Heat Equation on Unbounded Intervals
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1 Heat Equation on Unbounded Intervals MATH 467 Partial Differential Equations J. Robert Buchanan Department of Mathematics Fall 28
2 Objectives In this lesson we will learn about: the fundamental solution to the heat equation, solutions to the heat equation for x <, and solutions to the heat equation for < x <.
3 Fundamental Solution For t > define the function Ux, t = 4πkt e x 2 /4kt. Remarks: Ux, t is related to the probability density function for a normally distributed random variable. While defined only for t >, the limit as t + exists. Ux, t solves the heat equation.
4 Connection to Normal Distribution A normally distributed, continuous random variable X with mean µ and standard deviation σ has a probability distribution of f X x = σ 2π e x µ2 /2σ2 for < x <.
5 Connection to Normal Distribution A normally distributed, continuous random variable X with mean µ and standard deviation σ has a probability distribution of f X x = σ 2π e x µ2 /2σ2 for < x <. Consider the fundamental solution to the heat equation, Ux, t = 4πk t e x 2 /4kt = 2k t 2π e x 2 /2 2k t 2. For every t > the heat energy is distributed normally with mean µ = and standard deviation σ = 2k t.
6 Graph
7 lim Ux, t t + If x then lim Ux, t = lim e x 2 /4kt =. t + t + 4πkt
8 lim Ux, t t + If x then lim Ux, t = lim e x 2 /4kt =. t + t + 4πkt If x = then lim U, t = lim =. t + t + 4πkt
9 Justification Suppose x then lim Ux, t = lim e x 2 /4kt t + t + 4πkt t = lim indeterminate / 4πke x 2 /4kt t + / /2t 3/2 = lim t + x 2 4kt 4πke x 2 /4kt 2 = lim =. t + x 2 4kt /2 πke x 2 /4kt
10 Area Under the Curve Assume that for fixed t > the improper integral Ux, t dx converges. Find the value of the integral.
11 Solution If S = S 2 = = = = Ux, t dx Ux, t dx 4πkt 4πkt 4πkt = 2kt 2π S 2 = = S =. e x 2 /4kt dx re r 2 /4kt dr Uy, t dy e y 2 /4kt dy e x 2 +y 2 /4kt dx dy re r 2 /4kt dr dθ
12 Dirac Delta Function Since for x, lim Ux, t = and + then t Ux, t dx = for all t > the Dirac delta function. lim e x 2 /4kt = δx t + 4πkt
13 Logarithmic Differentiation of 2 Ux, t = 4πkt e x 2 /4kt ln U = 2 ln4πkt x 2 4kt t [ln U] = ] 2 x 2 [ ln4πkt t 4kt
14 Logarithmic Differentiation of 2 Ux, t = 4πkt e x 2 /4kt ln U = 2 ln4πkt x 2 4kt t [ln U] = ] 2 x 2 [ ln4πkt t 4kt U t U = 2t + x 2 4kt 2 U t = e x 2 /4kt 2t + x 2 4πkt 4kt 2
15 Logarithmic Differentiation 2 of 2 Ux, t = [ln U] = x x 4πkt e x 2 /4kt U x U = x 2kt x U x = U 2kt x U xx = U x U xx = ] 2 x 2 [ ln4πkt 4kt 2kt x 2 = U 4k 2 t 2 U 2kt U 4πkt e x 2 /4kt 2kt x 2 4k 2 t 2 2kt
16 Solution to the Heat Equation U t x, t = U xx x, t = 4 kπt e x 2 /4kt x 2 2kt 2 t 4k kπt e x 2 /4kt x 2 2kt 2 t and thus U t = ku xx. Remark: since the fundamental solution is defined for < x <, no boundary conditions need be considered.
17 Solution to the Heat Equation U t x, t = U xx x, t = 4 kπt e x 2 /4kt x 2 2kt 2 t 4k kπt e x 2 /4kt x 2 2kt 2 t and thus U t = ku xx. Remark: since the fundamental solution is defined for < x <, no boundary conditions need be considered. If ux, = f x is an initial condition defined on < x <, how do we form a solution to the IVP?
18 Solving the IVP Theorem Consider the initial value problem u t = k u xx for < x < and t > ux, = f x, for < x <. If f x is continuous and if f x dx converges, then the piecewise defined function Ux y, tf y dy if t >, ux, t = f x if t = solves the heat equation and satisfies the initial condition in the sense that lim x,t x, + ux, t = f x.
19 Uniqueness Theorem Consider the initial value problem with conditions imposed as x ±, u t = ku xx, for < x < and t > ux, = f x, for < x < lim max ux, t = x ± t T for < x <, t >, and T >. If f x is continuous, if lim f x =, and if f x dx converges, then x ± Ux y, tf y dy if t >, ux, t = f x if t = is the unique, continuous solution to the initial value problem above on, [,.
