A rectangular plate has zero temperature on three sides and a sinusoidal temperature along the top. The interior is steady and source-free, so Δu=0. Unlike the heat equation, there is no time variable. The second coordinate must supply spatial growth or decay that balances the sinusoidal curvature in the first.
Take the rectangle (0,π)×(0,H). Prescribe sin(nx) at y=H and zero on the other sides, with n a positive integer.
Predict
If X(x)=sin(nx), should the y factor satisfy Y″+n²Y=0 or Y″−n²Y=0? Substitute the product into the Laplacian before choosing sine or hyperbolic sine. Predict which function vanishes at y=0.
Experiment
The solution is u=sin(nx)sinh(ny)/sinh(nH).
Check
Calculation
Vertical sides
sin(0)=sin(nπ)=0
Bottom
sinh(0)=0
Top
Ratio of identical hyperbolic sines is 1
PDE
−n²u+n²u=0
At the center height y=H/2, compare the attenuation for n=1 and n=4. Higher spatial frequencies imposed at the top decay more strongly into the interior. This is spatial filtering, not time evolution.
Understand
In a disk, polar coordinates give Δu=uᵣᵣ+(1/r)uᵣ+(1/r²)uθθ away from r=0. For an angular mode cos(nθ), the radial equation is r²R″+rR′−n²R=0. Its positive-frequency solutions are rⁿ and r^(−n). Regularity at the origin excludes the negative power. At n=0 the alternatives are a constant and log r, and regularity excludes the logarithm.
Thus unit-disk boundary data cos(nθ) produce u=rⁿcos(nθ). A general Fourier expansion on the boundary gives a constant plus positive powers multiplying sine and cosine modes. On every smaller disk r≤ρ with ρ strictly below 1, exponential decay of ρⁿ dominates polynomial factors introduced by differentiation. Under standard bounded or L² boundary assumptions, this yields smooth interior solutions. It does not make discontinuous boundary data continuous at the boundary.
For continuous disk data, summing the modes leads to the Poisson kernel Pᵣ(φ)=(1−r²)/(1−2r cosφ+r²), with average normalization 1/(2π). Its positivity and unit integral show that interior values are weighted averages of boundary values. As r approaches 1, the kernel concentrates near the matching angle, giving boundary recovery at continuity points.
Rectangles introduce corner compatibility: if adjacent side data approach different values at a corner, no solution continuous on the entire closure can satisfy both. A separated interior solution may still be meaningful with weaker boundary traces, but the regularity statement must acknowledge the conflict.
Apply
Solve a disk problem with boundary 2+3cos(2θ)−sinθ and check its center value. Explain why the center depends only on the boundary average. Then formulate a rectangle problem with incompatible corner values and state exactly which continuity claim fails. Keep the origin and corners in your verification rather than checking only generic interior points.
Continue
Green functions add interior forcing to boundary problems. Their normalization and signs are fixed by a point-source identity, so checking the derivative jump or flux is essential.
Apply · explain · revise
Practice and reasoning
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For disk boundary data cos(3θ), what is the radial amplitude at r=1/2?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For sin(4x) on a rectangle of width π, what separation eigenvalue occurs?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Solve Δu=0 on (0,π)×(0,H), zero on three sides and u(x,H)=sin(nx).
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Why discard r^(−n) for a regular harmonic mode of order n≥1 in the disk?
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Explain why rough disk boundary data become smoother strictly inside the disk.
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Can arbitrary discontinuous data prescribed on meeting rectangle sides produce a solution continuous on the entire closure?
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Reflect and continue
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