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Theorem 3. Let \( f\left( x\right) \) and \( g\left( x\right) \) be functions defined for \( 0 \leq x \leq L \) and let \( {\widetilde{f}}_{0}\left( x\right) \) and \( {\widetilde{g}}_{0}\left( x\right) \) be the periodic extensions of the odd extensions of \( f\left( x\right) \) and \( g\left( x\right) \) . Assume tha... | Proof. By Theorem 1, with \( f\left( x\right) \) and \( g\left( x\right) \) replaced by \( {f}_{o}\left( x\right) \) and \( {\widetilde{g}}_{o}\left( x\right) \), we know that (22) solves the D.E. with I.C. \( \;u\left( {x,0}\right) = {\overset{\alpha }{f}}_{o}\left( x\right) \; \) and \( \;{u}_{t}\left( {x,0}\right) =... | Yes |
Theorem 4. If \( {f}_{o}\left( x\right) \) is \( {C}^{2} \) and \( {g}_{o}\left( x\right) \) is \( {C}^{1} \), then the solution of problem (21) is given by the following formula (which is equivalent to (22) by Theorem 1, Section 5.1)\n\n\[ u\left( {x, t}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\left\lbrack {... | Proof. By Property 3, \( u\left( {x, t}\right) \), given by (22), is a \( {C}^{2} \) periodic function in \( x \) for each fixed \( t \) , since \( {f}_{o}\left( x\right) \) is \( {C}^{2} \) and \( {\widetilde{g}}_{o}\left( x\right) \) is \( {C}^{1} \) . Also, \( u\left( {0, t}\right) = u\left( {L, t}\right) = 0 \) . T... | Yes |
Theorem 5 (A maximum magnitude principle). If \( {f}_{o}\left( x\right) \) is \( {C}^{2} \) and \( {\widetilde{g}}_{o}\left( x\right) \) is \( {C}^{1} \), then the solution \( u\left( {x, t}\right) \) of the problem\n\nD.E. \( \;{u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq L,\; - \infty < t < \infty ,\n\nB.C. \( \mathrm{... | Proof. We know from (22) in Theorem 3, that\n\n\[ u\left( {x, t}\right) = \frac{1}{2}\left\lbrack {{\widetilde{f}}_{o}\left( {x + {at}}\right) + {\widetilde{f}}_{o}\left( {x - {at}}\right) }\right\rbrack + \frac{1}{2a}{\int }_{x - {at}}^{x + {at}}{\widetilde{g}}_{o}\left( r\right) {dr}. \]\n\n(29)\n\n\n\nClearly, \( \l... | Yes |
Solve the problem in Example 1, by the method of images and D'Alembert's formula. | The method of images for problem (3) with free ends differs only in one respect from the corresponding treatment for the case of fixed ends (cf. (21) in Section 5.2). Indeed, all we do is replace the odd periodic extensions \( {\widetilde{f}}_{o}\left( x\right) \) and \( {\widetilde{g}}_{o}\left( x\right) \) by the eve... | Yes |
Example 3. Let \( u\left( {x, t}\right) \) solve the problem\n\nD.E. \( {u}_{tt} = {u}_{xx},\;0 \leq x \leq 2, - \infty < t < \infty \) ,\n\nB.C. \( u\left( {0, t}\right) = 0,{u}_{x}\left( {2, t}\right) = 0 \) ,\n\nI.C. \( \;u\left( {x,0}\right) = f\left( x\right) = \frac{1}{2}{x}^{3}{\left( 2 - x\right) }^{3},\;{u}_{t... | Solution. The graph of \( u\left( {x,0}\right) = \frac{1}{2}{x}^{3}{\left( 2 - x\right) }^{3} \) is shown in Figure 2.\n\nThe method of images dictates that we extend this initial profile to all \( \;x,\; - \infty < x < \infty ,\; \) in such a way that the extension is odd about \( \;x = 0\; \) (to insure \( \;u\left( ... | Yes |
Solve the problem\n\nD.E. \( {u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq \pi ,\; - \infty < t < \infty \) ,\n\nB.C. \( u\left( {0, t}\right) = - 1,\;{u}_{x}\left( {\pi, t}\right) = 2 \) ,\n\n(17)\n\nI.C. \( u\left( {x,0}\right) = \sin \left( {x/2}\right) + {2x} - 1,\;{u}_{t}\left( {x,0}\right) = - 2\sin \left( {{3x}/2}\... | Solution. As in Section 3.3, we choose a particular solution of the D.E. and B.C.. The simplest choice is the steady-state (time-independent) function \( {u}_{p}\left( {x, t}\right) = {2x} - 1 \) . The solution of (17) is then \( \;u\left( {x, t}\right) = {u}_{p}\left( {x, t}\right) + v\left( {x, t}\right) \), where \(... | Yes |
Solve D.E. \( {u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq L, - \infty < t < \infty \) ,\n\nB.C. \( {u}_{x}\left( {0, t}\right) = c,{u}_{x}\left( {L, t}\right) = d \) ,\n\n\[ \n\text{ I.C. }u\left( {x,0}\right) = f\left( x\right) ,{u}_{t}\left( {x,0}\right) = g\left( x\right) .\n\] | Solution. Note that the B.C. mean that the end \( x = 0 \) is subject to a downward force \( c/{T}_{0} \) and the end \( x = L \) is subject to an upward force \( d/{T}_{0} \) (cf. equations (1) and (2)). Thus, we expect the string to drift vertically if \( c \neq d \) . Indeed, there is no steady-state particular solu... | Yes |
Theorem 1 (Duhamel’s principle for the wave equation). Let \( h\left( {x, t}\right) \) be a \( {C}^{1} \) function, \( - \infty < \mathrm{x},\mathrm{t} < \infty \) . Then (27) is the unique solution of the problem\n\nD.E. \( {u}_{tt} - {a}^{2}{u}_{xx} = h\left( {x, t}\right) \; - \infty < x, t < \infty \)\n\nI.C. \( u\... | Proof. We know from (26) that \( \widetilde{\mathrm{v}}\left( {\mathrm{x},\mathrm{t};\mathrm{s}}\right) \) is \( {\mathrm{C}}^{2} \), since \( \mathrm{h}\left( {\mathrm{x},\mathrm{t}}\right) \) is assumed to be \( {\mathrm{C}}^{1} \) . We can then twice apply Lemma 1 of Section 3.4, once with \( \;g\left( {t, s}\right)... | Yes |
Solve D.E. \( {u}_{tt} - {u}_{xx} = x - t \) , \( \; - \infty < x, t < \infty \) , I.C. \( \;u\left( {x,0}\right) = {x}^{2},\;{u}_{t}\left( {x,0}\right) = \sin \left( x\right) . | Solution. We split the problem up into two familiar problems for functions \( {u}_{1}\left( {x, t}\right) \) and \( {u}_{2}\left( {x, t}\right) \) : D.E. \( \;{\left( {\mathrm{u}}_{1}\right) }_{\mathrm{{tt}}} - {\left( {\mathrm{u}}_{1}\right) }_{\mathrm{{XX}}} = 0 \) , D.E. \( {\left( {\mathrm{u}}_{2}\right) }_{\mathrm... | Yes |
Example 1. Suppose a uniform electrical charge density is applied to the z-axis. Find the most general form for the resulting harmonic electrostatic potential \( u\left( {x, y, z}\right) \) . | Solution. Since the physical situation is unchanged by translations in the z-direction, we deduce that \( u\left( {x, y, z}\right) \) does not depend on \( z \), say \( u\left( {x, y, z}\right) = u\left( {x, y}\right) \) . Thus, we seek appropriate solutions of (1). Note also that the physical situation is unchanged by... | Yes |
Show that, if the temperature \( u\left( {x, y, t}\right) \) in a flat, homogeneous heat conducting plate (without heat sources) obeys a second-order linear PDE, then this PDE must be of the form\n\n\[ \n{u}_{t} = k\left( {{u}_{xx} + {u}_{yy}}\right) , \n\]\n\nfor some constant \( \mathrm{k} > 0 \) . | Solution. The general second-order linear PDE for \( u = u\left( {x, y, t}\right) \) is\n\n\[ \n{q}_{1}{u}_{xx} + {q}_{2}{u}_{yy} + r{u}_{xy} + {r}_{1}{u}_{xt} + {r}_{2}{u}_{yt} + s{u}_{tt} + {a}_{1}{u}_{x} + {a}_{2}{u}_{y} + b{u}_{t} + {cu} = f.\n\]\n\nAll of the coefficients and \( f \) must be constants, because the... | Yes |
Reduce the Dirichlet problem for Poisson's equation,\n\n\[ \n\\text{D.E.}{u}_{xx} + {u}_{yy} = q\\left( {x, y}\\right) \\text{on}D \n\]\n\n(14)\n\n\[ \n\\text{B.C.}u\\left( {x, y}\\right) = g\\left( {x, y}\\right) \\text{for}\\left( {x, y}\\right) \\text{on}C\\text{,}\n\]\n\nto a related Dirichlet problem for Laplace's... | Solution. Let \( {u}_{p}\\left( {x, y}\\right) \) be a particular solution (e.g.,(13)) of the D.E., and let \( v \) be a solution of the related Dirichlet problem :\n\nD.E. \( {\\mathrm{v}}_{\\mathrm{{xx}}} + {\\mathrm{v}}_{\\mathrm{{yy}}} = 0 \) on \( \\mathrm{D} \n\n(15)\n\nB.C. \( v\\left( {x, y}\\right) = g\\left( ... | Yes |
