Solutions of differential equations using transforms

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Solutions of differential equations using transforms Process: Take transform of equation and boundary/initial conditions in one variable. Derivatives are turned into multiplication operators. Solve (hopefully easier) problem in k variable. Inverse transform to recover solution, often as a convolution integral.

Ordinary differential equations: example 1 u + u = f (x), lim u(x) = 0. x

Ordinary differential equations: example 1 u + u = f (x), lim u(x) = 0. x Transform using the derivative rule, giving k 2 û(k) + û(k) = ˆf (k). Just an algebraic equation, whose solution is û(k) = ˆf (k) 1 + k 2.

Ordinary differential equations: example 1 u + u = f (x), lim u(x) = 0. x Transform using the derivative rule, giving k 2 û(k) + û(k) = ˆf (k). Just an algebraic equation, whose solution is û(k) = ˆf (k) 1 + k 2. Inverse transform of product of ˆf (k) and 1/(1 + k 2 ) is convolution: ( ) 1 u(x) = f (x) 1 + k 2 = 1 e x y f (y)dy. 2 But where was far field condition used?

Ordinary differential equations: example 2 Example 2. The Airy equation is u xu = 0, lim u(x) = 0. x

Ordinary differential equations: example 2 Example 2. The Airy equation is u xu = 0, lim u(x) = 0. x Transform leads to k 2 û(k) iû (k) = 0.

Ordinary differential equations: example 2 Example 2. The Airy equation is u xu = 0, lim u(x) = 0. x Transform leads to k 2 û(k) iû (k) = 0. Solve by separation of variables: dû/û = ik 2 dk integrates to û(k) = Ce ik 3 /3.

Ordinary differential equations: example 2 Example 2. The Airy equation is u xu = 0, lim u(x) = 0. x Transform leads to k 2 û(k) iû (k) = 0. Solve by separation of variables: dû/û = ik 2 dk integrates to û(k) = Ce ik 3 /3. Inverse transform is u(x) = C 2π exp(i[kx + k 3 /3])dk. With the choice C = 1 get the Airy function.

Partial differential equations, example 1 Laplace equation in upper half plane: u xx + u yy = 0, < x <, y > 0, u(x, 0) = g(x), lim u(x, y) = 0. y

Partial differential equations, example 1 Laplace equation in upper half plane: u xx + u yy = 0, < x <, y > 0, u(x, 0) = g(x), Transform in the x variable only: U(k, y) = lim u(x, y) = 0. y e ikx u(x, y)dx. Note y-derivatives commute with the Fourier transform in x. k 2 U + U yy = 0, U(k, 0) = ĝ(k), lim y U(k, y) = 0.

Partial differential equations, example 1, cont. Now solve ODEs k 2 U + U yy = 0, U(k, 0) = ĝ(k), lim y U(k, y) = 0.

Partial differential equations, example 1, cont. Now solve ODEs k 2 U + U yy = 0, U(k, 0) = ĝ(k), lim y U(k, y) = 0. General solution is U = c 1 (k)e + k y + c 2 (k)e k y. Using boundary conditions, U(k, y) = ĝ(k)e k y.

Partial differential equations, example 1, cont. Now solve ODEs k 2 U + U yy = 0, U(k, 0) = ĝ(k), lim y U(k, y) = 0. General solution is U = c 1 (k)e + k y + c 2 (k)e k y. Using boundary conditions, U(k, y) = ĝ(k)e k y. Inverse transform using convolution and exponential formulas u(x, y) =g(x) (e k y) ( ) y = g(x) π(x 2 + y 2 ) = 1 yg(x 0 ) π (x x 0 ) 2 + y 2 dx 0. Same formula as obtained by Green s function methods!

Partial differential equations, example 2 Transport equation" u t + cu x = 0, < x <, t > 0, u(x, 0) = f (x).

Partial differential equations, example 2 Transport equation" u t + cu x = 0, < x <, t > 0, u(x, 0) = f (x). As before, U(k, t) = therefore transform in x variables is e ikx u(x, t)dx. U t + ikcu = 0, U(k, 0) = ˆf (k).

Partial differential equations, example 2 Transport equation" u t + cu x = 0, < x <, t > 0, u(x, 0) = f (x). As before, U(k, t) = therefore transform in x variables is e ikx u(x, t)dx. U t + ikcu = 0, U(k, 0) = ˆf (k). Simple differential equation with solution U(k, t) = e icktˆf (k).

