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The first sine mode with matched velocity data: exact solution and the numerical tail bound $8 (N + 1)^{- 4}$

Example

Let $L > 0$ , $c > 0$ and define the odd $2 L$ -periodic $C^{\infty}$ datum

$F (x) : = \sin (\frac{π x}{L}), G (x) : = \frac{c π}{L} \sin (\frac{π x}{L}) .$

Then the sine coefficients of Sine coefficients, sine series and the odd periodic extension are $b_{1} (F) = 1$ and $b_{n} (F) = 0$ for $n \geq 2$ , and $b_{1} (G) = c π / L$ , $b_{n} (G) = 0$ for $n \geq 2$ ; the series solution of the string theorem is the finite expression

$u (t, x) = [\cos (ω_{1} t) + \sin (ω_{1} t)] \sin (\frac{π x}{L}), ω_{1} = \frac{c π}{L},$

which solves $u_{t t} = c^{2} u_{x x}$ with $u (0, \cdot) = F$ , $u_{t} (0, \cdot) = G$ and $u (t, 0) = u (t, L) = 0$ ; its tail beyond the first mode vanishes. For each $N \geq 1$ define the matching partial sum $u_{N} (t, x) : = \sum_{n = 1}^{N} [a_{n} \cos (ω_{n} t) + d_{n} \sin (ω_{n} t)] \sin (n π x / L)$ , where $a_{n} = b_{n} (F)$ , $d_{n} = b_{n} (G) / ω_{n}$ and $ω_{n} = c n π / L$ . The numerical bound of the string theorem at these data is

$∣ u (t, x) - u_{N} (t, x) ∣ \leq 8 (N + 1)^{- 4} (N \geq 1),$

so the general bound is valid but far from sharp on this datum.

Facts & Assumptions

Given: The period $L > 0$ , wave speed $c > 0$ , the data $F, G$ above, and $ω_{1} = c π / L$ .

[L1]

Sine coefficients equal $(2 / L) \int_{0}^{L} (datum) \sin (n π x / L) d x$ ; for a single sine mode $\sin (π x / L)$ with the orthogonality of Sine orthogonality on a half period and the full-period integrals of odd $2 L$ -periodic functions one gets $b_{1} = 1$ , $b_{n} = 0$ for $n \geq 2$ , and the function is odd and $2 L$ -periodic (Sine coefficients, sine series and the odd periodic extension, Sine and cosine defined by their real power series, The zero sets of sine and cosine and the least positive common period 2 pi).

[L2]

The string theorem: the series (S) with $a_{n} = b_{n} (F)$ , $d_{n} = b_{n} (G) / ω_{n}$ is $C^{2}$ , solves $u_{t t} = c^{2} u_{x x}$ with the fixed-end conditions and the stated initial data, and obeys the tail bound with $C = 2 (\frac{2 L^{5} M_{5}}{π^{5}} + \frac{2 L^{5} M_{4}}{c π^{5}})$ (The classical $C^{2}$ fixed-end string: the sine series solves the wave equation with the given data, uniquely, with an explicit tail bound).

[L3]

Derivatives and values: $(\sin)^{'} = \cos$ , $(\cos)^{'} = - \sin$ , $\cos 0 = 1$ , $\sin 0 = 0$ , and the modes solve the wave equation for every $n$ (d'Alembert's formula solves the wave equation, is the unique $C^{2}$ solution, and the fixed-end sine modes solve it exactly, The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, Parity and the Pythagorean identity for sine and cosine).

Verification

technique · direct; identify the coefficients and evaluate the constants of the theorem

1.1givenL1algebra

By [L1], $b_{1} (F) = 1$ , $b_{n} (F) = 0$ for $n \geq 2$ ; similarly $b_{1} (G) = c π / L$ and $b_{n} (G) = 0$ for $n \geq 2$ . Hence $a_{1} = 1$ , $d_{1} = b_{1} (G) / ω_{1} = 1$ , and all other coefficients vanish, so the series (S) collapses to $u (t, x) = [\cos ω_{1} t + \sin ω_{1} t] \sin (π x / L)$ .

2.1step 1.1L3algebra

By [L3] and step 1.1, $\partial_{t}^{2} u = - ω_{1}^{2} u$ and $\partial_{x}^{2} u = - (π / L)^{2} u$ , so $u_{t t} = c^{2} u_{x x}$ ; $u (t, 0) = u (t, L) = 0$ ; $u (0, x) = \sin (π x / L) = F (x)$ ; and $u_{t} (0, x) = ω_{1} \sin (π x / L) = G (x)$ .

3.1step 1.1L2algebra∎

For the constant of the theorem, $M_{5} = \sup ∣ F^{(5)} ∣ = (π / L)^{5}$ and $M_{4} = \sup ∣ G^{(4)} ∣ = (c π / L) (π / L)^{4} = c π^{5} / L^{5}$ . Hence $C = 2 (\frac{2 L^{5} (π / L)^{5}}{π^{5}} + \frac{2 L^{5} c π^{5} / L^{5}}{c π^{5}}) = 2 (2 + 2) = 8$ . Since the exact tail vanishes for $N \geq 1$ , the bound $∣ u - u_{N} ∣ \leq 8 (N + 1)^{- 4}$ holds trivially for every $N \geq 1$ .

Remarks

The finite datum exercises the theorem without any infinite tail; the numerical bound $8 (N + 1)^{- 4}$ is valid because the general constant is valid, and the example shows explicitly that the general bound is not claimed to be sharp.

The first sine mode with matched velocity data: exact solution and the numerical tail bound $8 (N + 1)^{- 4}$

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

The first sine mode with matched velocity data: exact solution and the numerical tail bound 8(N+1)−48(N+1)−4

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

The first sine mode with matched velocity data: exact solution and the numerical tail bound $8 (N + 1)^{- 4}$