Theorem 2. Let \( {a}_{n},{b}_{n},{c}_{n} \), and \( {d}_{n} \) be the Fourier sine coefficients (assumed to vanish for \( n > N) \) of \( f\left( x\right), g\left( x\right), h\left( y\right) \), and \( k\left( y\right) \) . Then the solution of the Dirichlet problem\n\nD.E. \( \;{u}_{xx} + {u}_{yy} = 0\;0 < x < L,\;0 ... | is\n\n\[ u\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 1}}^{N}\left\lbrack {{A}_{n}\sin \left( {{n\pi x}/L}\right) \sinh \left\lbrack {{n\pi }\left( {M - y}\right) /L}\right\rbrack }\right.\n\n\[ + {\mathrm{B}}_{\mathrm{n}}\sin \left( {\mathrm{n}\pi \mathrm{x}/\mathrm{L}}\right) \sinh \left( {\mathrm{n}\pi \mathr... | Yes |
Find the unique harmonic function of the form\n\n\[ \mathrm{U}\left( {\mathrm{x},\mathrm{y}}\right) = \mathrm{a} + \mathrm{b}\mathrm{x} + \mathrm{c}\mathrm{y} + \mathrm{d}\mathrm{x}\mathrm{y}, \]\n\nwhere \( a, b, c \) and \( d \) are constants, such that \( U\left( {0,0}\right) = A, \) \( U\left( {L,0}\right) = B, \) ... | Solution. Note that \( \;U\left( {0,0}\right) = A\; \) implies \( \;a = A.\; \) Then \( \;U\left( {L,0}\right) = B\; \) implies \( \;a + {bL} = B\; \) or \( b = \left( {B - a}\right) /L.\; \) Similarly, \( \;U\left( {0, M}\right) = C\; \) implies \( \;c = \left( {C - A}\right) /M.\; \) Also, \( \;U\left( {L, M}\right) ... | Yes |
D.E. \( \;{u}_{xx} + {u}_{yy} = 0\;0 < x < \pi ,\;0 < y < \pi \)\n\nB.C. \( \left\{ \begin{array}{lll} u\left( {x,0}\right) = 0, & u\left( {x,\pi }\right) = 5\sin \left( {2x}\right) - 7\sin \left( {8x}\right) & 0 \leq x \leq \pi \\ u\left( {0, y}\right) = \sin \left( y\right) , & u\left( {\pi, y}\right) = 0 & 0 \leq y ... | Solution. Method 1 (By inspection). Note that the three product solutions of the D.E. which are relevant to the B.C. are \( \sin \left( {2x}\right) \sinh \left( {2y}\right) ,\;\sin \left( {8x}\right) \sinh \left( {8y}\right) \; \) and \( \;\sin \left( y\right) \sinh \left( {\pi - x}\right) .\; \) By forming a superposi... | Yes |
Find the points where the harmonic function \( \;u\left( {x, y}\right) = \frac{{x}^{3} - {3x}}{{y}^{2} + {xy} - x}\; \) achieves its maximum and minimum values in the square region \( 0 \leq x \leq 1,0 \leq y \leq 1 \) . | Solution. To find the maximum and minimum for \( u, \) one would usually compute \( {u}_{x} \) and \( {u}_{y}, \) and solve the equations \( \;{u}_{x} = 0\; \) and \( \;{u}_{y} = 0\; \) simultaneously for \( \;x\; \) and \( \;y,\; \) in order to find the critical points in the interior of the square. Then one would hav... | Yes |
Find the formal solution of the problem\n\nD.E. \( \;{u}_{xx} + {u}_{yy} = 0\;0 < x < \pi ,\;0 < y < \pi \)\n\nB.C. \( \left\{ \begin{array}{lll} u\left( {x,0}\right) = {x}^{3}\left( {x - \pi }\right) , & u\left( {x,\pi }\right) = 0 & 0 \leq x \leq \pi \\ u\left( {0, y}\right) = 0, & u\left( {\pi, y}\right) = 0 & 0 \le... | Solution. We have found (cf. equation (4)) that the product solutions of the D.E. which satisfy the homogeneous B.C. are the multiples of \( {u}_{n}\left( {x, y}\right) = \sin \left( {nx}\right) \sinh \left\lbrack {n\left( {\pi - y}\right) }\right\rbrack, n = 1,2,3,\ldots \) .\n\nSince \( \;{x}^{3}\left( {x\; - \;\pi }... | Yes |
Show that problem (23) has no solution, unless the following compatibility condition holds\n\n\[ \n{\int }_{0}^{L}g\left( x\right) {dx} - {\int }_{0}^{L}f\left( x\right) {dx} + {\int }_{0}^{M}k\left( y\right) {dy} - {\int }_{0}^{M}h\left( y\right) {dy} \n\]\n\n(24)\n\ni.e., the integral of the outward unit normal compo... | Solution. If \( u\left( {x, y}\right) \) is a solution of (23), then\n\n\[ \n0 = {\int }_{0}^{M}{\int }_{0}^{L}\left( {{u}_{xx} + {u}_{yy}}\right) {dxdy} = {\int }_{0}^{M}{\int }_{0}^{L}{u}_{xx}{dxdy} + {\int }_{0}^{L}{\int }_{0}^{M}{u}_{yy}{dydx} \n\]\n\n\[ \n= {\int }_{0}^{M}\left( {{u}_{x}\left( {L, y}\right) - {u}_... | Yes |
Find the product solutions \( \mathrm{X}\left( \mathrm{x}\right) \mathrm{Y}\left( \mathrm{y}\right) \) of the D.E. and homogeneous B.C. of the problem. | Solution. Recall that any product solution of the D.E. must be of one of the forms (1),(2) or (3). We first show that there is no nonzero Case 2 product solution which meets the last two B.C. . In Case 2, we have \( {u}_{x}\left( {0, y}\right) = b\left( {{c}_{1} - {c}_{2}}\right) \left( {{d}_{1}\cos \left( {by}\right) ... | Yes |
Solve problem (25) when \( f\left( x\right) = {\sum }_{n = 1}^{N}{a}_{n}\cos \left( {{n\pi x}/L}\right) \), a finite cosine series without a constant term \( {}_{2}^{1}{\mathrm{a}}_{0} \). Why must the constant term be zero in order that a solution should exist? | Solution. Let \( u\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{A}_{n}{u}_{n}\left( {x, y}\right) \), where \( {u}_{n} \) is given by (26). Then, by the superposition principle, u satisfies the D.E. and the homogeneous B.C. of (25). Note that\n\n\[ {u}_{y}\left( {x,0}\right) = \mathop{\sum }\limits_{{n = 1... | Yes |
Proposition 1. With the above notation,\n\n\[ \n{u}_{xx} + {u}_{yy} = {U}_{rr} + {r}^{-1}{U}_{r} + {r}^{-2}{U}_{\theta \theta }\;\left( {r > 0}\right) .\n\] | Proof. By the chain rule, we have\n\n\[ \n{U}_{r} = {u}_{x}\;{x}_{r} + {u}_{y}\;{y}_{r} = {u}_{x}\cos \theta + {u}_{y}\sin \theta \;\text{ and }\;{U}_{\theta } = {u}_{x}\;{x}_{\theta } + {u}_{y}\;{y}_{\theta } = - {u}_{x}\;r\sin \theta + {u}_{y}\;r\cos \theta .\n\]\n\nHence,\n\n\[ \n{U}_{rr} = {u}_{xx}\;{co}{s}^{2}\the... | Yes |
Example 1 (A steady-state temperature problem for the annulus). Solve the problem\n\nD.E. \( {\mathrm{U}}_{\mathrm{{rr}}} + {\mathrm{r}}^{-1}{\mathrm{U}}_{\mathrm{r}} + {\mathrm{r}}^{-2}{\mathrm{U}}_{\theta \theta } = 0,\;1 < \mathrm{r} < 2 \)\n\nB.C. \( \left\{ \begin{array}{l} U\left( {1,0}\right) = 3 + 4\cos \left( ... | Solution. As we have seen, separation of variables and the superposition principle lead us to a solution of the form (9). We could then simply use the formulas (8) with \( {A}_{0} = 6,{A}_{2} = 4 \) , \( {\mathrm{C}}_{1} = 5 \) and all other \( {\mathrm{A}}_{\mathrm{n}},{\mathrm{B}}_{\mathrm{n}},{\mathrm{C}}_{\mathrm{n... | Yes |
Proposition 3. In the Dirichlet problem (11), if\n\n\[ f\left( \theta \right) = \frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}\cos \left( {n\theta }\right) + {b}_{n}\sin \left( {n\theta }\right) \]\n\nthen the solution of (11) is\n\n\[ U\left( {r,\theta }\right) = \frac{1}{2}{a}_{0} + \mathop{\sum }\li... | Proof. It follows from the above remark and the superposition principle, that (14) defines a harmonic function throughout the disk. Note that if we set \( r = {r}_{0} \) in the right side of (14), then the result is \( f\left( \theta \right) \), whence the B.C. of (11) is met. By the uniqueness theorem for the Dirichle... | Yes |
Solve the Dirichlet problem for the disk of radius 1:\n\nD.E. \( \;{\mathrm{u}}_{\mathrm{{xx}}} + {\mathrm{u}}_{\mathrm{{yy}}} = 0\;\mathrm{r} < 1\; \cdot \)\n\nB.C. \( \mathrm{U}\left( {1,0}\right) = - 1 + 8{\cos }^{2}\left( \theta \right) \)\n\nP.C. \( \mathrm{U}\left( {\mathrm{r},\theta + {2\pi }}\right) = \mathrm{U... | Solution. The Fourier series of \( - 1 + 8{\cos }^{2}\left( \theta \right) \) is \( 3 + 4\cos \left( {2\theta }\right) \), which of the same as the function of the first B.C. in Example 1. In spite of the fact that the solution of (10) in Example 1 satisfies the B.C., it is not the solution of (15), because the terms i... | Yes |
