Introduction to Energy Methods

date2026-07-15document_iddoc_43a6bec521a85d16ea805c0223b35e4fdescriptionenergy method を、PDE の解の大きさや保存量を積分量で制御する方法として導入する。prerequisitesheat・wave・Laplace 方程式 / 多重積分と変数変換type講義content_typelecturestatusactive

mathpartial-differential-equationsenergy-methodlecture

1Introduction

This lecture explains how integral quantities over an entire domain can control a PDE solution without tracking it pointwise. The preceding maximum principle controls pointwise extrema, whereas an energy method evaluates the overall magnitude of a solution through quadratic integrals.

2Terminology and Definition

An energy methodEnergy method defines an integral involving squares of a solution and its derivatives as an energy and evaluates its evolution in time.

3Strategy

For the wave equation, kinetic and strain energy are conserved. For the heat equation, a quadratic integral decays. These integral estimates remain useful even when an explicit solution is unavailable.

4Connection to Earlier Material

Energy methods repeatedly use integration by parts. In several dimensions, Green's identities are the corresponding formulas obtained from Gauss's divergence theorem by converting domain integrals into boundary integrals. Boundary conditions determine whether the resulting boundary terms vanish and therefore form part of the estimate rather than an incidental assumption.

5Assumptions for the Calculations

Throughout this lecture, let $c > 0$ , $κ > 0$ , $0 < x < L$ , and $0 [PARSE ERROR: Undefined("Command(\"le\")")] t [PARSE ERROR: Undefined("Command(\"le\")")] T$ . We consider classical solutions sufficiently smooth to interchange integration and time differentiation and to apply integration by parts. A classical solution has enough differentiability to satisfy the equation and its conditions pointwise. Extending the estimates to weak solutions requires an additional approximation-and-limit argument.

6Example: The Wave Equation

Consider

u_{t t} = c^{2} u_{x x}, u (0, t) = u (L, t) = 0.

Define the energy by

E (t) = \frac{1}{2} \int_{0}^{L} (u_{t^{2}} + c^{2} u_{x^{2}}) d x .

Differentiation, integration by parts, and the boundary conditions give $E^{'} (t) = 0$ . The calculation begins with

E^{'} (t) = \int_{0}^{L} (u_{t} u_{t t} + c^{2} u_{x} u_{x t}) d x .

Substituting $u_{t t} = c^{2} u_{x x}$ yields

E^{'} (t) = c^{2} \int_{0}^{L} (u_{t} u_{x x} + u_{x} u_{x t}) d x .

Integration by parts in the first term gives

\int_{0}^{L} u_{t} u_{x x} d x = {[u_{t} u_{x}]}_{0^{L}} - \int_{0}^{L} u_{t x} u_{x} d x .

The two remaining integrals cancel, so

E^{'} (t) = c^{2} {[u_{t} u_{x}]}_{0^{L}} .

This boundary term represents energy flux through the endpoints. Differentiating the fixed-end conditions in time gives $u_{t} = 0$ at both endpoints, so the boundary term vanishes and $E^{'} (t) = 0$ . Conservation therefore depends both on the equation and on the absence of energy flux through the boundary.

7Uniqueness for the Wave Equation

Suppose that two solutions $u$ and $v$ have the same initial data and fixed-end boundary conditions. Their difference $w = u - v$ satisfies the homogeneous wave equation with zero initial and boundary data. Its initial energy is zero, and conservation implies

\int_{0}^{L} (w_{t^{2}} + c^{2} w_{x^{2}}) d x = 0

for every $t$ . Under the classical-solution assumptions, the integrand is continuous and nonnegative; hence $w_{t} = w_{x} = 0$ at every point. Thus $w$ is constant in both time and space, and the fixed-end conditions force $w = 0$ . Therefore $u = v$ .

8Example: The Heat Equation

Consider

u_{t} = κ u_{x x}, u (0, t) = u (L, t) = 0.

Differentiating the squared norm gives

\frac{d}{d t} \int_{0}^{L} u^{2} d x = 2 \int_{0}^{L} u u_{t} d x = 2 κ \int_{0}^{L} u u_{x x} d x .

Integration by parts yields

2 κ \int_{0}^{L} u u_{x x} d x = 2 κ {[u u_{x}]}_{0^{L}} - 2 κ \int_{0}^{L} u_{x^{2}} d x .

