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Theorem 1 (Existence and Uniqueness). For the PDE \( {a}_{x} + {b}_{y} + {cu} = f\left( {x, y}\right) \), suppose that we are given a regular side condition curve which intersects each characteristic line of the PDE exactly once, and transversely. Assume also that the values of \( u \) are specified in a \( {C}^{1} \) ... | Remark. Uniqueness follows easily from the fact that the values of a solution on a characteristic line is determined by its given value at the point of intersection with the side condition curve. The regularity of the side condition curve and the transversality of the intersections enter into the proof that the solutio... | No |
Solve the PDE \( {u}_{x} - {u}_{y} + u = 0 \), subject to the condition \( u\left( {x,{x}^{3}}\right) = {e}^{-x}\left( {x + {x}^{3}}\right) \) . | Solution. Here \( u \) is specified on the curve \( y = {x}^{3} \) . This curve intersects each characteristic line \( \mathrm{x} + \mathrm{v} = \mathrm{d} \) exactly once, and transversely (since the slope of the curve \( {\mathrm{x}}^{3} \) is \( 3{\mathrm{x}}^{2} \) which is never the same as the slope \( \left( {-1... | Yes |
What must C equal, so that there will be about 300 avocados on the shelf, in the long run? | Here \( \mathrm{D}\left( {\mathrm{y},\mathrm{t}}\right) \equiv \mathrm{y}/{25} \), for \( 0 \leq \mathrm{y} \) (and \( \mathrm{D}\left( \mathrm{y}\right) = 0 \), for \( \mathrm{y} < 0 \) ). For part (a), formula (26) implies that, in the long run, the population density is\n\n\[ \n{P}_{\infty }\left( y\right) = C\exp \... | Yes |
Find the general solution of\n\n\[ - y{u}_{x} + x{u}_{y} = 0 \] | Solution. The characteristic equation is \( \mathrm{{dy}}/\mathrm{{dx}} = - \mathrm{x}/\mathrm{y} \) . This is a separable equation which is readily solved by separating the variables and integrating :\n\n\[ {ydy} = - {xdx} \Rightarrow {\frac{1}{2}}^{2} = - {\frac{1}{2}}^{2} + \frac{1}{2}d. \]\n\nThus, the characterist... | Yes |
Show that the problem \( - y{u}_{x} + x{u}_{y} = 0, u\left( {x,0}\right) = {3x} \) has no solution. | The side condition is given on the \( \mathrm{x} \) -axis which intersects each of the characteristic circles \( {\mathrm{x}}^{2} + {\mathrm{y}}^{2} = {\mathrm{a}}^{2} \) twice, at \( \left( {\mathrm{a},0}\right) \) and \( \left( {-\mathrm{a},0}\right) \left\lbrack {\mathrm{a} \neq 0}\right\rbrack \) . We saw in Exampl... | Yes |
Find the parametric form of the solution of the problem\n\n\[ - y{u}_{x} + x{u}_{y} = 0, u\left( {s,{s}^{2}}\right) = {s}^{3}\;\left( {s > 0}\right) . \] | Solution. By (15), the family of characteristic curves \( \left( {\mathrm{X}\left( {\mathrm{s},\mathrm{t}}\right) ,\mathrm{Y}\left( {\mathrm{s},\mathrm{t}}\right) }\right) \) are found by solving\n\n\[ \frac{d}{dt}X\left( {s, t}\right) = - Y\left( {s, t}\right) ,\;\frac{d}{dt}Y\left( {s, t}\right) = X\left( {s, t}\righ... | Yes |
Find the general solution of the PDE\n\n\[ x{u}_{x} - y{u}_{y} + {yu} = 0. \] | Solution. The characteristic curves are found from\n\n\[ \frac{\mathrm{d}y}{\mathrm{\;d}x} = - \frac{y}{x}\;\text{ or }\;\frac{\mathrm{d}x}{x} + \frac{\mathrm{d}y}{y} = 0. \]\n\nIntegrating, we obtain \( \log \left( \left| x\right| \right) + \log \left( \left| y\right| \right) = \log \left( \left| d\right| \right) \) .... | Yes |
Find the general solution of the PDE\n\n\[ 2{u}_{x} + 3{u}_{y} + 5{u}_{z} - u = 0,\;u = u\left( {x, y, z}\right) . \] | Solution. The characteristic curves are found by solving the system\n\n\[ \frac{\mathrm{{dy}}}{\mathrm{{dx}}} = \frac{3}{2},\;\frac{\mathrm{{dz}}}{\mathrm{{dx}}} = \frac{5}{2}. \]\n\nWe obtain \( y = \frac{3}{2}x + \frac{\alpha }{2}, z = \frac{5}{2}x + \frac{\beta }{2} \) . Alternatively, the characteristic curves, are... | Yes |
Find the general solution of\n\n\[ \n{u}_{x} + z{u}_{y} + {6x}{u}_{z} = 0,\;u = u\left( {x, y, z}\right) .\n\] | Solution. The characteristic curves are found by solving the system\n\n\[ \n\frac{dy}{dx} = z,\;\frac{dz}{dx} = {6x}\n\]\n\nfor \( y = y\left( x\right) \) and \( z = z\left( x\right) \) . Note that the first equation cannot be integrated to give \( y = {zx} \) , because \( z \) is an unknown function of \( x \) . The s... | No |
Find a solution of the following quasi-linear PDE with the given side condition\n\n\\[ \n{u}_{x} + u \cdot {u}_{y} = {6x},\;u\\left( {0, y}\\right) = {3y}.\n\\]\n\n(16) | Solution. The associated linear PDE in dimension 3, is\n\n\\[ \n{\\varphi }_{\\mathrm{x}} + \\mathrm{z} \cdot {\\varphi }_{\\mathrm{y}} + 6\\mathrm{x}{\\varphi }_{\\mathrm{z}} = 0.\n\\]\n\n(17)\n\nThis is the same PDE which was solved in Example 2. By (13), the general solution is \\( \\varphi \\left( {\\mathrm{x},\\ma... | Yes |
Find a parametric solution of the following quasi-linear PDE with side condition\n\n\\[ \n{u}_{x} + u \cdot {u}_{y} = {6x},\;u\\left( {0, y}\\right) = G\\left( y\\right) ,\n\\]\n\nwhere \\( \\mathrm{G}\\left( \\mathrm{y}\\right) \\) is an arbitrary \\( {\\mathrm{C}}^{1} \\) function. | Solution. The characteristic system (cf. (2)) for the characteristic curves \\( \\left( {x\\left( t\\right), y\\left( t\\right), z\\left( t\\right) }\\right) \\)\n\nassociated with the linear PDE \\( {\\varphi }_{x} + z{\\varphi }_{y} + {6x}{\\varphi }_{z} = 0 \\) is\n\n\\[\n\\frac{\\mathrm{{dx}}}{\\mathrm{{dt}}} = 1,\... | Yes |
With the above notation, let the initial density of cars be \( \rho \left( {x,0}\right) = a{\left( 1 + {x}^{2}\right) }^{-1} \), for some positive constant \( a < d \) . Initially, the point of maximum density is at \( x = 0 \) . Where is the point of maximum density at time \( t \) ? When and where does the first shoc... | Solution. For each \( {x}_{0} \), the solution has the constant value \( f\left( {x}_{0}\right) = a{\left( 1 + {x}_{0}^{2}\right) }^{-1} \) on the line \( \mathrm{x} = \mathrm{M}\left( {1 - \frac{2}{\mathrm{d}} \cdot \mathrm{f}\left( {\mathrm{x}}_{0}\right) }\right) \mathrm{t} + {\mathrm{x}}_{0} \) (cf. (28)). Setting ... | Yes |
Assuming the equation of state \( p = A{\rho }^{\gamma }\left( {\gamma > 1}\right) \), find \( v\left( {x, t}\right) \) and \( \rho \left( {x, t}\right) \) for the above problem (41) - (44) when \( v\left( {x,0}\right) = {\alpha x}\left( {\alpha > 0\text{, constant}}\right) \), assuming \( v = 0 \) when \( \rho = {\rho... | Solution. We take \( g\left( x\right) = {\alpha x} \), and we need to determine \( R\left( v\right) \) and \( c\left( v\right) \). Since \( f\left( \rho \right) = A{\rho }^{\gamma } \), (37) yields\n\n\[ \mathrm{V}\left( \rho \right) = \pm {\int }_{{\rho }_{0}}^{\rho }{\left\lbrack \gamma \mathrm{A}{\rho }^{\gamma - 1}... | Yes |
Solve the PDE \( {u}_{x} + u \cdot {u}_{y} = {6x} \), subject to the side condition \( u\left( {0, s}\right) = G\left( s\right) \) . | Solution. Since the side condition curve is the \( \mathrm{y} \) -axis traced out by \( \left( {0,\mathrm{\;s}}\right) \), we have \( \mathrm{f}\left( \mathrm{s}\right) = 0 \) and \( \mathrm{g}\left( \mathrm{s}\right) = \mathrm{s} \) . Here \( \mathrm{F}\left( {\mathrm{x},\mathrm{y},\mathrm{u},\mathrm{p},\mathrm{q}}\ri... | Yes |
