For the wave equation u_tt = c²Δu, the total energy E(t) = ½∫(u_t² + c²|∇u|²)dx satisfies:
AdE/dt = 0 (energy is conserved)
BdE/dt < 0 (energy decreases)
CdE/dt > 0 (energy increases)
DdE/dt depends on the initial data
Differentiating and using the wave equation: dE/dt = ∫[u_t u_tt + c²∇u·∇u_t]dx = ∫[u_t c²Δu + c²∇u·∇u_t]dx. Integration by parts gives c²∫[u_t Δu + ∇u·∇u_t]dx = c²∫[u_t Δu - Δu·u_t]dx = 0 (with appropriate boundary conditions).
Question 2 True / False
Energy methods require finding explicit solution formulas.
TTrue
FFalse
Answer: False
The great advantage of energy methods is that they work without explicit solutions. By multiplying the PDE by the solution (or another test function), integrating, and using integration by parts, one derives inequalities on energy-like quantities that yield uniqueness, stability, and existence results.
Question 3 Short Answer
What is Gronwall's inequality used for in energy methods?
Think about your answer, then reveal below.
Model answer: It converts a differential inequality E'(t) ≤ C·E(t) into an exponential bound E(t) ≤ E(0)e^(Ct)
When an energy functional satisfies a differential inequality rather than an equality, Gronwall's lemma provides the crucial bound showing the energy cannot grow faster than exponentially. If C = 0, the energy is non-increasing; if E(0) = 0, then E(t) = 0 for all t, giving uniqueness.
Question 4 True / False
For the heat equation u_t = kΔu with homogeneous Dirichlet conditions, the integral ∫u²dx is non-increasing in time.
TTrue
FFalse
Answer: True
Multiplying by u and integrating: d/dt ∫½u²dx = ∫u·u_t dx = k∫u·Δu dx = -k∫|∇u|²dx ≤ 0 by integration by parts (with u = 0 on the boundary). This shows that the heat equation dissipates the L² energy, confirming the physical intuition that diffusion smooths out temperature variations.