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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 |
Example 10.2 By Newton's law, the differential equation of the second order\n\n\[ \nm{u}^{\prime \prime } = f\left( {t, u}\right) \n\]\n\ndescribes the motion of an object of mass \( m \) subject to the external force \( f\left( {t, u}\right) \) depending on the location \( u \) of the object and the time \( t \) . Giv... | Null | No |
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 |
Consider the initial value problem\n\n\\[ \n{u}^{\\prime } = {x}^{2} + {u}^{2},\\;u\\left( 0\\right) = 0, \n\\]\n\nfrom Example 10.7. Table 10.1 gives the difference between the exact solution as computed by the Picard-Lindelöf iterations in Example 10.7 and the approximate solution obtained by Euler's method for vario... | Null | No |
Example 10.13 Consider again the initial value problem from Example 10.7. Table 10.2 gives the difference between the exact solution as computed by the Picard-Lindelöf iterations and the approximate solution obtained by the improved Euler method for various step sizes \( h \) . We observe quadratic convergence as \( h ... | Null | No |
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) }\... | Yes |
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. | Yes |
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) +... | No |
Theorem 11.10 Assume that the solution to the boundary value problem (11.17)-(11.18) is four-times continuously differentiable. Then the error of the finite difference approximation can be estimated by\n\n\\[ \n\\left| {u\\left( {x}_{ij}\\right) - {u}_{ij}\\right| \\leq \\frac{{h}^{2}}{96}\\left\\lbrack {{\\begin{Vmatr... | Null | No |
Theorem 11.11 (Riesz) Let \( X \) be a Hilbert space. Then for each bounded linear function \( F : X \rightarrow \mathbb{C} \) there exists a unique element \( f \in X \) such that\n\n\[ F\\left( u\\right) = \\left( {u, f}\\right) \]\n\nfor all \( u \in X \) . The norms of the element \( f \) and the linear function \(... | Proof. Uniqueness follows from the observation that because of the positive definiteness of the scalar product, \( f = 0 \) is the only element representing the zero function \( F = 0 \) in the sense of (11.22). For \( F \neq 0 \) choose \( w \in X \) with \( F\\left( w\\right) \neq 0 \) . Since \( F \) is continuous, ... | 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 12.2 Let \( A : X \rightarrow X \) be a compact operator in a normed space \( X \) . Then \( I - A \) is surjective if and only if it is injective. If the inverse operator \( {\left( I - A\right) }^{-1} : X \rightarrow X \) exists, it is bounded. | Null | No |
Theorem 12.3 (Arzelà-Ascoli) Each sequence from a subset \( U \subset C\left\lbrack {a, b}\right\rbrack \) contains a uniformly convergent subsequence; i.e., \( U \) is relatively sequentially compact, if and only if it is bounded and equicontinuous, i.e., if there exists a constant \( C \) such that\n\n\[ \left| {\var... | Null | No |
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. | 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\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 ... | No |
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 12.24 Under the assumptions of Theorem 12.16, for the collocation method the condition number of the linear system satisfies\n\n\[ \n\\operatorname{cond}\\left( {{E}_{n} - {\\widetilde{A}}_{n}}\\right) \\leq C{\\begin{Vmatrix}{L}_{n}\\end{Vmatrix}}_{\\infty }^{2}\\operatorname{cond}{E}_{n} \n\]\n\nfor all suffi... | Null | 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 |
