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Lemma 8.28 Let \( m = 2\ell - 1 \) with \( \ell \in \mathbb{N} \) and \( \ell \geq 2 \), and let \( f \in {C}^{\ell }\left\lbrack {a, b}\right\rbrack \) . Assume that the spline \( s \in {S}_{m}^{n} \) interpolates \( f \), i.e., \[ s\left( {x}_{j}\right) = f\left( {x}_{j}\right) ,\;j = 0,\ldots, n, \] (8.26) and that ... | Proof. We have that \[ {\int }_{a}^{b}{\left\lbrack {f}^{\left( \ell \right) }\left( x\right) - {s}^{\left( \ell \right) }\left( x\right) \right\rbrack }^{2}{dx} = {\int }_{a}^{b}{\left\lbrack {f}^{\left( \ell \right) }\left( x\right) \right\rbrack }^{2}{dx} - {\int }_{a}^{b}{\left\lbrack {s}^{\left( \ell \right) }\lef... | Yes |
Lemma 8.29 Under the assumptions of Lemma 8.28 let \( f = 0 \) . Then \( s = 0 \) . | Proof. For \( f = 0 \), from (8.28) it follows that\n\n\[ \n{\int }_{a}^{b}{\left\lbrack {s}^{\left( \ell \right) }\left( x\right) \right\rbrack }^{2}{dx} = 0 \n\]\n\nThis implies that \( {s}^{\left( \ell \right) } = 0 \), and therefore \( s \in {P}_{\ell - 1} \) on \( \left\lbrack {a, b}\right\rbrack \) . Now the boun... | Yes |
Theorem 8.30 Let \( m = 2\ell - 1 \) with \( \ell \in \mathbb{N} \) and \( \ell \geq 2 \) . Then, given \( n + 1 \) values \( {y}_{0},\ldots ,{y}_{n} \) and \( m - 1 \) boundary data \( {a}_{1},\ldots ,{a}_{\ell - 1} \) and \( {b}_{1},\ldots ,{b}_{\ell - 1} \) , there exists a unique spline \( s \in {S}_{m}^{n} \) sati... | Proof. Representing the spline in the form (8.25), i.e.,\n\n\[ s\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{m}{\alpha }_{k}{u}_{k} + \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}{\beta }_{k}{v}_{k} \]\n\n(8.31)\n\n\nit follows that the interpolation conditions (8.29) and boundary conditions (8.30) are satisfied if a... | Yes |
Theorem 8.31 For \( m \in \mathbb{N} \cup \{ 0\} \) the B-splines\n\n\[ \n{B}_{m}\left( {\cdot - k}\right) ,\;k = 0,\ldots, m \n\]\n\n(8.37)\n\nare linearly independent on the interval \( {I}_{m} \mathrel{\text{:=}} \left\lbrack {\frac{m - 1}{2},\frac{m + 1}{2}}\right\rbrack \) . | Proof. This is trivial for \( m = 0 \), and we assume that it has been proven for degree \( m - 1 \) for some \( m \geq 1 \) . Let\n\n\[ \n\mathop{\sum }\limits_{{k = 0}}^{m}{\alpha }_{k}{B}_{m}\left( {x - k}\right) = 0,\;x \in {I}_{m} \n\]\n\n(8.38)\n\nThen, with the aid of (8.33), differentiating (8.38) yields\n\n\[ ... | Yes |
Corollary 8.32 Let \( {x}_{k} = a + {hk}, k = 0,\ldots, n \), be an equidistant subdivision of the interval \( \left\lbrack {a, b}\right\rbrack \) of step size \( h = \left( {b - a}\right) /n \) with \( n \geq 2 \), and let \( m = 2\ell - 1 \) with \( \ell \in \mathbb{N} \) . Then the B-splines\n\n\[ \n{B}_{m, k}\left(... | Proof. The \( n + m \) splines (8.39) belong to \( {S}_{m}^{n} \), and by the preceding Theorem 8.31 they can be shown to be linearly independent on \( \left\lbrack {a, b}\right\rbrack \) . Hence, the statement follows from Theorem 8.27. | Yes |
Theorem 8.33 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be twice continuously differentiable and let \( s \in {S}_{3}^{n} \) be the uniquely determined cubic spline satisfying the interpolation and boundary conditions of Lemma 8.28. Then\n\n\[ \parallel f - s{\parallel }_{\infty } \leq \frac{... | Proof. The error function \( r \mathrel{\text{:=}} f - s \) has \( n + 1 \) zeros \( {x}_{0},\ldots ,{x}_{n} \) . Hence, the distance between two consecutive zeros of \( r \) is less than or equal to \( h \) . By Rolle’s theorem, the derivative \( {r}^{\prime } \) has \( n \) zeros with distance less than or equal to \... | Yes |
Theorem 8.34 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be four-times continuously differentiable and let \( s \in {S}_{3}^{n} \) be the uniquely determined cubic spline satisfying the interpolation and boundary conditions of Lemma 8.28 for an equidistant subdivision with step width \( h \) .... | Proof. By \( {L}_{1} : C\left\lbrack {a, b}\right\rbrack \rightarrow {S}_{1}^{n} \) we denote the interpolation operator mapping \( g \in C\left\lbrack {a, b}\right\rbrack \) onto its uniquely determined piecewise linear interpolation. From Example 8.12 we obtain that\n\n\[ \parallel r{\parallel }_{\infty } = {\begin{V... | Yes |
Theorem 8.37 The Bernstein polynomials are nonnegative on \( \\left\\lbrack {0,1}\\right\\rbrack \) and provide a partition of unity; i.e., \[ {B}_{k}^{n}\\left( t\\right) \\geq 0,\\;t \\in \\left\\lbrack {0,1}\\right\\rbrack \] and \[ \\mathop{\\sum }\\limits_{{k = 0}}^{n}{B}_{k}^{n}\\left( t\\right) = 1,\\;t \\in \\m... | Proof. The first five properties are obvious. The statement on the maximum of \( {B}_{k}^{n} \) is a consequence of \[ \\frac{d}{dt}{B}_{k}^{n}\\left( t\\right) = \\left( \\begin{array}{l} n \\\\ k \\end{array}\\right) {t}^{k - 1}{\\left( 1 - t\\right) }^{n - k - 1}\\left( {k - {nt}}\\right) ,\\;k = 0,\\ldots n. \] The... | Yes |
Theorem 8.39 Let\n\n\\[ \np\\left( t\\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{n}{b}_{k}{B}_{k}^{n}\\left( t\\right) ,\\;t \\in \\left\\lbrack {0,1}\\right\\rbrack \n\\]\n\nbe a Bézier polynomial on \\( \\left\\lbrack {0,1}\\right\\rbrack \\) . Then\n\n\\[ \n{p}^{\\left( j\\right) }\\left( t\\right) = \\frac{n!}{\\... | Proof. Obviously, the statement is true for \\( j = 0 \\) . We assume that it has been proven for some \\( 0 \\leq j < n \\) . Then with the aid of (8.54) we obtain\n\n\\[ \n{p}^{\\left( j + 1\\right) }\\left( t\\right) = \\frac{n!}{\\left( {n - j}\\right) !}\\mathop{\\sum }\\limits_{{k = 0}}^{{n - j}}{\\bigtriangleup ... | Yes |
Corollary 8.40 The polynomial from Theorem 8.39 has the derivatives\n\n\\[ \n{p}^{\\left( j\\right) }\\left( 0\\right) = \\frac{n!}{\\left( {n - j}\\right) !}{\\bigtriangleup }^{j}{b}_{0},\\;{p}^{\\left( j\\right) }\\left( 1\\right) = \\frac{n!}{\\left( {n - j}\\right) !}{\\bigtriangleup }^{j}{b}_{n - j} \n\\]\n\n at t... | From Corollary 8.40 we note that \\( {p}^{\\left( j\\right) }\\left( 0\\right) \\) depends only on \\( {b}_{0},\\ldots ,{b}_{j} \\) and that \\( {p}^{\\left( j\\right) }\\left( 1\\right) \\) depends only on \\( {b}_{n - j},\\ldots ,{b}_{n} \\) . In particular, we have that\n\n\\[ \n{p}^{\\prime }\\left( 0\\right) = n\\... | Yes |
Theorem 8.41 The subpolynomials \( {b}_{i}^{k} \) of a Bézier polynomial \( p \) of degree \( n \) satisfy the recursion formulae\n\n\[ \n{b}_{i}^{k}\left( t\right) = \left( {1 - t}\right) {b}_{i}^{k - 1}\left( t\right) + t{b}_{i + 1}^{k - 1}\left( t\right) \n\]\n\n(8.57)\n\nfor \( i = 0,\ldots, n - k \) and \( k = 1,\... | Proof. We insert the recursion formulae (8.51) and (8.52) for the Bernstein polynomials into the definition (8.56) for the subpolynomials and obtain\n\n\[ \n{b}_{i}^{k}\left( t\right) = {b}_{i}{B}_{0}^{k}\left( t\right) + \mathop{\sum }\limits_{{j = 1}}^{{k - 1}}{b}_{i + j}{B}_{j}^{k}\left( t\right) + {b}_{i + k}{B}_{k... | Yes |