Show that, given a solution \( u\left( {x, y}\right) \) of \[ {u}_{x}^{2} + {u}_{y}^{2} = c{\left( x, y\right) }^{-2} \] (known as the eikonal equation or the Hamilton-Jacobi equation, depending on the context) the curve \( u\left( {x, y}\right) = \tau \) defines a wave front curve at time \( \tau \). | Solution. Let \( \;\left( {x\left( \tau \right), y\left( \tau \right) }\right) \; \) be the position of a point on the curve \( \;u\left( {x, y}\right) = \tau \; \) at time \( \;\tau \; \) (i.e., \( \mathbf{u}\left( {\mathrm{x}\left( \tau \right) ,\mathrm{y}\left( \tau \right) }\right) = \tau \), and suppose that its v... | Yes |
A very important solution of \( {u}_{t} = k{u}_{xx} \) is\n\n\[ \mathrm{u}\left( {\mathrm{x},\mathrm{t}}\right) = {\left( 4\pi \mathrm{{kt}}\right) }^{-\frac{1}{2}}{\mathrm{e}}^{-{\mathrm{x}}^{2}/\left( {4\mathrm{{kt}}}\right) },\mathrm{t} > 0, - \infty < \mathrm{x} < \infty . \] | For now, we check that (7) is in fact a solution, using logarithmic differentiation. Note that\n\n\[ \log \left( \mathrm{u}\right) = - \frac{1}{2}\log \left( {{4\pi }\mathrm{k}}\right) - \frac{1}{2}\log \left( \mathrm{t}\right) - \frac{{\mathrm{x}}^{2}}{4\mathrm{{kt}}}. \]\n\n\[ \text{Thus,}\frac{{u}_{x}}{u} = - \frac{... | Yes |
Proposition 1. Let \( {b}_{1},\ldots ,{b}_{N} \) be given constants. A solution of the problem\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0 \)\n\nB.C. \( u\left( {0, t}\right) = 0,\;u\left( {L, t}\right) = 0 \)\n\n(19)\n\nI.C. \( u\left( {x,... | is given by\n\n\[ u\left( {x, t}\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{b}_{n}{e}^{-{\left( n\pi /L\right) }^{2}{kt}}\sin \left( {{n\pi x}/L}\right) . \]\n\n(20) | Yes |
Find a solution of the problem\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} = 2{\mathrm{u}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq \pi \) ,且 \( \mathrm{t} \geq 0 \)\n\nB.C. \( u\left( {0, t}\right) = 0\;u\left( {\pi, t}\right) = 0 \)\n\nI.C. \( u\left( {x,0}\right) = 5\sin \left( {2x}\right) - {30}\sin \left( {3x}\right) \). | Solution. Given Proposition 1, we can immediately write down a solution. Indeed, here \( L = \pi \) , \( \mathrm{k} = 2,{\mathrm{\;b}}_{1} = 0,{\mathrm{\;b}}_{2} = \dot{5},{\mathrm{\;b}}_{3} = - {30} \) and \( \mathrm{N} = 3 \) . Substituting these values into (20), we obtain\n\n\[ u\left( {x, t}\right) = 5{e}^{-{8t}}{... | Yes |
Proposition 2. Let \( {a}_{0},{a}_{1},\ldots ,{a}_{N} \) and \( {b}_{1},\ldots ,{b}_{N} \) be given constants. A solution of the problem\n\nD.E. \( \;{u}_{t} = k{u}_{xx}\; - L \leq x \leq L\;, t \geq 0 \)\n\nB.C. \( \;u\left( {-L, t}\right) = u\left( {L, t}\right) \;{u}_{x}\left( {-L, t}\right) = {u}_{x}\left( {L, t}\r... | Observe that (29) tends to the constant \( {a}_{0} \) as \( t \rightarrow \infty \) . This constant is the average temperature in the circular wire. Indeed, integrating both sides of (29), we obtain\n\n\[ \frac{1}{2L}{\int }_{-L}^{L}u\left( {x, t}\right) {dx} = \frac{1}{2L}{\int }_{-L}^{L}{a}_{0}{dx} = {a}_{0}, \]\n\ns... | Yes |
Proposition 1. Let \( {b}_{1},\ldots ,{b}_{N} \) be given constants. A solution of the problem\n\nD.E. \( {u}_{t} = k{u}_{xx}\;0 \leq x \leq L, t \geq 0 \)\n\nB.C. \( u\left( {0, t}\right) = 0\;u\left( {L, t}\right) = 0 \)\n\nI.C. \( u\left( {x,0}\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{b}_{n}\sin \left( {{n\pi x}... | is given by\n\n\[ u\left( {x, t}\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{b}_{n}{e}^{-{\left( n\pi /L\right) }^{2}{kt}}\sin \left( {{n\pi x}/L}\right) . \] | Yes |
Proposition 2. Let \( {a}_{0},{a}_{1},\ldots ,{a}_{N} \) and \( {b}_{1},\ldots ,{b}_{N} \) be given constants. A solution of the problem\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{\mathrm{{xx}}}\; - \mathrm{L} \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0 \)\n\nB.C. \( u\left( {-L, t}\right) = u\left( ... | is given by\n\n\[ u\left( {x, t}\right) = {a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{N}{e}^{-{\left( n\pi /L\right) }^{2}{kt}}\left\lbrack {{a}_{n}\cos \left( {{n\pi x}/L}\right) + {b}_{n}\sin \left( {{n\pi x}/L}\right) }\right\rbrack . \] | Yes |
Theorem 3 (Continuous Dependence on the I.C. and the B.C.). Let \( {u}_{1}\left( {x, t}\right) \) and \( {u}_{2}\left( {x, t}\right) \) be \( {C}^{2} \) solutions of the respective problems \( \left( {0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0}\right) \)\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{... | Proof. Let \( v\left( {x, t}\right) = {u}_{1}\left( {x, t}\right) - {u}_{2}\left( {x, t}\right) \) . Then \( {v}_{t} = k{v}_{xx} \) and we have\n\n\[ \left| {v\left( {x,0}\right) }\right| = \left| {{f}_{1}\left( x\right) - {f}_{2}\left( x\right) }\right| \leq \epsilon ,\;0 \leq x \leq L, \]\n\n\[ \left| {v\left( {0, t}... | Yes |
Show that the no matter how small the constant \( \alpha \) is, the solution will become large as \( t \rightarrow {1}^{ - }. \) | Solution. For any fixed \( x \), the PDE is a separable ODE, namely \( {du}/u = {\left( 1 - t\right) }^{-1}{dt} \), and the general solution is \( u\left( {x, t}\right) = f\left( x\right) /\left( {1 - t}\right) \) for a \( {C}^{1} \) function \( f\left( x\right) \) . Since, \( f\left( x\right) = u\left( {x,0}\right) = ... | Yes |
For suitable initial distributions \( f\left( x\right) \), derive the solution of the problem\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0 \)\n\n\[ \n\text{B.C.}{u}_{x}\left( {0, t}\right) = 0\;u\left( {L, t}\right) = 0 \n\]\n\n(9)\n\nI.C. \... | Solution. Again, we find the product solutions of the D.E. that obey the B.C. . In Case 1 (cf. (11) of Section 3.1), we have (3) and (4) of Example 1, and the first B.C. yields \( {c}_{1} = 0 \) as before. However, the second B.C. yields\n\n\[ \n0 = u\left( {L, t}\right) = {e}^{-{\lambda }^{2}{kt}}{c}_{2}\cos \left( {\... | Yes |
For arbitrary real constants, a and \( b \), and suitable \( g\left( x\right) \), solve the problem\n\n\[ \n\text{ D.E. }{\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0 \n\]\n\n\[ \n\text{B.C.}u\left( {0, t}\right) = a\;u\left( {L, t}\right) = b \n\]\n\n(... | Solution. We first seek a particular solution \( {u}_{D}\left( {x, t}\right) \) of the D.E. and B.C. . Since any particular solution will do, we may as well strive for simplicity. Indeed, a Case 3 product solution (cf. (13) of Section 3.1) \( {u}_{D}\left( {x, t}\right) = {cx} + {dwill} \) do, if \( c \) and \( d \) ar... | Yes |
Example 4. Solve D.E. \( {\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0 \)\n\nB.C. \( u\left( {0, t}\right) = 0\;u\left( {L, t}\right) = L \)\n\n(18)\n\n\\[ \n\\text{I.C.}u\\left( {x,0}\\right) = x + 3\\sin \\left( {{2\\pi x}/L}\\right) \\text{.} \n\\] | Solution. We can find a particular solution of the D.E. and B.C. of the form \( {u}_{p}\left( {x, t}\right) = {cx} + d \) . From the B.C., \( 0 = u\left( {0, t}\right) = c \cdot 0 + d = d \) and \( L = u\left( {L, t}\right) = {cL} \) . Thus, \( d\overset{r}{ = }0, c = 1 \) and \( {u}_{p}\left( {x, t}\right) = x \) . Th... | Yes |
Example 5. Solve D.E. \( {\mathrm{u}}_{\mathrm{t}} = 2{\mathrm{u}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq 1,\mathrm{t} \geq 0 \)\n\nB.C. \( {u}_{x}\left( {0, t}\right) = 1\;u\left( {1, t}\right) = - 1 \)\n\nI.C. \( u\left( {x,0}\right) = x + {\cos }^{2}\left( {{3\pi x}/4}\right) - \frac{5}{2} \). | Solution. We try a particular solution of the form \( {u}_{p}\left( {x, t}\right) = {cx} + d \). The first B.C. yields \( c = 1 \), while \( {u}_{p}\left( {1, t}\right) = 1 + d \) yields \( d = - 2 \) by the second B.C. . Thus, \( {u}_{p}\left( {x, t}\right) = x - 2 \). The related homogeneous problem is\n\nD.E. \( {v}... | Yes |
