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Example 6. Consider\n\[ \n\\text{D.E.}{u}_{t} = {u}_{xx},\\;0 \leq x \leq 1, t \geq 0\\text{,}\n\]\n\n\[ \n\\text{B.C.}u\\left( {0, t}\\right) = 0, u\\left( {1, t}\\right) = 1\\text{,}\n\]\n\n(46)\n\n\[ \n\\text{I.C.}u\\left( {x,0}\\right) = x + {\\pi }^{-4}\\sin \\left( {\\pi x}\\right) \\text{.}\n\]\n\nUse the explic... | Solution. We must find the difference approximation \( {v}_{4,{64}} \), since \( M = L/{\\Delta x} = 8 \) and \( N = \) \( \\mathrm{T}/\\Delta \\mathrm{t} = {64} \) . With the help of a suitable computer program, we find that \( {v}_{4,{64}}^{\\prime } = {0.500064682}\\ldots \), while \( \\;u\\left( {{0.5},\\;{0.5}}\\r... | Yes |
Find the general solution of the difference equation\n\n\[ \n{y}_{n + 2} - 6{y}_{n + 1} + 8{y}_{n} = 0,\;n = 0,1,2,\ldots .\n\] | The auxiliary equation \( {r}^{2} - {6r} + 8 = 0 \) has roots \( {r}_{1} = 2 \) and \( {r}_{2} = 4 \) . Hence, by (7), the general solution of (11) is\n\n\[ \n{y}_{n} = {c}_{1}{2}^{n} + {c}_{2}{4}^{n},\;n = 0,1,2,\ldots .\n\] | Yes |
Find the general solution of the difference equation\n\n\[ \n{y}_{n + 2} - 4{y}_{n + 1} + 4{y}_{n} = 0,\;n = 0,1,2,\ldots \n\] | Solution. The auxiliary equation \( {r}^{2} - {4r} + 4 = 0 \) has roots \( {r}_{1} = {r}_{2} = 2 \) . Hence, by (8), the general solution of (13) is \( {y}_{n} = \left( {{c}_{1} + {c}_{2} \cdot n}\right) {2}^{n} \) . | Yes |
Find the general solution of the difference equation\n\n\[ \n{y}_{n + 2} - 2{y}_{n + 1} + 4{y}_{n} = 0,\;n = 0,1,2,\ldots .\n\] | Solution. The auxilliary equation \( {r}^{2} - {2r} + 4 = 0 \) has roots \( 1 \pm i\sqrt{3} \) . In (9), \( R = 2 \) and \( \theta = \pi /3 \) . Thus, by (10), the general solution of (14) is \( {y}_{n} = {2}^{n}({c}_{1}\cos \left( {{n\pi }/3}\right) + {c}_{2}\sin \left( {{n\pi }/3}\right) \) . | Yes |
Find the general solution of the difference equation\n\n\\[ \n{y}_{n + 2} - 2\\cos \\alpha {y}_{n + 1} + {y}_{n} = 0,\\;n = 0,1,2,\\ldots \n\\]\n\n(15) | Solution. The roots of \\( {r}^{2} - 2\\cos {\\alpha r} + 1 = 0 \\) are \\( \\cos \\alpha \\pm i \\cdot \\sin \\alpha \\) . If \\( \\sin \\alpha \\neq 0 \\), then the roots are complex. By (9), \\( \\mathrm{R} = 1,\\cos \\alpha = \\cos \\theta \\), and the general solution of (15) is\n\n\\[ \n{\\mathrm{y}}_{\\mathrm{n}... | Yes |
Example 5. Consider D.E. \( {u}_{t} = {u}_{xx} \) \( 0 \leq \mathrm{x} \leq 1,0 \leq \mathrm{t} \leq {0.5} \)\n\nB.C. \( u\left( {0, t}\right) = 0, u\left( {1, t}\right) = 0 \)\n\nI.C. \( u\left( {x,0}\right) = \sin \left( {\pi x}\right) \). | Solution. We use \( \lambda = 1/{12},\mathrm{k} = \mathrm{p} = 1 \), and \( {\mathrm{b}}_{\mathrm{p}} = 1 \), note that \( \alpha \) in (45) is given by\n\n\( \alpha = - \frac{1}{4}{\pi }^{4}{\mathrm{e}}^{-\frac{1}{2}{\pi }^{2}} \), which has the same sign as \( \mathrm{r} \approx - {10}^{-7} \) . Thus,(48) gives the o... | Yes |
Consider what happens in Example 5, when \( \lambda \) is changed from \( 1/{12} \) to \( 1/4 \) . | In this case, \( \alpha = \frac{{\pi }^{4}}{12}{\mathrm{e}}^{-\frac{1}{2}{\pi }^{2}} \) which has the opposite sign of \( \mathrm{r} \approx - {10}^{-7} \) . Using (48), the optimal value of \( \mathrm{M} \) at \( \mathrm{x} = {0.5} \) is approximately given by\n\n\[ \nM \approx {\left| \frac{\alpha {\lambda }^{2}}{rT}... | Yes |
Theorem 1 (A Convergence Result). Suppose that the iteration matrix \( T \) has the property,\n\nthat\n\n\[ \text{for some fixed}\mathrm{r} < 1,\parallel \mathrm{T}\mathbf{y}\parallel \leq \mathrm{r}\parallel \mathbf{y}\parallel \text{, for every vector}\mathbf{y} \]\n\n(i.e., \( \mathbf{T} \) contracts the lengths of ... | Proof. First we establish the uniqueness of the solution. Suppose that \( \;A\mathbf{y} = \mathbf{b}\; \) and \( \;A\mathbf{z} = \mathbf{b}. \) Then \( A\left( {\mathbf{y} - \mathbf{z}}\right) = \mathbf{b} - \mathbf{b} = \mathbf{0} \) . Thus, \( \mathbf{0} = {\mathbf{N}}^{-1}\left( {\mathbf{M} - \mathbf{N}}\right) \lef... | Yes |
Example 2. Solve\n\nD.E. \( \;{u}_{t} = k\left( {{u}_{xx} + {u}_{yy}}\right) ,\;0 \leq x \leq L,\;0 \leq y \leq M,\;t \geq 0 \)\n\nB.C. \( \left\{ \begin{array}{l} u\left( {x,0, t}\right) = 0,{u}_{y}\left( {x, M, t}\right) = 0 \\ {u}_{x}\left( {0, y, t}\right) = 0,{u}_{x}\left( {L, y, t}\right) = 0 \end{array}\right. \... | Solution. Here, the edge \( y = 0\;\left( {0 \leq x \leq L}\right) \) is maintained at zero, while the other three edges are insulated. As usual, we seek the product solutions of the D.E. which satisfy the B.C., and then consider the I.C. \( .\; \) Separation of variables leads to equations (4) for \( \check{\mathrm{X}... | Yes |
Find the value of the constant \( c \) such that the following Neumann problem has a solution, and find such a solution. Is the solution unique ? | Solution. One quickly verifies that the product solutions of the D.E. which satisfy the homogeneous B.C. are of the form\n\n\[ {u}_{n, m}\left( {x, y, z}\right) = {a}_{n, m}\;{cosh}\lbrack \sqrt{{n}^{2} + {m}^{2}}\;z\rbrack {cos}\left( {nx}\right) {cos}\left( {my}\right) \;,\;n, m = 0,\;1,\;2,\;...\;. \]\n\nWe apply th... | Yes |
Show that if \( {\left( \mathrm{L}/\mathrm{M}\right) }^{2} \) is not rational, then no two of the harmonics (30) can have the same frequency. For the square drum, show that there are infinitely many pairs of harmonics which have the same frequency. | Suppose that \( {\nu }_{n, m} = {\nu }_{p, q} \) and \( \left( {n, m}\right) \neq \left( {p, q}\right) \) . Then \( n \neq p \) and \( m \neq q \) (Why ?). Thus, \( {\mathrm{q}}^{2} - {\mathrm{m}}^{2} \neq 0 \), and so\n\n\[ \n{\left( n/L\right) }^{2} + {\left( m/M\right) }^{2} = {\left( p/L\right) }^{2} + {\left( q/M\... | Yes |
Show that if \( g\left( {x, y}\right) \) is any \( {C}^{2} \) eigenfunction of \( \Delta \), with eigenvalue \( \lambda \geq 0 \), then\n\n(a) \( v\left( {x, y, t}\right) = b{e}^{-{\lambda kt}}g\left( {x, y}\right) \) is a solution of the heat equation \( {v}_{t} = {k\Delta v} \), and\n\n\( \left( b\right) \;u\left( {x... | Solution. Note that \( {\Delta v} = \Delta \left( {b{e}^{-{\lambda kt}}g}\right) = b{e}^{-{\lambda kt}}{\Delta g} = - b{e}^{-{\lambda kt}}{\lambda g} = - {\lambda v} \), and similarly \( {\Delta u} = \n\n- \( {\lambda u} \) . Thus, \( {v}_{t} = - {\lambda bk}{e}^{-{\lambda kt}}g = - {k\lambda v} = {k\Delta v} \) and \(... | Yes |
Find an eigenfunction of \( \Delta \) which is zero on the boundary of a square, but which is not a product of the form \( f\left( x\right) h\left( y\right) \) . | Solution. Note that if \( {g}_{1}\left( {x, y}\right) \) and \( {g}_{2}\left( {x, y}\right) \) are two eigenfunctions of \( \Delta \) with the same eigenvalue \( \lambda \), then any linear combination \( {c}_{1}{g}_{1} + {c}_{2}{g}_{2} \) is also an eigenfunction with eigenvalue \( \lambda \) . Indeed, using the linea... | Yes |
