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MathidiaMathematics Academy
Partial differential equations II
Lesson 30 of 75
Review

Energy methods

Objective. Prove uniqueness and stability estimates.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Energy methods prove uniqueness and stability for PDEs by tracking a quadratic quantity, usually an integral of the square of the solution or its gradient, and showing it cannot grow. For the heat equation uₜ = uₓₓ with homogeneous boundary conditions, define E(t) = ∫u² dx; differentiating and integrating by parts gives E'(t) = −2∫uₓ² dx ≤ 0, so the energy is non-increasing. If two solutions share the same initial and boundary data, their difference w solves the same equation with zero initial energy, forcing E(t) = 0 for all t and hence w ≡ 0, proving uniqueness. Example: this same argument shows small perturbations in initial data lead to controlled, non-growing differences in the solution, giving stability.

Worked example

Use an energy method to show uniqueness for uₜ = uₓₓ on (0,1) with u(0,t) = u(1,t) = 0 and prescribed initial data.

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  1. 1Let w = u₁ − u₂ be the difference of two solutions; w solves wₜ = wₓₓ with w(0,t) = w(1,t) = 0 and w(x,0) = 0.
  2. 2Define E(t) = ∫₀¹ w² dx and differentiate: E'(t) = 2∫₀¹ wwₜ dx = 2∫₀¹ wwₓₓ dx.
  3. 3Integrate by parts: 2∫₀¹ wwₓₓ dx = [2wwₓ]₀¹ − 2∫₀¹ wₓ² dx = −2∫₀¹ wₓ² dx, since the boundary term vanishes.
  4. 4So E'(t) ≤ 0 and E(0) = 0, forcing E(t) = 0 for all t, hence w ≡ 0 and the solution is unique.

Answer: w ≡ 0, so the solution is unique

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Common mistake. A common mistake is differentiating ∫u² dx under the integral sign without substituting the PDE for uₜ, so the boundary terms from integration by parts are lost and the sign of E'(t) cannot be determined.

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