Theorem 8.42 The Bézier polynomials\n\n\\[ \n{p}_{1}\\left( x\\right) \\mathrel{\\text{:=}} \\mathop{\\sum }\\limits_{{k = 0}}^{n}{b}_{0}^{k}\\left( t\\right) {B}_{k}^{n}\\left( {x;0, t}\\right) \\;\\text{ and }\\;{p}_{2}\\left( x\\right) \\mathrel{\\text{:=}} \\mathop{\\sum }\\limits_{{k = 0}}^{n}{b}_{k}^{n - k}\\left... | Proof. Inserting the equivalent definition (8.56) of the subpolynomials and reordering the summation, we find that\n\n\\[ \n{p}_{1}\\left( x\\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{n}\\mathop{\\sum }\\limits_{{j = 0}}^{k}{b}_{j}{B}_{j}^{k}\\left( t\\right) {B}_{k}^{n}\\left( {x;0, t}\\right) = \\mathop{\\sum }\\l... | Yes |
Theorem 9.1 The polynomial interpolatory quadrature of order \( n \) defined by\n\n\[ \n{Q}_{n}\left( f\right) \mathrel{\text{:=}} {\int }_{a}^{b}\left( {{L}_{n}f}\right) \left( x\right) {dx} \n\]\n\n(9.3)\n\nis of the form (9.2) with the weights given by\n\n\[ \n{a}_{k} = \frac{1}{{q}_{n + 1}^{\prime }\left( {x}_{k}\r... | Proof. From (8.2) we obtain\n\n\[ \n{\int }_{a}^{b}\left( {{L}_{n}f}\right) \left( x\right) {dx} = \mathop{\sum }\limits_{{k = 0}}^{n}f\left( {x}_{k}\right) {\int }_{a}^{b}{\ell }_{k}\left( x\right) {dx} \n\]\n\nwith\n\[ \n{a}_{k} = {\int }_{a}^{b}{\ell }_{k}\left( x\right) {dx} = {\int }_{a}^{b}\mathop{\prod }\limits_... | Yes |
Theorem 9.2 Given \( n + 1 \) distinct quadrature points \( {x}_{0},\ldots ,{x}_{n} \in \left\lbrack {a, b}\right\rbrack \) , the interpolatory quadrature (9.3) of order \( n \) is uniquely determined by its property of integrating all polynomials \( p \in {P}_{n} \) exactly, i.e., by the property\n\n\[ \n\mathop{\sum ... | Proof. From (9.3) and \( {L}_{n}p = p \) for all \( p \in {P}_{n} \) it follows that\n\n\[ \n\mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}p\left( {x}_{k}\right) = {\int }_{a}^{b}\left( {{L}_{n}p}\right) \left( x\right) {dx} = {\int }_{a}^{b}p\left( x\right) {dx} \n\]\n\ni.e., the quadrature is exact for all \( p \in {P}_{... | Yes |
Theorem 9.3 The polynomial interpolatory quadrature of order \( n \) with equidistant quadrature points\n\n\[ \n{x}_{k} = a + {kh},\;k = 0,\ldots, n, \]\n\nand step width \( h = \left( {b - a}\right) /n \) is called the Newton-Cotes quadrature formula of order \( n \) . Its weights are given by\n\n\[ \n{a}_{k} = h\frac... | Proof. The weights are obtained from (9.4) by substituting \( x = {x}_{0} + {hz} \) and observing that\n\n\[ \n{q}_{n + 1}\left( x\right) = {h}^{n + 1}\mathop{\prod }\limits_{{j = 0}}^{n}\left( {z - j}\right) \]\n\nand\n\n\[ \n{q}_{n + 1}^{\prime }\left( {x}_{k}\right) = {\left( -1\right) }^{n - k}k!\left( {n - k}\righ... | Yes |
Theorem 9.4 Let \( f : C\left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be twice continuously differentiable. Then the error for the trapezoidal rule can be represented in the form\n\n\[{\int }_{a}^{b}f\left( x\right) {dx} - \frac{b - a}{2}\left\lbrack {f\left( a\right) + f\left( b\right) }\right\rbrack = - ... | Proof. Let \( {L}_{1}f \) denote the linear interpolation of \( f \) at the interpolation points \( {x}_{0} = a \) and \( {x}_{1} = b \) . By construction of the trapezoidal rule we have that the error\n\n\[{E}_{1}\left( f\right) \mathrel{\text{:=}} {\int }_{a}^{b}f\left( x\right) {dx} - \frac{b - a}{2}\left\lbrack {f\... | Yes |
Theorem 9.5 Let \( f : C\left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be four-times continuously differentiable. Then the error for Simpson's rule can be represented in the form\n\n\[{\int }_{a}^{b}f\left( x\right) {dx} - \frac{b - a}{6}\left\lbrack {f\left( a\right) + {4f}\left( \frac{a + b}{2}\right) + f... | Proof. Let \( {L}_{2}f \) denote the quadratic interpolation polynomial for \( f \) at the interpolation points \( {x}_{0} = a,{x}_{1} = \left( {a + b}\right) /2 \), and \( {x}_{2} = b \) . By construction of Simpson's rule we have that the error\n\n\[{E}_{2}\left( f\right) \mathrel{\text{:=}} {\int }_{a}^{b}f\left( x\... | Yes |
Example 9.6 The approximation of\n\n\\[ \ln 2 = {\\int }_{0}^{1}\\frac{dx}{1 + x} \\]\n\nby the trapezoidal rule yields\n\n\\[ \ln 2 \\approx \\frac{1}{2}\\left\\lbrack {1 + \\frac{1}{2}}\\right\\rbrack = {0.75} \\] | For \\( f\\left( x\\right) \\mathrel{\\text{:=}} 1/\\left( {1 + x}\\right) \\) we have\n\n\\[ \\frac{{h}^{3}}{12}{\\begin{Vmatrix}{f}^{\\prime \\prime }\\end{Vmatrix}}_{\\infty } = \\frac{1}{6} \\]\n\nand hence, from Theorem 9.4, we obtain the estimate \\( \\left| {\\ln 2 - {0.75}}\\right| \\leq {0.167} \\) as compared... | Yes |
Theorem 9.7 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be twice continuously differentiable. Then the error for the composite trapezoidal rule is given by\n\n\[{\int }_{a}^{b}f\left( x\right) {dx} - {T}_{h}\left( f\right) = - \frac{b - a}{12}{h}^{2}{f}^{\prime \prime }\left( \xi \right)\]\n\n... | Proof. By Theorem 9.4 we have that\n\n\[{\int }_{a}^{b}f\left( x\right) {dx} - {T}_{h}\left( f\right) = - \frac{{h}^{3}}{12}\mathop{\sum }\limits_{{k = 1}}^{n}{f}^{\prime \prime }\left( {\xi }_{k}\right)\]\n\nwhere \( a \leq {\xi }_{1} \leq {\xi }_{2} \leq \cdots \leq {\xi }_{n} \leq b \). From\n\n\[n\mathop{\min }\lim... | Yes |
Theorem 9.10 (Szegö) Let\n\n\[ \n{Q}_{n}\left( f\right) = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}^{\left( n\right) }f\left( {x}_{k}^{\left( n\right) }\right) \n\]\n\nbe a sequence of quadrature formulae that converges for all polynomials, i.e,\n\n\[ \n\mathop{\lim }\limits_{{n \rightarrow \infty }}{Q}_{n}\left( p\ri... | Proof. Let \( f \in C\left\lbrack {a, b}\right\rbrack \) and \( \varepsilon > 0 \) be arbitrary. By the Weierstrass approximation theorem (see [16]) there exists a polynomial \( p \) such that\n\n\[ \n\parallel f - p{\parallel }_{\infty } \leq \frac{\varepsilon }{2\left( {C + b - a}\right) }.\n\]\n\nThen, since by (9.1... | Yes |
Corollary 9.11 (Steklov) Assume that the sequence \( \left( {Q}_{n}\right) \) of quadrature formulae converges for all polynomials and that all the weights are nonnegative. Then the sequence \( \left( {Q}_{n}\right) \) is convergent. | Proof. This follows from\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{n}\left| {a}_{k}^{\left( n\right) }\right| = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}^{\left( n\right) } = {Q}_{n}\left( 1\right) \rightarrow {\int }_{a}^{b}{dx} = b - a,\;n \rightarrow \infty ,\]\n\nand the preceding Theorem 9.10. | No |
Lemma 9.13 Let \( {x}_{0},\ldots ,{x}_{n} \) be the \( n + 1 \) distinct quadrature points of a Gaussian quadrature formula. Then\n\n\[ \n{\int }_{a}^{b}w\left( x\right) {q}_{n + 1}\left( x\right) q\left( x\right) {dx} = 0 \n\]\n\n(9.16)\n\nfor \( {q}_{n + 1}\left( x\right) \mathrel{\text{:=}} \left( {x - {x}_{0}}\righ... | Proof. Since \( {q}_{n + 1}q \in {P}_{{2n} + 1} \) and \( {q}_{n + 1}\left( {x}_{k}\right) = 0 \), we have that\n\n\[ \n{\int }_{a}^{b}w\left( x\right) {q}_{n + 1}\left( x\right) q\left( x\right) {dx} = \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}{q}_{n + 1}\left( {x}_{k}\right) q\left( {x}_{k}\right) = 0 \n\]\n\nfor all... | Yes |
Lemma 9.14 Let \( {x}_{0},\ldots ,{x}_{n} \) be \( n + 1 \) distinct points satisfying the condition (9.16). Then the corresponding polynomial interpolatory quadrature is a Gaussian quadrature formula. | Proof. Let \( {L}_{n} \) denote the polynomial interpolation operator for the interpolation points \( {x}_{0},\ldots ,{x}_{n} \) . By construction, for the interpolatory quadrature we have\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}f\left( {x}_{k}\right) = {\int }_{a}^{b}w\left( x\right) \left( {{L}_{n}f}\right) \... | Yes |