Show that for \( r < 1 \), we have\n\n\[\n\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\frac{\left( {1 - {r}^{2}}\right) \sin \left( t\right) }{1 - {2r}\cos \left( {\theta - t}\right) + {r}^{2}}{dt} = r\sin \theta .\n\] | Solution. Both sides of (29) are solutions of the Dirichlet problem for the disk with radius \( {\mathrm{r}}_{\mathrm{O}} = 1 \) with B.C. \( \mathrm{f}\left( \theta \right) = \sin \left( \theta \right) \) . By uniqueness for the solution of this problem, the two sides must be equal for \( r < 1 \) . Incidentally,(29) ... | No |
Theorem 2 (The Mean–Value Theorem). Let \( u \) be a harmonic function on some open region R. Then the value of \( u \) at the center of any closed disk \( D \) contained in \( R \) is the average (or mean) of the values of \( u \) on the circular boundary of \( D \) . | Proof. By introducing polar coordinates with pole at the center of the disk, we may assume that \( \mathrm{D} \) is the disk \( \mathrm{r} \leq {\mathrm{r}}_{\mathrm{o}} \) . On the boundary of \( \mathrm{D} \), u is \( \mathrm{U}\left( {{\mathrm{r}}_{\mathrm{o}},\theta }\right) \) . Thus, the Poisson Integral Formula ... | Yes |
Theorem 3 (Regularity of harmonic functions). If \( u \) is harmonic on an open region \( R, \) then \( u \) is \( {C}^{\infty } \) on \( R \) . | Proof. Let \( p \) be any point in \( \mathbb{R} \) and choose polar coordinates \( \left( {r,\theta }\right) \) with \( p \) as the pole. Suppose that \( {r}_{0} \) is chosen small enough so that the disk \( r \leq {r}_{0} \) is contained in \( R \), and let \( \mathrm{x} = \mathrm{r}\cos \theta \) and \( \mathrm{y} =... | Yes |
Theorem 4 (Infinite series solutions). Let \( f\left( \theta \right) \) be a continuous periodic function of period \( {2\pi } \) , with \( \mathrm{{FS}}f\left( \theta \right) = \frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}\cos \left( {n\theta }\right) + {b}_{n}\sin \left( {n\theta }\right) \) .... | Proof. By the uniqueness theorem (cf. Theorem 1 of Section 6.2) and Theorem 1, the solution \( U\left( {r,\theta }\right) \) of problem (32) is given by (20) for \( r < {r}_{0} \) . Thus, we need only to show that the right sides of \( \left( {20}\right) \) and \( \left( {33}\right) \) are equal. It follows from Theore... | Yes |
Theorem 1 (The Maximum/Minimum Principle). Let \( u = u\left( {x, y}\right) \) be a continuous function on \( \overline{\mathrm{D}} \), for some open, bounded set \( \mathrm{D} \) . If \( \mathrm{u} \) is harmonic on \( \mathrm{D} \), then the maximum and the minimum values of \( u \) are achieved on the boundary of \(... | Proof. We know (cf. Appendix A.4) that the maximum of \( u \) is achieved at some point in \( \overline{D} \) , since \( u \) is continuous on the closed, bounded set \( \overline{D} \) . We prove the theorem by contradiction. Suppose that the maximum is not achieved on the boundary. Then the maximum is achieved at som... | Yes |
Theorem 2 (The uniqueness theorem for the Dirichlet problem). For some open, bounded set \( \mathrm{D} \), let \( {\mathrm{u}}_{1} \) and \( {\mathrm{u}}_{2} \) be continuous functions on \( \overline{\mathrm{D}} \) which are harmonic on \( \mathrm{D} \). If \( {\mathrm{u}}_{1} \) and \( {u}_{2} \) are equal at all bou... | Proof. The difference \( v \equiv {u}_{1} - {u}_{2} \) is a continuous function on \( \bar{D} \) which is harmonic on \( D \). Since \( v \equiv 0 \) on the boundary of \( D, \) the Maximum/Minimum Principle implies that \( v \leq 0 \) and \( v \geq 0 \) on \( \overline{\mathrm{D}} \). Hence, \( \mathrm{v} \equiv 0 \) ... | Yes |
Theorem 3 (Continuous dependence of solutions on boundary data). Let \( {u}_{i} \) (for \( i = 1,2 \) ) be the solution (if it exists) of the Dirichlet problem\n\nD.E. \( \Delta {\mathrm{u}}_{\mathrm{i}} = 0\; \) on \( \mathrm{D} \)\n\nB.C. \( {u}_{i} = {f}_{i}\; \) on \( \;C \) ,\n\nwhere \( \mathrm{C} \) is the bound... | Proof. Let \( \mathrm{v} \equiv {\mathrm{u}}_{1} - {\mathrm{u}}_{2} \) . For all \( \left( {\mathrm{x},\mathrm{y}}\right) \) in \( \overline{\mathrm{D}} \), we have\n\n\[- \mathop{\max }\limits_{\mathrm{C}}\left( \left| {{\mathrm{f}}_{1} - {\mathrm{f}}_{2}}\right| \right) \leq \mathop{\min }\limits_{\mathrm{C}}\left( {... | Yes |
Let \( u\left( {x, y}\right) \) be any nonconstant harmonic function on the entire \( {xy} \) -plane. By Problem 12(d) of Section 6.3, we know that the zero level set of \( u,\{ \left( {x, y}\right) \mid u\left( {x, y}\right) = 0\} \), is nonempty. Show that this level set cannot contain any circle. | Solution. If this were possible, the harmonic function \( u\left( {x, y}\right) \; \) would yield a solution of the Dirichlet problem for the disk \( \mathrm{D} \) enclosed by the circle, where the boundary data is zero. By the uniqueness theorem, \( u \) would have to be \( 0 \) throughout the disk \( D. \) ’We show t... | Yes |
Example 3. Suppose that \( u\left( {x, y}\right) \) is a continuous function on the closed disk \( r \leq 1, \) and assume that \( \bar{u} \) is harmonic on the open disk \( r < 1 \) . If \( u\left( {\cos \theta ,\sin \theta }\right) \leq \sin \theta + \cos \left( {2\theta }\right) \), then show that we have \( u\left(... | Solution. Note that \( \;v\left( {x, y}\right) \; \equiv \;y\; + \;{x}^{2}\; - \;{y}^{2}\; \) is a harmonic function with \( \;v\left( {\cos \theta \;,\;\sin \theta }\right) \) \( = \sin \theta + \cos \left( {2\theta }\right) \) . By assumption, \( \mathrm{u} \leq \mathrm{v} \) on the boundary of the disk \( \mathrm{r}... | Yes |
For any \( {C}^{2} \) function \( u\left( {x, y}\right) \) defined on the disk \( D\left( {r \leq R}\right) \), we have the formula\n\n\[{\iint }_{D}\left( {{u}_{xx} + {u}_{yy}}\right) {dxdy} = {\int }_{0}^{2\pi }{U}_{r}\left( {R,\theta }\right) {Rd\theta }.\]\n\nIn other words, the integral of the Laplacian of \( u \)... | Proof. Computing the left side of (4) in terms of polar coordinates,\n\n\[{\int }_{0}^{2\pi }{\int }_{0}^{R}\left( {{U}_{rr} + {r}^{-1}{U}_{r} + {r}^{-2}{U}_{\theta \theta }}\right) r\operatorname{drd}\theta = {\int }_{0}^{2\pi }{\int }_{0}^{R}\left( {r{U}_{rr} + {U}_{r}}\right) {drd\theta } + {\int }_{0}^{2\pi }{\int ... | Yes |
Let \( u\left( {x, y}\right) \) be harmonic in the disk \( {x}^{2} + {y}^{2} < {r}_{0}{}^{2} \) . If \( u \) achieves its maximum at the point \( \left( {0,0}\right) \), then show that \( u \) must be constant throughout this disk. | Solution. By the mean-value theorem, we have (for any \( r < {r}_{0} \) )\n\n\[ \n{\int }_{0}^{2\pi }u\left( {0,0}\right) \mathrm{d}\theta = {2\pi u}\left( {0,0}\right) = {\int }_{0}^{2\pi }U\left( {R,\theta }\right) \mathrm{d}\theta .\n\]\n\nSubtracting, we obtain \( {\int }_{0}^{2\pi }\left\lbrack {u\left( {0,0}\righ... | Yes |
Theorem 4 (The Strong Maximum/Minimum Principle). Let \( u \) be a harmonic function on the open connected set D. Suppose that the maximum or minimum of \( u \) is achieved at some point in D. Then \( u \) must be constant throughout D. | Proof. Let \( p \) be the point in \( D \) where \( u \) achieves its maximum, say \( M, \) and let \( q \) be any other point in \( D. \) Since \( D \) is connected, we can join \( p \) to \( \mathring{q} \) by a curve, say with parametrization \( \left( {x\left( t\right), y\left( t\right) }\right) \), where \( x\left... | Yes |