Proposition 1. A solution of problem (1) is given by\n\n\[ u\left( {x, t}\right) = w\left( {x, t}\right) + {u}_{1}\left( {x, t}\right) + {u}_{2}\left( {x, t}\right) ,\] | where \( w\left( {x, t}\right) \) is the particular solution (2) of the B.C. and \( {u}_{1}\left( {x, t}\right) \) solves (6a) with \( g\left( x\right) \) \( = f\left( x\right) - w\left( {x,0}\right) \texttt{ and }{u}_{2}\left( {x, t}\right) \texttt{ solves (6b) with }h\left( {x, t}\right) = - \left( {{w}_{t} - k{w}_{x... | Yes |
Reduce the problem\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} = {\mathrm{{ku}}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{t} \geq 0 \)\n\n\[ \text{B.C.}u\left( {0, t}\right) = a\left( t\right) \;{u}_{x}\left( {L, t}\right) = b\left( t\right) \]\n\n(7)\n\nI.C. \( u\left( {x,0}\right) = f\left( x\right) \)\n\n... | Solution. Note that \( w\left( {x, t}\right) = b\left( t\right) x + a\left( t\right) \) satisfies the B.C. of (7). Setting \( v\left( {x, t}\right) = u\left( {x, t}\right) \) \( - w\left( {x, t}\right) , \) we have \( {v}_{t} - k{v}_{xx} = {u}_{t} - k{u}_{xx} - \left( {{w}_{t} - k{w}_{xx}}\right) = - {b}^{\prime }\left... | Yes |
Example 2. Solve the following problem with heat source distribution \( \;h\left( {x, t}\right) = t \cdot \sin \left( x\right) \;. \)\n\nD.E. \( {u}_{t} - k{u}_{xx} = t \cdot \sin \left( x\right) \;0 \leq x \leq \pi, t \geq 0 \)\n\nB.C. \( u\left( {0, t}\right) = 0\;u\left( {\pi, t}\right) = 0 \)\n\nI.C. \( u\left( {x,... | Solution. We solve the related problem (14)\n\nD.E. \( {\widetilde{v}}_{t} = k{\widetilde{v}}_{xx}\;0 \leq x \leq \pi, t \geq 0 \)\n\nB.C. \( \widetilde{v}\left( {0, t;s}\right) = 0\;\widetilde{v}\left( {\pi, t;s}\right) = 0 \)\n\nI.C. \( \widetilde{v}\left( {x,0;s}\right) = h\left( {x, s}\right) = s \cdot \sin \left( ... | Yes |
Example 3. Solve the following problem, where the heat source distribution is \( {e}^{-{ct}}\sin \left( x\right) \) for an arbitrary constant c. Does anything interesting happen when \( \mathrm{c} \approx 1 \) ?\n\nD.E. \( {\mathrm{u}}_{\mathrm{t}} - {\mathrm{u}}_{\mathrm{{xx}}} = {\mathrm{e}}^{-\mathrm{{ct}}}\sin \lef... | Solution. We solve the related problem (14) with \( \;h\left( {x, s}\right) \; = \;{e}^{-{cs}}{sin}\left( x\right) \;,\; \) obtaining \( \;\widetilde{v}\left( {x, t;s}\right) \; = \) \( {\mathrm{e}}^{-\mathrm{{cs}}}{\mathrm{e}}^{-\mathrm{t}}\sin \left( \mathrm{x}\right) \) . Then, by (15) the solution should be\n\n\[ u... | Yes |
Lemma 1. Suppose \( g\left( {t, s}\right) \) and \( {g}_{t}\left( {t, s}\right) \) are continuous functions. Then\n\n\[ \frac{d}{dt}\left\lbrack {{\int }_{0}^{t}g\left( {t, s}\right) {ds}}\right\rbrack = g\left( {t, t}\right) + {\int }_{0}^{t}{g}_{t}\left( {t, s}\right) {ds} \] | Proof. Let \( \mathrm{H}\left( {\mathrm{t},\mathrm{y}}\right) \) be defined by\n\n\[ H\left( {t, y}\right) = {\int }_{0}^{y}g\left( {t, s}\right) {ds} \]\n\nWe compute \( \frac{\mathrm{d}}{\mathrm{{dt}}}\mathrm{H}\left( {\mathrm{t},\mathrm{t}}\right) \), since this is the left-hand side of (17). Let \( \mathrm{y}\left(... | Yes |
Theorem 1 (Duhamel’s principle). Suppose that \( \;h\left( {x, t}\right) \; \) is a given \( \;{C}^{2} \) function for \( \;0 \leq x \leq L\;, \) \( \mathrm{t} \geq 0 \) . Assume that for each \( \mathrm{s} \geq 0 \) the problem\n\nD.E. \( {\mathrm{v}}_{\mathrm{t}} = {\mathrm{{kv}}}_{\mathrm{{xx}}}\;0 \leq \mathrm{x} \... | Proof. The function \( u\left( {x, t}\right) \) defined by (21) satisfies the I.C. \( u\left( {x,0}\right) = 0 \) . It also satisfies the B.C. of \( \left( {20}\right) \), since \( v\left( {x, t;s}\right) \) ’satisfies the B.C. of \( \left( {19}\right) . \) Now use Lemma 1, with \( g\left( {t, s}\right) = v\left( {x, t... | Yes |
Example 5. Solve the problem\n\n\\[ \n\\text{D.E.}{u}_{t} - 8{u}_{xx} = \\cos \\left( t\\right) + {e}^{t}\\sin \\left( {x/2}\\right) \\;0 \\leq x \\leq \\pi, t \\geq 0 \n\\]\n\nB.C. \\( u\\left( {0, t}\\right) = \\sin \\left( t\\right) \\;{u}_{x}\\left( {\\pi, t}\\right) = 0 \\)\n\n(26)\n\nI.C. \\( u\\left( {x,0}\\righ... | Solution. Note that \\( w\\left( {x, t}\\right) = \\sin \\left( t\\right) \\) satisfies the B.C. . Letting \\( v\\left( {x, t}\\right) = u\\left( {x, t}\\right) - w\\left( {x, t}\\right) \\), we obtain the related problem\n\nD.E. \\( {v}_{t} - 8{v}_{xx} = {u}_{t} - 8{u}_{xx} - \\left( {{w}_{t} - 8{w}_{xx}}\\right) = {e... | Yes |
Example 6. Solve the following inhomogeneous problem with insulated ends.\n\nD.E. \( {u}_{t} - {u}_{xx} = \left( {{2t} + 1}\right) \cos \left( {3x}\right) \;0 \leq x \leq \pi, t \geq 0 \)\n\nB.C. \( {u}_{x}\left( {0, t}\right) = 0\;{u}_{x}\left( {\pi, t}\right) = 0 \)\n\nI.C. \( u\left( {x,0}\right) = 0 \) . | Solution. We apply Duhamel's principle when the ends are insulated, to obtain \( u\left( {x, t}\right) = {\int }_{0}^{t}\widetilde{v}\left( {x, t - s;s}\right) {ds} \), where \( \widetilde{v}\left( {x, t;s}\right) \) solves the problem\n\nD.E. \( {\widetilde{v}}_{t} - {\widetilde{v}}_{xx} = 0\;0 \leq x \leq \pi, t \geq... | Yes |
Proposition 1. The family of functions \( {s}_{n}\left( x\right) \equiv \sin \left( {{n\pi x}/L}\right) \;\left( {n = 1,2,3\ldots }\right) \) is orthogonal of norm-square \( \mathrm{L} \) on the interval \( \left\lbrack {-\mathrm{L},\mathrm{L}}\right\rbrack \) (cf. (8)). | Proof. We apply (10) with \( f\left( x\right) = {s}_{n}\left( x\right) \) and \( g\left( x\right) = {s}_{m}\left( x\right) \), noting that \( {s}_{n}^{\prime \prime } = - {\left( n\pi /L\right) }^{2}{s}_{n} \) and\n\n\[{\mathrm{s}}_{\mathrm{m}}{}^{\prime \prime } = - {\left( \mathrm{m}\pi /\mathrm{L}\right) }^{2}{\math... | Yes |
Show that the constant function \( {c}_{0}\left( x\right) = \cos \left( {{0\pi x}/L}\right) \equiv 1 \) is orthogonal to each member of the family \( {\mathrm{s}}_{1},{\mathrm{c}}_{1},{\mathrm{s}}_{2},{\mathrm{c}}_{2},\ldots \) . However, note that \( {\mathrm{c}}_{0} \) does not have norm-square \( \mathrm{L} \) . Rem... | Solution. Note that \( < {\mathrm{c}}_{0},{\mathrm{\;s}}_{\mathrm{n}} > = {\int }_{-\mathrm{L}}^{\mathrm{L}}1 \cdot \sin \left( {\mathrm{n}\pi \mathrm{x}/\mathrm{L}}\right) \mathrm{{dx}} = 0 \), and similarly \( < {\mathrm{c}}_{0},{\mathrm{c}}_{\mathrm{n}} > = 0,\mathrm{n} = 1,2 \) , 3,... . Thus, \( {\mathrm{c}}_{0} \... | Yes |
Theorem 1. Suppose that \( f\left( x\right) \) is of the form\n\n\[ f\left( x\right) = \frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}\cos \left( {{n\pi x}/L}\right) + {b}_{n}\sin \left( {{n\pi x}/L}\right) .\n\]\n\nThen, the coefficients \( {a}_{n} \) and \( {b}_{n} \) are uniquely determined by the fo... | Proof. In terms of \( {s}_{n} \) and \( {c}_{n} \), we can write (13) in the form\n\n\[ f = \frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}{c}_{n} + {b}_{n}{s}_{n} \]\n\n\( \left( {13}^{\prime }\right) \)\n\nTaking the inner product of both sides of \( \left( {13}^{\prime }\right) \) with \( {\mathrm{c}... | Yes |
Find the Fourier series of the function \( f\left( x\right) = x \) for \( - L \leq x \leq L \) . | Solution. We compute the Fourier coefficients \( {a}_{n} \) first for \( n \geq 1 \) ,\n\n\[ \n{a}_{n} = \frac{1}{L}{\int }_{-L}^{L}x \cdot \cos \left( {{n\pi x}/L}\right) {dx} = {\left. \frac{x}{n\pi }\sin \left( n\pi x/L\right) \right| }_{-L}^{L} - \frac{1}{n\pi }{\int }_{-L}^{L}\sin \left( {{n\pi x}/L}\right) {dx} \... | Yes |
