted to simplify the algebra
• B. The heat equation requires all spatial derivatives to vanish at the boundary
• C. X and T were assumed to be separable, so their ratio is necessarily zero
• D. x and t are independent variables, so a function of x alone can equal a function of t alone only if both are constant

Answer: (D)

This is the key logical step in separation of variables. The left side X''/X depends only on x; the right side T'/(κT) depends only on t. For these to be equal for every pair (x, t), neither side can actually depend on its variable — they must both equal the same constant. If the x-side varied with x, you could hold t fixed and find an x that breaks equality, and vice versa. The independence of x and t forces the separation constant.

Question 2

For the heat equation on a rod [0, L] with boundary conditions X(0) = 0 and X(L) = 0, which statement about the separation constant λ is correct?

• A. Any real value of λ is valid since ODEs always have solutions
• B. Only λ = 0 is permitted because the boundary conditions force X to zero everywhere
• C. Any positive λ is permitted; negative values are excluded because they produce exponential growth
• D. Only the discrete values λ_n = (nπ/L)² for positive integers n are permitted, as these are the only values compatible with both boundary conditions

Answer: (D)

For λ > 0, the spatial ODE has solution X = A sin(√λ x) + B cos(√λ x). Applying X(0) = 0 forces B = 0. Applying X(L) = 0 then requires A sin(√λ L) = 0, and since we need non-trivial solutions (A ≠ 0), we need sin(√λ L) = 0, meaning √λ L = nπ for positive integers n. These eigenvalues λ_n = (nπ/L)² are the only allowed values. Boundary conditions, not initial conditions, select the eigenvalues.

Question 3

A single separated solution u_n(x,t) = sin(nπx/L)e^{-κ(nπ/L)²t} is the general solution to the heat equation with zero boundary conditions, valid for any initial condition.

• T. True
• F. False

Answer: False

u_n satisfies the PDE and the boundary conditions, but it can only match initial conditions of the specific form f(x) = sin(nπx/L). A general initial condition f(x) — say, a temperature spike in the middle of the rod — cannot be represented by a single sine mode. The general solution requires superposing infinitely many modes: u = Σ b_n sin(nπx/L)e^{-κ(nπ/L)²t}, with the Fourier coefficients b_n chosen to match f(x).

Question 4

In separation of variables, the allowed values of the separation constant (eigenvalues) are determined by the boundary conditions on X(x), while the Fourier coefficients of the solution are determined by the initial condition u(x,0) = f(x).

• T. True
• F. False

Answer: True

These are two distinct steps. First, boundary conditions on the spatial variable restrict λ to the discrete eigenvalues λ_n — this produces the basis functions sin(nπx/L). Second, the initial condition f(x) is expanded in those basis functions via a Fourier sine series, which determines the coefficients b_n. Boundary conditions fix the 'what modes exist'; initial conditions fix 'how much of each mode'.

Question 5

Why does solving a PDE by separation of variables require superposing infinitely many separated solutions rather than just using a single product solution u = X(x)T(t)?

Model answer: A single product solution u_n = X_n(x)T_n(t) satisfies the PDE and boundary conditions but corresponds to only one spatial frequency mode. It can match an initial condition only if f(x) happens to be proportional to that mode. For an arbitrary initial condition f(x), we need to represent f as a sum of all eigenfunctions — a Fourier series. Because the PDE is linear, the superposition Σ b_n X_n(x)T_n(t) is also a solution, and choosing the coefficients b_n via Fourier expansion allows us to satisfy any initial condition that can be represented in that basis.

The separated solutions form a complete orthogonal basis for the space of functions satisfying the boundary conditions (e.g., {sin(nπx/L)} on [0,L]). Superposition exploits the linearity of the PDE and the completeness of this basis to represent arbitrary initial data.