Theorem 1 (Convergence Theorem for Double Fourier Series). Let \( f\left( {x, y}\right) \) be a \( {C}^{k} \) function \( \left( {\mathrm{k} \geq 3}\right) \) on the rectangle \( - \mathrm{L} \leq \mathrm{x} \leq \mathrm{L}, - \mathrm{M} \leq \mathrm{y} \leq \mathrm{M} \) . Let \( \mathrm{K} \) be the largest of the ma... | Proof. Integrating by parts repeatedly with respect to \( \mathrm{x} \) and using the equations (9) to eliminate the endpoint evaluations, we have\n\n\[ \n{c}_{n, m} = \frac{1}{4LM}{\int }_{-M}^{M}{\int }_{-L}^{L}f\left( {x, y}\right) \overline{{E}_{n, m}\left( {x, y}\right) }{dxdy} = \ldots\n\]\n\n\[ \n= \frac{1}{4LM}... | Yes |
The function \( u\left( {x, y, z}\right) = {xyz} \) is clearly harmonic (i.e., \( {\Delta u} = 0 \) ), but rewrite \( u\left( {x, y, z}\right) \) as a function \( \mathrm{U}\left( {\rho ,\varphi ,\theta }\right) \) in spherical coordinates and verify that the right side of (2) is 0. | Solution. Using \( \left( 1\right) ,\;u\left( {x, y, z}\right) = U\left( {\rho ,\varphi ,\theta }\right) = {\rho }^{3}{\sin }^{2}\varphi \cos \varphi \cos \theta \sin \theta = \frac{1}{2}{\rho }^{3}{\sin }^{2}\varphi \cos \varphi \sin {2\theta }.\; \) Thus,\n\n\[ \n{\mathrm{U}}_{\rho \rho } + 2{\rho }^{-1}{\mathrm{U}}_... | Yes |
For \( \left( {x, y, z}\right) \) on the unit sphere \( {x}^{2} + {y}^{2} + {z}^{2} = 1\;\left( {\text{or}\;\rho = 1}\right) \), let \( \;u\left( {x, y, z}\right) = {xyz}. \) Write this function in terms of the coordinates \( \;\left( {\varphi ,\theta }\right) \; \) on the sphere, say \( \;u\left( {x, y, z}\right) \; =... | Solution. From Example 1, we know that for \( \;\left( {x, y, z}\right) \; \) unrestricted, \( \;u\left( {x, y, z}\right) \; = \;U\left( {\rho ,\varphi ,\theta }\right) \) \( = \frac{1}{2}{\rho }^{3}{\sin }^{2}\varphi \;\cos \varphi \;\sin {2\theta }.\; \) Since \( \;\rho = 1\; \) on the unit sphere, \( \;f\left( {\var... | Yes |
Theorem 1. \( \;F\left( {\rho ,\varphi ,\theta }\right) = R\left( \rho \right) f\left( {\varphi ,\theta }\right) \; \) is an eigenfunction of the Laplace operator \( \;\Delta \; \) (on space) with eigenvalue \( c \) (i.e., \( \Delta \mathrm{F} + c\mathrm{F} = 0 \) ) if and only if \( f\left( {\varphi ,\theta }\right) \... | Recall that \( 1/\rho \) defines a harmonic function for \( \rho > 0 \) . Thus, \( 1/\rho \) must be a solution of (10), when \( \mathrm{c} = 0 \) and \( \lambda = 0 \) . Thus, it is natural to try a solution of the form \( \mathrm{R}\left( \rho \right) = \mathrm{g}\left( \rho \right) /\rho \) , when \( \mathrm{c} \) i... | Yes |
Find all solutions (valid for \( \rho > 0 \) ), of the three-dimensional heat and wave equations, which are of the form \( \mathrm{R}\left( \rho \right) \mathrm{T}\left( \mathrm{t}\right) \), and which are bounded as \( t \) and \( \rho \) approach \( + \infty \) . | Substituting \( R\left( \rho \right) T\left( t\right) \; \) into the heat and wave equations \( \;{u}_{t} = {k\Delta u}\; \) and \( \;{u}_{tt} = {a}^{2}{\Delta u} \) respectively, and separating variables, we obtain\n\n\[ \frac{{\mathrm{T}}^{\prime }}{\mathrm{{kT}}} = \Delta \mathrm{R}/\mathrm{R} = - \mathrm{c}\;\text{... | Yes |
Consider a solid, homogeneous ball of diameter \( 1 \), and with heat diffusivity constant \( k \) . The spherical surface of the ball is maintained at temperature 0 for \( t > 0, \) and it has a constant initial temperature distribution \( \;U\left( {\rho ,0}\right) = {100}^{ \circ }\; \) C . Tind a formal series solu... | Solution. Since there is no angular dependence in the boundary condition or in the initial\ntemperature, the formal solution \( \mathrm{U}\left( {\rho ,\mathrm{t}}\right) \) does not involve \( \varphi \) and \( \theta \) and it formally solves\n\nD.E. \( {U}_{t} = k{\rho }^{-2}{\left( {\rho }^{2}{U}_{\rho }\right) }_{... | Yes |
Example 6 (D'Alembert's formula in space). By means of formal computations, find the analog of D'Alembert's formula for the solution of the following wave problem in space.\n\nD.E. \( {u}_{tt} = {a}^{2}\left( {{u}_{xx} + {u}_{yy} + {u}_{zz}}\right) \; - \infty < x, y, z, t < \infty \)\n\nI.C. \( \begin{aligned} u\left(... | Solution. We proceed formally, by applying Fourier transform methods. Taking the three-dimensional Fourier transform of both sides of the D.E., we obtain\n\n\[ {\widehat{u}}_{tt}\left( {\xi ,\eta ,\zeta, t}\right) = - {a}^{2}\left( {{\xi }^{2} + {\eta }^{2} + {\zeta }^{2}}\right) \widehat{u}\left( {\xi ,\eta ,\zeta, t}... | Yes |
Show that if \( \;U\left( {\rho ,\varphi ,\theta }\right) = R\left( \rho \right) f\left( {\varphi ,\theta }\right) \; \) is a solution of Laplace’s equation \( \;{\Delta U} = 0\; \) in space, then \( \mathrm{f}\left( {\varphi ,\theta }\right) \) must be a spherical harmonic. | Solution. If \( \;U\left( {\rho ,\varphi ,\theta }\right) \; = \;R\left( \rho \right) f\left( {\varphi ,\theta }\right) \; \) solves Laplace’s equation \( \;{\Delta U}\; = \;0,\; \) then \( \;U\; \) is an eigenfunction of \( \Delta \) with eigenvalue \( c = 0 \) (i.e., \( {\Delta U} + 0 \cdot U = 0 \) ). Thus, \( f\lef... | Yes |
Find a harmonic polynomial of degree 3 in \( x, y \) and \( z \), such that its restriction to the sphere \( \rho = 1 \), is a spherical harmonic \( \mathrm{f}\left( {\varphi ,\theta }\right) \) which is independent of \( \theta \) . | Solution. We will first find the appropriate spherical harmonic, and then construct the polynomial. The product spherical harmonics are of the form \( \mathrm{L}\left( \varphi \right) {\mathrm{e}}^{\mathrm{{im}}\theta } \), where \( \mathrm{L}\left( \varphi \right) \) is a solution of the associated Legendre equation (... | Yes |
Consider the familiar harmonic potential \( u\left( {x, y, z}\right) = {\left\lbrack {x}^{2} + {y}^{2} + {z}^{2}\right\rbrack }^{-\frac{1}{2}} = {\rho }^{-1} \), and let \( {\mathrm{u}}_{\mathrm{c}}\left( {\mathrm{x},\mathrm{y},\mathrm{z}}\right) \equiv \mathrm{u}\left( {\mathrm{x},\mathrm{y},\mathrm{z} - \mathrm{c}}\r... | Solution. Note that \( \;{x}^{2}\; + \;{y}^{2}\; + \;{\left( z–c\right) }^{2}\; = \;{\rho }^{2}\; - \;{2cz}\; + \;{c}^{2}\; = \;{\rho }^{2}\; - \;{2c\rho cos\varphi }\; + \;{c}^{2} \)\n\n\( = {\rho }^{2}\left( {1 - 2\cos \left( \varphi \right) \mathrm{c}/\rho + {\left( \mathrm{c}/\rho \right) }^{2}}\right) \) . Thus, l... | Yes |
Theorem 2. For any integer \( \mathrm{n} \geq 0 \), the following \( 2\mathrm{n} + 1 \) functions are eigenfunctions of \( {\Delta }_{\mathrm{S}} \) (i.e., spherical harmonics) with common eigenvalue \( \mathrm{n}\left( {\mathrm{n} + 1}\right) \) :\n\n\[ \n{S}_{n, m}\left( {\varphi ,\theta }\right) \equiv {e}^{im\theta... | Proof of Theorem 2. Each function in (12) is a product of the function \( {M}_{m}\left( \theta \right) \) (cf. (4)) and a solution of (5), and hence is a solution of \( {\Delta }_{\mathrm{s}}\mathrm{f} = - \mathrm{n}\left( {\mathrm{n} + 1}\right) \mathrm{f} \) (cf. (3)). It easily follows that the real\n\nand imaginary... | No |
Theorem 3 (Orthogonality of spherical harmonics). For all integers \( n,{n}^{\prime } \geq 0\; \) and \( \; - n \leq m \leq n \) and \( - {\mathrm{n}}^{\prime } \leq {\mathrm{m}}^{\prime } \leq {\mathrm{n}}^{\prime } \), using the notation in (12), | Proof. For continuous complex-valued functions \( \mathrm{f}\left( {\varphi ,\theta }\right) \) and \( \mathrm{g}\left( {\varphi ,\theta }\right) \) on the unit sphere, we define\n\n\[ \langle f, g\rangle = {\int }_{0}^{2\pi }{\int }_{0}^{\pi }f\left( {\varphi ,\theta }\right) \overline{g\left( {\varphi ,\theta }\right... | Yes |