Proposition 1 (The Cauchy–Riemann Equations). If \( f\left( {x + {iy}}\right) = u\left( {x, y}\right) + {iv}\left( {x, y}\right) \; \) is analytic on an open set \( D \), then the real and imaginary parts ( \( u \) and \( v \), respectively) of \( f \) obey the Cauchy-Riemann equations in \( \mathrm{D} \n\n\[ \n{u}_{x}... | Proof. If we take \( h \) to be real in (1), then we get (for \( z \) in D)\n\n\[ \n{f}^{\prime }\left( z\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{u\left( {x + h, y}\right) + {iv}\left( {x + h, y}\right) - \left\lbrack {u\left( {x, y}\right) + {iv}\left( {x, y}\right) }\right\rbrack }{h} = {u}_{x} + i{v}... | Yes |
Proposition 2. Let \( P\left( {x, y}\right) \) and \( Q\left( {x, y}\right) \) be \( {C}^{1} \) functions on an open rectangular region \( R \) (possibly with one or more sides of infinite length, so that R may be a strip). Then there is a \( {\mathrm{C}}^{2} \) function \( \mathrm{f}\left( {\mathrm{x},\mathrm{y}}\righ... | Proof. If \( \mathrm{f} \) satisfies (3), then \( {\mathrm{P}}_{\mathrm{y}} = {\mathrm{f}}_{\mathrm{{xy}}} = {\mathrm{f}}_{\mathrm{{yx}}} = {\mathrm{Q}}_{\mathrm{x}} \) . Conversely, we assume that \( {\mathrm{P}}_{\mathrm{y}} \equiv {\mathrm{Q}}_{\mathrm{x}} \) on \( \mathrm{R} \), and construct \( \mathrm{f} \) satis... | Yes |
Proposition 3. Any harmonic function \( u, \) defined on an open rectangular region \( R, \) has a harmonic conjugate \( v \) defined on \( R \) . | Proof. By the remarks before Proposition 2, we need only to show that for the given harmonic function \( u \), we can solve the Cauchy-Riemann equations \( {v}_{x} = - {u}_{y} \) and \( {v}_{y} = {u}_{x} \) . We apply Proposition 2 with \( P = - {u}_{y} \) and \( Q = {u}_{x} \) . Since \( u \) is harmonic, we have the ... | Yes |
Find a harmonic conjugate of the harmonic function \( u\left( {x, y}\right) = \sin \left( x\right) \cosh \left( y\right) + y \) , defined on the whole plane. | Solution. Proceeding as in the proof of Proposition 2, we integrate the equation \( {v}_{y} = {u}_{x} = {cos}\left( x\right) {cosh}\left( y\right) \;{with}\;{respect}\;{to}\;y,\;{and}\;{v}_{x}\dot{ = } - {u}_{y} = \dot{-{sin}\left( x\right) }{sinh}\left( y\right) - 1\;{with}\;{respect} \) to \( x \) . Then \( v\left( {... | Yes |
Show that if \( v \) is a harmonic conjugate of the harmonic function \( u \), then at any point, the gradients \( \nabla u \) and \( \nabla u \) are of equal length and are perpendicular. Conclude that at a point where \( \nabla u \neq \mathbf{0} \), the level curves of \( u \) and \( v \) are orthogonal. | Solution. Using the Cauchy-Riemann equations, for the analytic function \( f = u + {iv} \), we have \( {\left| \nabla u\right| }^{2} = {\left( {u}_{x}\right) }^{2} + {\left( {u}_{y}\right) }^{2} = {\left( {v}_{y}\right) }^{2} + {\left( -{v}_{x}\right) }^{2} = {\left| \nabla v\right| }^{2} \), whence the gradients of \(... | Yes |
Show that the function \( u\left( {x, y}\right) = \frac{1}{2}\log \left( {{x}^{2} + {y}^{2}}\right) \), or \( U\left( {r,\theta }\right) = \log r \), which is harmonic on the punctured plane \( \left( {r > 0}\right) \), has no harmonic conjugate defined on the punctured plane. However, if the negative x-axis is deleted... | Solution. Since the level curves of \( u = \log r \) are circles centered at the pole, we know by Example 2 that the level curves of a harmonic conjugate \( v\;{of}\;u\; \) must be the rays issuing from the origin. (Note that the gradient of \( u \) does not vanish, and hence all level curves are indeed curves, and \( ... | Yes |
Show that the analytic function \( f\left( z\right) = {z}^{2} \) maps the wedge \( 0 \leq r \leq 3,0 \leq \theta \leq \alpha \) to the wedge \( 0 \leq \mathrm{r} \leq 9,0 \leq \theta \leq {2\alpha } \) . | Solution. The function \( f\left( z\right) = {z}^{2} \) maps the point \( \left( {x, y}\right) \) to \( \left( {{x}^{2} - {y}^{2},{2xy}}\right) \), but it is much easier to see the mapping geometrically, by using polar coordinates. Indeed, by De Moivre's formula, we have \( \mathrm{f}\left( {\mathrm{{re}}}^{\mathrm{i}\... | Yes |
Suppose that \( f\left( z\right) \) is an analytic function and \( {f}^{\prime }\left( {z}_{0}\right) = {\operatorname{Me}}^{i\tau } \neq 0 \) . As \( \theta \) varies, the point \( {z}_{0} + r{e}^{i\theta }\left( {r > 0}\right) \) traces out a circle \( C \) of radius \( r \) about \( {z}_{0} \) . Show that for small ... | Solution. From the definition (1) of \( {f}^{\prime }\left( z\right) \), we know that \( f\left( {{z}_{0} + h}\right) - f\left( {z}_{0}\right) \approx {f}^{\prime }\left( {z}_{0}\right) h \), for small \( \left| h\right| \) . Taking \( h = r{e}^{i\theta } \) for small \( r \), we then obtain\n\n\[ f\left( {{z}_{0} + r{... | Yes |
Proposition 4. If \( h\left( {u, v}\right) \) is a harmonic function on an open set \( E \) of the uv-plane (i.e., the \( w \) -plane, \( w = u + {iv}) \) and if \( f\left( z\right) = u\left( {x, y}\right) + {iv}\left( {x, y}\right) \) is an analytic function on the open set \( \mathrm{D} \) in the \( \mathrm{{xy}} \) ... | Proof. Using the chain rule, \( {g}_{x} = {h}_{u}{u}_{x} + {h}_{v}{v}_{x} \) and \( {g}_{xx} = {h}_{uu}{\left( {u}_{x}\right) }^{2} + 2{h}_{uv}{u}_{x}{v}_{x} + {h}_{vv}{\left( {v}_{x}\right) }^{2} \) , and we have a similar expression for \( {\mathrm{g}}_{\mathrm{{yy}}} \) . Thus, using the solution of Example 2, we ob... | Yes |
Consider a heat-conducting plate \( D \) which is the first quadrant of the xy-plane minus the quarter disk \( \left( {\mathrm{r} < 1,0 < \theta < \pi /2}\right) \), as in Figure 2. Assume that the circular arc is insulated and the edge \( y = 0\left( {x > 1}\right) \) is held at temperature 0, while the remaining edge... | Solution. The region \( D \) is defined by \( r > 1 \) and \( 0 < \theta < \pi /2 \) . Since \( u = \log r \) and \( v = \theta \), the image, say \( E, \) of \( D \) under the conformal map \( f \), is defined by \( u > 0 \) and \( \breve{0} < v < \pi /2 \), which is the strip (cf. Figure 3) in the uv-plane (or w-plan... | No |
Let \( R \) be a positive constant and define \( F\left( z\right) = {R}^{2}{z}^{-1}, \) for \( z \neq 0. \) Show that \( F \) maps the exterior \( r > R \) of the circle \( r = R \) onto the interior of this circle minus the pole (i.e., onto the punctured disk \( 0 < \mathrm{r} < \mathrm{R} \) ). | Solution. In terms of polar coordinates, \( F\left( {r{e}^{i\theta }}\right) = {R}^{2}{\left( r{e}^{i\theta }\right) }^{-1} = {R}^{2}{r}^{-1}{e}^{-{i\theta }} \) . In other words, \( F \) maps the point \( \left( {r,\theta }\right) \) to \( \left( {{R}^{2}/r, - \theta }\right) \) . If \( r \geq R \), then \( {R}^{2}/r ... | Yes |
Proposition 5. Let \( v = {v}_{1}i + {v}_{2}j \) be a \( {C}^{1} \) velocity vector field of an irrotational fluid flow of an incompressible fluid. \( \frac{\text{ Suppose that }v\text{ is defined on a simply-connected open set }}{R} \)\n\n(cf. the remark following Proposition 3). Then there are \( {C}^{2} \) functions... | Proof. The functions \( \Phi \) and \( \Omega \) exist, by Proposition 2 and the remark following Proposition 3. The integrability condition for (6) is the irrotationality condition \( \left\lbrack {{\left( {\mathrm{v}}_{2}\right) }_{\mathrm{x}} - {\left( {\mathrm{v}}_{1}\right) }_{\mathrm{v}}\overset{ \circ }{ = }0}\r... | Yes |