Lemma 9.15 There exists a unique sequence \( \left( {q}_{n}\right) \) of polynomials of the form \( {q}_{0} = 1 \) and\n\n\[ \n{q}_{n}\left( x\right) = {x}^{n} + {r}_{n - 1}\left( x\right) ,\;n = 1,2,\ldots ,\n\]\n\nwith \( {r}_{n - 1} \in {P}_{n - 1} \) satisfying the orthogonality relation\n\n\[ \n{\int }_{a}^{b}w\le... | Proof. This follows by the Gram-Schmidt orthogonalization procedure from Theorem 3.18 applied to the linearly independent functions \( {u}_{n}\left( x\right) \mathrel{\text{:=}} {x}^{n} \) for \( n = 0,1,\ldots \) and the scalar product\n\n\[ \n\left( {f, g}\right) \mathrel{\text{:=}} {\int }_{a}^{b}w\left( x\right) f\... | Yes |
Lemma 9.16 Each of the orthogonal polynomials \( {q}_{n} \) from Lemma 9.15 has \( n \) simple zeros in \( \left( {a, b}\right) \) . | Proof. For \( m = 0 \), from (9.18) we have that\n\n\[ \n{\int }_{a}^{b}w\left( x\right) {q}_{n}\left( x\right) {dx} = 0 \n\]\n\nfor \( n > 0 \) . Hence, since \( w \) is positive on \( \left( {a, b}\right) \), the polynomial \( {q}_{n} \) must have at least one zero in \( \left( {a, b}\right) \) where the sign of \( {... | Yes |
Theorem 9.17 For each \( n = 0,1,\ldots \) there exists a unique Gaussian quadrature formula of order \( n \) . Its quadrature points are given by the zeros of the orthogonal polynomial \( {q}_{n + 1} \) of degree \( n + 1 \) . | Proof. This is a consequence of Lemmas 9.13-9.16. | Yes |
Theorem 9.18 The weights of the Gaussian quadrature formulae are all positive. | Proof. Define\n\n\[ \n{f}_{k}\left( x\right) \mathrel{\text{:=}} {\left\lbrack \frac{{q}_{n + 1}\left( x\right) }{x - {x}_{k}}\right\rbrack }^{2},\;k = 0,\ldots, n. \]\n\nThen\n\n\[ \n{a}_{k}{\left\lbrack {q}_{n + 1}^{\prime }\left( {x}_{k}\right) \right\rbrack }^{2} = \mathop{\sum }\limits_{{j = 0}}^{n}{a}_{j}{f}_{k}\... | Yes |
Corollary 9.19 The sequence of Gaussian quadrature formulae is convergent. | Proof. For each polynomial \( p \) we have\n\n\[ \n{Q}_{n}\left( p\right) = {\int }_{a}^{b}w\left( x\right) p\left( x\right) {dx} \]\n\nprovided that \( {2n} + 1 \) is greater than or equal to the degree of \( p \) . From their proofs it is obvious that Theorem 9.10 and its Corollary 9.11 remain valid for the integral ... | No |
Theorem 9.20 Let \( f \in {C}^{{2n} + 2}\left\lbrack {a, b}\right\rbrack \) . Then the error for the Gaussian quadrature formula of order \( n \) is given by\n\n\[{\int }_{a}^{b}w\left( x\right) f\left( x\right) {dx} - \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}f\left( {x}_{k}\right) = \frac{{f}^{\left( 2n + 2\right) }\... | Proof. Recall the Hermite interpolation polynomial \( {H}_{n}f \in {P}_{{2n} + 1} \) for \( f \) from Theorem 8.18. Since \( \left( {{H}_{n}f}\right) \left( {x}_{k}\right) = f\left( {x}_{k}\right), k = 0,\ldots, n \), for the error\n\n\[{E}_{n}\left( f\right) \mathrel{\text{:=}} {\int }_{a}^{b}w\left( x\right) f\left( ... | Yes |
Example 9.21 We consider the Gaussian quadrature formulae for the weight function\n\n\[ \nw\left( x\right) = \frac{1}{\sqrt{1 - {x}^{2}}},\;x \in \left\lbrack {-1,1}\right\rbrack .\n\] | The Chebyshev polynomial \( {T}_{n} \) of degree \( n \) is defined by\n\n\[ \n{T}_{n}\left( x\right) \mathrel{\text{:=}} \cos \left( {n\arccos x}\right) ,\; - 1 \leq x \leq 1.\n\]\n\nObviously \( {T}_{0}\left( x\right) = 1 \) and \( {T}_{1}\left( x\right) = x \) . From the addition theorem for the cosine function, \( ... | Yes |
The Legendre polynomial \( {L}_{n} \) of degree \( n \) is defined by \[ {L}_{n}\left( x\right) \mathrel{\text{:=}} \frac{1}{{2}^{n}n!}\frac{{d}^{n}}{d{x}^{n}}{\left( {x}^{2} - 1\right) }^{n}. \] | Obviously, \( {L}_{n} \in {P}_{n} \) . If \( m < n \), by repeated partial integration we see that \[ {\int }_{-1}^{1}{x}^{m}\frac{{d}^{n}}{d{x}^{n}}{\left( {x}^{2} - 1\right) }^{n}{dx} = 0 \] since \( {\left( {x}^{2} - 1\right) }^{n} \) has zeros of order \( n \) at the endpoints -1 and 1 . Therefore, \[ {\int }_{-1}^... | No |
Lemma 9.24 The Bernoulli polynomials have the symmetry property\n\n\\[ \n{B}_{n}\\left( x\\right) = {\\left( -1\\right) }^{n}{B}_{n}\\left( {1 - x}\\right) ,\\;x \\in \\mathbb{R},\\;n = 0,1,\\ldots \n\\]\n\n(9.24) | Proof. Obviously (9.24) holds for \\( n = 0 \\) . Assume that (9.24) has been proven for some \\( n \\geq 0 \\) . Then, integrating (9.24), we obtain\n\n\\[ \n{B}_{n + 1}\\left( x\\right) = {\\left( -1\\right) }^{n + 1}{B}_{n + 1}\\left( {1 - x}\\right) + {\\beta }_{n + 1} \n\\]\n\nfor some constant \\( {\\beta }_{n + ... | Yes |
Lemma 9.25 The Bernoulli polynomials \( {B}_{{2m} + 1}, m = 1,2,\ldots \), of odd degree have exactly three zeros in \( \left\lbrack {0,1}\right\rbrack \), and these zeros are at the points \( 0,1/2 \), and 1 . The Bernoulli polynomials \( {B}_{2m}, m = 0,1,\ldots \), of even degree satisfy \( {B}_{2m}\left( 0\right) \... | Proof. From (9.23) and (9.24) we conclude that \( {B}_{{2m} + 1} \) vanishes at the points \( 0,1/2 \), and 1 . We prove by induction that these are the only zeros of \( {B}_{{2m} + 1} \) in \( \left\lbrack {0,1}\right\rbrack \) . This is true for \( m = 1 \), since \( {B}_{3} \) is a polynomial of degree three. Assume... | Yes |
Theorem 9.26 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be \( m \) times continuously differentiable for \( m \geq 2 \) . Then we have the Euler-Maclaurin expansion\n\n\[{\int }_{a}^{b}f\left( x\right) {dx} = {T}_{h}\left( f\right) - \mathop{\sum }\limits_{{j = 1}}^{\left\lbrack \frac{m}{2}\r... | Proof. Let \( g \in {C}^{m}\left\lbrack {0,1}\right\rbrack \) . Then, by \( m - 1 \) partial integrations and using (9.23) we find that\n\n\[{\int }_{0}^{1}{B}_{1}\left( z\right) {g}^{\prime }\left( z\right) {dz} = \mathop{\sum }\limits_{{j = 2}}^{m}{\left( -1\right) }^{j}{B}_{j}\left( 0\right) \left\lbrack {{g}^{\left... | Yes |
Corollary 9.27 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be \( \left( {{2m} + 1}\right) \) -times continuously differentiable and \( {2\pi } \) -periodic for \( m \in \mathbb{N} \) and let \( n \in \mathbb{N} \) . Then for the error of the rectangular rule we have\n\n\[ \left| {{E}_{n}\left( f\right) }\right| \le... | Proof. From Theorem 9.26 we have that\n\n\[ {E}_{n}\left( f\right) = - {\left( \frac{2\pi }{n}\right) }^{{2m} + 1}{\int }_{0}^{2\pi }{\widetilde{B}}_{{2m} + 1}\left( \frac{2\pi x}{n}\right) {f}^{\left( 2m + 1\right) }\left( x\right) {dx} \]\n\nand the estimate follows from the inequality\n\n\[ \left| {{\widetilde{B}}_{... | Yes |
Theorem 9.28 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be analytic and \( {2\pi } \) -periodic. Then there exists a strip \( D = \mathbb{R} \times \left( {-a, a}\right) \subset \mathbb{C} \) with \( a > 0 \) such that \( f \) can be extended to a holomorphic and \( {2\pi } \) -periodic bounded function \( f : D \... | Proof. Since \( f : \mathbb{R} \rightarrow \mathbb{R} \) is analytic, at each point \( x \in \mathbb{R} \) the Taylor expansion provides a holomorphic extension of \( f \) into some open disk in the complex plane with radius \( r\left( x\right) > 0 \) and center \( x \) . The extended function again has period \( {2\pi... | Yes |