Find a formal solution of D.E. \( {v}_{tt} = {a}^{2}{\Delta }_{s}v - {\omega }^{2}v\;0 \leq \varphi ,\;\theta /2 \leq \pi ,\; - \infty < t < \infty \) (26) I.C. \( v\left( {\varphi ,\theta }\right) = f\left( {\varphi ,\theta }\right) ,{v}_{t}\left( {\varphi ,\theta }\right) = g\left( {\varphi ,\theta }\right) \). | Solution. Separation of variables, with \( \mathrm{v} = \mathrm{T}\left( \mathrm{t}\right) \mathrm{F}\left( {\varphi ,\theta }\right) \), yields \[ \frac{1}{{a}^{2}}\left\lbrack {{\frac{T}{T}}^{\prime \prime } + {\omega }^{2}}\right\rbrack = \frac{{\Delta }_{S}F}{F} = - \lambda . \] Thus, \( \mathrm{F}\left( {\varphi ,... | Yes |
Proposition 1 (The Cauchy–Riemann operator in polar coordinates). Let \( g\left( {x, y}\right) \) be a \( {C}^{\infty } \) function on the punctured plane \( {\mathbb{R}}^{2} - \left( {0,0}\right) \), and let \( g\left( {x, y}\right) = G\left( {r,\theta }\right) \) in terms of polar coordinates. Then \[ \left( {{\parti... | Proof. On the right side of (40), replace \( {G}_{r} \) by the equivalent expression \( {g}_{x}{x}_{r} + {g}_{y}{y}_{r} = \) \( {g}_{x}\cos \theta + {g}_{y}\sin \theta \), and replace \( {G}_{\theta } \) by \( {g}_{x}{x}_{\theta } + {g}_{y}{y}_{\theta } = - {g}_{x}r\sin \theta + {g}_{y}r\cos \theta \) . Then algebraic ... | Yes |
Proposition 2. If \( F\left( \rho \right) \) is a \( {C}^{\infty } \) function depending only on \( \rho = {\left( {x}^{2} + {y}^{2} + {z}^{2}\right) }^{1/2} \), then\n\n\[ \n{\left( {\partial }_{x} + i{\partial }_{y}\right) }^{m}\left\lbrack {F\left( \rho \right) }\right\rbrack = {e}^{im\theta }{\left( \rho sin\varphi... | Proof. Fix \( z, \) and consider the function \( g\left( r\right) = F\left( \sqrt{{r}^{2} + {z}^{2}}\right) \) with \( r = \sqrt{{x}^{2} + {y}^{2}}. \) For \( \rho = \sqrt{{r}^{2} + {z}^{2}} \) , we have \( {r}^{-1}{\partial }_{r}g = {r}^{-1}{F}^{\prime }\left( \rho \right) {\rho }_{r} = {r}^{-1}{F}^{\prime }\left( \rh... | Yes |
Theorem 5. For \( n = 0 \) and \( c = \pm 1 \), let \( {R}_{0}^{ \pm }\left( \rho \right) \) be the nonzero solutions of\n\n\[ \n{\rho }^{2}{R}^{\prime \prime }\left( \rho \right) + {2\rho }{R}^{\prime }\left( \rho \right) + \left( {c{\rho }^{2} - n\left( {n + 1}\right) }\right) R\left( \rho \right) = 0.\n\]\n\n(43)\n\... | Proof. Equation (44) is immediate from (42). By Theorem 1 of Section 9.3, we know that \( \Delta \left\lbrack {{\mathrm{R}}_{0}^{ \pm }\left( \rho \right) }\right\rbrack + \pm {\mathrm{R}}_{0}^{ \pm }\left( \rho \right) = 0 \) . Since the order of differentiation, with respect to \( \mathrm{x},\mathrm{y} \), and \( \ma... | Yes |
Theorem 6. Let \( \;{u}_{c}\left( {x, y, z}\right) \equiv {\left\lbrack {x}^{2} + {y}^{2} + {\left( z - c\right) }^{2}\right\rbrack }^{-1/2} = {\left\lbrack {\rho }^{2} - 2c\rho cos\varphi + {c}^{2}\right\rbrack }^{-1/2} \equiv {U}_{c}\left( {\rho ,\varphi ,\theta }\right) \; \) be the shift of the standard harmonic po... | Proof. Let \( {T}_{c} \) be the translation operator which assigns to each function \( f\left( {x, y, z}\right) \) the new function \( f\left( {x, y, z - c}\right) \) (i.e., \( {T}_{c}\left\lbrack f\right\rbrack \left( {x, y, z}\right) = f\left( {x, y, z - c}\right) \) ). This operator commutes with the operators \( {\... | Yes |
Example 2 (Steady-state temperature distribution in a cylinder). \( \; \) Consider the problem of determining the steady-state temperature distribution of a homogeneous solid cylinder which is insulated on the top and bottom faces with a prescribed temperature on the lateral surface:\n\nD.E. \( \;{u}_{xx} + {u}_{yy} + ... | Solution. Separation of variables for the product solution \( \mathrm{U}\left( {\mathrm{r},\theta ,\mathrm{z}}\right) = \mathrm{R}\left( \mathrm{r}\right) \mathrm{H}\left( \theta \right) \mathrm{Z}\left( \mathrm{z}\right) \), leads to\n\n\[ {r}^{2}{R}^{\prime \prime }\left( r\right) + r{R}^{\prime }\left( r\right) + \l... | Yes |
Show that the spherical Bessel function \( {j}_{n} \) (cf. (29) and (30) of Section 9.4) is related to \( {\mathrm{J}}_{\mathrm{n} + \frac{1}{2}} \) by the formula\n\n\[ \n{j}_{n}\left( \rho \right) = \sqrt{\frac{1}{2}\pi /\rho }\;{J}_{n + \frac{1}{2}}\left( \rho \right) \;. \n\] | Solution. First we show that \( \mathrm{R}\left( \rho \right) \) is a solution of the three-dimensional radial equation (3), if and only if \( \sqrt{\mathrm{r}}\mathrm{R}\left( \mathrm{r}\right) \) solves Bessel’s equation (5) with \( \mathrm{m} = \left( {\mathrm{n} + \frac{1}{2}}\right) \) :\n\n\[ \n{r}^{2}{\left( \sq... | Yes |
Verify explicitly that the quarter sphere \( Q \) defined by \( \rho = 1, z \geq 0, x > 0 \) is a 2-manifold in \( {\mathbb{R}}^{3} \), whose boundary is the half equator \( \mathrm{E} \) defined by \( \rho = 1,\mathrm{z} \geq 0,\mathrm{x} > 0 \) . | Solution. Let \( F \) be the mapping which assigns to each point \( \left( {x, y, z}\right) \) in the open rectangular solid \( \;B,\; - \pi /2 < x < \pi /2\;,\; - \pi /2\; < \;y\; < \;\pi /2\;,\;\frac{1}{2}\; < \;z\; < \;2,\; \) the point \( \;F\left( {x, y, z}\right) \) \( \equiv \left( {\mathrm{{zcos}}\left( \mathrm... | Yes |
Show the validity of Weyl’s formula directly, in the simple case when \( M \) is the 1-manifold consisting of the interval \( \left\lbrack {0,\mathrm{\;L}}\right\rbrack \) . Consider all possible B.C. (2). | Solution. When \( \mathrm{M} \) is an interval \( \left\lbrack {0,\mathrm{\;L}}\right\rbrack \), and the B.C. are given by \( \mathrm{u}\left( 0\right) = 0 \) and \( \mathrm{u}\left( \mathrm{L}\right) = 0 \), we have the eigenfunctions \( \sin \left( \frac{n\pi x}{L}\right) \) of \( \Delta \equiv \operatorname{Vol}\lef... | Yes |
Theorem 4 (The homogeneous heat problem). For a compact \( k \) -manifold \( M \), the solution of\n\n\[ \n\\text{D.E.}{u}_{t} = \\kappa {\\Delta }_{M}u\\text{on}M\\text{for}t \\geq 0 \n\]\n\nB.C. \( \\left\\{ \\begin{array}{ll} u \\equiv 0 & \\text{ on }D \\\\ {\\partial }_{\\nu }u \\equiv 0 & \\text{ on }N \\end{arra... | \[ \nu \\left( {P, t}\\right) = \\mathop{\\sum }\\limits_{{q = 0}}^{Q}{c}_{q}{e}^{-\\lambda {q}^{\\kappa t}}{u}_{q}\\left( P\\right) . \n\]\n\n(12) | Yes |
Theorem 5 (The homogeneous wave problem). For a compact \( \mathrm{k} \) -manifold \( \mathrm{M} \), the solution of\n\nD.E. \( {u}_{tt} = {a}^{2}{\Delta }_{{M}^{u}} \) on \( M \) for \( - \infty < t < \infty \)\n\nB.C. \( \left\{ \begin{array}{ll} u \equiv 0 & \text{ on }D \\ {\partial }_{\nu }u \equiv 0 & \text{ on }... | \[ u\left( {P, t}\right) = \mathop{\sum }\limits_{{q = 0}}^{Q}\left\lbrack {{a}_{q}\cos \left( {\sqrt{{\lambda }_{q}}\;{at}}\right) + {\left( a\sqrt{{\lambda }_{q}}\;\right) }^{-1}{b}_{q}\sin \left( {\sqrt{{\lambda }_{q}}\;{at}}\right) }\right\rbrack {u}_{q}\left( P\right) \;,\]\n\n(14)\n\nwhere it is understood that t... | Yes |
