Solve one dimensional heat equation

The one dimensional heat equation ∂u/∂t = c² ∂²u/∂x² gives the temperature u(x, t) in a thin rod of length L. It is solved by separation of variables, with the standard conditions u(0, t) = 0, u(L, t) = 0 (ends kept at zero temperature) and u(x, 0) = f(x).

Step 1: separate. Let u = X(x)T(t). Then X T′ = c² X″ T, so

X″/X = T′/(c²T) = −p² (a constant)

The constant must be negative, otherwise u would grow without bound in time or fail the boundary conditions. This gives

X″ + p²X = 0 and T′ + c²p²T = 0

Step 2: solutions. X = A cos px + B sin px and T = C e^(−c²p²t). So

u = (A cos px + B sin px) e^(−c²p²t)

Step 3: boundary conditions.

• u(0, t) = 0 gives A = 0.
• u(L, t) = 0 gives B sin pL = 0, so p = nπ/L (n = 1, 2, 3, …).

Hence uₙ = Bₙ sin(nπx/L) e^(−c²n²π²t/L²).

Step 4: superpose and use the initial condition.

u(x, t) = Σ (n=1 to ∞) Bₙ sin(nπx/L) e^(−c²n²π²t/L²)

Putting t = 0: f(x) = Σ Bₙ sin(nπx/L), a half range sine series, so

Bₙ = (2/L) ∫₀ᴸ f(x) sin(nπx/L) dx

This is the complete solution. Each term decays exponentially, and the higher harmonics decay faster, so u → 0 as t → ∞.