Theorem 9.29 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be \( {2m} \) -times continuously differentiable. Then for the Romberg quadratures we have the error estimate\n\n\[ \n\left| {{\int }_{a}^{b}f\left( x\right) {dx} - {T}_{k}^{m}\left( f\right) }\right| \leq {C}_{m}{\begin{Vmatrix}{f}^{\le... | Proof. By induction, we show that there exist constants \( {\gamma }_{j, i} \) such that\n\n\[ \n\left| {{\int }_{a}^{b}f\left( x\right) {dx} - {T}_{k}^{i}\left( f\right) - \mathop{\sum }\limits_{{j = i}}^{{m - 1}}{\gamma }_{j, i}\left\lbrack {{f}^{\left( 2j - 1\right) }\left( b\right) - {f}^{\left( 2j - 1\right) }\lef... | Yes |
Theorem 9.30 The quadrature weights of the Romberg formulae are positive. | Proof. We define recursively \( {Q}_{k}^{1} \mathrel{\text{:=}} 4{T}_{k + 1}^{1} - 2{T}_{k}^{1} \) and\n\n\[ {Q}_{k}^{m + 1} \mathrel{\text{:=}} \frac{1}{{4}^{m} - 1}\left\lbrack {{2}^{{2m} + 1}{T}_{k + 1}^{m} + 2{T}_{k}^{m} + {4}^{m + 1}{Q}_{k + 1}^{m}}\right\rbrack \]\n\n(9.30)\n\nfor \( k = 1,2,\ldots \) and \( m = ... | Yes |
Corollary 9.31 For the Romberg quadratures we have convergence:\n\n\[ \n\mathop{\lim }\limits_{{m \rightarrow \infty }}{T}_{k}^{m}\left( f\right) = {\int }_{a}^{b}f\left( x\right) {dx}\;\text{ and }\;\mathop{\lim }\limits_{{k \rightarrow \infty }}{T}_{k}^{m}\left( f\right) = {\int }_{a}^{b}f\left( x\right) {dx} \n\]\n\... | Proof. This follows from Theorems 9.29 and 9.30 and Corollary 9.11. | Yes |
Theorem 9.32 Denote by \( {L}_{k}^{m} \) the uniquely determined polynomial in \( {h}^{2} \) of degree less than or equal to \( m \) with the interpolation property\n\n\[ \n{L}_{k}^{m}\left( {h}_{j}^{2}\right) = {T}_{j}^{1}\left( f\right) ,\;j = k,\ldots, k + m.\n\]\n\nThen the Romberg quadratures satisfy\n\n\[ \n{T}_{... | Proof. Obviously,(9.32) is true for \( m = 0 \) . Assume that it has been proven for \( m - 1 \) . Then, using the Neville scheme from Theorem 8.9, we obtain\n\n\[ \n{L}_{k}^{m}\left( 0\right) = \frac{1}{{h}_{k + m}^{2} - {h}_{k}^{2}}\left\lbrack {-{h}_{k}^{2}{L}_{k + 1}^{m - 1}\left( 0\right) + {h}_{k + m}^{2}{L}_{k}^... | Yes |
Let \( p = p\left( t\right) \) describe the population of a species of animals or plants at time \( t \) . If \( r\left( {t, p}\right) \) denotes the growth rate given by the difference between the birth and death rate depending on the time \( t \) and the size \( p \) of the population, then an isolated population sat... | with the explicit solution \( p\left( t\right) = {p}_{0}{e}^{a\left( {t - {t}_{0}}\right) } \) . Such an exponential growth is realistic only if the population is not too large. | Yes |
Corollary 10.6 Under the assumptions of Theorem 10.5, the sequence \( \left( {u}_{\nu }\right) \) defined by \( {u}_{0}\left( x\right) = {u}_{0} \) and\n\n\[ \n{u}_{\nu + 1}\left( x\right) \mathrel{\text{:=}} {u}_{0} + {\int }_{{x}_{0}}^{x}f\left( {\xi ,{u}_{\nu }\left( \xi \right) }\right) {d\xi },\;\left| {x - {x}_{0... | Proof. This follows from Theorem 3.46. | No |
Example 10.7 Consider the initial value problem\n\n\\[ \n{u}^{\prime } = {x}^{2} + {u}^{2},\;u\\left( 0\\right) = 0 \n\\] \n\non \\( G = \\left( {-{0.5},{0.5}}\\right) \\times \\left( {-{0.5},{0.5}}\\right) \\) . For \\( f\\left( {x, u}\\right) \\mathrel{\\text{:=}} {x}^{2} + {u}^{2} \\) we have\n\n\\[ \n\\left| {f\\le... | Here, the iteration (10.6) reads\n\n\\[ \n{u}_{\\nu + 1}\\left( x\\right) = {\\int }_{0}^{x}\\left\\lbrack {{\\xi }^{2} + {u}_{\\nu }^{2}\\left( \\xi \\right) }\\right\\rbrack {d\\xi }.\n\\] \n\nStarting with \\( {u}_{0}\\left( x\\right) = 0 \\) we first compute\n\n\\[ \n{u}_{1}\\left( x\\right) = {\\int }_{0}^{x}{\\xi... | Yes |
Theorem 10.17 A single-step method is consistent if and only if\n\n\[ \mathop{\lim }\limits_{{h \rightarrow 0}}\varphi \left( {x, u;h}\right) = f\left( {x, u}\right) \]\n\nuniformly for all \( \left( {x, u}\right) \in G \) . | Proof. Since we assume \( f \) to be bounded, we have\n\n\[ \eta \left( {x + t}\right) - \eta \left( x\right) = {\int }_{0}^{t}{\eta }^{\prime }\left( {x + s}\right) {ds} = {\int }_{0}^{t}f\left( {x + s,\eta \left( {x + s}\right) }\right) {ds} \rightarrow 0,\;t \rightarrow 0, \]\n\nuniformly for all \( \left( {x, u}\ri... | Yes |
Theorem 10.17 A single-step method is consistent if and only if\n\n\\[ \n\\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\varphi \\left( {x, u;h}\\right) = f\\left( {x, u}\\right) \n\\]\n\nuniformly for all \\( \\left( {x, u}\\right) \\in G \\) . | Proof. Since we assume \\( f \\) to be bounded, we have\n\n\\[ \n\\eta \\left( {x + t}\\right) - \\eta \\left( x\\right) = {\\int }_{0}^{t}{\\eta }^{\\prime }\\left( {x + s}\\right) {ds} = {\\int }_{0}^{t}f\\left( {x + s,\\eta \\left( {x + s}\\right) }\\right) {ds} \\rightarrow 0,\\;t \\rightarrow 0, \n\\]\n\nuniformly... | Yes |
Theorem 10.18 The Euler method is consistent. If \( f \) is continuously differentiable in \( G \), then the Euler method has consistency order one. | Proof. Consistency is a consequence of Theorem 10.17 and the fact that \( \varphi \left( {x, u;h}\right) = f\left( {x, u}\right) \) for Euler’s method. If \( f \) is continuously differentiable, then from the differential equation \( {\eta }^{\prime } = f\left( {\xi ,\eta }\right) \) it follows that \( \eta \) is twice... | Yes |
Theorem 10.19 The improved Euler method is consistent. If \( f \) is twice continuously differentiable in \( G \), then the improved Euler method has consistency order two. | Proof. Consistency follows from Theorem 10.17 and\n\n\[ \varphi \left( {x, u;h}\right) = \frac{1}{2}\left\lbrack {f\left( {x, u}\right) + f\left( {x + h, u + {hf}\left( {x, u}\right) }\right) }\right\rbrack \rightarrow f\left( {x, u}\right) ,\;h \rightarrow 0. \]\n\nIf \( f \) is twice continuously differentiable, then... | Yes |
Lemma 10.21 Let \( \\left( {\\xi }_{j}\\right) \) be a sequence in \( \\mathbb{R} \) with the property\n\n\[\\left| {\\xi }_{j + 1}\\right| \\leq \\left( {1 + A}\\right) \\left| {\\xi }_{j}\\right| + B,\\;j = 0,1,\\ldots ,\]\n\nfor some constants \( A > 0 \) and \( B \\geq 0 \) . Then the estimate\n\n\[\\left| {\\xi }_... | Proof. We prove this by induction. The estimate is true for \( j = 0 \) . Assume that it has been proven for some \( j \\geq 0 \) . Then, with the aid of the inequality \( 1 + A < {e}^{A} \), which follows from the power series for the exponential function, we obtain\n\n\[\\left| {\\xi }_{j + 1}\\right| \\leq \\left( {... | Yes |
Theorem 10.23 Assume that the single-step method satisfies the assumptions of the previous Theorem 10.22 and that it has consistency order \( p \) ; i.e., \( \left| {\Delta \left( {x, u;h}\right) }\right| \leq K{h}^{p} \) . Then\n\n\[ \left| {e}_{j}\right| \leq \frac{K}{M}\left( {{e}^{M\left( {{x}_{j} - {x}_{0}}\right)... | Proof. This follows from (10.16) with the aid of \( c\left( h\right) \leq K{h}^{p} \) . | No |
