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Lesson 12 of 30

Separate forcing from homogeneous evolution

Derive Duhamel and lift inhomogeneous boundaries.

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A rod has different prescribed temperatures at its endpoints, so a sine expansion alone cannot match them. First construct a simple function that satisfies the boundary values, then solve for the remaining deviation. This is boundary lifting: it changes the unknown, not the physical boundary condition.

On [0,L], let the endpoint values be a(t) and b(t). The affine lift w(x,t)=a(t)+(b(t)−a(t))x/L has exactly those values. Define v=u−w, so v has zero endpoints.

Predict

If the endpoint temperatures vary in time, can subtracting w leave the same forcing term? Differentiate u=v+w before answering. In particular, a spatially affine lift has wₓₓ=0 but may still have nonzero wₜ.

Experiment

For uₜ=κuₓₓ+F, substitution gives vₜ=κvₓₓ+F−wₜ+κwₓₓ. With L=2, a(t)=t and b(t)=3t, the lift is w=t(1+x), so the corrected forcing is F−(1+x).

QuantityFormula in this example
Lift wt(1+x)
Lift time derivative1+x
Lift second spatial derivative0
Corrected initial profilef(x)−w(x,0)=f(x)

Check the transformed equation by adding w back. Ignoring the time derivative would predict the wrong interior evolution despite matching both boundary values.

Understand

For homogeneous boundary conditions, write the heat solution operator as S(t). Duhamel’s formula for a source is v(t)=S(t)v(0)+∫₀ᵗ S(t−s)G(s)ds, where G is the corrected forcing. A source added at time s evolves only for the remaining duration t−s. This explains both the integration limits and the argument of S.

For a single eigenmode with eigenvalue λ, the coefficient satisfies z′+κλz=g(t). Multiplying by e^(κλt) and integrating yields z(t)=e^(−κλt)z(0)+∫₀ᵗ e^(−κλ(t−s))g(s)ds. For constant g=4 and κλ=2 with z(0)=0, this becomes 2(1−e^(−2t)). The steady coefficient is 2, while the transient decays. Differentiate this result to verify the source balance rather than merely checking its limit.

On the whole line, replace S(t−s) by convolution with the heat kernel. On a bounded interval, use the boundary-adapted eigenmodes. Interchanging sums, integrals and derivatives requires regularity or a weak-solution framework; a formal mode calculation is not by itself a proof for arbitrary rough sources.

Corner compatibility remains important. If f(0) differs from a(0), no solution continuous on the full closed space-time cylinder can satisfy both values. A weaker solution can still approach its data in a suitable sense, but the regularity claim must change.

Apply

Choose linearly changing endpoint temperatures and an internal source. Calculate the lift, corrected source and corrected initial profile separately. Then solve one forced sine coefficient using an integrating factor. Check the reconstructed endpoints and explain whether the initial corners are compatible. A correct boundary fit does not excuse a missing source term.

Continue

Wave equations also use eigenmodes and source integration, but each mode has a second-order time equation. The whole-line d’Alembert formula first makes the roles of displacement, velocity and finite propagation explicit.

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Practice and reasoning

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Boundary values are u(0,t)=2,u(4,t)=6. What is the linear steady lift at x=1?

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For a mode amplitude a′+3a=6 with a(0)=0, what is its limiting value?

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

If uₜ−κuₓₓ=F and u=v+w, derive the equation for v.

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State and verify the heat Duhamel formula using a solution operator S(t).

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Derive the forced sine-mode equation on a homogeneous Dirichlet interval.

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Why does subtracting a boundary lift not remove an incompatibility between initial and boundary data?

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