Strichartz estimates and admissible exponents
Match spacetime integrability to free Schrödinger scaling, respect the excluded endpoint, and close a cubic nonlinear estimate with a small time factor.
On this page
The question for today
What you’ll understand
- Check Schrödinger admissibility with the correct dimension and endpoint exclusion.
- Distinguish homogeneous estimates from a stated nonendpoint retarded forcing estimate.
- Derive the one-dimensional cubic time factor in the L⁸-time/L⁴-space contraction framework.
A nonlinear wave proof often needs to control where and when a solution is large, rather than a separate supremum at every time. Strichartz norms combine time and space integrability so the free dispersive evolution gains exactly the balance allowed by scaling. The exponents are not interchangeable labels. A single excluded endpoint can invalidate an argument, and the forcing term uses dual exponents that must be checked independently from the solution norm.
Admissibility couples time and space
For free Schrödinger on R to the d, call a pair (q,r) admissible when both exponents are at least two, they satisfy two over q plus d over r equals d over two, and the triple (q,r,d) is not (2,infinity,2). The homogeneous estimate bounds the L to the q time, L to the r space norm of U(t)f over the full real-time line by a constant times the L² norm of f. The pair (infinity,2) is the energy estimate and belongs to this family.
The exceptional two-dimensional pair satisfies scaling but fails the general estimate, showing that dimensional balance alone is insufficient. In three dimensions the endpoint (2,6) does hold; it must not be discarded merely because q equals two. In one dimension, (4,infinity) is admissible and valid. Different dispersive equations have different admissibility relations, and compact spatial domains need their own statements, sometimes with losses or time restrictions. Every application should name its equation, spatial domain, and endpoint convention.
The source belongs to a dual spacetime space
For two admissible pairs, the retarded Duhamel operator maps an appropriate dual source space into the solution spacetime space. A sufficient version used here takes both time exponents strictly greater than two, allowing the energy time exponent infinity, and excludes the exceptional pair already stated. If the source pair is (q-tilde,r-tilde), its Hölder conjugates q-tilde-prime and r-tilde-prime describe the forcing norm. The time integral runs from the initial time to t, not over all times without order.
The proof combines the dispersive bound, unitarity, duality, and a time-ordering argument. Endpoint forcing estimates require their own justified theorem; they are not obtained by simply substituting q=2 into every nonendpoint proof. To use the estimate nonlinearly, the source |u|²u must land in the required dual norm and its difference must obey a corresponding Lipschitz bound. The linear theorem is a tool for that nonlinear accounting, not a statement that every power nonlinearity is automatically controlled.
Subcritical time smallness closes a nonlinear map
For one-dimensional cubic NLS, work in the intersection of continuous L²-valued functions and L⁸-time/L⁴-space functions on [0,T]. The displayed source bound, with its difference version, makes the Duhamel correction at most a constant times square root of T times the cube of a ball radius. The Lipschitz coefficient has the same time factor times radius squared. Choosing a short interval gives a local L² construction, rougher than the H¹ algebra construction used earlier on the torus.
At the one-dimensional mass-critical quintic power, the admissible pair (6,6) gives the source L to the six fifths norm directly from five factors in L⁶, with no extra interval-length power. A critical local proof instead exploits smallness of the free evolution's L⁶ spacetime norm on a sufficiently short interval for the particular datum, together with a carefully chosen contraction domain. One cannot manufacture the missing subcritical time power by relabeling exponents. This contrast makes scaling's warning concrete in an actual nonlinear estimate.
Work it through
Account for the cubic time gain
In one dimension, verify the pair (8,4) and the time power in the cubic source estimate. If T=1/16 and the L⁸L⁴ norm is at most two, evaluate the resulting source bound before any Strichartz constant.
Admissibility gives 2/8+1/4=1/2, exactly d/2 for d=1. The spatial cube of an L⁴ function lies in L^(4/3).
The three time factors in L⁸ give their product in L^(8/3). On an interval of length T, embedding into L^(8/7) costs T^(7/8−3/8)=T^(1/2).
At the stated values, the source norm is bounded by sqrt(1/16)·2³=(1/4)·8=2. The Duhamel bound then multiplies this by its linear estimate constant and the absolute nonlinear coefficient.
The source bound is two. A complete contraction still checks image size and a difference estimate; this single product estimate is only one component.
The living notebook
Explore the model
Interactive lab · heat
Notice that different norms scale differently
Pause and reset, then set both controls to n = 1 and κ = 0.5. At time one, predict heat amplitude exp(−0.5) and unnormalized L² norm sqrt(pi) exp(−0.5); use two Steps from reset. For a fresh comparison, pause and reset, then set n = 2 and κ = 0.5, predicting the two readings with exp(−2) instead. These are spatial norms of one analytic heat mode, not a Strichartz integral or a Schrödinger solver. Use the explicit readings to distinguish spatial amplitude from the additional time integration a mixed norm requires.
- 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 forA mixed spacetime norm needs a specified temporal exponent and interval, not just a sequence of peak readings.
Pause & reason
Check your understanding
Your turn
Try it yourself
Check whether (q,r)=(4,3) is admissible in three dimensions. Then verify that (6,6) is admissible in one dimension and explain why the quintic source has no interval-length gain when estimated in L^(6/5) in both time and space.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compute the reciprocal-exponent sums, and multiply five L⁶ factors before considering time embedding.
Compare with the solution
In three dimensions, 2/4+3/3=1/2+1=3/2, so (4,3) is admissible and is not an excluded endpoint. In one dimension, 2/6+1/6=1/2, so (6,6) is admissible. Five factors of an L⁶ spacetime function already lie in L^(6/5), exactly the dual source exponent. No change to a different temporal exponent occurs, so this Hölder estimate produces no positive power of T. A critical proof must obtain smallness through another part of its solution-space argument.
Keep the idea
Take it with you
- Admissibility combines a scaling relation with dimension-specific exponent restrictions.
- The forcing estimate uses dual norms and an explicitly justified retarded-time theorem.
- A positive time power can close subcritical iterations; critical problems require a more refined smallness mechanism.
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