Corollary 10.24 The Euler method and the improved Euler method are convergent. For continuously differentiable \( f \) the Euler method has convergence order one. For twice continuously differentiable \( f \) the improved Euler method has convergence order two. | Proof. By Theorems 10.18, 10.19, 10.22, and 10.23 it remains only to verify the Lipschitz condition of the function \( \varphi \) for the improved Euler method given by (10.11). From the Lipschitz condition for \( f \) we obtain\n\n\[ \left| {\varphi \left( {x, u;h}\right) - \varphi \left( {x, v;h}\right) }\right| \]\n... | Yes |
Theorem 10.26 The Runge-Kutta method is consistent. If \( f \) is four-times continuously differentiable, then it has consistency order four and hence convergence order four. | Proof. The function \( \varphi \) describing the Runge-Kutta method is given recursively by\n\n\[ \varphi = \frac{1}{6}\left( {{\varphi }_{1} + 2{\varphi }_{2} + 2{\varphi }_{3} + {\varphi }_{4}}\right) \]\n\nwhere\n\n\[ {\varphi }_{1}\left( {x, u;h}\right) = f\left( {x, u}\right) \]\n\n\[ {\varphi }_{2}\left( {x, u;h}... | No |
Theorem 10.29 If \( f \) is \( \left( {s + 1}\right) \) -times continuously differentiable, then the multistep methods (10.21) are consistent of order \( s + 1 \) . | Proof. By construction we have that\n\n\[ \Delta \left( {x, u;h}\right) = \frac{1}{h}{\int }_{x + \left( {r - k}\right) h}^{x + {rh}}\left\lbrack {f\left( {\xi, u\left( \xi \right) }\right) - p\left( \xi \right) }\right\rbrack {d\xi } \]\n\nwhere \( p \) denotes the polynomial satisfying the interpolation condition\n\n... | Yes |
Let \( p \) be the quadratic interpolation polynomial satisfying\n\n\[ p\left( {x}_{j}\right) = u\left( {x}_{j}\right) ,\;j = 0,1,2, \]\n\nand approximate\n\n\[ {u}^{\prime }\left( {x}_{0}\right) \approx {p}^{\prime }\left( {x}_{0}\right) \] | Using the fact that the approximation for the derivative is exact for polynomials of degree less than or equal to two, simple calculations show that (see Problem 10.15)\n\n\[ {p}^{\prime }\left( {x}_{0}\right) = \frac{1}{2h}\left\lbrack {-u\left( {x}_{2}\right) + {4u}\left( {x}_{1}\right) - {3u}\left( {x}_{0}\right) }\... | No |
For \( k = 0,1,\ldots, r - 1 \), let \( {u}_{j, k} \) denote the unique solutions to the homogeneous difference equation (10.31) with initial values\n\n\[ \n{u}_{j, k} = {\delta }_{j, k},\;j = 0,1,\ldots, r - 1.\n\]\n\nThen for a given right-hand side \( {c}_{r},{c}_{r + 1},\ldots \), the unique solution to the inhomog... | Proof. Setting \( {u}_{m, r - 1} = 0 \) for \( m = - 1, - 2,\ldots \), we can rewrite (10.37) in the form\n\n\[ \n{z}_{j} = \mathop{\sum }\limits_{{k = 0}}^{{r - 1}}{z}_{k}{u}_{j, k} + {w}_{j},\;j = 0,1,\ldots ,\n\]\n\nwhere\n\n\[ \n{w}_{j} \mathrel{\text{:=}} \mathop{\sum }\limits_{{k = 0}}^{\infty }{c}_{k + r}{u}_{j ... | Yes |
Lemma 10.37 Let \( \\left( {\\xi }_{j}\\right) \) be a sequence in \( \\mathbb{R} \) with the property\n\n\[ \n\\left| {\\xi }_{j}\\right| \\leq A\\mathop{\\sum }\\limits_{{m = 0}}^{{j - 1}}\\left| {\\xi }_{m}\\right| + B,\\;j = 1,2,\\ldots ,\n\]\n\nfor some constants \( A > 0 \) and \( B \\geq 0 \) . Then the estimate... | Proof. We prove by induction that\n\n\[ \n\\left| {\\xi }_{j}\\right| \\leq \\left( {A\\left| {\\xi }_{0}\\right| + B}\\right) {\\left( 1 + A\\right) }^{j - 1},\\;j = 1,2,\\ldots \n\]\n\n(10.38)\n\nThen the assertion follows by using the estimate \( 1 + A \\leq {e}^{A} \) . The inequality (10.38) is true for \( j = 1 \... | Yes |
Consider the boundary value problem\n\n\[ \n{u}^{\prime \prime } = {u}^{3},\;u\left( 1\right) = \sqrt{2},\;u\left( 2\right) = \frac{1}{2}\sqrt{2}, \n\]\n\nwith the exact solution \( u\left( x\right) = \sqrt{2}/x \) . | We solve numerically the associated initial value problem\n\n\[ \n{u}^{\prime \prime } = {u}^{3},\;u\left( 1\right) = \sqrt{2},\;{u}^{\prime }\left( 1\right) = s, \n\]\n\nby the improved Euler method of Section 10.2 with step sizes \( h = {0.1} \), \( h = {0.01} \), and \( h = {0.001} \). For this we transform the init... | Yes |
The linear boundary value problem\n\n\[ \n{u}^{\prime \prime } - {u}^{\prime } - {110u} = 0,\;u\left( 0\right) = u\left( {10}\right) = 1, \n\] | has the unique solution\n\n\[ \nu\left( x\right) = \frac{1}{{e}^{110} - {e}^{-{100}}}\left\{ {\left( {{e}^{110} - 1}\right) {e}^{-{10x}} + \left( {1 - {e}^{-{100}}}\right) {e}^{11x}}\right\} .\n\] | Yes |
Theorem 11.4 Assume that \( q, r \in C\left\lbrack {a, b}\right\rbrack \) and \( q \geq 0 \) . Then the boundary value problem for the linear differential equation\n\n\[ - {u}^{\prime \prime } + {qu} = r\;\text{ on }\left\lbrack {a, b}\right\rbrack \]\n\nwith homogeneous boundary conditions\n\n\[ u\left( a\right) = u\l... | Proof. Assume that \( {u}_{1} \) and \( {u}_{2} \) are two solutions to the boundary value problem. Then the difference \( u = {u}_{1} - {u}_{2} \) solves the homogeneous boundary value problem\n\n\[ - {u}^{\prime \prime } + {qu} = 0,\;u\left( a\right) = u\left( b\right) = 0.\]\n\nBy partial integration we obtain\n\n\[... | Yes |
Theorem 11.5 For each \( h > 0 \) the difference equations (11.10)-(11.11) have a unique solution. | Proof. The tridiagonal matrix \( A \) is irreducible and weakly row-diagonally dominant. Hence, by Theorem 4.7, the matrix \( A \) is invertible, and the Jacobi iterations converge. | No |
Lemma 11.6 Denote by \( A \) the matrix of the finite difference method for \( q \geq 0 \) and by \( {A}_{0} \) the corresponding matrix for \( q = 0 \) . Then\n\n\[ 0 \leq {A}^{-1} \leq {A}_{0}^{-1} \]\n\ni.e., all components of \( {A}^{-1} \) are nonnegative and smaller than or equal to the corresponding components o... | Proof. The columns of the inverse \( {A}^{-1} = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) satisfy \( A{a}_{j} = {e}_{j} \) for \( j = 1,\ldots, n \) with the canonical unit vectors \( {e}_{1},\ldots ,{e}_{n} \) in \( {\mathbb{R}}^{n} \) . The Jacobi iterations for the solution of \( {Az} = {e}_{j} \) starting with \( ... | Yes |
Lemma 11.7 Assume that \( u \in {C}^{4}\left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \left| {{u}^{\prime \prime }\left( x\right) - \frac{1}{{h}^{2}}\left\lbrack {u\left( {x + h}\right) - {2u}\left( x\right) + u\left( {x - h}\right) }\right\rbrack }\right| \leq \frac{{h}^{2}}{12}{\begin{Vmatrix}{u}^{\left( 4\right) }... | Proof. By Taylor's formula we have that\n\n\[ u\left( {x \pm h}\right) = u\left( x\right) \pm h{u}^{\prime }\left( x\right) + \frac{{h}^{2}}{2}{u}^{\prime \prime }\left( x\right) \pm \frac{{h}^{3}}{6}{u}^{\prime \prime \prime }\left( x\right) + \frac{{h}^{4}}{24}{u}^{\left( 4\right) }\left( {x \pm {\theta }_{ \pm }h}\r... | Yes |
Theorem 11.8 Assume that the solution to the boundary value problem (11.7)-(11.8) is four-times continuously differentiable. Then the error of the finite difference approximation can be estimated by | Proof. By Lemma 11.7, for\n\n\[ {z}_{j} \mathrel{\text{:=}} {u}^{\prime \prime }\left( {x}_{j}\right) - \frac{1}{{h}^{2}}\left\lbrack {u\left( {x}_{j + 1}\right) - {2u}\left( {x}_{j}\right) + u\left( {x}_{j - 1}\right) }\right\rbrack \]\n\nwe have the estimate\n\n\[ \left| {z}_{j}\right| \leq \frac{{h}^{2}}{12}{\begin{... | Yes |