20 Example Find the unique solution to the following initial boundary value problem. u t = u xx, for < x < and t > ux, = e x 2 cos x, for < x < lim max ux, t = x ± t T
21 Solution of 4 Since k = then ux, t = = Ux y, te y 2 cos y dy e x y2 /4t e y 2 Re e iy dy
22 Solution of 4 Since k = then ux, t = = = Ux y, te y 2 cos y dy e x y2 /4t e y 2 Re e x 2 /4t e iy dy e 2xy y 2 /4t e y 2 Re e iy dy
23 Solution of 4 Since k = then ux, t = = = Ux y, te y 2 cos y dy e x y2 /4t e y 2 Re e x 2 /4t = Ux, t e iy dy e 2xy y 2 /4t e y 2 Re e iy dy e 2xy [+4t]y 2 /4t Re e iy dy
24 Solution of 4 Since k = then ux, t = = = Ux y, te y 2 cos y dy e x y2 /4t e y 2 Re e x 2 /4t = Ux, t = Ux, t e iy dy e 2xy y 2 /4t e y 2 Re e iy dy e 2xy [+4t]y 2 /4t Re e iy dy Re e 2xy [+4t]y 2 /4t e iy dy
25 Solution of 4 Since k = then ux, t = = = Ux y, te y 2 cos y dy e x y2 /4t e y 2 Re e x 2 /4t = Ux, t e iy dy e 2xy y 2 /4t e y 2 Re e iy dy e 2xy [+4t]y 2 /4t Re e iy dy Re e 2xy [+4t]y 2 /4t e iy dy = Ux, t ux, t = Ux, t Re e [2x+i4t]y [+4t]y 2 /4t dy
26 Solution 2 of 4 ux, t = Ux, t Re e [2x+i4t]y [+4t]y 2 /4t dy Complete the square in the exponent. ux, t = Ux, t Re e +4t 4t y 2 2x+i4t +4t y dy = Ux, t Re e +4t 4t = Ux, t Re e +4t 4t [ x+i2t +4t ] 2 y 2 2x+i4t +4t y+[ x+i2t +4t ] 2 [ x+i2t +4t ] 2 dy e +4t 4t y x+i2t +4t 2 dy
27 Solution 3 of 4 ux, t = Ux, t Re e x+i2t 2 4t+4t + 4t Substitute z = y x + i2t. 2t + 4t ux, t = Ux, t Re e x+i2t 2 4t+4t 2t + 4t = Ux, t Re e x+i2t 2 4t+4t = Ux, t Re + 4t e +4t 2t y x+i2t +4t 2 /2 dy + 4t e x2 +i4xt 4t 2 4t+4t e z2 /2 dz
28 Solution 4 of 4 ux, t = = = = + 4t + 4t + 4t Ux, t Re e x2 4t 2 4t+4t e i4xt 4t+4t x 2 4t 2 Ux, te 4t+4t x cos e x2 4t e x 2 4t 2 4t+4t cos + 4t e x2 +t +4t cos x + 4t + 4t x + 4t
29 Illustration x 2 +t ux, t = e +4t cos + 4t x + 4t t x 2 4 6
30 Semi-Infinite Intervals Theorem Suppose f x is continuous on [,, f =, lim f x =, and problem x f x dx converges. The initial boundary value u t = k u xx for < x < and t > u+, t = ux, + = f x lim max ux, t = x t T has a unique, continuous solution defined for t >, ux, t = Ux y, t Ux + y, t f y dy.
31 Example Solve the initial, boundary value problem: u+, t = ux, + = x e x lim x max ux, t t T =. u t = u xx for < x < and t >
32 Solution of 3 For simplicity k = and thus ux, t = = = = [e ] x y2 /4t e x+y2 /4t ye y dy [ ] y e x2 2x 2ty+y 2 4t e x2 +2x+2ty+y 2 4t dy [ ] e x2 4t y e y2 2x 2ty 4t e y2 +2x+2ty 4t dy e x2 x 2t 2 4t e 4t e x2 x+2t 2 4t e 4t y e y x 2t2 4t dy y e y+x+2t2 4t dy
33 Solution 2 of 3 ux, t = e x+t y x 2t + x 2te y x 2t2 4t dy e x+t y + x + 2t x + 2te y+x+2t2 4t dy = π e x+t x 2te x+t + e x+t π + x + 2tex+t y x 2t 2 t e 2 y + x + 2t 2 t e 2 e y x 2t2 4t y x 2t 2 2t dy e y+x+2t2 4t y+x+2t 2 2t dy dy dy
34 Solution 3 of 3 ux, t = t π e x 2 4t t + π e x 2 4t x 2te x+t 2π + x + 2tex+t 2π x 2t/ e z2 2 2t x+2t/ e z2 2 2t x 2t = x 2te x+t Φ 2t x + 2t 2t + x + 2te x+t Φ dz dz where Φ y = 2π y e y 2 /2 dy.
35 t Illustration ux, t = x 2te x+t Φ x 2t/ 2t + x + 2te x+t Φ x + 2t/ 2t x
36 Semi-Infinite Interval, Neumann BCs Theorem Suppose f x is continuous on [,, f =, lim f x =, and problem x f x dx converges. The initial boundary value u t = k u xx for < x < and t > u x +, t = ux, + = f x lim max ux, t = x t T has a unique, continuous solution defined for t >, ux, t = Ux y, t + Ux + y, t f y dy.
37 Example Solve the initial, boundary value problem: u t = u xx for < x < and t > u x +, t = ux, + = e x 2 cos x lim max ux, t =. x t T
38 t Solution ux, t = e x2 +t x +4t cos + 4t + 4t x
39 Homework Read Section 4.4 Exercises: 7, 8, 9
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