Compute the Fourier series of \( f\left( x\right) = x \) defined on the interval \( \left\lbrack {0,{2L}}\right\rbrack \) . | Solution. The integrals for the Fourier coefficients are now from 0 to \( 2\mathrm{\;L} \) . Thus, \( {a}_{0} = \frac{1}{L}{\int }_{0}^{2L}x\mathrm{\;d}x \) \( = 2\mathrm{\;L} \), while for \( \mathrm{n} = 1,2,3,\ldots \), we have\n\n\[ \n{a}_{n} = \frac{1}{L}{\int }_{0}^{2L}x \cdot \cos \left( {{n\pi x}/L}\right) \;{d... | Yes |
Compute \( \mathrm{{FS}}\mathrm{f}\left( \mathrm{x}\right) \), noting that \( \mathrm{L} = \pi \) . | Solution. Since \( \mathrm{f}\left( \mathrm{x}\right) = 0 \) for \( - \pi \leq \mathrm{x} \leq 0 \), we have (for \( \mathrm{n} > 0 \) )\n\n\[ \n{a}_{n} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left( x\right) \cos \left( {nx}\right) {dx} = \frac{1}{\pi }{\int }_{0}^{\pi }x \cdot \cos \left( {nx}\right) {dx} \n\]\n\n\[ \... | Yes |
Theorem 2 (same as Theorem 2 of Section 4.2). Let \( f\left( x\right) \) be a \( {C}^{2} \) function on the interval \( \left\lbrack {-L, L}\right\rbrack \), such that \( f\left( {-L}\right) = f\left( L\right) \) and \( {f}^{\prime }\left( {-L}\right) = {f}^{\prime }\left( L\right) \) . Let \( {a}_{n} \) and \( {b}_{n}... | \[ \left| {f\left( x\right) - \left\lbrack {\frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}\cos \left( {{n\pi x}/L}\right) + {b}_{n}\sin \left( {{n\pi x}/L}\right) }\right\rbrack }\right| \leq \frac{4{L}^{2}M}{{\pi }^{2}N}, \] for all \( \mathrm{x} \) in \( \left\lbrack {-\mathrm{L},\mathrm{L}}\right\rb... | Yes |
Example 5. Take \( L = 1 \) and \( f\left( x\right) = {x}^{3} - x, - 1 \leq x \leq 1 \) . Apply Theorem 2 to get an estimate on the number of terms of \( {FSf}\left( x\right) \) needed to approximate \( f\left( x\right) \) within an error of .01 . | Solution. First, we check that \( f\left( x\right) \) satisfies the hypotheses of Theorem 2. Note that \( f\left( {-1}\right) = f\left( 1\right) \) \( = 0 \) and \( {f}^{\prime }\left( {-1}\right) = {f}^{\prime }\left( 1\right) = 2 \), since \( {f}^{\prime }\left( x\right) = 3{x}^{2} - 1 \) ; also, \( {f}^{\prime \prim... | Yes |
Compute the Fourier series of \( f\left( x\right) = {x}^{3}, - L \leq x \leq L \) . | Solution. We have \( {a}_{n} = {L}^{-1} < {c}_{n}, f > = \frac{1}{L}{\int }_{-L}^{L}{x}^{3}\cos \left( {{n\pi x}/L}\right) {dx} = 0 \), because the integrand is odd (i.e., it is changed to its negative under replacing \( x \) by \( - x \) ). To compute \( {b}_{n} = {L}^{-1} < {s}_{n}, f > \), we could integrate by part... | Yes |
Show that for \( f\left( x\right) = {x}^{3} - x \), defined on \( \left\lbrack {-1,1}\right\rbrack \), we have\n\n\[\n\mathrm{{FS}}f\left( x\right) = \frac{12}{{\pi }^{3}}\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n}\frac{1}{{n}^{3}}\sin \left( {n\pi x}\right) .\n\] | Solution. We computed FS x in Example 2, and FS \( {x}^{3} \) in Example 6. Taking \( L = 1 \) in these examples, we have \( \operatorname{FS}f\left( x\right) = \operatorname{FS}\left( {{x}^{3} - x}\right) = \operatorname{FS}{x}^{3} - \operatorname{FS}x \), where the last equation follows from \( < {x}^{3} - x,{s}_{n} ... | Yes |
Theorem 2 (same as Theorem 2 of Section 4.2). Let \( f\left( x\right) \leq \) be a \( {C}^{2} \) function on the interval \( \left\lbrack {-L, L}\right\rbrack \), such that \( f\left( {-L}\right) = f\left( L\right) \) and \( {f}^{\prime }\left( {-L}\right) = {f}^{\prime }\left( L\right) \) . Let \( {a}_{n} \) and \( {b... | \[ \left| {f\left( x\right) - \left\lbrack {\frac{1}{2}{a}_{0} + \mathop{\sum }\limits_{{n = 1}}^{N}{a}_{n}\cos \left( {{n\pi x}/L}\right) + {b}_{n}\sin \left( {{n\pi x}/L}\right) }\right\rbrack }\right| \leq \frac{4{L}^{2}M}{{\pi }^{2}N}, \] for all \( x \) in \( \left\lbrack {-L, L}\right\rbrack \) . | Yes |
Proposition 1. If \( g\\left( x\\right) \) is a periodic function of period \( 2 \) L, then the integral (if it exists) of \( g\\left( x\\right) \) over an interval of length \( {2L} \) is the same as the integral of \( g\\left( x\\right) \) over any other interval of length \( 2\\mathrm{\\;L} \) . In other words, for ... | Proof. Since the interval \( \\;\\left\\lbrack {L, L + c}\\right\\rbrack \) is obtained from \( \\;\\left\\lbrack {-L, - L + c}\\right\\rbrack \) by shifting to the right by \( \\;{2L} \) and \( g\\left( x\\right) \) is periodic of period \( {2L} \), we have\n\n\[{\\int }_{-L}^{-L + c}g\\left( x\\right) {dx} = {\\int }... | Yes |
Compute \( {\int }_{0}^{2\pi }{\sin }^{5}\left( x\right) \cdot {\cos }^{100}\left( x\right) {dx} \). | The integrand is periodic of period \( {2\pi } \). Thus, according to Proposition 1, the integral is\n\n\[ \n{\int }_{-\pi }^{\pi }{\sin }^{5}\left( \mathrm{x}\right) \cdot {\cos }^{100}\left( \mathrm{x}\right) \mathrm{{dx}}. \n\]\n\nThe integrand is odd (i.e., it is changed to its negative, by replacing \( x \) by \( ... | Yes |
Use Bessel’s inequality for \( f\left( x\right) = x\left( {-L \leq x \leq L}\right) \) to prove that\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{2}} \leq \frac{{\pi }^{2}}{6} \] | Solution. We computed FS f(x) in Example 2 of Section 4.1, and found that\n\n\[ {FSf}\left( x\right) = \frac{2L}{\pi }\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( -1\right) }^{n + 1}\frac{1}{n}\sin \left( {{n\pi x}/L}\right) ,\]\n\n(i.e., \( {a}_{n} = 0, n \geq 0 \) and \( {b}_{n} = {\left( -1\right) }^{n + 1}{2L}/... | Yes |
Proposition 2. Let \( f\left( x\right) \) be a \( {C}^{2} \) function on \( \left\lbrack {-L, L}\right\rbrack \) such that \( f\left( {-L}\right) = f\left( L\right) \) and \( {f}^{\prime }\left( {-L}\right) \) \( = {f}^{\prime }\left( L\right) \) . Let \( M \) be the maximum of \( \left| {{f}^{\prime \prime }\left( x\r... | Proof. We use Green’s formula (cf. Example 6 in Section 4.1 ) to get, for \( \mathrm{n} \geq 1 \) ,\n\n\[ = \frac{-L}{{n}^{2}{\pi }^{2}}{\int }_{-L}^{L}{f}^{\prime \prime }\left( x\right) \cos \left( {{n\pi x}/L}\right) {dx} \]\n\nwhere the endpoint evaluations cancel by the assumptions \( f\left( {-L}\right) = f\left(... | Yes |
Proposition 3. For any real \( \\theta \) such that \( \\sin \\left( {\\theta /2}\\right) \\neq 0 \) , \n\n\[ \n\\frac{1}{2} + \\cos \\left( \\theta \\right) + \\cos \\left( {2\\theta }\\right) + \\ldots + \\cos \\left( {\\mathrm{n}\\theta }\\right) = \\frac{\\sin \\left( {\\left\\lbrack {\\mathrm{n} + \\frac{1}{2}}\\r... | Proof. Multiplying the left side of (12) by \( 2\\sin \\left( {\\theta /2}\\right) \), we obtain \n\n\[ \n\\mathrm{{sin}}\\left( {\\theta /2}\\right) + 2\\mathrm{{sin}}\\left( {\\theta /2}\\right) \\mathrm{{cos}}\\left( \\theta \\right) + 2\\mathrm{{sin}}\\left( {\\theta /2}\\right) \\mathrm{{cos}}\\left( {2\\theta }\\... | Yes |
Theorem 1 (Pointwise Convergence of Fourier Series). Let \( f\left( x\right) \) be \( {C}^{1} \) on \( \left\lbrack {-L, L}\right\rbrack \), and assume that \( f\left( {-L}\right) = f\left( L\right) \) and \( {f}^{\prime }\left( {-L}\right) = {f}^{\prime }\left( L\right) \) [so that \( f\left( x\right) \) may be regard... | Proof. We demonstrate that the limit (22) holds by virtue of (15) in Proposition 4. We write \( {S}_{N}\left( x\right) \) in terms of an integral involving \( {D}_{N}^{\prime } \) . Using the definition (20) of \( {\widehat{S}}_{N}\left( x\right) \) and keeping \( x \) fixed, so that \( \cos \left( {\mathrm{n}\pi \math... | No |