Theorem 6 (The heat problem with a source). For a compact \( k \) -manifold \( M \), the solution of\n\nD.E. \( {u}_{t}\left( {P, t}\right) = \kappa {\Delta }_{M}u\left( {P, t}\right) + \varphi \left( {P, t}\right) \; \) on \( M \) with \( t \geq 0 \)\n\nB.C. \( \left\{ \begin{array}{ll} u \equiv 0 & \text{ on }D \\ {\... | is\n\[ u\left( {P, t}\right) = \mathop{\sum }\limits_{{q = 0}}^{Q}\left\lbrack {\left. \left\lbrack {{e}^{-{\lambda }_{q}{\kappa t}}{\int }_{0}^{t}{e}^{\lambda {q}^{\kappa s}}{\alpha }_{q}\left( s\right) {ds}}\right\rbrack \right| \;{u}_{q}\left( P\right) }\right\rbrack .\n\]\n(16) | Yes |
Theorem 7 (The wave problem with a source). For a compact \( k \) -manifold \( M \), the solution of\n\nD.E. \( {u}_{tt}\left( {P, t}\right) = {a}^{2}{\Delta }_{M}u\left( {P, t}\right) + \varphi \left( {P, t}\right) \)\n\nB.C. \( \left\{ \begin{array}{ll} u \equiv 0 & \text{ on }D \\ {\partial }_{\nu }u \equiv 0 & \tex... | with source\n\[ \varphi \left( {P, t}\right) = \mathop{\sum }\limits_{{q = 0}}^{Q}{\alpha }_{q}\left( t\right) {u}_{q}\left( P\right) \]\n\nis\n\[ u\left( {P, t}\right) = \mathop{\sum }\limits_{{q = 0}}^{Q}\left\lbrack {\left\lbrack {{\int }_{0}^{t}\frac{\sin \left\lbrack {a\sqrt{{\lambda }_{q}}\left( {t - s}\right) }\... | Yes |
Formally, find integral formulas for the solutions of the heat problem (11). | In the solution (12) of problem (11), we write \( {c}_{q} = < f,{u}_{q} > \) (cf. (8) and (9)) in terms of an integral, and we interchange the sum and integrals in the following computation\n\n\[\n\begin{aligned} u\left( {P, t}\right) & = {\int }_{q = 0}^{Q}{c}_{q}{e}^{-{\lambda }_{q}{\kappa t}}{u}_{q}\left( P\right) =... | Yes |
Express the solution (16) of problem (15) in terms of an integral formula. | The solution (16) can be written in the form\n\n\[ u\left( {P, t}\right) = \mathop{\sum }\limits_{{q = 0}}^{Q}{e}^{-{\lambda }_{q}{\kappa t}}{\int }_{0}^{t}{\alpha }_{q}\left( s\right) {e}^{{\lambda }_{q}{\kappa s}}{ds}{u}_{q}\left( P\right) \]\n\n\[ = {\int }_{M}{\int }_{0}^{t}\left\lbrack {\mathop{\sum }\limits_{{q =... | Yes |
Formally find an integral solution for the Poisson problem\n\n\[ \text{D.E.}{\Delta }_{\mathrm{M}}\mathrm{u}\left( \mathrm{P}\right) = \mathrm{h}\left( \mathrm{P}\right) \;\text{on}\mathrm{M} \]\n\n\[ \text{B.C.}\left\{ \begin{array}{ll} u \equiv 0 & \text{ on }D \\ {\partial }_{\nu }u \equiv 0 & \text{ on }N \end{arra... | Solution. Taking the inner product of both sides of the D.E. with \( {u}_{q} \), we obtain\n\n\[ < h,{u}_{q} > = < {\Delta }_{M}{}^{u, u}{}_{q} > = < u,{\Delta }_{M}{}^{u}{}_{q} > = - < u,{\lambda }_{q}{}^{u}{}_{q} > = - {\lambda }_{q} < u,{u}_{q} > .\n\nThus, in order that there be a solution, it is necessary that\n\n... | Yes |
By means of formal manipulations and Green's formula (7), find a hypothetical solution of the steady-state heat problem\n\n\\[ \n\\text{D.E.}{\\Delta }_{\\mathrm{M}}\\mathrm{u} = 0\\;\\text{on}\\mathrm{M} \n\\]\n\n\\[ \n\\text{B.C.}\\left\\{ \\begin{array}{ll} u = f & \\text{ on }D \\\\ {\\partial }_{\\nu }u \\equiv g ... | Solution. We consider first the case \\( {\\lambda }_{0} \\neq 0 \\) . According to (42), we formally have\n\n\\[ \nh\\left( P\\right) = {\\Delta }_{M}{\\int }_{M}G\\left( {P,{P}^{\\prime }}\\right) h\\left( {P}^{\\prime }}\\right) {dM}\\left( {P}^{\\prime }}\\right) = {\\int }_{M}{\\Delta }_{M}G\\left( {P,{P}^{\\prime... | Yes |
Theorem 1 (The Classification Theorem). Consider the second-order, linear PDE\n\n\[ a{U}_{\xi \xi } + b{U}_{\xi \tau } + c{U}_{\tau \tau } + d{U}_{\xi } + e{U}_{\tau } + {kU} = F\left( {\xi ,\tau }\right) ,\;\left( {{a}^{2} + {b}^{2} + {c}^{2} \neq 0}\right) \]\n\nwhere the unknown function \( \mathrm{U} = \mathrm{U}\l... | Proof. Our strategy is to write the left side of (1) in terms of \( {\text{ u }}_{xx},{\text{ u }}_{xt},{\text{ u }}_{tt},{\text{ u }}_{x},{\text{ u }}_{t} \) and u, and choose the constants \( \alpha ,\beta ,\gamma ,\delta \) and \( \rho \), such that the left side becomes (after multiplication by \( \exp \left\lbrack... | Yes |
We show that \[ {\int }_{0}^{1}{\int }_{0}^{1}f\left( {x, y}\right) {dydx} \neq {\int }_{0}^{1}{\int }_{0}^{1}f\left( {x, y}\right) {dxdy} \] | The left side of (2) is \[ {\int }_{0}^{1}{\left\lbrack \frac{y}{{x}^{2} + {y}^{2}}\right\rbrack }_{y = 0}^{y = 1}{dx} = {\int }_{0}^{1}\frac{1}{{x}^{2} + 1}{dx} = \arctan \left( 1\right) = \frac{\pi }{4}. \] However, the right side of (2) is \[ {\int }_{0}^{1}{\left( \frac{-x}{{x}^{2} + {y}^{2}}\right) }_{x = 0}^{x = ... | Yes |
Let \( f\left( {x, y}\right) = \left( {{2xy} - {x}^{2}{y}^{2}}\right) {e}^{-{xy}} \) . This function is \( {C}^{\infty } \) throughout the \( {xy} \) -plane, and yet we will show that\n\n\[{\int }_{0}^{1}{\int }_{0}^{\infty }f\left( {x, y}\right) \mathrm{d}y\mathrm{\;d}x \neq {\int }_{0}^{\infty }{\int }_{0}^{1}f\left(... | For \( x > 0 \), we have\n\n\[{\int }_{0}^{\infty }f\left( {x, y}\right) {dy} = \mathop{\lim }\limits_{{R \rightarrow \infty }}{\int }_{0}^{R}\left( {{2xy} - {x}^{2}{y}^{2}}\right) {e}^{-{xy}}{dy} = {\left. \mathop{\lim }\limits_{{R \rightarrow \infty }}\left\lbrack x{y}^{2}{e}^{-{xy}}\right\rbrack \right| }_{y = 0}^{y... | Yes |
For each positive value of \( x \), we have\n\n\[ \mathrm{I}\left( \mathrm{x}\right) \equiv {\int }_{0}^{\infty }\frac{\sin \left( \mathrm{{xy}}\right) }{\mathrm{y}}\mathrm{{dy}} \equiv \mathop{\lim }\limits_{{\mathrm{R} \rightarrow \infty }}{\int }_{0}^{\mathrm{R}}\frac{\sin \left( \mathrm{{xy}}\right) }{\mathrm{y}}\m... | Indeed, the change of variables from \( y \) to \( z = {xy} \) shows that \( I\left( x\right) \) is independent of the positive value chosen for \( \mathrm{x} \), and the result (i) when \( \mathrm{x} = 1 \) is shown in the Supplement at the end of Chapter 7. Thus, we have\n\n\[ \frac{d}{dx}{\int }_{0}^{\infty }\frac{\... | No |
Let \( f\left( {x, y}\right) = \left\{ \begin{matrix} \frac{x - y}{\left| x - y\right| } & \text{if}x \neq y \\ 0 & \text{if}x = y \end{matrix}\right. \) , (i.e., \( f\left( {x, y}\right) = 1 \), if \( x > y;f\left( {x, y}\right) = - 1 \), if \( x < y \) ; and \( f\left( {x, y}\right) = 0 \), if \( x = y \) ). | The graph of \( f\left( {x, y}\right) \) over the region \( x > y \) is a horizontal half-plane. Thus, \( {f}_{x}\left( {x, y}\right) = 0 \) for \( x > y \) . Similarly, \( {f}_{x}\left( {x, y}\right) = 0 \), for \( x < y \) . When \( y = x,{f}_{x} \) does not exist. Recall that a definite integral of a function (of a ... | Yes |
Example 3. Let \( f\left( {0,0}\right) = 0 \) and \( f\left( {x, y}\right) = \frac{x}{{x}^{2} + {y}^{2}} \), for \( \left( {x, y}\right) \neq \left( {0,0}\right) \) . We have | \[ {\int }_{0}^{1}f\left( {x, y}\right) {dy} = {\int }_{0}^{1}\frac{x}{{x}^{2} + {y}^{2}}{dy} = {\int }_{0}^{1/x}\frac{1}{1 + {z}^{2}}{dz} = \arctan \left( \frac{1}{x}\right) \] Thus, \[ \frac{d}{dx}{\int }_{0}^{1}f\left( {x, y}\right) {dy} = \frac{1}{1 + {x}^{-2}}\left( {-{x}^{-2}}\right) = \frac{-1}{1 + {x}^{2}}. \] ... | Yes |
