Prove uniqueness and bounds without summing the solution

Try

Suppose two smooth temperature profiles satisfy the same heat equation, initial condition and endpoint temperatures. Must they coincide? Instead of subtracting two complicated Fourier series, subtract the solutions themselves. Their difference has zero initial and boundary data. A nonnegative measure of its size can prove that it stays zero.

For a rod, define E(t) = (1/2)∫u(x,t)²dx. This mathematical energy measures squared temperature relative to a reference; it is not automatically thermodynamic internal energy.

Predict

With homogeneous Dirichlet endpoints, predict the sign of E′. Then consider endpoints held at nonzero temperatures. Could the boundary supply energy? Identify the term that must be retained before deciding.

Experiment

Differentiate and integrate by parts:

E′ = κ∫u uₓₓ = κ[u uₓ] at the endpoints − κ∫uₓ².

Boundary situation	Boundary contribution	Consequence
u = 0 at endpoints	Zero	E cannot increase
uₓ = 0 at endpoints	Zero	E cannot increase
General imposed data	May be nonzero	Account for flux

For u = Ae^(−κt)sin x on [0,π], E = πA²e^(−2κt)/4. Differentiate this expression and compare it with −κ∫uₓ². The agreement checks both the factor 1/2 and the integration-by-parts sign.

Understand

For the difference w of two sufficiently regular solutions with the same homogeneous boundary type, E′≤0 and E(0)=0. Since E≥0, E remains zero; continuity then gives w=0 everywhere. More general weak uniqueness arguments use approximation and energy-space identities, not unjustified differentiation of rough functions.

The maximum principle gives a different kind of information. On a bounded space-time cylinder, a classical heat solution continuous on its closure cannot exceed the maximum on the parabolic boundary: the initial spatial interval plus the lateral spatial boundaries. The final-time face is not prescribed data.

To see the mechanism, set v=u−εt. Then vₜ−κvₓₓ=−ε. If v achieved a maximum larger than all parabolic-boundary values at an interior spatial point and positive time, the spatial second derivative would be nonpositive and the time derivative at a maximum up to that time would be nonnegative. Their combination could not equal −ε. Letting ε decrease to zero proves the bound. Applying the same reasoning to −u supplies the lower bound.

Energy controls an integral size; the maximum principle controls pointwise extremes. Neither should be casually substituted for the other. A profile can have small L² norm but a narrow, tall peak, while a maximum bound alone says little about spatial gradients.

Apply

Suppose initial and boundary temperatures lie between −1 and 3. State the resulting pointwise bounds and prove uniqueness for two solutions sharing those data by taking their difference. Next introduce a positive internal source. Explain why the homogeneous maximum bound no longer follows unchanged and how subtracting a suitable time-dependent comparison function can restore a bound.

Continue

On the whole line, the heat kernel constructs solutions directly. Its positivity connects to the maximum principle, while its nonzero tails reveal the model’s infinite propagation speed.