Theorem 11.9 For each \( h > 0 \) the difference equations (11.19)-(11.20) have a unique solution. | From the proof of Lemma 11.6 it can be seen that its statement also holds for the corresponding matrices of the system (11.19)-(11.20). Lemma 11.7 implies that\n\n\[ \n{\Delta u}\left( {{x}_{1},{x}_{2}}\right) - \frac{1}{{h}^{2}}\left\lbrack {u\left( {{x}_{1} + h,{x}_{2}}\right) + u\left( {{x}_{1} - h,{x}_{2}}\right) +... | Yes |
Theorem 11.13 (Lax-Milgram) In a Hilbert space \( X \) a bounded and strictly coercive linear operator \( A : X \rightarrow X \) has a bounded inverse \( {A}^{-1} : X \rightarrow X \) . | Proof. Using the Cauchy-Schwarz inequality, we can estimate\n\n\[ \parallel {Au}\parallel \parallel u\parallel \geq \operatorname{Re}\left( {{Au}, u}\right) \geq c\parallel u{\parallel }^{2}. \]\n\nHence\n\n\[ \parallel {Au}\parallel \geq c\parallel u\parallel \]\n\n(11.25)\n\nfor all \( u \in X \) . From (11.25) we ob... | Yes |
Theorem 11.15 Let \( S \) be a bounded and strictly coercive sesquilinear function on a Hilbert space \( X \) . Then there exists a uniquely determined bounded and strictly coercive linear operator \( A : X \rightarrow X \) such that\n\n\[ S\left( {u, v}\right) = \left( {u,{Av}}\right) \] for all \( u, v \in X \) . | Proof. For each \( v \in X \) the mapping \( u \mapsto S\left( {u, v}\right) \) clearly defines a bounded linear function on \( X \), since \( \left| {S\left( {u, v}\right) }\right| \leq C\parallel u\parallel \parallel v\parallel \) . By the Riesz Theorem 11.11 we can write \( S\left( {u, v}\right) = \left( {u, f}\righ... | Yes |
Corollary 11.16 Let \( S \) be a bounded and strictly coercive sesquilinear function and \( F \) a bounded linear function on a Hilbert space \( X \) . Then there exists a unique \( u \in X \) such that\n\n\[ S\left( {v, u}\right) = F\left( v\right) \]\n\nfor all \( v \in X \) . | Proof. By Theorem 11.15 there exists a uniquely determined bounded and strictly coercive linear operator \( A \) such that\n\n\[ S\left( {v, u}\right) = \left( {v,{Au}}\right) \]\n\nfor all \( u, v \in X \), and by Theorem 11.11 there exists a uniquely determined element \( f \) such that\n\n\[ F\left( v\right) = \left... | Yes |
Theorem 11.17 For a bounded and strictly coercive linear operator \( A \) the Galerkin equations (11.30) have a unique solution. It satisfies the error estimate\n\n\[ \begin{Vmatrix}{{u}_{n} - u}\end{Vmatrix} \leq M\mathop{\inf }\limits_{{v \in {X}_{n}}}\parallel v - u\parallel \] \n\n(11.32) \n\nwhere \( M \) is some ... | Proof. Since \( {A}_{n} : {X}_{n} \rightarrow {X}_{n} \) is strictly coercive with coercitivity constant \( c \), by the Lax-Milgram Theorem 11.13 we conclude that \( {A}_{n} \) is bijective; i.e., the Galerkin equations (11.30) have a unique solution \( {u}_{n} \in {X}_{n} \) . The estimate (11.26) applied to the oper... | Yes |
Theorem 11.19 The linear space\n\n\\[ \n{H}^{1}\\left\\lbrack {a, b}\\right\\rbrack \\mathrel{\\text{:=}} \\left\\{ {u \\in {L}^{2}\\left\\lbrack {a, b}\\right\\rbrack : {u}^{\\prime } \\in {L}^{2}\\left\\lbrack {a, b}\\right\\rbrack }\\right\\} \n\\]\n\nendowed with the scalar product\n\n\\[ \n{\\left( u, v\\right) }_... | Proof. It is readily checked that \\( {H}^{1}\\left\\lbrack {a, b}\\right\\rbrack \\) is a linear space and that (11.44) defines a scalar product. Let \\( \\left( {u}_{n}\\right) \\) denote an \\( {H}^{1} \\) Cauchy sequence. Then \\( \\left( {u}_{n}\\right) \\) and \\( \\left( {u}_{n}^{\\prime }\\right) \\) are both \... | Yes |
Theorem 11.20 \( {C}^{1}\left\lbrack {a, b}\right\rbrack \) is dense in \( {H}^{1}\left\lbrack {a, b}\right\rbrack \) . | Proof. Since \( C\left\lbrack {a, b}\right\rbrack \) is dense in \( {L}^{2}\left\lbrack {a, b}\right\rbrack \), for each \( u \in {H}^{1}\left\lbrack {a, b}\right\rbrack \) and \( \varepsilon > 0 \) there exists \( w \in C\left\lbrack {a, b}\right\rbrack \) such that \( {\begin{Vmatrix}{u}^{\prime } - w\end{Vmatrix}}_{... | Yes |
Theorem 11.21 \( {H}^{1}\left\lbrack {a, b}\right\rbrack \) is contained in \( C\left\lbrack {a, b}\right\rbrack \) . | Proof. From (11.43) we have\n\n\[ u\left( x\right) - u\left( y\right) = {\int }_{y}^{x}{u}^{\prime }\left( \xi \right) {d\xi } \]\n\n(11.45)\n\nwhence by the Cauchy-Schwarz inequality,\n\n\[ \left| {u\left( x\right) - u\left( y\right) }\right| \leq {\left| x - y\right| }^{1/2}{\begin{Vmatrix}{u}^{\prime }\end{Vmatrix}}... | Yes |
Theorem 11.22 The space\n\n\[ \n{H}_{0}^{1}\left\lbrack {a, b}\right\rbrack \mathrel{\text{:=}} \left\{ {u \in {H}^{1}\left\lbrack {a, b}\right\rbrack : u\left( a\right) = u\left( b\right) = 0}\right\} \n\]\n\nis a complete subspace of \( {H}^{1}\left\lbrack {a, b}\right\rbrack \) . | Proof. Since the \( {H}^{1} \) norm is stronger than the maximum norm, each \( {H}^{1} \) convergent sequence of elements of \( {H}_{0}^{1}\left\lbrack {a, b}\right\rbrack \) has its limit in \( {H}_{0}^{1}\left\lbrack {a, b}\right\rbrack \) . Therefore \( {H}_{0}^{1}\left\lbrack {a, b}\right\rbrack \) is a closed subs... | Yes |
Theorem 11.24 Assume that \( p > 0 \) and \( q \geq 0 \) . Then there exists a unique weak solution to the boundary value problem (11.36)-(11.37). | Proof. The sesquilinear function \( S : {H}_{0}^{1}\left\lbrack {a, b}\right\rbrack \times {H}_{0}^{1}\left\lbrack {a, b}\right\rbrack \) is bounded, since\n\n\[ \left| {S\left( {u, v}\right) }\right| \leq \max \left\{ {\parallel p{\parallel }_{\infty },\parallel q{\parallel }_{\infty }}\right\} \parallel u{\parallel }... | Yes |
Theorem 11.25 Each weak solution to the boundary value problem (11.36)- (11.37) is also a classical solution; i.e., it is twice continuously differentiable. | Proof. Define\n\n\[ f\left( x\right) \mathrel{\text{:=}} {\int }_{a}^{x}\left\lbrack {q\left( \xi \right) u\left( \xi \right) - r\left( \xi \right) }\right\rbrack {d\xi },\;x \in \left\lbrack {a, b}\right\rbrack . \]\n\nThen \( f \in {C}^{1}\left\lbrack {a, b}\right\rbrack \) . From (11.38), by partial integration we o... | Yes |
Lemma 11.26 Let \( f\left\lbrack {a, b}\right\rbrack \in {C}^{2}\left\lbrack {a, b}\right\rbrack \) . Then the remainder \( {R}_{1}f \mathrel{\text{:=}} f - {L}_{1}f \) for the linear interpolation at the two endpoints \( a \) and \( b \) can be estimated \( {by} \)\n\n\[ \n{\begin{Vmatrix}{R}_{1}f\end{Vmatrix}}_{{L}^{... | Proof. For each function \( g \in {C}^{1}\left\lbrack {a, b}\right\rbrack \) satisfying \( g\left( a\right) = 0 \), from\n\n\[ \ng\left( x\right) = {\int }_{a}^{x}{g}^{\prime }\left( \xi \right) {d\xi } \]\n\nby using the Cauchy-Schwarz inequality we obtain\n\n\[ \n{\left| g\left( x\right) \right| }^{2} \leq \left( {b ... | Yes |
Theorem 11.27 The error in the finite element approximation by linear splines for the boundary value problem (11.36)-(11.37) can be estimated by\n\n\\[ \n{\\begin{Vmatrix}{u}_{n} - u\\end{Vmatrix}}_{{H}^{1}} \\leq C{\\begin{Vmatrix}{u}^{\\prime \\prime }\\end{Vmatrix}}_{{L}^{2}}h\n\\]\n\n(11.51)\n\nfor some positive co... | Proof. By summing up the inequalities (11.49), applied to each of the subintervals of length \\( h \\), for the interpolating linear spline \\( {w}_{n} \\in {X}_{n} \\) with \\( {w}_{n}\\left( {x}_{j}\\right) = u\\left( {x}_{j}\\right) \\) for \\( j = 0,\\ldots, n \\) we find that\n\n\\[ \n{\\begin{Vmatrix}{w}_{n}^{\\p... | Yes |