Example 8. Let \( f\left( z\right) = {z}^{2} \) . Sketch the level curves of the real and imaginary parts of \( f \) and interpret these curves physically. | Solution. Since \( f\left( z\right) = \left( {{x}^{2} - {y}^{2}}\right) + {i2xy} \), we have \( \Phi \left( {x, y}\right) = \operatorname{Re}\left( {f\left( z\right) }\right) = {x}^{2} - {y}^{2} \), and \( \Omega \left( {x, y}\right) \) \( = \) Im \( \left( {f\left( z\right) }\right) = {2xy}. \) in fluid mechanics, the... | Yes |
Example 9. Let \( f\left( z\right) = \log \left( z\right) = \log \left( {r{e}^{i\theta }}\right) = \log r + {i\theta } \) for \( r > 0 \) and \( - \pi < \theta < \pi \) (i.e., \( f \) is the principal branch of \( \log \mathrm{z} \) ; cf. Example 3). Analyze and interpret this function as was done in Example 8. Also, c... | Solution. Here the streamlines for the fluid flow associated with \( f\left( z\right) \) are the rays \( \theta = \) constant (cf. Figure 5). The gradient of the velocity potential, \( \log r \), is \( {r}^{-1}{\mathbf{e}}_{r} \), where \( {\mathbf{e}}_{r} \) is the unit radial vector field. Thus, the fluid appears to ... | Yes |
In the same way as in Examples 8 and 9, supply interpretations of the function \( \mathrm{f}\left( \mathrm{z}\right) = {\mathrm{V}}_{0}\left( {\mathrm{z} + {\mathrm{R}}^{2}{\mathrm{z}}^{-1}}\right) \), where \( {\mathrm{V}}_{0} \) and \( \mathrm{R} \) are positive constants and \( \mathrm{z} \neq 0 \) . | Solution. For \( \left| z\right| \) large, we have \( f\left( z\right) \approx {V}_{0}z = {V}_{0}\left( {x + {iy}}\right) \) . Thus, the velocity of the associated fluid flow is nearly \( {\mathrm{V}}_{0}\mathrm{i} \), far away from the pole. In terms of polar coordinates, we have\n\n\[ \begin{aligned} f\left( z\right)... | Yes |
Compute the complex Fourier series of the \( f\left( x\right) = {e}^{ax}, - L \leq x \leq L \), where a is a real constant. | We have\n\n\[ \n{c}_{m} = \frac{1}{2L}{\int }_{-L}^{L}{e}^{ax}{e}^{-{im\pi x}/L}{dx} = \frac{1}{2L}{\int }_{-L}^{L}{e}^{\left( {a - \left( {{im\pi }/L}\right) }\right) x}{dx} \n\]\n\n\[ \n= {\left. \frac{1}{2L}\frac{1}{a - \operatorname{im}\pi /L}{e}^{\left( {a - \left( {{im\pi }/L}\right) }\right) x}\right| }_{-L}^{L}... | No |
What does Parseval’s equality say if \( f\left( x\right) = {e}^{ax}, - L \leq x \leq L\;\left( {a \neq 0\;{and}\;{real}}\right) \;? \) | We have\n\n\[ \n{\int }_{-L}^{L}{\left| f\left( x\right) \right| }^{2}{dx} = {\int }_{-L}^{L}{e}^{2ax}{dx} = {\left. \frac{1}{2a}{e}^{2ax}\right| }_{-L}^{L} = \frac{1}{2a}\left( {{e}^{2aL} - {e}^{-{2aL}}}\right) \n\]\n\n\[ \n= \frac{1}{2a}\left( {{e}^{aL} + {e}^{-{aL}}}\right) \left( {{e}^{aL} - {e}^{-{aL}}}\right) = \... | Yes |
Compute the Fourier transform of \( f\left( x\right) = {e}^{-a\left| x\right| } \), where \( a > 0 \) and \( - \infty < x < \infty \) . | \[ f\left( \xi \right) = {\int }_{-\infty }^{\infty }{e}^{-a\left| x\right| }{e}^{-{i\xi x}}{d}^{\prime }x = {\int }_{0}^{\infty }{e}^{-{ax}}{e}^{-{i\xi x}}{d}^{\prime }x + {\int }_{-\infty }^{0}{e}^{ax}{e}^{-{i\xi x}}{d}^{\prime }x \]\n\[ = {\int }_{0}^{\infty }{\mathrm{e}}^{-\left( {a + {i\xi }}\right) x}{d}^{\prime ... | Yes |
Find the Fourier transform of the function\n\n\[ f\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ for }\\left| x\\right| \\leq L \\\\ 0 & \\text{ for }\\left| x\\right| > L \\end{array}\\right. \] | Solution.\n\n\[ \\widehat{f}\\left( \\xi \\right) = {\\int }_{-\\infty }^{\\infty }f\\left( x\\right) {e}^{-{i\\xi x}}{d}^{\\prime }x = {\\int }_{-L}^{L}{e}^{-{i\\xi x}}{d}^{\\prime }x = {\\left. \\frac{1}{\\sqrt{2\\pi }}\\frac{{e}^{-{i\\xi x}}}{-{i\\xi }}\\right| }_{-L}^{L} \]\n\n\[ = \\frac{1}{\\sqrt{2\\pi }}\\frac{{... | Yes |
By computing the Fourier transform of the function\n\n\[ f\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ for }0 \\leq x \\leq L \\\\ 0 & \\text{ otherwise } \\end{array}\\right. \]\n\nshow that the Fourier transform of a real-valued function need not be real-valued itself. | Solution.\n\n\[ \\widehat{\\mathrm{f}}\\left( \\xi \\right) = {\\int }_{0}^{\\mathrm{L}}{\\mathrm{e}}^{-\\mathrm{i}\\xi \\mathrm{x}}{\\mathrm{d}}^{\\prime }\\mathrm{x} = {\\left. \\frac{1}{\\sqrt{2\\pi }}\\frac{{\\mathrm{e}}^{-\\mathrm{i}\\xi \\mathrm{x}}}{-\\mathrm{i}\\xi }\\right| }_{0}^{\\mathrm{L}} = \\frac{1}{\\sq... | Yes |
Example 6. Let \( f\left( x\right) = {e}^{-a{x}^{2}/2}, a > 0, - \infty < x < \infty \) . Show that\n\n\[ \n\widehat{f}\left( \xi \right) = {\int }_{-\infty }^{\infty }{\mathrm{e}}^{-a{x}^{2}/2 - {i\xi x}}{d}^{\prime }x = \frac{1}{\sqrt{a}}{\mathrm{e}}^{-{\xi }^{2}/{2a}}.\n\] | Solution. Completing the square, the exponent in the integrand equals \( - \frac{a}{2}{\left( x + i\frac{\xi }{a}\right) }^{2} - \frac{{\xi }^{2}}{2a} \) . Thus,\n\n\[ \n\widehat{f}\left( \xi \right) = {\mathrm{e}}^{-{\xi }^{2}/{2a}}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-a{\left( x + i\xi /a\right) }^{2}/2}{d}^{\pr... | Yes |
Example 1. Show that \( {\mathrm{e}}^{-{x}^{2}} \) is rapidly decreasing. | Solution. For any \( \mathrm{k} \geq 0 \) and for any real \( \mathrm{x} \), the following estimate holds :\n\n\[ \n{e}^{-{x}^{2}} = {\left\lbrack {e}^{{x}^{2}}\right\rbrack }^{-1} = \frac{1}{1\; + \;{x}^{2}\; + \;\frac{1}{2!}\;{x}^{4}\; + \ldots \; + \;\frac{1}{k!}\;{x}^{2k}\; + \;\ldots } \leq \frac{1}{\frac{1}{k!}\;... | Yes |
Proposition 1. Let \( f\left( x\right) \) have decay order \( \left( {1,2}\right) \), i.e., \( f \) is \( {C}^{1} \) and \( \left| {f\left( x\right) }\right| + \left| {{f}^{\prime }\left( x\right) }\right| \nleq \mathrm{K}{\left| \mathrm{x}\right| }^{-2} \), for \( \left| \mathrm{x}\right| \geq 1 \) and some constant \... | Proof. In order to show that \( \widehat{\mathrm{f}}\left( \xi \right) \) exists, it suffices to show that \( \mathrm{f}\left( \mathrm{x}\right) \) is absolutely integrable (cf. Problem 10 of Exercises 7.1). We have\n\n\[{\int }_{-\infty }^{\infty }\left| {f\left( x\right) }\right| {dx} = {\int }_{-1}^{1}\left| {f\left... | Yes |
Corollary 1. If \( f\left( x\right) \) has decay order \( \left( {m,2}\right) \), then for all real \( \xi \) ,\n\n\[ \n{\left\lbrack {f}^{\left( m\right) }\left( x\right) \right\rbrack }^{\widehat{}}\left( \xi \right) = {i}^{m}{\xi }^{m}\widehat{f}\left( \xi \right) .\n\] | Proof. Since \( f\left( x\right) \) has decay order \( \left( {m,2}\right) \) we have that \( {f}^{\prime }\left( x\right) \) is of decay order \( \left( {m - 1,2}\right) \), \( {f}^{\prime \prime }\left( x\right) \) is of decay order \( \left( {\mathrm{m} - 2,2}\right) ,\ldots \), and \( {\mathrm{f}}^{\left( \mathrm{m... | Yes |
Proposition 2. Suppose \( f\left( x\right) \) has decay order \( \left( {0,3}\right) \), i.e., \( f\left( x\right) \) is continuous and\n\n\( \left| {f\left( x\right) }\right| \leq K{\left| x\right| }^{-3} \) for \( \left| x\right| \geq 1, K > 0 \) . Then, for all real \( \xi \)\n\n\[ i\frac{d\widehat{f}}{d\xi }\left( ... | Proof. Since \( f\left( x\right) \) has decay order \( \left( {0,3}\right) \), both \( f\left( x\right) {e}^{-{i\xi x}} \) and its derivative with respect to \( \xi \) , namely, \( \; - \mathrm{{ixf}}\left( \mathrm{x}\right) {\mathrm{e}}^{-\mathrm{i}\xi \mathrm{x}} \), are absolutely integrable and continuous. Thus, Le... | Yes |
Corollary 2. If \( f\left( x\right) \) has decay order \( \left( {0, n + 2}\right) \), then for all real \( \xi \) | \[ {i}^{n}\frac{{d}^{n}\widehat{f}}{d{\xi }^{n}}\left( \xi \right) = {\left\lbrack {x}^{n}f\left( x\right) \right\rbrack }^{\widehat{}}\left( \xi \right) \] (5) In particular, both sides of equation (5) exist ! | Yes |