Show that the sequences \( {f}_{1}\left( x\right) ,{f}_{2}\left( x\right) ,{f}_{3}\left( x\right) ,\ldots \) and \( {g}_{1}\left( x\right) ,{g}_{2}\left( x\right) ,{g}_{3}\left( x\right) ,\ldots \) both converge pointwise to the zero function \( h\left( x\right) \equiv 0. \) However, show that the sequence \( {g}_{1}\l... | Solution. We have \( {g}_{n}\left( x\right) = \left( {x + 2}\right) /\left( {4n}\right) \), and so \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{g}_{n}\left( x\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {\left( {x + 2}\right) /\left( {4n}\right) }\right\rbrack = 0 = h\left( x\right) \) fo... | Yes |
Theorem 2. Let \( f\left( x\right) \) be defined and \( {C}^{2} \) on the interval \( \left\lbrack {-L, L}\right\rbrack \), with \( f\left( {-L}\right) = f\left( L\right) \) and \( {f}^{\prime }\left( {-L}\right) = {f}^{\prime }\left( L\right) \) . Then FS \( f\left( x\right) \) converges uniformly to \( f\left( x\righ... | Proof. We know from Theorem 1 that \( \operatorname{FS}f\left( x\right) = f\left( x\right) \) . Thus we have\n\n\[ f\left( x\right) - {S}_{N}\left( x\right) = {FS}\;f\left( x\right) - {S}_{N}\left( x\right) = \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\left\lbrack {{a}_{n}{cos}\left( {{n\pi x}/L}\right) + {b}_{n}{sin... | Yes |
Find two functions \( f\left( x\right) \) and \( g\left( x\right) \) whose graphs are congruent, such that \( f\left( x\right) \) is \( {C}^{\infty } \) and \( \mathrm{g}\left( \mathrm{x}\right) \) is not even piecewise \( {\mathrm{C}}^{1} \) . | Solution. Let \( f\left( x\right) = {x}^{3} \) and let \( g\left( x\right) = \sqrt[3]{x} \) . The graphs are congruent, since the the graphs of inverse functions are reflections of each other in the line \( y = x \) . Clearly, \( f\left( x\right) \) is \( {C}^{\infty } \), but note that \( {\mathrm{g}}^{\prime }\left( ... | Yes |
Theorem 3. Let \( f\left( x\right) \) be a piecewise \( {C}^{1} \) function on \( \left\lbrack {-L, L}\right\rbrack \) and let \( f\left( x\right) \) be the adjusted function in (28). Then \( \mathrm{{FS}}f\left( x\right) = f\left( x\right) \), for all \( x \) in \( \left\lbrack {-L, L}\right\rbrack \) . Indeed, we hav... | Proof. The proof proceeds in the same way as the proof of Theorem 1, until we reach (23) which is now replaced by the following limit that must be established for each \( \mathrm{x} \) in \( \left\lbrack {-\mathrm{L},\mathrm{L}}\right\rbrack \) : \[ \mathop{\lim }\limits_{{N \rightarrow \infty }}\frac{1}{L}{\int }_{-L}... | Yes |
Proposition 5. Let \( h\left( x\right) \) be a piecewise \( {C}^{1} \) function on \( \left\lbrack {-L, L}\right\rbrack \) . If \( {D}_{n} \) denotes the \( n - \) th Dirichlet kernel (14), then \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{L}{\int }_{-L}^{L}{D}_{n}\left( x\right) h\left( x\right) {dx} = \... | Proof. The result (32) follows from adding the two results : \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{L}{\int }_{-L}^{0}{D}_{n}\left( x\right) h\left( x\right) {dx} = \frac{1}{2}h\left( {0}^{ - }\right) \text{and}\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{L}{\int }_{0}^{L}{D}_{n}\left( x\... | Yes |
Verify directly that \( \mathrm{{FS}}f\left( x\right) \) converges to the adjusted function at the discontinuities of \( f\left( x\right) \), but show that the convergence of \( \mathrm{{FS}}f\left( x\right) \) to \( f\left( x\right) \) on \( \left\lbrack {-L, L}\right\rbrack \) is not uniform. | Solution. It is not necessary to compute the Fourier coefficients to draw the graph of \( \mathrm{{FS}}f\left( x\right) \) . According to Theorem 3, \( {FSf}\left( x\right) \) is the periodic extension \( \overset{ \sim }{\bar{f}}\left( x\right) \) which is graphed in Figure 8.\n\n![b1d8443a-d96a-44a5-a3a1-2e99c7819d0d... | Yes |
Theorem 4. Let \( f\left( x\right) \) be a continuous piecewise \( {C}^{1} \) function on \( \left\lbrack {-L, L}\right\rbrack \), such that \( f\left( {-L}\right) = f\left( L\right) \). Then’FS \( f\left( x\right) \) converges uniformly to \( f\left( x\right) \) on \( \left\lbrack {-L, L}\right\rbrack \). In other wor... | Proof. From Theorem 3, we already know that \( {FSf}\left( x\right) = f\left( x\right) \) for \( x \) in \( \left\lbrack {-L, L}\right\rbrack \), since the adjusted function \( f\left( x\right) \) is the same as \( f\left( x\right) \) by the continuity assumptions on \( f\left( x\right) \). Thus, \[ f\left( x\right) - ... | No |
Proposition 1. Let \( f\left( x\right) \) be a function, defined for \( - L \leq x \leq L \), with Fourier coefficients\n\n\[ \n{a}_{n} = \frac{1}{L}{\int }_{-L}^{L}f\left( x\right) \cos \left( {{n\pi x}/L}\right) {dx}\;\text{ and }\;{b}_{n} = \frac{1}{L}{\int }_{-L}^{L}f\left( x\right) \sin \left( {{n\pi x}/L}\right) ... | Proof. If \( f\left( x\right) \) is even, then \( {b}_{n} = 0 \) by facts \( \left( C\right) \) and \( \left( D\right) \), since \( \sin \left( {{n\pi x}/L}\right) \) is odd. Formula (3) follows from (A) and (E). The case when \( f\left( x\right) \) is odd is handled similarly. \( ▱ \) | Yes |
Proposition 2. Let \( f\left( x\right) \) be defined for \( 0 \leq x \leq L \), and suppose that the integrals in (5) and (6) exist. Then (redefining \( f\left( 0\right) \) to be 0 ) the Fourier sine series of \( f\left( x\right) \) is the Fourier series of the odd extension \( {f}_{0}\left( x\right) \) defined on \( \... | Proof. We simply check that the Fourier coefficients of \( {f}_{o}\left( x\right) \) are given by \( {a}_{n} = 0 \) and \( {b}_{n} \) as in (5). Indeed, \( {a}_{n} = 0 \) (for all \( n = 0,1,2,\ldots \) ) by Proposition 1, and \[ {b}_{n} = \frac{1}{L}{\int }_{-L}^{L}{f}_{o}\left( x\right) \sin \left( {{n\pi x}/L}\right... | Yes |
Theorem 1. Let \( f\left( x\right) \) be a piecewise \( {C}^{1} \) function defined on \( \left\lbrack {0, L}\right\rbrack \) . Then\n\n\[ \operatorname{FSS}f\left( x\right) = \left\{ {\begin{array}{ll} \frac{1}{2}\left\lbrack {f\left( {x}^{ - }\right) + f\left( {x}^{ + }\right) }\right\rbrack & 0 < x < L \\ 0 & x = 0\... | Proof. If necessary, redefine \( f\left( 0\right) \) to be 0, and let \( {f}_{0}\left( x\right) \) be the odd extension of \( f\left( x\right) \) . Then \( {f}_{o}\left( x\right) \) is a piecewise \( {C}^{1} \) function defined on \( \left\lbrack {-L, L}\right\rbrack . \) By Theorem \( 3 \) of Section \( {4.2} \), we k... | Yes |
Theorem 2. Let \( f\left( x\right) \) be a piecewise \( {C}^{1} \) function on \( \left\lbrack {0, L}\right\rbrack \) . Then\n\n\[ \n{FCSf}\left( x\right) = \left\{ \begin{matrix} \frac{1}{2}\left\lbrack {f\left( {x}^{ - }\right) + f\left( {x}^{ + }\right) }\right\rbrack & 0 < x < L \\ f\left( {0}^{ + }\right) & x = 0 ... | Proof. We apply Theorem 3 of Section 4.2 to the (piecewise \( {C}^{1} \) ) even extension \( {f}_{e}\left( x\right) \) in order to obtain \( \begin{aligned} \operatorname{FCS}f\left( x\right) & = {\bar{f}}_{e}\left( x\right) . \\ \text{ Note that }{\bar{f}}_{e}\left( 0\right) & = \frac{1}{2}\left\lbrack {{f}_{e}\left( ... | Yes |