Example 1. \( {y}^{\prime } = x - 1/y\;\left( {y > 0}\right) \) . | In the first example, the special solution \( \phi \) is the only bounded global solution. The solutions above \( \phi \) tend to \( \infty \) as \( x \rightarrow \infty \), while every positive solution beneath \( \phi \) exists only in a finite interval \( \lbrack 0, b) \) and tends to 0 as \( x \rightarrow b - \) . | Yes |
Example 2. \( {y}^{\prime } = {x}^{3} + {y}^{3} \) . | Null | No |
The inequality \( {Pv} \geq 0 \) holds for \( v = {\mathrm{e}}^{-x} \), and the inequality \( {Pw} \leq 0 \) is satisfied by the function\n\n\[ w\left( x\right) = \left\{ \begin{array}{lll} 2 - x & \text{ for } & 0 \leq x \leq 1 \\ 1/x & \text{ for } & x > 1 \end{array}\right. \]\n\nThus there exists a global solution ... | Null | No |
One can chose \( v = - \left( {x + 1}\right), w = - x \), as can be easily seen. Thus there exists a global solution \( \phi \) that satisfies the inequality \( - \left( {x + 1}\right) < \) \( \phi \left( x\right) < - x \) . The reader should show that \( {v}_{1} = - x - 1/3{x}^{2} \) is a better lower bound. | Null | No |
Example 1. Here the global solutions \( \phi ,\psi \) are bounded; that is, there exists \( L > 0 \) such that \( 0 < \phi < \psi < L \) holds in \( \lbrack 0,\infty ) \) and hence \( {f}_{y}\left( {x, y}\right) = 1/{y}^{2} > \) \( 1/{L}^{2} \mathrel{\text{:=}} \alpha \) . | \[ {u}^{\prime } \geq {\alpha u},\;\text{ which implies that }\;u\left( x\right) \geq u\left( 0\right) {\mathrm{e}}^{\alpha x}. \] But \( u \) is bounded by assumption. This contradiction proves the assertion made at the beginning that there is only one bounded global solution. The estimate \( {v}_{1} < \phi < 1/x \) (... | Yes |
Let \( y \) be a solution and \( y\left( a\right) \geq - a \) for some \( a \geq 0 \). It is easy to see from the differential equation that there exists \( b > a \) with \( y\left( b\right) > 0 \). Since the solution of the initial value problem \( {v}^{\prime } = {v}^{3}, v\left( b\right) = y\left( b\right) \) is a l... | If \( \phi \) and \( \psi \) are global solutions with \( \phi < \psi \), then accordingly, \( \psi \left( x\right) < - x \). Thus in (14) we have \( {f}_{y}\left( {x,{y}^{ * }}\right) = 3{y}^{*2} > 3{x}^{2} \), and hence \( u = \psi - \phi \geq \delta \exp \left( {x}^{3}\right) \), where \( \delta = u\left( 0\right) >... | Yes |
Corollary 1. If the sets \( A \) and \( B \) are homeomorphic and if \( A \) has the fixed point property, then \( B \) also has the fixed point property. | The proof is very simple. Let \( h : A \rightarrow B \) be a homeomorphism and \( f \) : \( B \rightarrow B \) a continuous mapping. Then \( F = {h}^{-1} \circ f \circ h \) is a continuous mapping of \( A \) to itself. If \( x \) is a fixed point of \( F \), then the image point \( \xi = h\left( x\right) \) is a fixed ... | Yes |
Corollary 2. Let the set \( A \subset {\mathbb{R}}^{n} \) be compact, and let there exist a continuous mapping \( P : {\mathbb{R}}^{n} \rightarrow A \) with \( {\left. P\right| }_{A} = {\operatorname{id}}_{A} \), i.e., \( P\left( x\right) = x \) for \( x \in A \) . Then \( A \) has the fixed point property. | For the proof let \( B \supset A \) be a closed ball and \( f : A \rightarrow A \) continuous. Then \( F = f \circ P \) is a continuous mapping of \( B \) into itself. By the Brouwer fixed point theorem, \( F \) has a fixed point \( \xi \), and because \( F\left( B\right) \subset A \), this fixed point belongs to \( A ... | Yes |
Corollary 3. A nonempty, convex, and compact set \( A \subset {\mathbb{R}}^{n} \) has the fixed point property. | Proof. For every \( x \in {\mathbb{R}}^{n} \) there exists, since \( A \) is convex and compact, exactly one \ | No |
