Duhamel's principle and forcing
Accumulate source contributions through the correct heat semigroup and distinguish mild from classical solutions.
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The question for today
What you’ll understand
- Construct a forced solution by a time convolution with the evolution operator.
- Solve a forced Fourier coefficient including a resonant time dependence.
- Explain why boundary lifting and forcing regularity matter to the formula's meaning.
A heat source acts at many different times. Material introduced early has more time to diffuse than material introduced just before observation. Duhamel's principle accounts for that age difference by evolving each source contribution for the remaining time and integrating the results. It is the infinite-dimensional variation-of-constants formula. Its compact notation still contains a choice of boundary operator, a function-space integral, and a regularity question.
Each source contribution starts its own diffusion clock
Let S(t) be the heat semigroup for a specified spatial domain and homogeneous boundary condition. For uₜ=κΔu+f with initial state u₀, the formula below adds the freely evolved initial state to source contributions S(t−s)f(s). On the whole space, each term is a Gaussian convolution in space. On a bounded Dirichlet domain or a periodic torus, S is different. The elapsed time t−s is essential; using S(t) for every source slice would incorrectly diffuse recently added heat for the full observation time.
For u₀ in L² and f integrable in time with values in L² on a finite interval, the Bochner integral defines a continuous L² mild solution. It satisfies the evolution equation in an appropriate weak or distributional sense, but f of this regularity need not make u classically differentiable at every point. Additional regularity supports differentiating the formula and recovering a classical PDE. Calling the integral a solution is meaningful only after specifying which of these senses is intended.
A Fourier coefficient turns the formula into a checked scalar calculation
If a spatial eigenfunction has negative Laplacian eigenvalue −λ, its coefficient obeys c′+κλc=g(t). Variation of constants gives an exponentially weighted time integral. A constant source produces a transient plus a stationary coefficient when λ>0. A zero mode behaves differently: with no damping, its coefficient accumulates the time integral of the source mean. Whether that mode exists depends on the boundary condition. A periodic or insulated problem may heat up indefinitely under a positive mean source even when every nonconstant mode remains bounded.
If g(t) has the same exponential decay as the homogeneous mode, the integral produces a factor t. This is sometimes called temporal resonance in the coefficient calculation, but it does not imply unbounded growth: t times a decaying exponential still tends to zero. Formulas obtained by dividing the difference of two exponential rates must be replaced by their limit when the rates coincide. Evaluating the scalar integral directly is often safer than memorizing a quotient whose denominator can vanish.
Boundary inputs and nonlinear forcing need their own bookkeeping
A semigroup built for zero Dirichlet data cannot directly propagate arbitrary nonzero boundary values. One method is to choose a sufficiently regular lifting h with the required boundary trace and write u=v+h. Then v has homogeneous boundary data and an effective source f−hₜ+κΔh, together with adjusted initial data. The lifting's regularity and compatibility at time zero must be checked. Ignoring those terms solves a different boundary problem even if the interior source integral looks familiar.
For a semilinear equation, replace f by a function of the unknown solution inside the integral. The resulting expression is an integral equation used in a fixed-point argument, not an explicit formula. Local Lipschitz control on bounded ranges can produce local existence and uniqueness, but does not automatically prevent later blow-up. Likewise, an a priori source estimate cannot be inserted before establishing it for the evolving solution. Duhamel organizes the analysis; it does not remove the nonlinear estimates or the need to identify a continuation criterion.
Work it through
A forced periodic mode and an undamped mean
On a periodic interval of length 2π, take κ=1/2, u₀=3+sin x, and time-independent forcing f=cos2x. Find u and its long-time limit.
The constant initial mean remains 3 because the forcing has zero spatial mean. The first sine coefficient solves c′+(1/2)c=0 with c(0)=1, giving exp(−t/2).
The second cosine coefficient solves d′+2d=1 with d(0)=0. Its Duhamel integral is (1−e⁻²ᵗ)/2.
Thus u=3+exp(−t/2)sin x+(1−e⁻²ᵗ)cos2x/2. As t tends to infinity it approaches 3+(1/2)cos2x, whose diffusion term exactly cancels the forcing.
The mean is conserved here because the source mean is zero. Adding a positive constant source would instead make that mean grow linearly.
The living notebook
Explore the model
Interactive lab · heat
Separate homogeneous decay from forced accumulation
Reset, set n=2 and κ=0.5, then Step once to t=0.5. The instrument's unforced amplitude is exp(−1). Predict the constant-unit-source coefficient from the worked formula as (1−exp(−1))/2≈0.3161, then compare the two numbers. The instrument displays only the homogeneous mode; the forced value is an analytical comparison, not a hidden forcing control.
- Current amplitude
- 1
- Spatial L² norm
- 1.772
- Decay rate · κn²
- 3.6 /s
The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe homogeneous decay factor becomes a time-weighting kernel for source contributions.
Pause & reason
Check your understanding
Your turn
Try it yourself
On the same periodic interval with κ=1/2, zero initial data, and forcing f(t,x)=e⁻²ᵗsin2x, find u and the time of its largest positive mode amplitude.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The forcing rate equals the homogeneous damping rate a=2, so evaluate the time integral before simplifying.
Compare with the solution
The second sine coefficient is ∫₀ᵗe⁻²⁽ᵗ⁻ˢ⁾e⁻²ˢds=te⁻²ᵗ. Thus u=te⁻²ᵗsin2x. Its amplitude derivative is e⁻²ᵗ(1−2t), so the maximum occurs at t=1/2 with value 1/(2e). The extra t factor comes from coincident exponential rates, yet the amplitude still decays to zero at long time.
Keep the idea
Take it with you
- Duhamel weights each source by its own remaining diffusion time.
- Mild-solution formulas do not automatically imply classical derivatives.
- Boundary lifting, zero modes, and nonlinear dependence require explicit additional analysis.
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