Find a function \( f\left( x\right) \) which is not rapidly decreasing, but is such that \( \left| {\widehat{\mathrm{f}}\left( \xi \right) }\right| \leq {\mathrm{K}}_{\mathrm{m}}{\left| \xi \right| }^{-\mathrm{m}} \) for all \( \left| \xi \right| \geq 1 \) and \( \mathrm{m} > 0 \) (i.e., \( \widehat{\mathrm{f}}\left( \... | Solution. Theorem 1 says that \( \widehat{\mathrm{f}}\left( \xi \right) \) will have decay order \( \left( {0,\mathrm{\;m}}\right) \) for all \( \mathrm{m} \geq 0 \), if \( \mathrm{f}\left( \mathrm{x}\right) \) has decay order \( \left( {m,2}\right) \) for all \( m \geq 0 \) (i.e., \( f\left( x\right) \) is a \( {C}^{\... | Yes |
Theorem 2 (The Convolution Theorem). Let \( f\left( x\right) \) and \( g\left( x\right) \) be piecewise continuous with \( \left| {f\left( x\right) }\right| ,\left| {g\left( x\right) }\right| \leq \) const. \( {\left| x\right| }^{-2},\left| x\right| \geq 1 \) . Then | \[ \widehat{\mathrm{f}}\left( \xi \right) \widehat{\mathrm{g}}\left( \xi \right) = \frac{1}{\sqrt{2\pi }}\left( {\mathrm{f} * \mathrm{g}}\right) \left( \xi \right) \] | Yes |
Compute the inverse Fourier transform of the function \( g\left( \xi \right) = {\left( {a}^{2} + {\xi }^{2}\right) }^{-1}\;\left( {a > 0}\right) . \) | Solution. If we had to do this directly, we would need to compute the integral\n\n\[ \check{g}\left( x\right) = {\int }_{-\infty }^{\infty }\frac{{e}^{i\xi x}}{{a}^{2} + {\xi }^{2}}{d}^{\prime }\xi \]\n\nWhile this is not hard to do using complex contour integration, we will proceed indirectly by using the Inversion Th... | Yes |
Find the Fourier transform of \( h\left( x\right) = {x}^{n}{e}^{-\frac{1}{2}{x}^{2}} \), for \( n = 0,1,2,\ldots \) . | Solution. A direct computation of \( \widehat{\mathrm{h}}\left( \xi \right) \), using the definition\n\n\[ \widehat{h}\left( \xi \right) = {\int }_{-\infty }^{\infty }{x}^{n}{e}^{-\frac{1}{2}{x}^{2}}{e}^{-{i\xi x}}{d}^{\prime }x \]\n\n\nis not very easy. One could differentiate \( {\mathrm{e}}^{-\frac{1}{2}{\xi }^{2}} ... | Yes |
Compute \( I \equiv {\int }_{-\infty }^{\infty }\frac{{x}^{2}}{{\left( {x}^{2} + 1\right) }^{4}}{dx} \), using Parseval’s equality and \( \left( {2}^{\prime }\right) \) . | Solution. Let \( g\left( x\right) = \frac{1}{2}{\left( {x}^{2} + 1\right) }^{-1} \) . By Parseval’s equality and Proposition 1 of Section 7.2,\n\n\[ I = {\int }_{-\infty }^{\infty }{\left| {g}^{\prime }\left( x\right) \right| }^{2}{dx} = {\int }_{-\infty }^{\infty }{\left| {\left( {g}^{\prime }\right) }^{2}\left( \xi \... | Yes |
By means of formal calculations express the Fourier transform of\n\n\\[ \nh\\left( x\\right) = {\\int }_{-\\infty }^{\\infty }\\left( {x - s}\\right) f\\left( {x - s}\\right) {f}^{\\prime }\\left( s\\right) {ds} \n\\]\n\nin terms of \\( \\widehat{\\mathrm{f}}\\left( \\xi \\right) \\) and \\( \\frac{\\mathrm{d}\\widehat... | Solution. Let \\( g\\left( x\\right) = {xf}\\left( x\\right) \\) . Then \\( h\\left( x\\right) = \\left( {g * {f}^{\\prime }}\\right) \\left( x\\right) \\) and hence by the Convolution Theorem\n\n\\[ \n\\widehat{\\mathrm{h}}\\left( \\xi \\right) = {\\left( \\mathrm{g} * {\\mathrm{f}}^{\\prime }\\right) }^{\\widehat{}}\... | Yes |
Solve for \( g\left( x\right) \) in the integral equation\n\n\[ \n{\int }_{-\infty }^{\infty }\frac{g\left( s\right) }{{\left( x - s\right) }^{2} + {b}^{2}}{ds} = \frac{1}{{x}^{2} + {a}^{2}}\;\left( {a > b > 0}\right) ,\n\] | Solution. If we set \( f\left( x\right) = \frac{1}{{x}^{2} + {b}^{2}} \), then (8) becomes \( \left( {f * g}\right) \left( x\right) = \frac{1}{{x}^{2} + {a}^{2}} \) .\n\nHence, by the Convolution Theorem,\n\n\[ \n\sqrt{2\pi }\widehat{f}\left( \xi \right) \widehat{g}\left( \xi \right) = \left\lbrack \frac{1}{{x}^{2} + {... | Yes |
Parseval’s equality : If \( f\left( x\right), f\left( \xi \right) \) and \( g\left( x\right) \) are absolutely integrable on \( \left( {-\infty ,\infty }\right) \) and \( f\left( x\right) \) is piecewise \( {\mathrm{C}}^{1} \) on \( \left( {-\infty ,\infty }\right) \), then\n\n\[ \n{\int }_{-\infty }^{\infty }f\left( x... | (S2) | No |
In 1935, the Russian mathematician A.N. Tychonov demonstrated that the problem\n\n\\[ \n\\text{D.E.}{u}_{t} = {u}_{xx}\\; - \\infty < x < \\infty, t > 0 \n\\]\n\n\\[ \n\\text{ I.C. }u\\left( {x,0}\\right) = 0\\text{,}\n\\]\n\nfor the infinite rod with a zero initial temperature, has a solution other than the obvious tr... | We sketch the construction of Tychonov’s solution.\n\nConstruction. Let \\( f\\left( t\\right) = \\left\\{ \\begin{matrix} {e}^{-1/{t}^{2}} & \\text{if}\\;t\\; \\neq \\;0 \\\\ 0 & \\text{if}\\;t = 0 \\end{matrix}\\right. ,\\)\n\nand let\n\\[ \nu\\left( {x, t}\\right) = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{f}^{... | Yes |
Find a formal solution of\n\n\\[ \n\\text{D.E.}{u}_{t} = k{u}_{xx} + q\\left( {x, t}\\right) \\; - \\infty < x < \\infty, t > 0 \n\\]\n\n\\[ \n\\text{I.C.}u\\left( {x,0}\\right) = 0\\text{.} \n\\] | Solution. The following formal manipulations serve only to provide a hypothetical solution whose validity could be justified directly, under certain assumptions concerning the source term \\( q\\left( {x, t}\\right) \\) . We first take the Fourier transform of both sides of the D.E. with respect to \\( x \\) . Thus, we... | Yes |
Verify the following relation and give a physical interpretation of it :\n\n\[ \n{\int }_{-\infty }^{\infty }u\left( {x, t}\right) {dx} = {\int }_{-\infty }^{\infty }f\left( y\right) {dy} \n\] | Solution. Since the integrand of (21) is absolutely integrable over the xy-plane, in the following calculations we may interchange the order of integration (cf. Appendix A.2).\n\n\[ \n{\int }_{-\infty }^{\infty }u\left( {x, t}\right) {dx} = {\int }_{-\infty }^{\infty }\left( {\frac{1}{\sqrt{4\pi kt}}{\int }_{-\infty }^... | Yes |
Show that \(\\left( {{}_{t}H * {}_{t}H}\\right) \\left( x\\right) = {}_{2t}H\\left( t\\right) \\;,\\;\\left( {t > 0}\\right) . | Solution. According to the definition of convolution (cf. Section 7.2), for \( t > 0 \)\n\n\[\\left( {{}_{t}H{ * }_{t}H}\\right) \\left( x\\right) = {\\int }_{-\\infty }^{\\infty }{}_{t}H\\left( {x - y}\\right) {}_{t}H\\left( y\\right) {dy} = {\\int }_{-\\infty }^{\\infty }\\frac{1}{\\sqrt{4\\pi kt}}{e}^{-{\\left( x - ... | Yes |
Example 7. Find a continuous solution of the problem\n\n\\[ \n\\text{D.E.}{u}_{t} - k{u}_{xx} = - {\\lambda u}\\; - \\infty < x < \\infty, t > 0 \n\\]\n\n(25)\n\n\\[ \n\\text{I.C.}u\\left( {x,0}\\right) = f\\left( x\\right) \\text{,}\n\\]\n\nwhere \\( \\;\\lambda \\; \\) is a constant and \\( \\;f\\left( x\\right) \\; ... | Solution. Formally applying the Fourier transform to the D.E., we obtain\n\n\\[ \n{\\widehat{u}}_{t}\\left( {\\xi, t}\\right) + \\left( {k{\\xi }^{2} + \\lambda }\\right) \\widehat{u}\\left( {\\xi, t}\\right) = 0\n\\]\n\nThus, formally \\( \\widehat{\\mathrm{u}}\\left( {\\xi ,\\mathrm{t}}\\right) = {\\mathrm{e}}^{-\\la... | Yes |
By formal calculations find an integral representation for the solution of\n\n\\[ \n\\text{D.E.}{u}_{tt} = {a}^{2}{u}_{xx} - u\\; - \\infty < x, t < \\infty \n\\]\n\n\\[ \n\\text{I.C.}u\\left( {x,0}\\right) = {e}^{-\\frac{1}{2}{x}^{2}},{u}_{t}\\left( {x,0}\\right) = 0\\text{.}\n\\] | Solution. If we take the Fourier transform of both sides of the D.E. with respect to \\( x \\), we obtain\n\n\\[ \n{\\widehat{u}}_{tt}\\left( {\\xi, t}\\right) = {a}^{2}{\\left( i\\xi \\right) }^{2}\\widehat{u}\\left( {\\xi, t}\\right) - \\widehat{u}\\left( {\\xi, t}\\right) \\text{ or }{\\widehat{u}}_{tt}\\left( {\\xi... | Yes |