Theorem 13. Given a separable Banach space \( E \) such that every sequence of elements of \( E \) that is bounded in norm has a subsequence weakly convergent to an element of \( E \), the space \( E \) is isometrically isomorphic to the space \( {E}^{* * } \) (the dual of \( {E}^{ * } \) ). | Though some of these terms have yet to be defined here, their meanings are not important at the moment. An examination of Banach's proof of this theorem shows that the isometric isomorphism he had in mind is the natural map from \( E \) into \( {E}^{* * } \), so the conclusion of Banach’s theorem is that \( E \) must b... | No |
Theorem 1 The edge set of a graph can be partitioned into cycles if, and only if, every vertex has even degree. | Proof. The condition is clearly necessary, since if a graph is the union of some edge disjoint cycles and isolated vertices, then a vertex contained in \( k \) cycles has degree \( {2k} \) .\n\nSuppose that every vertex of a graph \( G \) has even degree and \( e\left( G\right) > 0 \) . How can we find a single cycle i... | Yes |
Theorem 2 Every graph of order \( n \) and size greater than \( \left\lfloor {{n}^{2}/4}\right\rfloor \) contains a triangle. | Proof. Let \( G \) be a triangle-free graph of order \( n \) . Then \( \Gamma \left( x\right) \cap \Gamma \left( y\right) = \varnothing \) for every edge \( {xy} \in E\left( G\right) \), so\n\n\[ d\left( x\right) + d\left( y\right) \leq n. \]\n\nSumming these inequalities for all \( e\left( G\right) \) edges \( {xy} \)... | Yes |
Theorem 3 Let \( x \) be a vertex of a graph \( G \) and let \( W \) be the vertex set of a component containing \( x \) . Then the following assertions hold.\ni. \( W = \{ y \in G : G \) contains an \( x - y \) path \( \} \.\nii. \( W = \{ y \in G : G \) contains an \( x - y \) trail \( \} \.\niii. \( W = \{ y \in G :... | Null | No |
Theorem 4 A graph is bipartite iff it does not contain an odd cycle. | Proof. Suppose \( G \) is bipartite with vertex classes \( {V}_{1} \) and \( {V}_{2} \) . Let \( {x}_{1}{x}_{2}\cdots {x}_{l} \) be a cycle in \( G \) . We may assume that \( {x}_{1} \in {V}_{1} \) . Then \( {x}_{2} \in {V}_{2},{x}_{3} \in {V}_{1} \), and so on: \( {x}_{i} \in {V}_{1} \) iff \( i \) is odd. Since \( {x... | Yes |
Theorem 5 A graph is a forest iff for every pair \( \{ x, y\} \) of distinct vertices it contains at most one \( x - y \) path. | Proof. If \( {x}_{1}{x}_{2}\cdots {x}_{l} \) is a cycle in a graph \( G \), then \( {x}_{1}{x}_{2}\cdots {x}_{l} \) and \( {x}_{1}{x}_{l} \) are two \( {x}_{1} - {x}_{l} \) paths in \( G \) .\n\nConversely, let \( {P}_{1} = {x}_{0}{x}_{1}\cdots {x}_{l} \) and \( {P}_{2} = {x}_{0}{y}_{1}{y}_{2}\cdots {y}_{k}{x}_{l} \) b... | Yes |
Theorem 6 The following assertions are equivalent for a graph \( G \) .\ni. \( G \) is a tree.\nii. \( G \) is a minimal connected graph, that is, \( G \) is connected and if \( {xy} \in E\left( G\right) \), then \( G - {xy} \) is disconnected. [In other words, \( G \) is connected and every edge is a bridge.]\niii. \(... | Proof. Suppose \( G \) is a tree. For an edge \( {xy} \in E\left( G\right) \), the graph \( G - {xy} \) cannot contain an \( x - y \) path \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \), since otherwise \( G \) contains the cycle \( x{z}_{1}{z}_{2}\cdots {z}_{k}y \) . Hence \( G - {xy} \) is disconnected; and so \( G \) is a min... | Yes |
Corollary 7 Every connected graph contains a spanning tree, that is, a tree containing every vertex of the graph. | ## Proof. Take a minimal connected spanning subgraph. | No |