Find the Fourier sine and cosine series for the function \( f\left( x\right) = L - x\;\left( {0 \leq x \leq L}\right) \), and sketch the graphs of FSS f(x) and FCS f(x) in the interval \( \left\lbrack {-{3L},{3L}}\right\rbrack \) . | Solution. We compute FSS f(x), using Green’s formula (cf. (9) of Section 4.1). Recall that \( {s}_{n}\left( x\right) \) \( = \sin \left( {{n\pi x}/L}\right) \) . Using the inner product notation \( < g, h > = {\int }_{0}^{L}g\left( x\right) h\left( x\right) \mathrm{d}x \) (now on \( \left\lbrack {0, L}\right\rbrack ! \... | Yes |
Find the Fourier cosine series of \( f\left( x\right) = \sin \left( x\right) \left\lbrack {0 \leq x \leq \pi }\right\rbrack \), and sketch the graph of FCS \( f\left( x\right) , \) for \( - {2\pi } \leq \mathrm{x} \leq {2\pi } \) . | Solution. We use Green’s formula to compute \( {a}_{n} \), for \( n = 2,3,4,\ldots \) :\n\n\[ \n{a}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }f\left( x\right) \cos \left( {nx}\right) {dx} = \frac{2}{\pi } < f,{c}_{n} > = - \frac{2}{\pi {n}^{2}} < f,{c}_{n}^{\prime \prime } > \n\]\n\n\[ \n= - \frac{2}{\pi {n}^{2}}\left\lbra... | Yes |
Theorem 3. The Fourier sine series of \( {f}^{e}\left( x\right) \) on \( \left\lbrack {0,{2L}}\right\rbrack \) is given by\n\n\[ \operatorname{FSS}{f}^{e}\left( x\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}\sin \left\lbrack {\left( {n + \frac{1}{2}}\right) {\pi x}/L}\right\rbrack ,\] \n\nwhere \n\n\[ {c}_... | Proof. By definition, the Fourier sine series of \( {f}^{e}\left( x\right) \) defined on \( \left\lbrack {0,{2L}}\right\rbrack \) is given by\n\n\[ \mathrm{{FSS}}{f}^{e}\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }{b}_{k}\sin \left( {{k\pi x}/{2L}}\right) ,\;\text{ where }\;{b}_{k} = \frac{2}{2L}{\int }_{... | Yes |
Example 3. Let \( f\left( x\right) = x\left( {{2L} - x}\right) \), for \( 0 \leq x \leq L \) . Find a series representation for \( f\left( x\right) \) of the form\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}\sin \left\lbrack {\left( {n + \frac{1}{2}}\right) {\pi x}/L}\right\rbrack . \] | Solution. According to Theorem 3, the series (19) with \( {c}_{n} \) defined by (16) will converge uniformly to \( f\left( x\right) \) on \( \left\lbrack {0, L}\right\rbrack \), since \( f\left( x\right) \) is piecewise \( {C}^{1} \), continuous and \( f\left( 0\right) = 0 \) . We compute the coefficients \( {c}_{n} \)... | Yes |
Attempt to find an exact solution of the problem\n\n\\[ \n\\text{ D.E. }{\\mathrm{u}}_{\\mathrm{t}} = {\\mathrm{u}}_{\\mathrm{{xx}}}\\;0 \\leq \\mathrm{x} \\leq \\mathrm{L},\\mathrm{t} \\geq 0 \n\\]\n\n\\[ \n\\text{B.C.}u\\left( {0, t}\\right) = 0\\;u\\left( {L, t}\\right) = 0 \n\\]\n\n(20)\n\n\\[ \n\\text{ I.C. }u\\le... | Although \\( \\;f\\left( x\\right) = x\\left( {L - x}\\right) \\;\\left\\lbrack {0 \\leq x \\leq L}\\right\\rbrack \\) is not a finite linear combination of the functions \\( \\sin \\left( {{n\\pi x}/L}\\right) \\), we know from Theorem 1 that’ FSS f(x) \\( = f\\left( x\\right) \\) on [0, L]. The Fourier sine coefficie... | Yes |
Find the formal solution of the problem\n\nD.E. \( {u}_{t} = k{u}_{xx}\;0 \leq x \leq \pi, t \geq 0 \)\n\nB.C. \( {u}_{x}\left( {0, t}\right) = 0\;{u}_{x}\left( {\pi, t}\right) = 0 \)\n\nI.C. \( u\left( {x,0}\right) = \sin \left( x\right) \). | Solution. From the B.C., we know (cf. Example 1 of Section 3.3) that we should expand \( \;{sin}\left( x\right) \) \( \left( {0 \leq \mathrm{x} \leq \pi }\right) \) into a cosine series. In Example 2, we found\n\n\[ \sin \left( \mathrm{x}\right) = \frac{2}{\pi } - \frac{4}{\pi }\mathop{\sum }\limits_{{\mathrm{n} = 1}}^... | Yes |
Find a formal solution of the following problem :\n\n\\[ \n\\text{ D.E. }{\\mathrm{u}}_{\\mathrm{t}} = {\\mathrm{u}}_{\\mathrm{{xx}}}\\;0 \\leq \\mathrm{x} \\leq \\pi \\text{,且 }\\mathrm{t} \\geq 0 \n\\]\n\n\\[ \n\\text{ B.C. }u\\left( {0, t}\\right) = \\sin \\left( t\\right) \\;u\\left( {\\pi, t}\\right) = 0 \n\\]\n\n... | Solution. We apply the methods of Section 3.4 involving Duhamel's principle, and we treat the infinite sum which arises in a formal manner. A particular solution of the B.C. is easily found to be \\( \\;w\\left( {x, t}\\right) = \\left( {1 - \\frac{x}{\\pi }}\\right) \\sin \\left( t\\right) \\;.\\; \\) The related prob... | Yes |
Determine the eigenvalues and eigenfunctions of the Sturm-Liouville problem\n\n\[ \text{D.E.}{\mathrm{y}}^{\prime \prime } + \lambda \mathrm{y} = 0,\;0 \leq \mathrm{x} \leq \mathrm{L} \]\n\n\[ \text{B.C.}y\left( 0\right) = 0,\;y\left( L\right) = 0\text{.} \] | Solution. We first note that (9) reduces to (10), if \( \mathrm{K}\left( \mathrm{x}\right) \equiv 1,\mathrm{q}\left( \mathrm{x}\right) \equiv 0,\mathrm{\;g}\left( \mathrm{x}\right) \equiv 1,\mathrm{a} = 0,\mathrm{\;b} = \mathrm{L} \) , \( {\mathrm{c}}_{1} = 1,{\mathrm{c}}_{2} = 0,{\mathrm{c}}_{3} = 1,{\mathrm{c}}_{4} =... | Yes |
For the eigenvalue problem\n\nD.E. \( {\mathrm{y}}^{\prime \prime } + \lambda \mathrm{y} = 0,\;0 \leq \mathrm{x} \leq \mathrm{L},\mathrm{L} < \pi /2 \)\n\n(12)\n\nB.C. \( \mathrm{y}\left( 0\right) - {\mathrm{y}}^{\prime }\left( 0\right) = 0,\mathrm{y}\left( \mathrm{L}\right) + {\mathrm{y}}^{\prime }\left( \mathrm{L}\ri... | Solution. If \( \lambda = 0 \) or \( \lambda < 0 \), then it is easy to check that the only solution of (12) is the trivial solution. If \( \lambda > 0 \), then the general solution of the D.E. is\n\n\[ y\left( x\right) = A\cos \left( {x\sqrt{\lambda }}\right) + B\sin \left( {x\sqrt{\lambda }}\right) .\n\]\n\n(13)\n\nS... | Yes |
Consider the differential operators \( L \) and \( {L}^{ * } \) defined by \( \left( {17}\right) \) and \( \left( {21}\right) \) respectively. If \( {\mathrm{p}}_{2},{\mathrm{p}}_{1},{\mathrm{p}}_{0},\mathrm{y} \) and \( \mathrm{z} \) are \( {\mathrm{C}}^{2} \) functions, verify the following identity\n\n\[ \n{zL}\left... | Solution. By (20) we have\n\n\[ \n\int \left( {{zL}\left\lbrack y\right\rbrack - y{L}^{ * }\left\lbrack z\right\rbrack }\right) {dx} = \left( {z{p}_{2}}\right) {y}^{\prime } - {\left( z{p}_{2}\right) }^{\prime }y + \left( {z{p}_{1}}\right) y.\n\]\n\n(25)\n\nThus, if we differentiate both sides of (25) with respect to \... | No |
Let \( \mathrm{L} \) denote the linear, second-order differential operator defined by\n\n\[ L\left\lbrack y\right\rbrack = \frac{d}{dx}\left\lbrack {K\left( x\right) \frac{dy}{dx}}\right\rbrack + q\left( x\right) y \]\n\nwhere \( K\left( x\right) \) and \( q\left( x\right) \) satisfy the requirements in (9) and \( y \)... | Solution. By comparing the expression\n\n\[ L\left\lbrack y\right\rbrack = K\left( x\right) \frac{{d}^{2}y}{d{x}^{2}} + {K}^{\prime }\left( x\right) \frac{dy}{dx} + q\left( x\right) y \]\n\nwith (17), we see that \( {p}_{2}\left( x\right) = K\left( x\right) ,{p}_{1}\left( x\right) = {K}^{\prime }\left( x\right) \) and ... | Yes |
Show that the linear, second-order differential operator\n\n\[ L\left\lbrack y\right\rbrack = {p}_{2}\left( x\right) {y}^{\prime \prime } + {p}_{1}\left( x\right) {y}^{\prime } + {p}_{0}\left( x\right) y \]\n\n(31)\n\nis self-adjoint (i.e., \( \mathrm{L} = {\mathrm{L}}^{ * } \) ) if and only if \( {\mathrm{p}}_{2}^{\pr... | Solution. The adjoint \( {\mathrm{L}}^{ * } \) of \( \mathrm{L} \) is given by (cf. (21))\n\n\[ {L}^{ * }\left\lbrack y\right\rbrack = \frac{{d}^{2}}{d{x}^{2}}\left( {y{p}_{2}}\right) - \frac{d}{dx}\left( {y{p}_{1}}\right) + y{p}_{0} = {p}_{2}{y}^{\prime \prime } + \left( {2{p}_{2}^{\prime } - {p}_{1}}\right) {y}^{\pri... | Yes |