Theorem 1 (The Bolzano-Weierstrass Theorem for \( \left. {\mathbb{R}}^{2}\right) \) . Let \( {p}_{n} = \left( {{x}_{n},{y}_{n}}\right) \;\left( {n = 1,2,3,\ldots }\right) \) be a sequence of points which form a bounded subset \( B \) of \( {\mathbb{R}}^{2} \) . Then the sequence \( {p}_{1},{p}_{2},{p}_{3},\ldots \) has... | Proof. For convenience, we uniformly shrink and/or translate the bounded set \( B \), so that it lies in the unit square \( \{ \acute{\left( {x, y}\right) : 0 < x, y < 1}\} . \) Let \( \acute{D} \equiv \{ 0,1,2,\ldots ,9\} \) be the set of digits. There are digits \( {a}_{1} \) and \( {b}_{1} \) in \( D \), such that f... | Yes |
According to Itô's formula, the solution of the stochastic differential equation\n\n\[ \left\{ \begin{array}{l} {dY} = {YdW} \\ Y\left( 0\right) = 1 \end{array}\right. \]\n\n | is\n\n\[ Y\left( t\right) \mathrel{\text{:=}} {e}^{W\left( t\right) - \frac{t}{2}} \]\n\nand not what might seem the obvious guess, namely \( \widehat{Y}\left( t\right) \mathrel{\text{:=}} {e}^{W\left( t\right) } \) . | Yes |
Let \( P\left( t\right) \) denote the (random) price of a stock at time \( t \geq 0 \) . A standard model assumes that \( \frac{dP}{P} \), the relative change of price, evolves according to the SDE\n\n\[ \frac{dP}{P} = {\mu dt} + {\sigma dW} \]\n\nfor certain constants \( \mu > 0 \) and \( \sigma \), called respectivel... | Using once again Itô’s formula we can check that the solution is\n\n\[ P\left( t\right) = {p}_{0}{e}^{{\sigma W}\left( t\right) + \left( {\mu - \frac{{\sigma }^{2}}{2}}\right) t}. \] | Yes |
Suppose instead we fix a point \( z \in {\mathbb{R}}^{n} \), and now define\n\n\[ P\left( B\right) \mathrel{\text{:=}} \left\{ \begin{array}{ll} 1 & \text{ if }z \in B \\ 0 & \text{ if }z \notin B \\ 9 & \text{ if }z \notin B \end{array}\right. \]\n\nfor sets \( B \in \mathcal{B} \) . Then \( \left( {{\mathbb{R}}^{n},\... | We call \( P \) the Dirac mass concentrated at the point \( z \), and write \( P = {\delta }_{z} \). | No |
Example 4 (Buffon's needle problem). The plane is ruled by parallel lines 2 inches apart and a 1-inch long needle is dropped at random on the plane. What is the probability that it hits one of the parallel lines? | The first issue is to find some appropriate probability space \( \\left( {\\Omega ,\\mathcal{U}, P}\\right) \) . For this, let\n\n\[ \n\\left\\{ \\begin{array}{l} h = \\text{ distance from the center of needle to nearest line,} \\\\ \\theta = \\text{ angle }\\left( { \\leq \\frac{\\pi }{2}}\\right) \\text{ that the nee... | Yes |
LEMMA. Let \( \mathbf{X} : \Omega \rightarrow {\mathbb{R}}^{n} \) be a random variable. Then \[ \mathcal{U}\left( \mathbf{X}\right) \mathrel{\text{:=}} \left\{ {{\mathbf{X}}^{-1}\left( B\right) \mid B \in \mathcal{B}}\right\} \] is a \( \sigma \) -algebra, called the \( \sigma \) -algebra generated by \( \mathbf{X} \) ... | Proof. Check that \( \left\{ {{\mathbf{X}}^{-1}\left( B\right) \mid B \in \mathcal{B}}\right\} \) is a \( \sigma \) -algebra; clearly it is the smallest \( \sigma \) -algebra with respect to which \( \mathbf{X} \) is measurable. | No |
If \( X : \Omega \rightarrow \mathbb{R} \) has density\n\n\[ f\left( x\right) = \frac{1}{\sqrt{{2\pi }{\sigma }^{2}}}{e}^{-\frac{{\left| x - m\right| }^{2}}{2{\sigma }^{2}}}\;\left( {x \in \mathbb{R}}\right) ,\] | we say \( X \) has a Gaussian (or normal) distribution, with mean \( m \) and variance \( {\sigma }^{2} \) . In this case let us write\n\n\[ X\text{is an}N\left( {m,{\sigma }^{2}}\right) \text{random variable.} \] | No |
LEMMA. Let \( \mathbf{X} : \Omega \rightarrow {\mathbb{R}}^{n} \) be a random variable, and assume that its distribution function \( F = {F}_{\mathbf{X}} \) has the density \( f \) . Suppose \( g : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \), and \( Y = g\left( \mathbf{X}\right) \) is integrable. Then \( E\left( Y\right... | Proof. Suppose first \( g \) is a simple function on \( {\mathbb{R}}^{n} \) : \( g = \mathop{\sum }\limits_{{i = 1}}^{m}{b}_{i}{\chi }_{{B}_{i}}\;\left( {{B}_{i} \in \mathcal{B}}\right) \). Then \( E\left( {g\left( \mathbf{X}\right) }\right) = \mathop{\sum }\limits_{{i = 1}}^{m}{b}_{i}{\int }_{\Omega }{\chi }_{{B}_{i}}... | Yes |
Let \( X\left( \cdot \right) = W\left( \cdot \right), u\left( x\right) = {x}^{m} \) . Then \( {dX} = {dW} \) and thus \( F \equiv 0, G \equiv 1 \) . Hence Itô's formula gives | \[ d\left( {W}^{m}\right) = m{W}^{m - 1}{dW} + \frac{1}{2}m\left( {m - 1}\right) {W}^{m - 2}{dt}. \] In particular the case \( m = 2 \) reads \[ d\left( {W}^{2}\right) = {2WdW} + {dt} \] This integrated is the identity \[ {\int }_{s}^{r}{WdW} = \frac{{W}^{2}\left( r\right) - {W}^{2}\left( s\right) }{2} - \frac{\left( r... | Yes |
Example 2. Again take \( X\left( \cdot \right) = W\left( \cdot \right), u\left( {x, t}\right) = {e}^{{\lambda x} - \frac{{\lambda }^{2}t}{2}}, F \equiv 0, G \equiv 1 \) . Then | \[ d\left( {e}^{{\lambda W}\left( t\right) - \frac{{\lambda }^{2}t}{2}}\right) = \left( {-\frac{{\lambda }^{2}}{2}{e}^{{\lambda W}\left( t\right) - \frac{{\lambda }^{2}t}{2}} + \frac{{\lambda }^{2}}{2}{e}^{{\lambda W}\left( t\right) - \frac{{\lambda }^{2}t}{2}}}\right) {dt} + \lambda {e}^{{\lambda W}\left( t\right) - \... | Yes |
THEOREM (Stochastic calculus with Hermite polynomials). We have\n\n\[ {\int }_{0}^{t}{h}_{n}\left( {W, s}\right) {dW} = {h}_{n + 1}\left( {W\left( t\right), t}\right) \;\text{ for }t \geq 0\text{ and }n = 0,1,\ldots ; \]\n\nthat is,\n\n\[ d{h}_{n + 1}\left( {W, t}\right) = {h}_{n}\left( {W, t}\right) {dW}. \] | Proof. (from McKean [McK]) Since\n\n\[ {\left. \frac{{d}^{n}}{d{\lambda }^{n}}\left( {e}^{-\frac{{\left( x - \lambda t\right) }^{2}}{2t}}\right) \right| }_{\lambda = 0} = {\left( -t\right) }^{n}\frac{{d}^{n}}{d{x}^{n}}\left( {e}^{-{x}^{2}/{2t}}\right) ,\]\n\nwe have\n\n\[ {\left. \frac{{d}^{n}}{d{\lambda }^{n}}\left( {... | Yes |
Let \( m = n = 1 \) and suppose \( g \) is a continuous function (not a random variable). Then the unique solution of\n\n\[ \left\{ \begin{array}{l} {dX} = {gXdW} \\ X\left( 0\right) = 1 \end{array}\right. \]\n\nis\n\n\[ X\left( t\right) = {e}^{-\frac{1}{2}{\int }_{0}^{t}{g}^{2}{ds} + {\int }_{0}^{t}{gdW}} \] | To verify this, note that\n\n\[ Y\left( t\right) \mathrel{\text{:=}} - \frac{1}{2}{\int }_{0}^{t}{g}^{2}{ds} + {\int }_{0}^{t}{gdW} \]\n\nsatisfies\n\n\[ {dY} = - \frac{1}{2}{g}^{2}{dt} + {gdW}. \]\n\nThus Itô’s lemma for \( u\left( x\right) = {e}^{x} \) gives\n\n\[ {dX} = \frac{\partial u}{\partial x}{dY} + \frac{1}{2... | Yes |
Example 2. Similarly, the unique solution of\n\n\[ \left\\{ \begin{array}{l} {dX} = {fXdt} + {gXdW} \\ X\left( 0\right) = 1 \end{array}\right. \] | is\n\n\[ X\left( t\right) = {e}^{{\int }_{0}^{t}f - \frac{1}{2}{g}^{2}{ds} + {\int }_{0}^{t}{gdW}} \]\n\nfor \( 0 \leq t \leq T \) . | Yes |
Let \( P\left( t\right) \) denote the price of a stock at time \( t \) . We can model the evolution of \( P\left( t\right) \) in time by supposing that \( \frac{dP}{P} \), the relative change of price, evolves according to the SDE\n\n\[ \frac{dP}{P} = {\mu dt} + {\sigma dW} \]\n\nfor certain constants \( \mu > 0 \) and... | \[ d\left( {\log \left( P\right) }\right) = \frac{dP}{P} - \frac{1}{2}\frac{{\sigma }^{2}{P}^{2}{dt}}{{P}^{2}}\;\text{ by Itô’s formula } \]\n\n\[ = \left( {\mu - \frac{{\sigma }^{2}}{2}}\right) {dt} + {\sigma dW} \]\n\nConsequently\n\n\[ P\left( t\right) = {p}_{0}{e}^{{\sigma W}\left( t\right) + \left( {\mu - \frac{{\... | Yes |