Partial differential equations, example 2 Transport equation" u t + cu x = 0, < x <, t > 0, u(x, 0) = f (x). As before, U(k, t) = therefore transform in x variables is e ikx u(x, t)dx. U t + ikcu = 0, U(k, 0) = ˆf (k). Simple differential equation with solution U(k, t) = e icktˆf (k). Use translation formula f (x a) = e iatˆf (k) with a = ct, u(x, t) = f (x ct).

Partial differential equations, example 3 Consider the wave equation on the real line u tt = u xx, < x <, t > 0, u(x, 0) = f (x), u t (x, 0) = g(x).

Partial differential equations, example 3 Consider the wave equation on the real line u tt = u xx, < x <, t > 0, u(x, 0) = f (x), u t (x, 0) = g(x). Transforming as before, U tt + k 2 U = 0, U(k, 0) = ˆf (k), U t (k, 0) = ĝ(k).

Partial differential equations, example 3 Consider the wave equation on the real line u tt = u xx, < x <, t > 0, u(x, 0) = f (x), u t (x, 0) = g(x). Transforming as before, U tt + k 2 U = 0, U(k, 0) = ˆf (k), U t (k, 0) = ĝ(k). Solution of initial value problem U(k, t) = ˆf (k) cos(kt) + ĝ(k) k sin(kt). Need inverse transform formulas for cos,sin and ˆf /k.

Partial differential equations, example 3, cont. For cos,sin, write in terms of complex exponentials: [cos(ak)] = 1 2 [eiak + e iak ] = 1 [δ(x + a) + δ(x a)], 2 [sin(ak)] = i 2 [eiak e iak ] = i [δ(x + a) δ(x a)]. 2

Partial differential equations, example 3, cont. For cos,sin, write in terms of complex exponentials: [cos(ak)] = 1 2 [eiak + e iak ] = 1 [δ(x + a) + δ(x a)], 2 [sin(ak)] = i 2 [eiak e iak ] = i [δ(x + a) δ(x a)]. 2 If f (x) is integrable, int. by parts gives [ x ] f (x )dx = e ikx x f (x )dx dx = 1 e ikx f (x)dx. ik

Partial differential equations, example 3, cont. For cos,sin, write in terms of complex exponentials: [cos(ak)] = 1 2 [eiak + e iak ] = 1 [δ(x + a) + δ(x a)], 2 [sin(ak)] = i 2 [eiak e iak ] = i [δ(x + a) δ(x a)]. 2 If f (x) is integrable, int. by parts gives [ x ] f (x )dx = therefore e ikx x ] [ˆf (k) = ik f (x )dx dx = 1 e ikx f (x)dx. ik x f (x )dx. That is, integration in x gives a factor of 1/(ik) in the transform.

Partial differential equations, example 3, cont. Now for inverse transform of U(k, t) = ˆf (k) cos(kt) + ĝ(k) k sin(kt).

Partial differential equations, example 3, cont. Now for inverse transform of U(k, t) = ˆf (k) cos(kt) + ĝ(k) k First term: ] 1 [ˆf (k) cos(kt) = f (x) [δ(x + t) + δ(x t)] 2 = 1 2 sin(kt). f (x y)[δ(y + t) + δ(y t)]dy = 1 [f (x t) + f (x + t)]. 2

Partial differential equations, example 3, cont. Now for inverse transform of U(k, t) = ˆf (k) cos(kt) + ĝ(k) k First term: ] 1 [ˆf (k) cos(kt) = f (x) [δ(x + t) + δ(x t)] 2 Second term: = 1 2 [ĝ(k) sin(kt)/k] = i = 1 2 = 1 2 = 1 2 sin(kt). f (x y)[δ(y + t) + δ(y t)]dy = 1 [f (x t) + f (x + t)]. 2 x x [ĝ(k) sin(kt)] (x )dx x x g(x ) [δ(x + t) δ(x t)]dx g(x y)[δ(y + t) δ(y t)]dydx g(x + t) g(x t)dx.

Partial differential equations, example 3, cont. Finally, can change variables ξ = x + t or ξ = x t x g(x + t) g(x t)dx = = x+t x+t x t x t g(ξ)dξ g(ξ)dξ g(ξ)dξ.