The Dirichlet conditions eliminate the boundary term, so

\frac{d}{d t} \int_{0}^{L} u^{2} d x = - 2 κ \int_{0}^{L} u_{x^{2}} d x [PARSE ERROR: Undefined("Command(\"le\")")] 0.

Thus the quadratic integral decays for heat flow, whereas the wave energy is conserved.

This estimate alone proves only that the quadratic integral is nonincreasing. To obtain a decay rate, use the Poincare inequality for functions that vanish at both endpoints:

\int_{0}^{L} u^{2} d x [PARSE ERROR: Undefined("Command(\"le\")")] {(\frac{L}{π})}^{2} \int_{0}^{L} u_{x^{2}} d x .

The inequality controls the magnitude of a function by the magnitude of its derivative. Its constant follows from the sine eigenfunctions introduced in separation of variables. Expanding

u (x) = \sum_{n = 1}^{\infty} b_{n} \sin (\frac{n π x}{L})

and using orthogonality gives

\int_{0}^{L} u^{2} d x = \frac{L}{2} \sum_{n = 1}^{\infty} | b_{n} |^{2}, \int_{0}^{L} u_{x^{2}} d x = \frac{L}{2} \sum_{n = 1}^{\infty} {(\frac{n π}{L})}^{2} | b_{n} |^{2.}

Since $n^{2} [PARSE ERROR: Undefined("Command(\"ge\")")] 1$ , it follows that $\int_{0}^{L} u_{x^{2}} d x [PARSE ERROR: Undefined("Command(\"ge\")")] (π / L)^{2} \int_{0}^{L} u^{2} d x$ , which is the stated Poincare inequality. This derivation uses the sine expansion of a sufficiently smooth function that vanishes at both endpoints.

Substituting this inequality into the decay estimate gives

\frac{d}{d t} \int_{0}^{L} u^{2} d x [PARSE ERROR: Undefined("Command(\"le\")")] - 2 κ {(\frac{π}{L})}^{2} \int_{0}^{L} u^{2} d x,

and therefore

\int_{0}^{L} u (x, t)^{2} d x [PARSE ERROR: Undefined("Command(\"le\")")] e^{- 2 κ (π / L)^{2} t} \int_{0}^{L} u (x, 0)^{2} d x .

Hence heat flow with homogeneous boundary data decays exponentially in the quadratic-integral sense. The factor $π / L$ is the wave number of the first sine eigenfunction in separation of variables.

9The Case with External Forcing

For comparison, consider

u_{t t} = c^{2} u_{x x} + F (x, t), u (0, t) = u (L, t) = 0.

The boundary flux still vanishes, but now

E^{'} (t) = \int_{0}^{L} F (x, t) u_{t} (x, t) d x .

The right-hand side is the rate at which the external force performs work on the system. Define

∥ F (\cdot, t) ∥_{2} = {(\int_{0}^{L} | F (x, t) |^{2} d x)}^{1 / 2} .

The Cauchy--Schwarz inequality and $∥ u_{t} (\cdot, t) ∥_{2^{2}} [PARSE ERROR: Undefined("Command(\"le\")")] 2 E (t)$ imply

E^{'} (t) [PARSE ERROR: Undefined("Command(\"le\")")] ∥ F (\cdot, t) ∥_{2} \sqrt{2 E (t)} .

On intervals where $E (t) > 0$ , differentiating $\sqrt{E (t)}$ gives, provided $∥ F (\cdot, t) ∥_{2}$ is integrable on $0 [PARSE ERROR: Undefined("Command(\"le\")")] t [PARSE ERROR: Undefined("Command(\"le\")")] T$ ,

\sqrt{E (t)} [PARSE ERROR: Undefined("Command(\"le\")")] \sqrt{E (0)} + \frac{1}{\sqrt{2}} \int_{0}^{t} ∥ F (\cdot, s) ∥_{2} d s .

If $E$ vanishes at some time, the same result follows by estimating $\sqrt{E + ε}$ and then letting $ε ↓ 0$ . This stability estimate shows that small forcing produces a controlled energy response. Energy methods therefore describe not only conservation but also growth and decay through inequalities.

10Scope of Validity

For nonlinear PDEs, the energy being estimated may not control every term on the right-hand side of the resulting inequality. One must determine how dissipation, conservation, forcing, and nonlinear terms interact before closing an estimate.

Energy methods can establish uniqueness, decay, and stability without constructing an explicit solution. The next lecture instead constructs solutions of linear PDEs as integrals of point-source responses using Green functions.

11Related Lectures

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