The solution of the SDE\n\n\[ \left\{ \begin{array}{l} {dB} = - \frac{B}{1 - t}{dt} + {dW}\\ B\left( 0\right) = 0 \end{array}\right. \] | \[ B\left( t\right) = \left( {1 - t}\right) {\int }_{0}^{t}\frac{1}{1 - s}{dW} \] | Yes |
Consider first the linear stochastic differential equation\n\n\[ \left\{ \begin{array}{l} {dX} = d\left( t\right) {Xdt} + f\left( t\right) {XdW} \\ X\left( 0\right) = {X}_{0} \end{array}\right. \]\n\nfor \( m = n = 1 \) . We will try to find a solution having the product form\n\n\[ X\left( t\right) = {X}_{1}\left( t\ri... | where\n\n\[ \left\{ \begin{array}{l} d{X}_{1} = f\left( t\right) {X}_{1}{dW} \\ {X}_{1}\left( 0\right) = {X}_{0} \end{array}\right. \]\n\nand\n\n\[ \left\{ \begin{array}{l} d{X}_{2} = A\left( t\right) {dt} + B\left( t\right) {dW} \\ {X}_{2}\left( 0\right) = 1, \end{array}\right. \]\n\nwhere the functions \( A \) and \(... | Yes |
THEOREM. We have\n\n(5)\n\n\[ u\left( x\right) = E\left( {\tau }_{x}\right) \;\text{ for all }x \in U. \] | Proof. We employ formula (3), with \( {Lu} = \frac{1}{2}{\Delta u} \) . We have for each \( n = 1,2,\ldots \)\n\n\[ E\left( {u\left( {\mathbf{X}\left( {{\tau }_{x} \land n}\right) }\right) }\right) - E\left( {u\left( {\mathbf{X}\left( 0\right) }\right) }\right) = E\left( {{\int }_{0}^{{\tau }_{x} \land n}\frac{1}{2}{\D... | Yes |
We have for each point \( x \in U \)\n\n(7)\n\n\[ u\left( x\right) = E\left( {g\left( {\mathbf{X}\left( {\tau }_{x}\right) }\right) }\right) \] \n\nfor \( \mathbf{X}\left( \cdot \right) \mathrel{\text{:=}} \mathbf{W}\left( \cdot \right) + x \), Brownian motion starting at \( x \) . | Proof. As shown above,\n\n\[ E\left( {u\left( {\mathbf{X}\left( {\tau }_{x}\right) }\right) }\right) = E\left( {u\left( {\mathbf{X}\left( 0\right) }\right) }\right) + E\left( {{\int }_{0}^{{\tau }_{x}}\frac{1}{2}{\Delta u}\left( \mathbf{X}\right) {ds}}\right) = E\left( {u\left( {\mathbf{X}\left( 0\right) }\right) }\rig... | Yes |
For each point \( x \in U, u\left( x\right) \) is the probability that a Brownian motion starting at \( x \) hits \( {\Gamma }_{1} \) before hitting \( {\Gamma }_{2} \) . | Apply (7) for\n\n\[ g = \left\{ \begin{array}{ll} 1 & \text{ on }{\Gamma }_{1} \\ 0 & \text{ on }{\Gamma }_{2} \end{array}\right. \]\n\nThen\n\n\[ u\left( x\right) = E\left( {g\left( {\mathbf{X}\left( {\tau }_{x}\right) }\right) }\right) = \text{ probability of hitting }{\Gamma }_{1}\text{ before }{\Gamma }_{2}. \] | Yes |
Lemma 2.1. 问题: 久 AttaAz-0 血为 Pixt2= Pix-et5. \( \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}\right) \left( {x + 2}\right) \) | 证: \( {\left. \bar{u}\right| }_{tco} = v.\;2{\bar{u}}_{t} = w\left( {{xt}, t}\right) + {\int }_{0}^{t}\frac{\partial w}{\partial t}{ds} \)\n\n\( = f\left( {xt}\right) - f\frac{t}{2}P\left( {Dx}\right) .w\left( {x, t, s}\right) {ds}. \)\n\n\( = f\left( {x + }\right) - P\left( {Dx}\right) x \cdot {L}^{\prime } \) 零条件 \( ... | No |
Clairaut's equation from differential geometry is the PDE\n\n\[ \nx \cdot {Du} + f\left( {Du}\right) = u \]\n\nwhere \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is given. | A complete integral is\n\n\[ \nu\left( {x;a}\right) = a \cdot x + f\left( a\right) \;\left( {x \in U}\right) \]\n\nfor \( a \in {\mathbb{R}}^{n} \). | Yes |
The eikonal equation from geometric optics is the PDE\n\n\[ \left| {Du}\right| = 1\text{.} \] | A complete integral is\n\n\[ u\left( {x;a, b}\right) = a \cdot x + b\;\left( {x \in U}\right) \] \n\nfor \( x \in U, a \in \partial B\left( {0,1}\right), b \in \mathbb{R} \) . | Yes |
Consider the PDE\n\n\[ \n{u}^{2}\left( {1 + {\left| Du\right| }^{2}}\right) = 1 \n\] | A complete integral is\n\n\[ \nu\left( {x, a}\right) = \pm {\left( 1 - {\left| x - a\right| }^{2}\right) }^{1/2}\;\left( {\left| {x - a}\right| < 1}\right) .\n\]\n\nWe compute\n\n\[ \n{D}_{a}u = \frac{\mp \left( {x - a}\right) }{{\left( 1 - {\left| x - a\right| }^{2}\right) }^{1/2}} = 0\n\]\n\nprovided \( a = \phi \lef... | No |
An alternative form for a complete integral of the eikonal equation \( \\left| {Du}\\right| = 1 \) for \( n = 2 \) is\n\n(15)\n\n\[ u\\left( {x;a}\\right) = {x}_{1}\\cos {a}_{1} + {x}_{2}\\sin {a}_{1} + {a}_{2}\\;\\left( {x, a \\in {\\mathbb{R}}^{2}}\\right) . \] | We set \( h \\equiv 0 \), so that\n\n\[ {u}^{\\prime }\\left( {x;{a}_{1}}\\right) = {x}_{1}\\cos {a}_{1} + {x}_{2}\\sin {a}_{1} \]\n\nrepresents the subfamily of planar solutions of \( \\left| {Du}\\right| = 1 \), whose graphs pass through the point \( \\left( {0,0,0}\\right) \\in {\\mathbb{R}}^{3} \). We then compute ... | Yes |
Example 6. Let \( H\left( p\right) = {\left| p\right| }^{2}, h \equiv 0 \) in Example 3 above. Then \[ {u}^{\prime }\left( {x, t;a}\right) = x \cdot a - t{\left| a\right| }^{2}. \] | We calculate the envelope by setting \( {D}_{a}{u}^{\prime } = x - {2ta} = 0 \) . Hence \( a = \frac{x}{2t} \) , and so \[ {u}^{\prime }\left( {x, t}\right) = x \cdot \frac{x}{2t} - t{\left| \frac{x}{2t}\right| }^{2} = \frac{{\left| x\right| }^{2}}{4t}\;\left( {x \in {\mathbb{R}}^{n}, t > 0}\right) \] solves the Hamilt... | Yes |
We demonstrate the utility of equations (17) by explicitly solving the problem\n\n(18)\n\n\[ \n\\begin{cases} {x}_{1}{u}_{{x}_{2}} - {x}_{2}{u}_{{x}_{1}} & = u & & \\text{ in }U \\\\ u & = g & & \\text{ on }\\Gamma , \\end{cases} \n\]\n\nwhere \( U \) is the quadrant \( \\left\\{ {{x}_{1} > 0,{x}_{2} > 0}\\right\\} \) ... | The PDE in (18) is of the form (12), for \( \\mathbf{b} = \\left( {-{x}_{2},{x}_{1}}\\right) \) and \( c = - 1 \) . Thus the equations (17) read\n\n(19)\n\n\[ \n\\left\\{ \\begin{array}{l} {\\dot{x}}^{1} = - {x}^{2},{\\dot{x}}^{2} = {x}^{1} \\\\ \\dot{z} = z. \\end{array}\\right. \n\]\n\nAccordingly we have\n\n\[ \n\\l... | Yes |
\[ \begin{cases} {u}_{{x}_{1}} + {u}_{{x}_{2}} & = {u}^{2} & & \text{ in }U \\ u & = g & & \text{ on }\Gamma . \end{cases} \] | Now \( U \) is the half-space \( \left\{ {{x}_{2} > 0}\right\} \) and \( \Gamma = \left\{ {{x}_{2} = 0}\right\} = \partial U \) . Here \( \mathbf{b} = \left( {1,1}\right) \) and \( c = - {z}^{2} \) . Then (21) becomes \[ \begin{cases} {\dot{x}}^{1} & = 1,{\dot{x}}^{2} = 1 \\ \dot{z} & = {z}^{2}. \end{cases} \] Conseque... | Yes |
Can we solve the linear boundary-value problem\n\n(53)\n\n\\[ \n\\begin{cases} \\mathbf{b} \\cdot {Du} & = 0 & & \\text{ in }U \\\\ u & = g & & \\text{ on }\\Gamma ? \\end{cases} \n\\] | Invoking Theorem 2 we see that there exists a unique solution \( u \) defined near \( \\Gamma \), and indeed that \( u\\left( {\\mathbf{x}\\left( s\\right) }\\right) \\equiv u\\left( {\\mathbf{x}\\left( 0\\right) }\\right) = g\\left( {x}^{0}\\right) \) for each solution of the\n\n![f8ff4aae-474b-46d8-b2c2-c00cf33b0339_... | Yes |