Theorem 3. Let \( f\left( x\right) , - \infty < x < \infty \), be a bounded, continuous function. Then the problem\n\nD.E. \( {\mathrm{u}}_{\mathrm{{xx}}} + {\mathrm{u}}_{\mathrm{{yy}}} = 0\; - \infty < \mathrm{x} < \infty ,\mathrm{y} > 0 \)\n\n(39)\n\nB.C. \( u\left( {x,0}\right) = f\left( x\right) \; - \infty < x < \... | Formula (40) can be found using Fourier transforms. Proceeding formally, we take the Fourier transform of the D.E. with respect to \( \mathrm{x} \), treating \( \mathrm{y} \) as a constant :\n\n\[ {\left( \mathrm{i}\xi \right) }^{2}\widehat{\mathrm{u}}\left( {\xi ,\mathrm{y}}\right) + {\widehat{\mathrm{u}}}_{\mathrm{{y... | Yes |
Example 9. Solve the problem\n\n\\[ \n\\text{D.E.}{u}_{xx} + {u}_{yy} = 0\\; - \\infty < x < \\infty, y > 0 \n\\]\n\n\\[ \n\\text{B.C.}u\\left( {x,{0}^{ + }}\\right) = f\\left( x\\right) = \\left\\{ \\begin{array}{ll} 0 & \\text{ if }\\;x < - 1 \\\\ 1 & \\text{ if } - 1 < x < 1 \\\\ 2 & \\text{ if }\\;x > 1 \\end{array... | Solution. In this example, the boundary function \\( \\;f\\left( x\\right) \\; \\) has some jump discontinuities. Nevertheless, if we use the formula (40), we obtain\n\n\\[ \nu\\left( {x, y}\\right) = \\frac{1}{\\pi }{\\int }_{-\\infty }^{\\infty }\\frac{y}{{y}^{2} + {\\left( x - s\\right) }^{2}}f\\left( s\\right) {ds}... | Yes |
Consider the problem\n\nD.E. \( {u}_{xx} + {u}_{yy} = 0\; - \infty < x < \infty, y > 0 \)\n\nB.C. \( u\left( {x,0}\right) = {e}^{-{x}^{2}} \).\n\nIf \( u\left( {x, y}\right) \) is the unique solution (40) which is continuous and bounded, then show that\n\n\[ \n{\int }_{-\infty }^{\infty }u\left( {x, y}\right) {dx} = \s... | Solution. Since \( {\mathrm{e}}^{-{\mathrm{x}}^{2}} \) is bounded and continuous, formula (40) yields (for \( \mathrm{y} > 0 \) )\n\n\[ \nu\left( {x, y}\right) = \frac{1}{\pi }{\int }_{-\infty }^{\infty }\frac{y}{{y}^{2} + {\left( x - s\right) }^{2}}{e}^{-{s}^{2}}{ds} \]\n\n(43)\n\nIf we integrate both sides of (43) wi... | Yes |
Theorem 1. Let \( {f}_{0}\left( x\right) , - \infty < x < \infty \), be the periodic extension of the odd extension \( {f}_{0}\left( x\right) \) , \( - \mathrm{L} \leq \mathrm{x} \leq \mathrm{L} \), of \( \mathrm{f}\left( \mathrm{x}\right) \) . Then the only solution of problem (1), which meets (2), is \[ u\left( {x, t... | Proof. We already know from Theorem 1 of Section 7.4 that the function \( u \) defined by (3) is continuous and satisfies the D.E. for \( 0 \leq x \leq L \) and \( t > 0 \) . Since \( u\left( {0, t}\right) \) is the integral of an odd function, it is 0. Moreover, since \( {\widetilde{f}}_{o}\left( y\right) \) is also o... | Yes |
Theorem 3. The unique continuous solution of problem (6) is given by the following equivalent formulas, for \( t > 0 \) ,\n\n\[ u\left( {x, t}\right) = \left\{ {\begin{array}{l} \frac{1}{\sqrt{4\pi kt}}{\int }_{-\infty }^{\infty }{e}^{-{\left( x - y\right) }^{2}/{4kt}}f\left( y\right) {dy} \\ {\int }_{n = - \infty }^{\... | Proof. We know from Theorem 1 of Section 7.4 that the integral in (7) is \( {\mathrm{C}}^{\infty } \) for \( \mathrm{t} > 0 \), and solves the D.E.. It is also periodic. Indeed, replacing \( x \) by \( x + {2L} \) in the integral and changing variables by the formula \( z = y - {2L} \), we obtain the same result since ... | Yes |
Is the solution unique for the problem\n\n\[ \text{D.E.}{u}_{t} = k{u}_{xx}\;0 < x < \infty, t > 0 \]\n\n\[ \text{B.C.}\mathrm{u}\left( {{0}^{ + },\mathrm{t}}\right) = 0 \]\n\n(8)\n\n\[ \text{ I.C. }u\left( {x,{0}^{ + }}\right) = f\left( x\right) \text{,} \]\n\nwhere \( f\left( x\right) \) is continuous and absolutely ... | Solution. Let \( {f}_{0}\left( x\right) \) denote the odd extension of \( f\left( x\right) \) (i.e., \( {f}_{0}\left( x\right) = - f\left( {-x}\right) \), for \( x \leq 0 \) ).\n\nNote that \( {f}_{0}\left( x\right) \) is continuous on \( \left( {-\infty ,\infty }\right) \), since it is assumed that \( f\left( 0\right)... | Yes |
Attempt to solve the problem posed in Example 1, by using the even extension \( {f}_{e}\left( x\right) \) of \( f\left( x\right) \) (i.e., \( {f}_{e}\left( x\right) = f\left( {-x}\right) \) for \( x < 0 \) ). | Solution. Once again it is easy to verify that \( {f}_{e}\left( x\right) \) is continuous and absolutely integrable on \( \left( {-\infty ,\infty }\right) \) . Hence, by Theorem 1 of Section 7.4, the function given by\n\n\[ u\left( {x, t}\right) = \frac{1}{\sqrt{4\pi kt}}{\int }_{-\infty }^{\infty }{e}^{-{\left( x - y\... | Yes |
Example 3. Solve the problem\n\nD.E. \( {u}_{t} = k{u}_{xx}\;0 < x < \infty, t > 0 \)\n\nB.C. \( {u}_{x}\left( {{0}^{ + }, t}\right) = 0 \)\n\nI.C. \( u\left( {x,{0}^{ + }}\right) = f\left( x\right) \), \n\nwhere \( f\left( x\right) \) is continuous and absolutely integrable on \( \lbrack 0,\infty ) \) . | Solution. For this problem, we take the even extension \( {f}_{e}\left( x\right) \) of \( f\left( x\right) \) . Then, as in Example 1,\n\n\[ u\left( {x, t}\right) = \frac{1}{\sqrt{4\pi kt}}{\int }_{0}^{\infty }\left\lbrack {{e}^{-{\left( x - y\right) }^{2}/{4kt}} + {e}^{-{\left( x + y\right) }^{2}/{4kt}}}\right\rbrack ... | No |
Let \( g\left( x\right) \) be an odd function which is \( {C}^{2} \) on \( \left( {-\infty ,\infty }\right) \), except at \( x = 0 \) . Assume that \( \mathrm{g}\left( {0}^{ + }\right) ,{\mathrm{\;g}}^{\prime }\left( {0}^{ + }\right) \) and \( {\mathrm{g}}^{\prime \prime }\left( {0}^{ + }\right) \) exist, and suppose t... | Solution. Since \( {\mathrm{g}}^{\prime \prime }\left( \mathrm{x}\right) \) is odd, we have\n\n\[ \left( {g}^{\prime \prime }\right) \left( \xi \right) = {\int }_{-\infty }^{\infty }{g}^{\prime \prime }\left( x\right) {e}^{-{i\xi x}}{d}^{\prime }x = {\int }_{0}^{\infty }{g}^{\prime \prime }\left( x\right) \left( {{e}^{... | Yes |
For \( u\left( {x, t}\right) \) given by (16), show that, despite appearances, we have \( u\left( {{0}^{ + }, t}\right) = h\left( t\right) \) (not necessarily 0 ), for \( \mathrm{t} > 0 \) . Use the fact that\n\n\[ \n{\int }_{0}^{\infty }{y}^{-\frac{1}{2}}{e}^{-{ay}}{dy} = \sqrt{\pi /a} \n\]\n\n(17)\n\nwhich may be eas... | Solution. In (16), change the variable of integration from \( s \) to \( y = {x}^{2}/\left( {t - s}\right) \), so that \( \mathrm{{dy}} = {\mathrm{x}}^{2}/{\left( \mathrm{t} - \mathrm{s}\right) }^{2}\mathrm{{ds}} \) . We then obtain (cf. Appendix A.3)\n\n\[ \nu\left( {{0}^{ + }, t}\right) = \mathop{\lim }\limits_{{x \r... | Yes |
Assuming the validity of the solutions found in Examples 1 and 5, solve the problem\n\nD.E. \( {u}_{t} = {u}_{xx}\;x > 0, t > 0 \)\n\nB.C. \( u\left( {{0}^{ + }, t}\right) = h\left( t\right) \)\n\n(18)\n\nI.C. \( u\left( {x,{0}^{ + }}\right) = f\left( x\right) \) ,\n\nwhere \( h\left( t\right) \) and \( f\left( x\right... | Solution. Since the D.E., B.C. and I.C are linear, the superposition principle implies that a solution of (18) is simply the sum of the solutions which were found in Examples 1 and 5 :\n\n\[ u\left( {x, t}\right) = \frac{1}{\sqrt{4\pi t}}{\int }_{0}^{\infty }\left\lbrack {{e}^{-{\left( x - y\right) }^{2}/{4t}} - {e}^{-... | Yes |