Partial differential equations, example 3, cont. Finally, can change variables ξ = x + t or ξ = x t x g(x + t) g(x t)dx = = x+t x+t All together get d Alembert s formula x t u(x, t) = 1 2 [f (x t) + f (x + t)] + 1 2 x t g(ξ)dξ g(ξ)dξ g(ξ)dξ. x+t x t g(ξ)dξ.

Fundamental solutions Consider generic, linear, time-dependent equation u t (x, t) = Lu(x, t), < x <, u(x, 0) = f (x), lim u(x, t) = 0, x where L is some operator(e.g. L = 2 / x 2 ).

Fundamental solutions Consider generic, linear, time-dependent equation u t (x, t) = Lu(x, t), < x <, u(x, 0) = f (x), lim u(x, t) = 0, x where L is some operator(e.g. L = 2 / x 2 ). The fundamental solution S(x, x 0, t) is a type of Green s function, solving S t = L x S, < x <, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0.

Fundamental solutions Consider generic, linear, time-dependent equation u t (x, t) = Lu(x, t), < x <, u(x, 0) = f (x), lim u(x, t) = 0, x where L is some operator(e.g. L = 2 / x 2 ). The fundamental solution S(x, x 0, t) is a type of Green s function, solving S t = L x S, < x <, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Initial condition means S limits to a δ-function as t 0: lim t 0 S(x, x 0, t)φ(x)dx = δ(x x 0 )φ(x)dx = φ(x 0 ),

Fundamental solutions, cont. Claim that the initial value problem has solution u(x, t) = S(x, x 0, t)f (x 0 )dx 0,

Fundamental solutions, cont. Claim that the initial value problem has solution Check: u(x, 0) = lim t 0 u(x, t) = S(x, x 0, t)f (x 0 )dx 0 = S(x, x 0, t)f (x 0 )dx 0, δ(x x 0 )f (x 0 )dx 0 = f (x).

Fundamental solutions, cont. Claim that the initial value problem has solution Check: u(x, 0) = lim t 0 u(x, t) = S(x, x 0, t)f (x 0 )dx 0 = S(x, x 0, t)f (x 0 )dx 0, δ(x x 0 )f (x 0 )dx 0 = f (x). Plugging u into the equation and moving time derivative inside the integral u t = S t (x, x 0, t)f (x 0 )dx 0 = L x S(x, x 0, t)f (x 0 )dx 0.

Fundamental solutions, cont. Claim that the initial value problem has solution Check: u(x, 0) = lim t 0 u(x, t) = S(x, x 0, t)f (x 0 )dx 0 = S(x, x 0, t)f (x 0 )dx 0, δ(x x 0 )f (x 0 )dx 0 = f (x). Plugging u into the equation and moving time derivative inside the integral u t = S t (x, x 0, t)f (x 0 )dx 0 = Now move operator outside integral L x S(x, x 0, t)f (x 0 )dx 0. u t = L x S(x, x 0, t)f (x 0 )dx 0 = L x u.

Fundamental solutions using the Fourier transform, example 1 For diffusion equation on the real line, S solves S t = DS xx, < x <, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0.

Fundamental solutions using the Fourier transform, example 1 For diffusion equation on the real line, S solves S t = DS xx, < x <, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Take Fourier transform in x by letting Ŝ(k, x 0, t) = S(x, x 0, t)e ikx dx, giving

Fundamental solutions using the Fourier transform, example 1 For diffusion equation on the real line, S solves S t = DS xx, < x <, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Take Fourier transform in x by letting Ŝ(k, x 0, t) = S(x, x 0, t)e ikx dx, giving Ŝ t = Dk 2 Ŝ, Ŝ(k, 0) = e ix 0k.

Fundamental solutions using the Fourier transform, example 1 For diffusion equation on the real line, S solves S t = DS xx, < x <, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Take Fourier transform in x by letting Ŝ(k, x 0, t) = S(x, x 0, t)e ikx dx, giving Solution to this ODE Ŝ t = Dk 2 Ŝ, Ŝ(k, 0) = e ix 0k. Ŝ = e ix 0k Dk 2t.

Fundamental solutions using the Fourier transform, example 1 Inverse transform of Ŝ = e ix 0k Dk 2t. uses translation, dilation, and Gaussian formulas: S(x, x 0, t) = 1 4πDt e (x x 0) 2 /(4Dt).