As an instance of a quasilinear first-order PDE, we turn now to the scalar conservation law\n\n\\[ \nG\\left( {{Du},{u}_{t}, u, x, t}\\right) = {u}_{t} + \\operatorname{div}\\mathbf{F}\\left( u\\right) \n\\]\n\n(56)\n\n\\[ \n= {u}_{t} + {\\mathbf{F}}^{\\prime }\\left( u\\right) \\cdot {Du} = 0 \n\\]\nin \\( U = {\\math... | Since the direction \\( t = {x}_{n + 1} \\) plays a special role, we appropriately modify our notation. Writing now \\( q = \\left( {p,{p}_{n + 1}}\\right) \\) and \\( y = \\left( {x, t}\\right) \\), we have\n\n\\[ \nG\\left( {q, z, y}\\right) = {p}_{n + 1} + {\\mathbf{F}}^{\\prime }\\left( z\\right) \\cdot p \n\\]\n\n... | Yes |
Example 6 (Characteristics for the Hamilton-Jacobi equation). We look now at the general Hamilton-Jacobi PDE\n\n(62)\n\n\[ \nG\left( {{Du},{u}_{t}, u, x, t}\right) = {u}_{t} + H\left( {{Du}, x}\right) = 0, \n\]\n\nwhere \( {Du} = {D}_{x}u = \left( {{u}_{{x}_{1}},\ldots ,{u}_{{x}_{n}}}\right) \) . | Then writing \( q = \left( {p,{p}_{n + 1}}\right), y = \left( {x, t}\right) \) ,\n\nwe have\n\n\[ \nG\left( {q, z, y}\right) = {p}_{n + 1} + H\left( {p, x}\right) \n\]\n\nand so\n\n\[ \n{D}_{q}G = \left( {{D}_{p}H\left( {p, x}\right) ,1}\right) ,{D}_{y}G = \left( {{D}_{x}H\left( {p, x}\right) ,0}\right) ,{D}_{z}G = 0. ... | Yes |
Let us consider the initial-value problem for Burgers’ equation:\n\n\[ \n\\begin{cases} {u}_{t} + {\\left( \\frac{{u}^{2}}{2}\\right) }_{x} & = 0\\;\\text{in }\\mathbb{R} \\times \\left( {0,\\infty }\\right) \\\\ u & = g\\;\\text{on }\\mathbb{R} \\times \\{ t = 0\\} , \\end{cases} \n\]\n\nwith the initial data\n\n\[ \n... | According to the characteristic equations (cf. §3.2.5) any smooth solution \( u \) of (13),(14), takes the constant value \( {z}^{0} = g\\left( {x}^{0}\\right) \) along the projected characteristic\n\n\[ \n\\mathbf{y}\\left( s\\right) = \\left( {g\\left( {x}^{0}\\right) s + {x}^{0}, s}\\right) \\;\\left( {s \\geq 0}\\r... | Yes |
Again consider the initial-value problem (13), for which now we take\n\n\[ g\\left( x\\right) = \\left\\{ \\begin{array}{ll} 0 & \\text{ if }x < 0 \\\\ 1 & \\text{ if }x > 0 \\end{array}\\right. \] | The method of characteristics this time does not lead to any ambiguity in defining \( u \), but does fail to provide any information within the wedge \( \\{ 0 < x < t\\} \) . To illustrate this lack of knowledge, let us first set\n\n\[ {u}_{1}\\left( {x, t}\\right) \\mathrel{\\text{:=}} \\left\\{ \\begin{array}{ll} 0 &... | Yes |
Example 3. We again return to Burgers' equation (13), now for the initial function\n\n(20)\n\n\[ \n g\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x < 0 \\ 1 & \text{ if }0 \leq x \leq 1 \\ 0 & \text{ if }x > 1 \end{array}\right. \n\]\n\nFor \( 0 \leq t \leq 2 \), we may combine the analysis in Examples 1 ... | For times \( t \geq 2 \), we expect the shock wave parameterized by \( s\left( \cdot \right) \) to continue, with \( u = x/t \) to the left of \( s\left( \cdot \right), u = 0 \) to the right. This is compatible with the entropy condition (19). We calculate the behavior of the shock curve by applying the Rankine-Hugonio... | Yes |
Let \( U \subset {\mathbb{R}}^{n} \) be a bounded, open set with smooth boundary. We consider the initial/boundary-value problem for the heat equation\n\n\[\n\\begin{cases} {u}_{t} - {\Delta u} = 0 & \\text{ in }U \\times \\left( {0,\\infty }\\right) \\\\ u = 0 & \\text{ on }\\partial U \\times \\lbrack 0,\\infty ) \\\... | Will this work? To find out, we compute\n\n\[ {u}_{t}\\left( {x, t}\\right) = {v}^{\\prime }\\left( t\\right) w\\left( x\\right) ,{\\Delta u}\\left( {x, t}\\right) = v\\left( t\\right) {\\Delta w}\\left( x\\right) .\n\]\n\nHence\n\n\[ 0 = {u}_{t}\\left( {x, t}\\right) - {\\Delta u}\\left( {x, t}\\right) = {v}^{\\prime ... | Yes |
Let us next apply the separation of variables technique to discover a solution of the porous medium equation\n\n\[ \n{u}_{t} - \Delta \left( {u}^{\gamma }\right) = 0\;\text{ in }{\mathbb{R}}^{n} \times \left( {0,\infty }\right) ,\n\]\n\nwhere \( u \geq 0 \) and \( \gamma > 1 \) is a constant. | As in the previous example, we seek a solution of the form\n\n\[ \nu\left( {x, t}\right) = v\left( t\right) w\left( x\right) \;\left( {x \in {\mathbb{R}}^{n}, t \geq 0}\right) .\n\]\n\nInserting into (11), we discover that\n\n\[ \n\frac{{v}^{\prime }\left( t\right) }{v{\left( t\right) }^{\gamma }} = \mu = \frac{\Delta ... | Yes |
Let us turn once again to the Hamilton-Jacobi equation\n\n\[ \n{u}_{t} + H\left( {Du}\right) = 0\;\text{ in }{\mathbb{R}}^{n} \times \left( {0,\infty }\right) \n\]\n\nand look for a solution \( u \) having the form\n\n\[ \nu\left( {x, t}\right) = w\left( x\right) + v\left( t\right) \;\left( {x \in {\mathbb{R}}^{n}, t \... | Then\n\n\[ \n0 = {u}_{t}\left( {x, t}\right) + H\left( {{Du}\left( {x, t}\right) }\right) = {v}^{\prime }\left( t\right) + H\left( {{Dw}\left( x\right) }\right) \n\]\n\nif and only if\n\n\[ \nH\left( {{Dw}\left( x\right) }\right) = \mu = - {v}^{\prime }\left( t\right) \;\left( {x \in {\mathbb{R}}^{n}, t > 0}\right) \n\... | Yes |
We investigate first the PDE\n\n\[ \n- {\Delta u} + u = f\;\text{ in }{\mathbb{R}}^{n}, \]\n\nwhere \( f \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \) . | To find an explicit formula for \( u \), we take the Fourier transform, recalling Theorem 2,(ii) to obtain\n\n\[ \n\left( {1 + {\left| y\right| }^{2}}\right) \widehat{u}\left( y\right) = \widehat{f}\left( y\right) \;\left( {y \in {\mathbb{R}}^{n}}\right) . \]\n\nThe effect of the Fourier transform has been to convert t... | Yes |
Consider again the initial-value problem for the heat equation\n\n\[ \n\\begin{cases} {u}_{t} - {\\Delta u} & = 0 & & \\text{ in }{\\mathbb{R}}^{n} \\times \\left( {0,\\infty }\\right) \\\\\nu & = g & & \\text{ on }{\\mathbb{R}}^{n} \\times \\{ t = 0\\} . \\end{cases} \n\] | We establish a new method for solving (16) by computing \( \\widehat{u} \), the Fourier transform of \( u \) in the spatial variables \( x \) only. Thus\n\n\[ \n\\begin{cases} {\\widehat{u}}_{t} + {\\left| y\\right| }^{2}\\widehat{u} & = 0\\;\\text{ for }t > 0 \\\\\n\\widehat{u} & = \\widehat{g}\\;\\text{ for }t = 0 \\... | Yes |
Let us next look at the initial-value problem for Schrödinger's equation\n\n\[\n\\begin{cases} i{u}_{t} + {\\Delta u} & = 0\\;\\text{ in }{\\mathbb{R}}^{n} \\times \\left( {0,\\infty }\\right) \\\\ u & = g\\;\\text{ on }{\\mathbb{R}}^{n} \\times \\{ t = 0\\} . \\end{cases}\n\]\n\nHere \( u \) and \( g \) are complex-va... | If we formally replace \( t \) by \( {it} \) on the right hand side of (18), we obtain the formula\n\n\[ u\\left( {x, t}\\right) = \\frac{1}{{\\left( 4\\pi it\\right) }^{n/2}}{\\int }_{{\\mathbb{R}}^{n}}{e}^{\\frac{i{\\left| x - y\\right| }^{2}}{4t}}g\\left( y\\right) {dy}\\;\\left( {x \\in {\\mathbb{R}}^{n}, t > 0}\\r... | Yes |
We next analyze the initial-value problem for the wave equation\n\n\[ \left\{ \begin{matrix} {u}_{tt} - {\Delta u} = 0 & \text{in }{\mathbb{R}}^{n} \times \left( {0,\infty }\right) \\ u = g,\;{u}_{t} = 0 & \text{on }{\mathbb{R}}^{n} \times \{ t = 0\} , \end{matrix}\right. \] | where for simplicity we suppose the initial velocity to be zero. Take as before \( \widehat{u} \) to be the Fourier transform of \( u \) in the variable \( x \in {\mathbb{R}}^{n} \) . Then\n\n\[ \left\{ \begin{matrix} {\widehat{u}}_{tt} + {\left| y\right| }^{2}\widehat{u} = 0 & \text{ for }t > 0 \\ \widehat{u} = \wideh... | Yes |