Corollary 8 A tree of order \( n \) has size \( n - 1 \) ; a forest of order \( n \) with \( k \) components has size \( n - k \) . | Null | No |
Corollary 9 A tree of order at least 2 contains at least 2 vertices of degree 1. | Proof. Let \( {d}_{1} \leq {d}_{2} \leq \cdots \leq {d}_{n} \) be the degree sequence of a tree \( T \) of order \( n \geq 2 \) . Since \( T \) is connected, \( \delta \left( T\right) = {d}_{1} \geq 1 \) . Hence if \( T \) had at most one vertex of degree 1, by (1) and Corollary 8 we would have\n\n\[ \n{2n} - 2 = {2e}\... | Yes |
Theorem 10 Each of the four methods described above produces an economical spanning tree. If no two edges have the same cost, then there is a unique economical spanning tree. | Proof. Choose an economical spanning tree \( T \) of \( G \) that has as many edges in common with \( {T}_{1} \) as possible, where \( {T}_{1} \) is a spanning tree constructed by the first method.\n\nSuppose that \( E\left( {T}_{1}\right) \neq E\left( T\right) \) . The edges of \( {T}_{1} \) have been selected one by ... | Yes |
Theorem 11 For \( n \geq 3 \) the complete graph \( {K}_{n} \) is decomposable into edge disjoint Hamilton cycles iff \( n \) is odd. For \( n \geq 2 \) the complete graph \( {K}_{n} \) is decomposable into edge-disjoint Hamilton paths iff \( n \) is even. | Null | No |
Theorem 12 A non-trivial connected graph has an Euler circuit iff each vertex has even degree. | Proof. The conditions are clearly necessary. For example, if \( G \) has an Euler circuit \( {x}_{1}{x}_{2}\cdots {x}_{m} \), and \( x \) occurs \( k \) times in the sequence \( {x}_{1},{x}_{2},\ldots ,{x}_{m} \), then \( d\left( x\right) = {2k} \) .\n\nWe prove the sufficiency of the first condition by induction on th... | Yes |
Theorem 13 Let \( G \) be a directed multigraph with vertex set \( V\left( G\right) = \{ {v}_{1},\ldots ,{v}_{n}\} \) , such that \( {d}^{ + }\left( {v}_{i}\right) = {d}^{ - }\left( {v}_{i}\right) \) for every \( i \) . Denote by \( s\left( G\right) \) the number of Euler circuits of \( G \), and by \( {t}_{i}\left( G\... | Null | No |
Theorem 14 Let \( G = \left( {V, E}\right) \) be a connected multigraph with \( E \) infinite. Then \( G \) has a two-way infinite Euler trail if and only if the following conditions are satisfied:\n\n(i) \( E \) is countable,\n\n(ii) every degree is even or infinite,\n\n(iii) for every subgraph \( {G}^{\prime } \subse... | Null | No |
Theorem 15 If a connected plane graph \( G \) has \( n \) vertices, \( m \) edges, and \( f \) faces, then\n\n\[ n - m + f = 2\text{.} \] | Proof. Let us apply induction on the number of faces. If \( f = 1 \), then \( G \) does not contain a cycle, so it is a tree, and the result holds by Corollary 8.\n\nSuppose now that \( f > 1 \) and the result holds for smaller values of \( f \) . Let \( {ab} \) be an edge in a cycle of \( G \) . Since a cycle separate... | Yes |
Theorem 16 A planar graph of order \( n \geq 3 \) has at most \( {3n} - 6 \) edges. Furthermore, a planar graph of order \( n \) and girth at least \( g,3 \leq g < \infty \), has size at most\n\n\[ \max \left\{ {\frac{g}{g - 2}\left( {n - 2}\right), n - 1}\right\} . \] | Proof. The first assertion is the case \( g = 3 \) of the second, so it suffices to prove the second assertion. Let \( G \) be a planar graph of order \( n \), size \( m \), and girth at least \( g \) . If \( n \leq g - 1 \), then \( G \) is acyclic, so \( m \leq n - 1 \) . Assume now that \( n \geq g \) and the assert... | Yes |
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