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Partial differential equations
Lesson 62 of 75
Easy practice

Separation of variables

Objective. Build eigenfunction solutions on bounded domains.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

Separation of variables solves linear PDEs on bounded domains by assuming the solution factorises, u(x,t) = X(x)T(t), turning one PDE into two ordinary differential equations linked by a separation constant. Substituting into the PDE and dividing by XT isolates x-dependence on one side and t-dependence on the other, so both must equal a constant −λ; boundary conditions on X then select discrete eigenvalues λₙ and eigenfunctions, usually sines or cosines. Example: for uₜ=uₓₓ on [0,L] with u(0,t)=u(L,t)=0, separation gives X(x)=sin(nπx/L) with eigenvalues λₙ=(nπ/L)². The full solution is a superposition (Fourier series) of these eigenfunction modes.

Worked example

Using separation of variables on uₓₓ + λu = 0 with u(0)=0 and u(L)=0, state the eigenvalues λₙ.

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Illustrated card showing 0 … 0
  1. 1Assume solution u(x) = sin(kx) satisfying u(0)=0 automatically
  2. 2Apply u(L)=0: sin(kL) = 0, so kL = nπ for integer n
  3. 3Hence k = nπ/L
  4. 4Since λ = k², the eigenvalues are λₙ = (nπ/L)²

Answer: λₙ = (nπ/L)²

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Common mistake. Students assume u(x,t)=X(x)T(t) can represent any solution directly without summing over all eigenfunctions; the general solution is an infinite series Σcₙ Xₙ(x)Tₙ(t), not a single separated term.

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