Theorem 3 (A Uniqueness Theorem). Consider the Sturm–Liouville problem (36). If \( y\left( x\right) \) and \( \mathrm{Y}\left( \mathrm{x}\right) \) are two eigenfunctions corresponding to the same eigenvalue \( \lambda \), then \( \mathrm{y}\left( \mathrm{x}\right) = \alpha \mathrm{Y}\left( \mathrm{x}\right) \) , \( a ... | Proof. We consider the function\n\n\[ \omega \left( \mathrm{x}\right) = {\mathrm{Y}}^{\prime }\left( \mathrm{a}\right) \mathrm{y}\left( \mathrm{x}\right) - {\mathrm{y}}^{\prime }\left( \mathrm{a}\right) \mathrm{Y}\left( \mathrm{x}\right) ,\]\n\n(37)\n\nand suppose that\n\n\[ {\left\lbrack {\mathrm{Y}}^{\prime }\left( \... | Yes |
Theorem 4 (Green’s Formula for L). Let \( L \) be the Sturm–Liouville differential operator defined by (35). If \( y\left( x\right) \) and \( z\left( x\right) \) are \( {C}^{2} \) functions on \( \left\lbrack {a, b}\right\rbrack \), then \[ {\int }_{a}^{b}\left( {{zL}\left\lbrack y\right\rbrack - {yL}\left\lbrack z\rig... | Proof. Since L is self-adjoint (cf. Example 5), Lagrange's identity (cf. (24)) becomes \[ {zL}\left\lbrack y\right\rbrack - {yL}\left\lbrack z\right\rbrack = \frac{d}{dx}\left\lbrack {K\left( x\right) \left( {{y}^{\prime }z - y{z}^{\prime }}\right) }\right\rbrack . \] Thus, upon integrating both sides of (42) with resp... | Yes |
Theorem 5 (Orthogonality of Eigenfunctions). Let \( {\lambda }_{\mathrm{m}} \) and \( {\lambda }_{\mathrm{n}} \) be two distinct eigenvalues of the Sturm-Liouville problem (36). Then the corresponding eigenfunctions \( {y}_{m}\left( x\right) \) and \( {y}_{n}\left( x\right) \) are orthogonal on \( \left\lbrack {a, b}\r... | Proof. Since \( L\left\lbrack {y}_{m}\right\rbrack = - {\lambda }_{m}g{y}_{m} \) and \( L\left\lbrack {y}_{n}\right\rbrack = - {\lambda }_{n}g{y}_{n} \), we have\n\n\[ {y}_{n}L\left\lbrack {y}_{m}\right\rbrack - {y}_{m}L\left\lbrack {y}_{n}\right\rbrack = \left( {{\lambda }_{n} - {\lambda }_{m}}\right) {y}_{m}{y}_{n}g.... | Yes |
Theorem 6. All of the eigenvalues of the Sturm-Liouville problem (36) are real. | Proof. Let \( \;\lambda = \alpha + {i\beta }\;\left( {\alpha ,\beta \text{ real}}\right) \; \) be an arbitrary eigenvalue of the Sturm-Liouville problem (36), and let \( y\left( x\right) \) be a complex-valued eigenfunction corresponding to \( \lambda \) . Since \( K\left( x\right), q\left( x\right) \) and \( \mathrm{g... | Yes |
Theorem 7. In the Sturm–Liouville problem (36), suppose that \( \;q\left( x\right) \leq 0\; \) for \( \;a \leq x \leq b \)\n\nand that the real constants \( {c}_{j}\left( {j = 1,\ldots ,4}\right) \) satisfy the inequalities\n\n\[ \n{\mathrm{c}}_{1} \cdot {\mathrm{c}}_{2} \leq 0\;\text{ and }\;{\mathrm{c}}_{3} \cdot {\m... | Proof. Suppose that \( L\left\lbrack y\right\rbrack + {\lambda gy} = 0 \), where \( y\left( x\right) ≢ 0 \) . Then\n\n\[ \n0 = {\int }_{a}^{b}y \cdot \left( {L\left\lbrack y\right\rbrack + {\lambda gy}}\right) {dx} = {\int }_{a}^{b}y\left( x\right) \frac{d}{dx}\left( {K\frac{dy}{dx}}\right) {dx} + {\int }_{a}^{b}q\left... | Yes |
Example 7. Suppose that we have constants \( {\lambda }_{2} > {\lambda }_{1} > 0 \) . Let \( {y}_{1}\left( x\right) \) be any nonzero solution of \( {\mathrm{y}}^{\prime \prime } + {\lambda }_{1}\mathrm{y} = 0 \) and let \( {\mathrm{y}}_{2}\left( \mathrm{x}\right) \) be any nonzero solution of \( {\mathrm{y}}^{\prime \... | Solution. The general solution of \( {\mathrm{y}}^{\prime \prime } + {\lambda }_{1}\mathrm{y} = 0 \) can be written in the form \( \operatorname{Asin}\left( {\mathrm{x}\sqrt{{\lambda }_{1}} + \delta }\right) \) , where \( \mathrm{A} \) and \( \delta \) are arbitrary constants. Thus, any interval, say \( \mathrm{J} \), ... | Yes |
Theorem 8 (The Sturm Comparison Theorem). Let \( L\left\lbrack y\right\rbrack \equiv \frac{d}{dx}\left\lbrack {K\left( x\right) \frac{dy}{dx}}\right\rbrack + q\left( x\right) y \), where \( K\left( x\right) > 0 \) on \( \left\lbrack {a, b}\right\rbrack \), and \( {K}^{\prime }\left( x\right) \) and \( q\left( x\right) ... | Proof. By (55) and Green’s formula (41) for the operator \( \mathrm{L} \) on \( \left\lbrack {\alpha ,\beta }\right\rbrack \), we obtain \( {\int }_{\alpha }^{\beta }\left\lbrack {{\lambda }_{2}{\mathrm{g}}_{2} - {\lambda }_{1}{\mathrm{g}}_{1}}\right\rbrack {\mathrm{y}}_{1}{\mathrm{y}}_{2}\mathrm{\;d}\mathrm{x} = {\int... | Yes |
Theorem 9. Consider the Sturm-Liouville problem\n\nD.E. \( {y}^{\prime \prime } + q\left( x\right) y + {\lambda g}\left( x\right) y = 0\;a \leq x \leq b \)\n\n(57)\n\nB.C. \( y\left( a\right) = 0, y\left( b\right) = 0 \) ,\n\nwhere \( q\left( x\right) \) and \( g\left( x\right) \) are continuous, and \( g\left( x\right... | Proof (sketch). The eigenvalues of problem (57) are precisely those values of \( \lambda \) for which \( Y\left( {b,\lambda }\right) = 0 \) (Why ?). We must show that there is a sequence of such values of \( \lambda \) which tend to \( \infty \) . For \( \mathrm{c} > 0 \), let \( \lambda \left( \mathrm{c}\right) \) be ... | Yes |
Using the fact that \( {J}_{m}\left( x\right) \) satisfies the ODE (59), verify that \( {y}_{m}\left( x\right) = {x}^{\frac{1}{2}}{J}_{m}\left( x\right) \) \( \left( {\mathrm{x} > 0}\right) \), satisfies the ODE\n\n\[ \n{y}^{\prime \prime } + \left\lbrack {1 + \frac{\frac{1}{4} - {m}^{2}}{{x}^{2}}}\right\rbrack y = 0, ... | Solution. First compute \( {\mathrm{y}}_{\mathrm{m}}^{\mathrm{u}} \) in terms of \( {\mathrm{J}}_{\mathrm{m}}\left( \mathrm{x}\right) \) . Then for \( \mathrm{x} > 0 \), we have by (59)\n\n\[ \n{x}^{\frac{3}{2}}\left\lbrack {{y}_{m}^{\prime \prime } + \left\lbrack {1 + \frac{\frac{1}{4} - {m}^{2}}{{x}^{2}}}\right\rbrac... | Yes |
Deduce that the following estimates holds for the eigenvalues \( {\lambda }_{\mathrm{n}} \) :\n\n\[ \frac{1}{M}{\left\lbrack \frac{n\pi }{b - a}\right\rbrack }^{2} < {\lambda }_{n} < \frac{1}{m}{\left\lbrack \frac{n\pi }{b - a}\right\rbrack }^{2}, n = 1,2,\ldots . \] | Solution. By Theorem 2 (or Theorem 9), we know that (61) has an infinite number of eigenvalues \( {\lambda }_{1} < {\lambda }_{2} < \ldots < {\lambda }_{\mathrm{n}} < \ldots \) . Moreover, by Theorem 2 and the B.C. in (61), the eigenfunction \( {y}_{n}\left( x\right) \) associated to \( {\lambda }_{n} \) has \( n + 1 \... | Yes |
Proposition 1. A solution of the problem\n\nD.E. \( \;{u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq L, - \infty < t < + \infty \) ,\n\nB.C. \( u\left( {0, t}\right) = 0, u\left( {L, t}\right) = 0 \) ,\n\n(8)\n\nI.C. \( \left\{ \begin{array}{l} u\left( {x,0}\right) = f\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{N}{B... | \[ u\left( {x, t}\right) = \mathop{\sum }\limits_{{n = 1}}^{N}\left\lbrack {\frac{L}{n\pi a}{\bar{A}}_{n}\sin \left( \frac{n\pi at}{L}\right) + {B}_{n}\cos \left( \frac{n\pi at}{L}\right) }\right\rbrack \sin \left( \frac{n\pi x}{L}\right) . \] | Yes |
Solve the initial/boundary-value problem\n\nD.E. \( \;{u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq L,\; - \infty < t < + \infty \) ,\n\nB.C. \( \mathrm{u}\left( {0,\mathrm{t}}\right) = 0,\mathrm{u}\left( {\mathrm{L},\mathrm{t}}\right) = 0 \) ,\n\nI.C. \( \left\{ \begin{array}{l} u\left( {x,0}\right) = f\left( x\right) = ... | Solution. We simply apply Proposition 1 with \( {\mathrm{B}}_{3} = 2,{\overline{\mathrm{A}}}_{1} = 1,{\overline{\mathrm{A}}}_{5} = - 3 \), and with all of the other \( {\overline{\mathrm{A}}}_{\mathrm{n}} \) and \( {\mathrm{B}}_{\mathrm{n}} \) equal to zero. Then\n\n\[ u\left( {x, t}\right) = \frac{L}{\pi a}\left\lbrac... | Yes |