Theorem 11.28 The error in the finite element approximation by linear splines for the boundary value problem (11.36)-(11.37) can be estimated by\n\n\[ \n{\\begin{Vmatrix}{u}_{n} - u\\end{Vmatrix}}_{{L}^{2}} \\leq C{\\begin{Vmatrix}{u}^{\\prime \\prime }\\end{Vmatrix}}_{{L}^{2}}{h}^{2}\n\]\n\nwith some positive constant... | Proof. Denote by \( {z}_{n} \) the weak solution to the boundary value problem with the right-hand side \( u - {u}_{n} \) ; i.e.,\n\n\[ \nS\\left( {v,{z}_{n}}\\right) = {\\left( v, u - {u}_{n}\\right) }_{{L}^{2}}\n\]\n\nfor all \( v \\in {H}_{0}^{1}\\left\\lbrack {a, b}\\right\\rbrack \) . In particular, inserting \( v... | Yes |
Theorem 12.4 The integral operator (12.3) with continuous kernel is a compact operator on \( C\left\lbrack {a, b}\right\rbrack \) . | Proof. For all \( \varphi \in C\left\lbrack {a, b}\right\rbrack \) with \( \parallel \varphi {\parallel }_{\infty } \leq 1 \) and all \( x \in \left\lbrack {a, b}\right\rbrack \), we have that\n\n\[ \left| {\left( {A\varphi }\right) \left( x\right) }\right| \leq \left( {b - a}\right) \mathop{\max }\limits_{{x, y \in \l... | Yes |
Theorem 12.5 The norm of the integral operator \( A : C\left\lbrack {a, b}\right\rbrack \rightarrow C\left\lbrack {a, b}\right\rbrack \) with continuous kernel \( K \) is given by\n\n\[ \parallel A{\parallel }_{\infty } = \mathop{\max }\limits_{{a \leq x \leq b}}{\int }_{a}^{b}\left| {K\left( {x, y}\right) }\right| {dy... | Proof. For each \( \varphi \in C\left\lbrack {a, b}\right\rbrack \) with \( \parallel \varphi {\parallel }_{\infty } \leq 1 \) we have\n\n\[ \left| {\left( {A\varphi }\right) \left( x\right) }\right| \leq {\int }_{a}^{b}\left| {K\left( {x, y}\right) }\right| {dy},\;x \in \left\lbrack {a, b}\right\rbrack ,\]\n\nand thus... | Yes |
Theorem 12.6 Let \( A : X \rightarrow X \) be a compact linear operator on a Banach space \( X \) such that \( I - A \) is injective. Assume that the sequence \( {A}_{n} : X \rightarrow X \) of bounded linear operators is norm convergent, i.e., \( \begin{Vmatrix}{{A}_{n} - A}\end{Vmatrix} \rightarrow 0, n \rightarrow \... | Proof. By the Riesz Theorem 12.2, the inverse \( {\left( I - A\right) }^{-1} : X \rightarrow X \) exists and is bounded. Since \( \begin{Vmatrix}{{A}_{n} - A}\end{Vmatrix} \rightarrow 0, n \rightarrow \infty \), by Remark 3.25 we have \( \begin{Vmatrix}{{\left( I - A\right) }^{-1}\left( {{A}_{n} - A}\right) }\end{Vmatr... | Yes |
Lemma 12.9 Let \( X \) be a Banach space, let \( {A}_{n} : X \rightarrow X \) be a collectively compact sequence, and let \( {B}_{n} : X \rightarrow X \) be a pointwise convergent sequence with limit operator \( B : X \rightarrow X \) . Then\n\n\[ \n\begin{Vmatrix}{\left( {{B}_{n} - B}\right) {A}_{n}}\end{Vmatrix} \rig... | Proof. Assume that (12.7) is not valid. Then there exist \( {\varepsilon }_{0} > 0 \), a sequence \( \left( {n}_{k}\right) \) in \( \mathbb{N} \) with \( {n}_{k} \rightarrow \infty, k \rightarrow \infty \), and a sequence \( \left( {\varphi }_{k}\right) \) in \( X \) with \( \begin{Vmatrix}{\varphi }_{k}\end{Vmatrix} \... | Yes |
Theorem 12.10 Let \( A : X \rightarrow X \) be a compact linear operator on a Banach space \( X \) such that \( I - A \) is injective, and assume that the sequence \( {A}_{n} : X \rightarrow X \) of linear operators is collectively compact and pointwise convergent; i.e., \( {A}_{n}\varphi \rightarrow {A\varphi }, n \ri... | Proof. By the Riesz Theorem 12.2, the inverse \( {\left( I - A\right) }^{-1} : X \rightarrow X \) exists and is bounded. The identity\n\n\[ {\left( I - A\right) }^{-1} = I + {\left( I - A\right) }^{-1}A \]\n\nsuggests\n\n\[ {M}_{n} \mathrel{\text{:=}} I + {\left( I - A\right) }^{-1}{A}_{n} \]\n\nas an approximate inver... | Yes |
Theorem 12.11 Let \( {\varphi }_{n} \) be a solution of\n\n\[ \n{\varphi }_{n}\left( x\right) - \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}K\left( {x,{x}_{k}}\right) {\varphi }_{n}\left( {x}_{k}\right) = f\left( x\right) ,\;x \in \left\lbrack {a, b}\right\rbrack .\n\]\n\n(12.13)\n\nThen the values \( {\varphi }_{j}^{\le... | Proof. The first statement is trivial. For a solution \( {\varphi }_{j}^{\left( n\right) }, j = 0,\ldots, n \), of the system (12.14) the function \( {\varphi }_{n} \) defined by (12.15) has values\n\n\[ \n{\varphi }_{n}\left( {x}_{j}\right) = f\left( {x}_{j}\right) + \mathop{\sum }\limits_{{k = 0}}^{n}{a}_{k}K\left( {... | Yes |
Theorem 12.12 The norm of the quadrature operators \( {A}_{n} \) is given by\n\n\[ \n{\begin{Vmatrix}{A}_{n}\end{Vmatrix}}_{\infty } = \mathop{\max }\limits_{{a \leq x \leq b}}\mathop{\sum }\limits_{{k = 0}}^{n}\left| {{a}_{k}K\left( {x,{x}_{k}}\right) }\right| .\n\] | Proof. For each \( \varphi \in C\left\lbrack {a, b}\right\rbrack \) with \( \parallel \varphi {\parallel }_{\infty } \leq 1 \) we have\n\n\[ \n{\begin{Vmatrix}{A}_{n}\varphi \end{Vmatrix}}_{\infty } \leq \mathop{\max }\limits_{{a \leq x \leq b}}\mathop{\sum }\limits_{{k = 0}}^{n}\left| {{a}_{k}K\left( {x,{x}_{k}}\right... | Yes |
Consider the integral equation\n\n\[ \varphi \left( x\right) - \frac{1}{2}{\int }_{0}^{1}\left( {x + 1}\right) {e}^{-{xy}}\varphi \left( y\right) {dy} = {e}^{-x} - \frac{1}{2} + \frac{1}{2}{e}^{-\left( {x + 1}\right) },\;0 \leq x \leq 1, \] | with exact solution \( \varphi \left( x\right) = {e}^{-x} \) . For its kernel we have\n\n\[ \mathop{\max }\limits_{{0 \leq x \leq 1}}{\int }_{0}^{1}\frac{1}{2}\left( {x + 1}\right) {e}^{-{xy}}{dy} = \mathop{\sup }\limits_{{0 < x \leq 1}}\frac{x + 1}{2x}\left( {1 - {e}^{-x}}\right) < 1. \]\n\nTherefore, by the Neumann s... | Yes |
Consider the integral equation\n\n\[ \n\\varphi \\left( t\\right) + \\frac{ab}{\\pi }{\\int }_{0}^{2\\pi }\\frac{\\varphi \\left( \\tau \\right) {d\\tau }}{{a}^{2} + {b}^{2} - \\left( {{a}^{2} - {b}^{2}}\\right) \\cos \\left( {t + \\tau }\\right) } = f\\left( t\\right) ,\\;0 \\leq t \\leq {2\\pi },\n\]\n\nwhere \( a \\... | Any solution \( \\varphi \) to the homogeneous form of equation (12.20) clearly must be a \( {2\\pi } \) -periodic analytic function, since the kernel is a \( {2\\pi } \) -periodic analytic function with respect to the variable \( t \) . Hence, we can expand \( \\varphi \) into a uniformly convergent Fourier series\n\n... | Yes |
Theorem 12.16 Let \( A : C\left\lbrack {a, b}\right\rbrack \rightarrow C\left\lbrack {a, b}\right\rbrack \) be a compact linear operator such that \( I - A \) is injective, and assume that the interpolation operators \( {L}_{n} : C\left\lbrack {a, b}\right\rbrack \rightarrow {X}_{n} \) satisfy \( {\begin{Vmatrix}{L}_{n... | Proof. From Theorem 12.6 applied to \( {A}_{n} = {L}_{n}A \), we conclude that for all sufficiently large \( n \) the inverse operators \( {\left( I - {L}_{n}A\right) }^{-1} \) exist and are uniformly bounded. To verify the error bound, we apply the interpolation operator \( {L}_{n} \) to (12.22) and get\n\n\[ \varphi ... | Yes |
