CAL-16

1D Laplacian

Δu = u''. Second derivative as a 1D Laplacian.

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Vector calculusLaplacian

Governing equation

\Delta u=u''\approx(u_+-2u+u_-)/h^2

where

u_+: u(x+h) (—)
u: u(x) (—)
u_-: u(x−h) (—)
h: Spacing h (—)
\Delta u: 1D Laplacian (—)

Lecture brief

Historical brief

Newton and Leibniz (1670s), Taylor, the fundamental theorem and the trapezoid rule are how change became a number. The lab differentiates, integrates and linearises in one variable. This sheet (CAL-16 — 1D Laplacian) is the form associated with Laplacian. Working symbols: $u_+$ , $u$ , $u_-$ , $h$ $\rightarrow$ $\Delta u$ . In 1D, ∇² = d²/dx². Harmonic functions are linear. The 3-point stencil (u_{+} − 2u + u_{-})/h² is the discrete Laplacian.

Purpose

Purpose: compute

\Delta u

from

u_+

u

u_-

h

in Calculus via

\Delta u=u''\approx(u_+-2u+u_-)/h^2

Δu = u''. Second derivative as a 1D Laplacian. Use it when a real calculus question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

u_+ = 1.100\,\mathrm{—}

u = 1.000\,\mathrm{—}

u_- = 0.950\,\mathrm{—}

h = 0.100\,\mathrm{—}

, the governing relation

\Delta u=u''\approx(u_+-2u+u_-)/h^2

yields

\Delta u = 5.0000\,\mathrm{—}

. Enter u'' directly, or the 3-point stencil with spacing h. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

u(x+h) u_+1.1 —u(x) u1 —u(x−h) u_-0.95 —Spacing h h0.1 —

Outputs

1D Laplacian \Delta u5.0000 —

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Narration of this film

Enter u'' directly, or the 3-point stencil with spacing h.

In 1D, ∇² = d²/dx². Harmonic functions are linear. The 3-point stencil (u_{+} − 2u + u_{-})/h² is the discrete Laplacian.

Reading speed

Watch on YouTube