Theorem 1 (Uniqueness). Let \( {u}_{1}\left( {x, t}\right) \) and \( {u}_{2}\left( {x, t}\right) \) be \( {C}^{2} \) solutions of the following problem\n\nD.E. \( \;{u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq L,\; - \infty < t < + \infty ,\n\nB.C. \( u\left( {0, t}\right) = A\left( t\right), u\left( {L, t}\right) = B\le... | Proof. Let \( \;v\left( {x, t}\right) \; = \;{u}_{1}\left( {x, t}\right) \; - \;{u}_{2}\left( {x, t}\right) .\; \) Note that \( \;v\; \) satisfies the related problem with homogeneous B.C. and I.C. . In particular, \( v\left( {x,0}\right) = 0 \) and \( {v}_{t}\left( {x,0}\right) = 0 \) . We need to show that \( v\left(... | Yes |
Calculate the energy of the \( \mathrm{n} \) -th harmonic | Solution. The energy \( \mathrm{E}\left( \mathrm{t}\right) \) is defined by (13), and we know from the proof of Theorem 1 that \( \mathrm{E}\left( \mathrm{t}\right) \) is constant. Thus, \( \mathrm{E}\left( \mathrm{t}\right) = \mathrm{E}\left( 0\right) \), and so we only need to compute \( \mathrm{E}\left( 0\right) \):... | Yes |
Find the formal solution of the problem for the motion of the plucked string:\n\nD.E. \( \;{u}_{tt} = {a}^{2}{u}_{xx},\;0 \leq x \leq L,\; - \infty < t < \infty ,\)\n\nB.C. \( u\left( {0, t}\right) = 0, u\left( {L, t}\right) = 0 \)\n\nI.C. \( \;u\left( {x,0}\right) = f\left( x\right) ,{u}_{t}\left( {x,0}\right) = 0, \) | Solution. We have\n\n\[ f\left( x\right) = \left\{ \begin{matrix} \left( {{u}_{0}/{x}_{0}}\right) x, & 0 \leq x \leq {x}_{0}, \\ {u}_{0}\left( {x - L}\right) /\left( {{x}_{0} - L}\right) , & {x}_{0} \leq x \leq L. \end{matrix}\right. \]\n\nThe formal solution of problem (14) is obtained by computing the Fourier sine co... | Yes |
Example 1. In problem (1) take \( f\left( x\right) = 0 \) and \( g\left( x\right) = \sin \left( {{\pi x}/L}\right) \) . Show that the maximum of the solution \( \mathrm{u}\left( {\mathrm{x},\mathrm{t}}\right) \) does not occur when \( \mathrm{t} = 0,\mathrm{x} = 0 \) or \( \mathrm{x} = \mathrm{L} \) . | Solution. The solution is \( \;u\left( {x, t}\right) = \frac{L}{\pi a}{sin}\left( {{\pi at}/L}\right) {sin}\left( {{\pi x}/L}\right) .\; \) Note that \( \;u\left( {x, t}\right) \; \) vanishes at the ends and also initially (i.e., \( u\left( {0, t}\right) = 0, u\left( {L, t}\right) = 0, u\left( {x,0}\right) = 0 \) ). Ho... | Yes |
We know that \( \cos \left( {\lambda at}\right) \sin \left( {\lambda x}\right) \; \) is a (product) solution of \( \;{u}_{tt} = {a}^{2}{u}_{xx}.\; \) Hence, it must be possible to rewrite \( \cos \left( {\lambda \mathrm{{at}}}\right) \sin \left( {\lambda \mathrm{x}}\right) \) in the form \( \mathrm{F}\left( {\mathrm{x}... | Solution. Using the identity \( \cos \left( \beta \right) \sin \left( \alpha \right) = \frac{1}{2}\left\lbrack {\sin \left( {\alpha + \beta }\right) + \sin \left( {\alpha - \beta }\right) }\right\rbrack \), with \( \alpha = {\lambda x} \) and \( \beta = \lambda \) at,\n\n\[ \cos \left( {\lambda at}\right) \sin \left( {... | Yes |
Example 3. Solve D.E. \( {u}_{tt} = {a}^{2}{u}_{xx},\; - \infty < x, t < \infty \) , I.C. \( u\left( {x,0}\right) = \frac{1}{1 + {x}^{2}},\;{u}_{t}\left( {x,0}\right) = 0 \) . | Solution. This is problem (8) with \( f\left( x\right) = {\left( 1 + {x}^{2}\right) }^{-1} \) and \( g\left( x\right) = 0 \) . By (9), the solution is \[ u\left( {x, t}\right) = \frac{1}{2}\left\lbrack {\frac{1}{1 + {\left( x + at\right) }^{2}} + \frac{1}{1 + {\left( x - at\right) }^{2}}}\right\rbrack . \] | Yes |
Solve D.E. \( {u}_{tt} = {a}^{2}{u}_{xx},\; - \infty < x, t < \infty \) , \n\n\[ \n\text{I.C.}u\left( {x,0}\right) = 0,\;{u}_{t}\left( {x,0}\right) = \frac{2}{1 + {x}^{2}}\text{.} \n\] \n\nWhat is the limit of the amplitude \( u\left( {x, t}\right) \) at any fixed \( x \), as \( t \rightarrow \infty \) ? | Solution. Here the string is initially straight \( \left( {u\left( {x,0}\right) = 0}\right) \), but has a variable upward velocity at \( \mathrm{t} = 0 \) . The upward velocity at \( \mathrm{x} \) is \( 2{\left( 1 + {\mathrm{x}}^{2}\right) }^{-1} \) . By (9), \n\n\[ \n\begin{aligned} u\left( {x, t}\right) & = {\left. \... | Yes |
Theorem 2 (A maximum magnitude principle). Let \( f\left( x\right) \) be \( {C}^{2} \) and \( g\left( x\right) \) be \( {C}^{1} \) \( \left( {-\infty < x < \infty }\right) \) . Suppose that \( {M}_{f} = \mathop{\max }\limits_{{-\infty < x < \infty }}\left| {f\left( x\right) }\right| < \infty \), and that \( {I}_{g} \) ... | Proof. Using D'Alembert's formula (9), we obtain \[ \left| {u\left( {x, t}\right) }\right| \leq \frac{1}{2}\left( {\left| {f\left( {x + {at}}\right) }\right| + \left| {f\left( {x - {at}}\right) }\right| }\right) + \frac{1}{2a}\left| {{\int }_{x - {at}}^{x + {at}}g\left( r\right) {dr}}\right| \leq \frac{1}{2}2{M}_{f} + ... | Yes |
Property 1. Disturbances propagate with speed a. | The value \( u\left( {{x}_{0},{t}_{0}}\right) \) depends only on the values of \( g \) in the interval \( \left\lbrack {{x}_{0} - a{t}_{0},{x}_{0} + a{t}_{0}}\right\rbrack \) and on the values of \( f \) at the endpoints of this interval. Geometrically, this is the interval cut out by the characteristic lines that pass... | Yes |
Property 2. Odd/even initial data yield odd/even solutions. | If \( f\left( x\right) \) and \( g\left( x\right) \) are odd, then \( u\left( {x, t}\right) \) is odd in the \( x \) -variable, since\n\n\[ u\left( {-x, t}\right) = \frac{1}{2}\left\lbrack {f\left( {-x + {at}}\right) + f\left( {-x - {at}}\right) }\right\rbrack + \frac{1}{2a}{\int }_{-x - {at}}^{-x + {at}}g\left( r\righ... | Yes |
Property 3. Periodic initial data yield periodic solutions. | If \( f\left( x\right) \) and \( g\left( x\right) \) are periodic functions of period \( {2L} \), then \( u\left( {x, t}\right) \) is also periodic of period 2L in \( x. \) This follows easily from \( D \) ’Alembert’s formula, but there is also a uniqueness argument \( \left\lbrack {v\left( {x, t}\right) \equiv u\left(... | Yes |
Example 5. Let \( f\left( x\right) \) and \( g\left( x\right) \) be \( {C}^{2} \) functions which are \( 0 \) for \( \left| x\right| \geq {10} \) . Suppose that \( u\left( {x, t}\right) \) is the solution of the wave equation \( {u}_{tt} = 4{u}_{xx} \) with \( u\left( {x,0}\right) = f\left( x\right) \) and \( {u}_{t}\l... | Solution. Here \( a = 2 \) and intuitively we do not expect the initial disturbances \( f\left( x\right) \; \) and \( \;g\left( x\right) \;\bar{t}\bar{o} \) soread faster than with speed 2. Since these disturbances are confined to the interval \( \left( {-{10},{10}}\right) \) for \( \mathrm{t} = 0 \) (cf. Figure 4), we... | Yes |
Example 6 (The semi–infinite string). Let \( f\left( x\right) \) be a \( {C}^{2} \) function defined for \( x \geq 0 \) such that \( f\left( 0\right) = 0 \) and \( {f}^{\prime \prime }\left( 0\right) = 0 \). Solve the following problem for the semi-infinite string \( \left( {0 \leq x < \infty }\right) \), with fixed en... | Solution. We exploit Property 2. Note that \( f\left( x\right) \) is defined for \( x \geq 0 \), but we can consider (cf. Figure 5) the odd extension \( {f}_{0}\left( x\right) , - \infty < x < \infty \) (i.e., \( {f}_{0}\left( x\right) = f\left( x\right) \) for \( x \geq 0 \), and \( {f}_{0}\left( x\right) = - f\left( ... | Yes |
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