Corollary 12.17 Let \( A : C\left\lbrack {a, b}\right\rbrack \rightarrow C\left\lbrack {a, b}\right\rbrack \) be a compact linear operator such that \( I - A \) is injective, and assume that the interpolation operators \( {L}_{n} : C\left\lbrack {a, b}\right\rbrack \rightarrow {X}_{n} \) are pointwise convergent; i.e.,... | Proof. By Lemma 12.9 the pointwise convergence of the interpolation operators \( {L}_{n} \) and the compactness of \( A \) imply that \( {\begin{Vmatrix}{L}_{n}A - A\end{Vmatrix}}_{\infty } \rightarrow 0, n \rightarrow \infty \) . Now the statement follows from the preceding theorem. | No |
Lemma 12.20 Let \( f \in {C}^{1}\left\lbrack {0,{2\pi }}\right\rbrack \) . Then for the remainder in trigonometric interpolation we have\n\n\[ \n{\begin{Vmatrix}{L}_{n}f - f\end{Vmatrix}}_{\infty } \leq {c}_{n}{\begin{Vmatrix}{f}^{\prime }\end{Vmatrix}}_{2} \n\]\n\n(12.34)\n\nwhere \( {c}_{n} \rightarrow 0, n \rightarr... | Proof. Consider the trigonometric monomials \( {f}_{m}\left( t\right) = {e}^{imt} \) and write \( m = \) \( \left( {{2k} + 1}\right) n + q \) with \( k \in \mathbf{Z} \) and \( 0 \leq q < {2n} \) . Since \( {f}_{m}\left( {t}_{j}\right) = {f}_{q - n}\left( {t}_{j}\right) \) for \( j = 0,\ldots ,{2n} - 1 \), the trigonom... | Yes |
Lemma 12.20 Let \( f \in {C}^{1}\left\lbrack {0,{2\pi }}\right\rbrack \) . Then for the remainder in trigonometric interpolation we have\n\n\[ \n{\begin{Vmatrix}{L}_{n}f - f\end{Vmatrix}}_{\infty } \leq {c}_{n}{\begin{Vmatrix}{f}^{\prime }\end{Vmatrix}}_{2} \n\]\n\nwhere \( {c}_{n} \rightarrow 0, n \rightarrow \infty \... | Proof. Consider the trigonometric monomials \( {f}_{m}\left( t\right) = {e}^{imt} \) and write \( m = \) \( \left( {{2k} + 1}\right) n + q \) with \( k \in \mathbf{Z} \) and \( 0 \leq q < {2n} \) . Since \( {f}_{m}\left( {t}_{j}\right) = {f}_{q - n}\left( {t}_{j}\right) \) for \( j = 0,\ldots ,{2n} - 1 \), the trigonom... | Yes |
Theorem 12.21 The collocation method with trigonometric polynomials converges for integral equations of the second kind with continuously differentiable periodic kernels and right-hand sides. | One possibility for the implementation of the collocation method is to use the trigonometric monomials as basis functions. Then the integrals \( {\int }_{0}^{2\pi }K\left( {{t}_{j},\tau }\right) {e}^{ik\tau }{d\tau } \) have to be integrated numerically. Replacing the kernel by its trigonometric interpolation leads to ... | Yes |
For the integral equation (12.20) from Example 12.15, Table 12.5 gives the error between the exact solution and the collocation approximation. | TABLE 12.5. Collocation method for equation (12.20)\n\n<table><thead><tr><th></th><th>\\( n \\)</th><th>\\( t = 0 \\)</th><th>\\( t = \\pi /2 \\)</th><th>\\( t = \\pi \\)</th></tr></thead><tr><td></td><td>4</td><td>-0.10752855</td><td>-0.03243176</td><td>0.03961310</td></tr><tr><td>\\( a = 1 \\)</td><td>8</td><td>-0.00... | Yes |
Theorem 12.23 For the Nyström method the condition numbers for the linear system are uniformly bounded. | This theorem states that the Nyström method essentially preserves the stability of the original integral equation. | No |
Theorem 12.25 Let \( X \) and \( Y \) be normed spaces and let \( A : X \rightarrow Y \) be a compact linear operator. Then \( A \) has a bounded inverse if and only if \( X \) is finite-dimensional. | Proof. Assume that \( A \) has a bounded inverse \( {A}^{-1} : Y \rightarrow X \) . Then we have \( {A}^{-1}A = I \), and therefore the identity operator must be compact, since the product of a bounded and a compact operator is compact (see Problem 12.2). However, the identity operator on \( X \) is compact if and only... | Yes |
\[ {y}^{\prime } = - {2y} \] | Here \( D = {\mathbb{R}}^{2} \) . Using the procedure in (5) one obtains\n\n\[ \frac{dy}{y} = - {2dx} \Leftrightarrow \ln \left| y\right| = - {2x} + C \Leftrightarrow \left| y\right| = {\mathrm{e}}^{C - {2x}}.\]\n\nThe general solution (with \( \pm {\mathrm{e}}^{C} \) replaced with \( C \) ) is\n\n\[ y\left( {x;C}\righ... | Yes |
\[ {y}^{\prime } = \sqrt{\left| y\right| } \] | Again \( D = {\mathbb{R}}^{2} \) . Since the direction field is symmetric, it follows that if \( y\left( x\right) \) is a solution, then \( z\left( x\right) = - y\left( {-x}\right) \) is also a solution. Indeed, we have\n\n\[ {z}^{\prime }\left( x\right) = {y}^{\prime }\left( {-x}\right) = \sqrt{\left| y\left( -x\right... | Yes |
\[ {y}^{\prime } = - x\left( {\operatorname{sgn}y}\right) \sqrt{\left| y\right| } = \left\{ \begin{array}{lll} - x\sqrt{y} & \text{ for } & y \geq 0, \\ x\sqrt{-y} & \text{ for } & y < 0. \end{array}\right. \] | The direction field is symmetric to the \( x \) -axis; i.e., if \( y\left( x\right) \) is a solution, then so is \( - y\left( x\right) \) . Thus it is sufficient to calculate the positive solutions. From \[ \int \frac{dy}{\sqrt{y}} = - 2\sqrt{y} = - \int {xdx} = \frac{1}{2}\left( {C - {x}^{2}}\right) \] it follows that... | Yes |
\[ {y}^{\prime } = {\mathrm{e}}^{y}\sin x. \] | The direction field is symmetric with respect to the \( y \) -axis and periodic in \( x \) of period \( {2\pi } \), i.e., if \( y\left( x\right) \) is a solution, then so are \( u\left( x\right) = y\left( {-x}\right) \) and \( v\left( x\right) = y\left( {x + {2k\pi }}\right) \) . By separation of variables (7) one obta... | Yes |
Theorem 1. Let \( \mathop{\lim }\limits_{{t \rightarrow \infty }}B\left( t\right) = \infty \) . If \( u \) is a positive solution, then\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}u\left( t\right) = \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{b\left( t\right) }{c\left( t\right) } \]\n\nprovided that t... | Proof. This theorem is a substantial generalization of 1.XIII.(a). It can be proved by writing \( y \) as the quotient \( Z\left( t\right) /N\left( t\right) \) with \( N\left( t\right) = {\mathrm{e}}^{B\left( t\right) } \) . The result then follows using l’Hospital’s rule; since both \( B\left( t\right) \) and \( N\lef... | Yes |
Theorem 2. If the coefficients \( b \) and \( c \) are \( T \) -periodic, then there exists exactly one positive \( T \) -periodic solution of (14). | Proof. It is sufficient to show that there is exactly one solution with \( u\left( 0\right) = \) \( u\left( T\right) > 0 \) . Under this assumption \( v\left( t\right) \mathrel{\text{:=}} u\left( {t + T}\right) \) is a solution of (14) with \( v\left( 0\right) = u\left( 0\right) \) . Then \( y = 1/u \) and \( z = 1/v \... | Yes |
Theorem 3. Let the coefficients \( b, c \) be positively bounded. Then equation (13) has exactly one positively bounded solution \( {u}^{ * } \) on \( \mathbb{R} \) ; and if \( u \) is any positive solution, then \( u\left( t\right) - {u}^{ * }\left( t\right) \rightarrow 0 \) as \( t \rightarrow \infty \) . | Proof. Let \( \alpha ,\beta ,\gamma ,\delta \) be positive constants with \( \alpha < b < \beta ,\gamma < c/b < \delta \) in \( \mathbb{R} \) . The first set of these inequalities leads to the estimates\n\n\[ \n{\alpha t} < B\left( t\right) < {\beta t}\text{for}t > 0,{\alpha t} > B\left( t\right) > {\beta t}\text{for}t... | Yes |
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