Consider again the heat equation\n\n\[ \n\\begin{cases} {v}_{t} - {\\Delta v} = 0 & \\text{ in }U \\times \\left( {0,\\infty }\\right) \\\\ v = f & \\text{ on }U \\times \\{ t = 0\\} \\end{cases} \n\]\n\nand perform a Laplace transform with respect to time:\n\n\[ \n{v}^{\\# }\\left( {x, s}\\right) = {\\int }_{0}^{\\inf... | We compute\n\n\[ \n\\Delta {v}^{\\# }\\left( {x, s}\\right) = {\\int }_{0}^{\\infty }{e}^{-{st}}{\\Delta v}\\left( {x, t}\\right) {dt} = {\\int }_{0}^{\\infty }{e}^{-{st}}{v}_{t}\\left( {x, t}\\right) {dt} \n\]\n\n\[ \n= s{\\int }_{0}^{\\infty }{e}^{-{st}}v\\left( {x, t}\\right) {dt} + {\\left. {e}^{-{st}}v\\right| }_{... | Yes |
Suppose \( u \) is a bounded, smooth solution of the initial-value problem:\n\n\[ \left\{ \begin{matrix} {u}_{tt} - {\Delta u} = 0 & \text{in }{\mathbb{R}}^{n} \times \left( {0,\infty }\right) \\ u = g,\;{u}_{t} = 0 & \text{on }{\mathbb{R}}^{n} \times \{ t = 0\} \end{matrix}\right. \] | where \( n \) is odd and \( g \) is smooth, with compact support. We extend \( u \) to negative times by writing\n\n\[ u\left( {x, t}\right) = u\left( {x, - t}\right) \;\text{ if }x \in {\mathbb{R}}^{n}, t < 0. \]\n\nThen\n\n\[ {u}_{tt} - {\Delta u} = 0\;\text{ in }{\mathbb{R}}^{n} \times \mathbb{R}. \]\n\nNext define\... | Yes |
What happens to \( {u}^{\varepsilon } \) as \( \varepsilon \rightarrow 0 \) ? | LEMMA (Asymptotics). Suppose that \( k, l : \mathbb{R} \rightarrow \mathbb{R} \) are continuous functions, that \( l \) grows at most linearly and that \( k \) grows at least quadratically. Assume also there exists a unique point \( {y}_{0} \in \mathbb{R} \) such that\n\n\[ k\left( {y}_{0}\right) = \mathop{\min }\limit... | Yes |
Let us once more turn our attention to the wave equation\n\n\[ \n{u}_{tt} - {\Delta u} = 0\;\text{ in }{\mathbb{R}}^{n} \times \left( {0,\infty }\right) ,\n\]\n\nand we now regard the solution \( u \) as taking complex values. We fix \( \varepsilon > 0 \) and seek a solution \( u = {u}^{\varepsilon } \) of (27) having ... | Substituting (28) into (27), we find after some computations that\n\n\[ \n0 = {u}_{tt}^{\varepsilon } - \Delta {u}^{\varepsilon } = {e}^{i{p}^{\varepsilon }/\varepsilon }\left( {\frac{i{p}_{tt}^{\varepsilon }}{\varepsilon }{a}^{\varepsilon } - {\left( \frac{{p}_{t}^{\varepsilon }}{\varepsilon }\right) }^{2}{a}^{\vareps... | Yes |
Let \( U \) denote an open, bounded subset of \( {\mathbb{R}}^{n} \), with smooth boundary \( \partial U \), and consider this boundary-value problem for a divergence structure PDE:\n\n\[ \begin{cases} - \mathop{\sum }\limits_{{i, j = 1}}^{n}{\left( {a}^{ij}\left( \frac{x}{\varepsilon }\right) {u}_{{x}_{i}}^{\varepsilo... | In the following heuristic discussion let us assume\n\n\[ {u}^{\varepsilon } \rightarrow u\;\text{ as }\varepsilon \rightarrow 0 \]\n\nin some suitable sense and try to determine an equation which \( u \) satisfies. The trick is to suppose \( {u}^{\varepsilon } \) admits the following two-scale expansion:\n\n\[ {u}^{\v... | Yes |
Let \( n = 1, U = \left( {0,2}\right) \), and\n\n\[ u\left( x\right) = \left\{ \begin{array}{ll} x & \text{ if }0 < x \leq 1 \\ 1 & \text{ if }1 \leq x < 2 \end{array}\right. \]\n\nDefine\n\[ v\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }0 < x \leq 1 \\ 0 & \text{ if }1 < x < 2 \end{array}\right. \]\n\nLe... | To see this, choose any \( \phi \in {C}_{c}^{\infty }\left( U\right) \) . We must demonstrate\n\n\[ {\int }_{0}^{2}u{\phi }^{\prime }{dx} = - {\int }_{0}^{2}{v\phi dx} \]\n\nBut we easily calculate\n\n\[ {\int }_{0}^{2}u{\phi }^{\prime }{dx} = {\int }_{0}^{1}x{\phi }^{\prime }{dx} + {\int }_{1}^{2}{\phi }^{\prime }{dx}... | Yes |
We assert \( {u}^{\prime } \) does not exist in the weak sense. To check this, we must show there does not exist any function \( v \in {L}_{\mathrm{{loc}}}^{1}\left( U\right) \) satisfying\n\n(5)\n\n\[{\int }_{0}^{2}u{\phi }^{\prime }{dx} = - {\int }_{0}^{2}{v\phi dx}\]\n\nfor all \( \phi \in {C}_{c}^{\infty }\left( U\... | Suppose, to the contrary,(5) were valid for some \( v \) and all \( \phi \) . Then\n\n(6)\n\n\[- {\int }_{0}^{2}{v\phi dx} = {\int }_{0}^{2}u{\phi }^{\prime }{dx} = {\int }_{0}^{1}x{\phi }^{\prime }{dx} + 2{\int }_{1}^{2}{\phi }^{\prime }{dx}\]\n\n\[= - {\int }_{0}^{1}{\phi dx} - \phi \left( 1\right)\]\n\nChoose a sequ... | Yes |
For which values of \( \alpha > 0, n, p \) does \( u \) belong to \( {W}^{1, p}\left( U\right) \)? | To answer, note first \( u \) is smooth away from 0, with\n\n\[ {u}_{{x}_{i}}\left( x\right) = \frac{-\alpha {x}_{i}}{{\left| x\right| }^{\alpha + 2}}\;\left( {x \neq 0}\right) ,\]\n\nand so\n\n\[ \left| {{Du}\left( x\right) }\right| = \frac{\left| \alpha \right| }{{\left| x\right| }^{\alpha + 1}}\;\left( {x \neq 0}\ri... | Yes |
Example 1 (Dirichlet's principle). Take\n\n\[ L\left( {p, z, x}\right) = \frac{1}{2}{\left| p\right| }^{2}. \]\n\nThen \( {L}_{{p}_{i}} = {p}_{i}\left( {i = 1,\ldots, n}\right) ,{L}_{z} = 0 \) ; and so the Euler-Lagrange equation associated with the functional\n\n\[ I\left\lbrack w\right\rbrack \mathrel{\text{:=}} \fra... | \[ {\Delta u} = 0\;\text{ in }U. \] | No |
Let\n\n\\[ \nL\\left( {p, z, x}\\right) = {\\left( 1 + {\\left| p\\right| }^{2}\\right) }^{1/2};\n\\]\n\nso that\n\n\\[ \nI\\left\\lbrack w\\right\\rbrack = {\\int }_{U}{\\left( 1 + {\\left| Dw\\right| }^{2}\\right) }^{1/2}{dx}\n\\]\n\nis the area of the graph of the function \\( w : U \\rightarrow \\mathbb{R} \\) . Th... | The expression \\( \\operatorname{div}\\left( \\frac{Du}{{\\left( 1 + {\\left| Du\\right| }^{2}\\right) }^{1/2}}\\right) \\) on the left side of (10) is \\( n \\) times the mean curvature of the graph of \\( u \\) . Thus a minimal surface has zero mean curvature. | Yes |
Example 1 (Reaction-diffusion equations). Let us investigate the solvability of the initial/boundary-value problem for the reaction-diffusion system\n\n\[\n\\begin{cases} {\\mathbf{u}}_{t} - \\Delta \\mathbf{u} & = \\mathbf{f}\\left( \\mathbf{u}\\right) & & \\text{ in }{U}_{T} \\\\ \\mathbf{u} & = \\mathbf{0} & & \\tex... | THEOREM 2 (Existence). There exists a unique weak solution of (2).\n\nProof. 1. We will apply Banach's theorem in the space\n\n\[ X = C\\left( {\\left\\lbrack {0, T}\\right\\rbrack ;{L}^{2}\\left( {U;{\\mathbb{R}}^{m}}\\right) }\\right) \]\n\nwith the norm\n\n\[ \\parallel \\mathbf{v}\\parallel = \\mathop{\\max }\\limi... | No |
The \( p \) -system is this collection of two conservation laws:\n\n\[\n\begin{cases} {u}_{t}^{1} - {u}_{x}^{2} & = 0\;\text{ (compatibility condition) } \\ {u}_{t}^{2} - p{\left( {u}^{1}\right) }_{x} & = 0\;\text{ (Newton’s law),} \end{cases}\n\]\n\nin \( \mathbb{R} \times \left( {0,\infty }\right) \), where \( p : \m... | Taking \( {u}^{1} \mathrel{\text{:=}} {u}_{x},{u}^{2} \mathrel{\text{:=}} {u}_{t} \), we obtain the system (6), with the stated interpretations. | No |
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