Theorem 1. Let \( {\widetilde{f}}_{0}\left( x\right) ,\; - \infty < x < \infty \), be the periodic extension of the odd extension \( {f}_{0}\left( x\right) \) , \( - \mathrm{L} \leq \mathrm{x} \leq \mathrm{L} \), of \( \mathrm{f}\left( \mathrm{x}\right) \) . Then the only solution of problem (S1), which meets (S2), is | \[ u\left( {x, t}\right) = \frac{1}{\sqrt{4\pi kt}}{\int }_{-\infty }^{\infty }{e}^{-{\left( x - y\right) }^{2}/{4kt}}{\widetilde{f}}_{0}\left( y\right) {dy}\left( {\text{ for }t > 0}\right) \text{ and }u\left( {x,0}\right) = f\left( x\right) \left( {\text{ for }t = 0}\right) .\](S3) | Yes |
Verify that \( \sin \left( x\right) = x + O\left( {x}^{3}\right) ,\;\left( {x \rightarrow 0}\right) \). | Consider the Taylor series expansion of \( \sin \left( \mathrm{x}\right) \) about \( \mathrm{x} = 0 \):\n\n\[ \sin \left( x\right) = x - \frac{{x}^{3}}{3!} + \frac{{x}^{5}}{5!} - \frac{{x}^{7}}{7!} + \;...\; = x + {x}^{3}\left\lbrack {-\frac{1}{3!} + \frac{{x}^{2}}{5!} - \;...}\right\rbrack = x + {x}^{3}g\left( x\right... | Yes |
Verify that\n\n\\[ \n{f}^{\prime }\left( x\right) = \frac{f\left( {x + {\Delta x}}\right) - f\left( {x - {\Delta x}}\right) }{2\Delta x} + O\left( {\Delta x}\right) ,\;\left( {{\Delta x} \rightarrow 0}\right) ,\n\\]\n\nwhere \\( f\left( x\right) \\) is a \\( {C}^{2} \\) function in some open interval containing the poi... | Solution. By Taylor’s theorem we have, with \\( {\Delta x} > 0 \\) sufficiently small, that\n\n\\[ \n\\begin{aligned} f\left( {x + {\Delta x}}\right) & = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + {R}_{2}\left( {\Delta x}\right) \\ & = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + O\lef... | Yes |
Suppose that \( \;u\left( {x, y}\right) \; \) is \( \;{C}^{4}\; \) in a neighborhood of the point \( \;\left( {x, y}\right) .\; \) Obtain a second-order difference approximation for \( {u}_{xx}\left( {x, y}\right) \) . | Replacing \( \Delta \mathrm{x} \) by \( - \Delta \mathrm{x} \) in (16) yields (as \( \Delta \mathrm{x} \rightarrow 0 \) )\n\n\[ u\left( {x - {\Delta x}, y}\right) = u\left( {x, y}\right) - {\Delta x}\;\frac{\partial u}{\partial x}\left( {x, y}\right) + \frac{{\left( \Delta x\right) }^{2}}{2\;!}\frac{{\partial }^{2}u}{\... | Yes |
If \( u\left( {x, y}\right) \) is \( {C}^{k} \), for \( k \) large enough, obtain the following difference approximations :\n\n\[ \n{\left\lbrack \frac{\partial u}{\partial t}\right\rbrack }_{\left( i, j\right) } = \frac{{u}_{i, j + 1} - {u}_{i, j}}{\Delta t} + O\left( {\Delta t}\right) ,\n\] | Solution. Observe that formula (26) is similar to (23) and that using the notation in (22), formulae (27) and (28) follow from (21). In Problem 4, formula (29) is derived. | No |
Theorem 1 (Taylor’s theorem for functions of two variables). Let \( u\left( {x, y}\right) \) be a \( {C}^{n} \) function in some disk \( \mathrm{D} \) in the \( \mathrm{{xy}} \) -plane. Let \( \left( {\mathrm{x},\mathrm{y}}\right) \) and \( \left( {\mathrm{x} + \Delta \mathrm{x},\mathrm{y} + \widehat{\Delta }\mathrm{y}... | \[ u\left( {x + {\Delta x}, y + {\Delta y}}\right) = {P}_{n - 1}\left( {{\Delta x},{\Delta y}}\right) + {R}_{n}\left( {{\Delta x},{\Delta y}}\right) ,\] where \[ {P}_{n - 1}\left( {{\Delta x},{\Delta y}}\right) = u\left( {x, y}\right) + \left\lbrack {{\Delta x}\;{u}_{x}\left( {x, y}\right) + {\Delta y}\;{u}_{y}\left( {... | Yes |
Consider D.E. \( {u}_{t} = {u}_{xx},\;0 \leq x \leq 5,\;0 \leq t \leq {0.5} \) ,\n\nB.C. \( u\left( {0, t}\right) = {2t}, u\left( {5, t}\right) = {25} + {2t} \) ,\n\n\[ \n\text{ I.C. }u\left( {x,0}\right) = {x}^{2}\text{. } \n\]\n\nUse the explicit difference method with grid spacings \( \;{\Delta x}\; = \;1\; \) and \... | Solution. In this problem \( L = 5, T = {0.5} \), and so \( M = L/{\Delta x} = 5 \) and \( N = T/{\Delta t} = {0.5}/{0.1} \) \( = 5 \) . Moreover, since \( \lambda = \Delta \mathrm{t}/{\left( \Delta \mathrm{x}\right) }^{2} = {0.1} \), the discretization (10) of the D.E. is\n\n\[ \n{v}_{i, j + 1} = \left( {0.1}\right) {... | Yes |
Consider D.E. \( {u}_{t} = {u}_{xx},\;0 < x < 4, t > 0, \)\n\nB.C. \( u\left( {0, t}\right) = 5, u\left( {4, t}\right) = 5,\;t > 0 \) ,\n\nI.C. \( u\left( {x,0}\right) = 0,\;0 \leq x \leq 4 \) .\n\nUse the explicit difference method, with \( \;{\Delta x} = 1\; \) and \( \;{\Delta t} = {0.125} \), to find \( \;{v}_{1,8}... | Solution. The values \( {v}_{1,8},{v}_{2,8} \) and \( {v}_{3,8} \) are in the last row of Table 3, which was constructed by solving (10) and (9), with \( {v}_{0,0} = 0,{v}_{4,0} = 0,\mathrm{\;T} = 1 \) and \( \lambda = {0.125} \) (Why?).\n\n<table><thead><tr><th></th><th>\( \mathrm{x} \) t</th><th>B.C. 0</th><th>1</th>... | Yes |
Develop an explicit difference method for problem (14), using central difference approximations for the B.C. | Solution. (a) We know that the explicit difference approximations \( {v}_{i, j} \) satisfy\n\n\[ \n{v}_{i, j + 1} = \lambda {v}_{i + 1, j} + \left( {1 - {2\lambda }}\right) {v}_{i, j} + \lambda {v}_{i - 1, j}, \]\n\n(15)\n\n\( \mathrm{i} = 1,2,\ldots ,\mathrm{M} - 1,\mathrm{j} = 0,1,\ldots ,\mathrm{N} - 1 \), where \( ... | Yes |
Proposition 1. If \( u\left( {x, t}\right) \) is a \( {C}^{4} \) solution of the problem\n\nD.E. \( {u}_{t} = {u}_{xx},\;0 \leq x \leq L,\;0 \leq t \leq T, \)\n\nB.C. \( u\left( {0, t}\right) = A\left( t\right), u\left( {L, t}\right) = B\left( t\right) \) ,\n\n(31)\n\nI.C. \( u\left( {x,0}\right) = f\left( x\right) ,\)... | Proof. Let \( w\left( {x, t}\right) = {u}_{tt}\left( {x, t}\right) \) . Using the D.E. and the assumption that \( u \) is \( {C}^{4} \), we have \( {w}_{t} = \) \( {u}_{ttt} = {u}_{xxtt} = {u}_{ttxx} = {w}_{xx} \) . Thus, \( w \) satisfies the problem\n\nD.E. \( {w}_{t} = {w}_{xx},\;0 \leq x \leq L,0 \leq t \leq T \) ,... | Yes |
Theorem 1 (A Convergence Theorem for the Explicit Difference Method). Let \( \;u\left( {x, t}\right) \; \) be a \( \;{C}^{4} \) solution of the problem (31). Let \( M \) and \( N \) be positive integers, and define \( {\Delta x} = L/M \) , \( \Delta \mathrm{t} = \mathrm{T}/\mathrm{N} \) and \( \lambda = \frac{\Delta \m... | Proof. Using Taylor's theorem with Lagrange's form of the remainder, we obtain\n\n\[ \n{u}_{i + 1, j} = {u}_{i, j} + {\Delta x}{\left\lbrack \frac{\partial u}{\partial x}\right\rbrack }_{\left( i, j\right) } + \frac{{\left( \Delta x\right) }^{2}}{2\;!}{\left\lbrack \frac{{\partial }^{2}u}{\partial {x}^{2}}\right\rbrack... | Yes |
Calculate by hand \( u\left( {{0.5},{0.5}}\right) \) with \( T = {0.5}, M = 2 \) and \( N = 1\left( {\lambda = 2}\right) \), using the explicit difference method. | Here \( {\Delta x} = L/M = {0.5} \) and \( {\Delta t} = T/N = {0.5}\left( {\lambda = {\Delta t}/{\left( \Delta x\right) }^{2} = 2}\right) \). Since \( u\left( {{0.5},{0.5}}\right) \) \( = {\mathrm{u}}_{1,1} \), we must find \( {\mathrm{v}}_{1,1} \). The system of difference equations is given by\n\n\[ \n{v}_{i, j + 1} ... | Yes |
Use the explicit method with \( \lambda = 1/2 \) to approximate \( \mathrm{u}\left( {{0.5},{0.5}}\right) \), where \( \mathrm{u} \) is the solution of the problem (45) of the previous example. | Solution. As in the previous example, \( \mathrm{L} = 1,\mathrm{\;T} = {0.5},\Delta \mathrm{x} = 1/\mathrm{M} \) and \( \Delta \mathrm{t} = {0.5}/\mathrm{N} \) . Now we will select the positive integers \( M \) and \( N \) such that \( \lambda = 1/2 \) . Table 5 provides five different approximations \( {\mathrm{v}}_{\... | Yes |
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