Fundamental solutions using the Fourier transform, example 1 Inverse transform of Ŝ = e ix 0k Dk 2t. uses translation, dilation, and Gaussian formulas: S(x, x 0, t) = 1 4πDt e (x x 0) 2 /(4Dt). It follows that the solution to u t = Du xx and u(x, 0) = f (x) is u(x, t) = f (x 0 ) 4πDt e (x x 0) 2 /(4Dt) dx 0.

Fundamental solutions using the Fourier transform, example 2 Linearized Korteweg - de Vries (KdV) equation: u t = u xxx, u(x, 0) = f (x), lim u(x, t) = 0. x

Fundamental solutions using the Fourier transform, example 2 Linearized Korteweg - de Vries (KdV) equation: u t = u xxx, u(x, 0) = f (x), lim u(x, t) = 0. x Fundamental solution solves S t = S xxx, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0.

Fundamental solutions using the Fourier transform, example 2 Linearized Korteweg - de Vries (KdV) equation: u t = u xxx, u(x, 0) = f (x), lim u(x, t) = 0. x Fundamental solution solves S t = S xxx, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Transforming Ŝ t = ik 3 Ŝ, Ŝ(k, 0) = e ix 0k, whose solution is Ŝ(k, x 0, t) = e ix 0k e ik 3t.

Fundamental solutions using the Fourier transform, example 2 Linearized Korteweg - de Vries (KdV) equation: u t = u xxx, u(x, 0) = f (x), lim u(x, t) = 0. x Fundamental solution solves S t = S xxx, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Transforming Ŝ t = ik 3 Ŝ, Ŝ(k, 0) = e ix 0k, whose solution is Ŝ(k, x 0, t) = e ix0k e ik 3t. Recall transform of Airy function Ai(x) is e ik 3 /3, therefore S(x, x 0, t) = [e ] ix0k e ik 3 [ t = e i(k/a)3 /3 (x x0 ) ( ) = aai a(x x 0 ), a (3t) 1/3.

Fundamental solutions using the Fourier transform, example 2 Linearized Korteweg - de Vries (KdV) equation: u t = u xxx, u(x, 0) = f (x), lim u(x, t) = 0. x Fundamental solution solves S t = S xxx, S(x, x 0, 0) = δ(x x 0 ), lim x S(x, x 0, t) = 0. Transforming Ŝ t = ik 3 Ŝ, Ŝ(k, 0) = e ix 0k, whose solution is Ŝ(k, x 0, t) = e ix0k e ik 3t. Recall transform of Airy function Ai(x) is e ik 3 /3, therefore S(x, x 0, t) = [e ] ix0k e ik 3 [ t = e i(k/a)3 /3 (x x0 ) ( ) = aai a(x x 0 ), a (3t) 1/3. Solution to original equation: u(x, t) = 1 (3t) 1/3 ( ) x x0 Ai f (x (3t) 1/3 0 )dx 0.

The method of images for fundamental solutions For solutions on half-line x > 0, can t use Fourier transform directly.

The method of images for fundamental solutions For solutions on half-line x > 0, can t use Fourier transform directly. Fundamental solution must satisfy boundary condition at x = 0

The method of images for fundamental solutions For solutions on half-line x > 0, can t use Fourier transform directly. Fundamental solution must satisfy boundary condition at x = 0 Inspiration: method of images. If S (x; x 0, t) is the fundamental solution for the whole line, then: Odd reflection S = S (x; x 0, t) S (x; x 0, t) gives S(0, x 0, t) = 0. Even reflection S = S (x; x 0, t) + S (x; x 0, t) gives S x (0, x 0, t) = 0.

The method of images for fundamental solutions, example Consider diffusion equation on half line: u t = Du xx, u(x, 0) = f (x), u(0, t) = 0, lim x u(x, t) = 0.

The method of images for fundamental solutions, example Consider diffusion equation on half line: u t = Du xx, u(x, 0) = f (x), u(0, t) = 0, lim x u(x, t) = 0. Use odd reflection of fundamental solution for whole line S = e (x x 0) 2 /(4Dt) / 4πDt, S(x, x 0, t) = 1 4πDt [ e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) ]

The method of images for fundamental solutions, example Consider diffusion equation on half line: u t = Du xx, u(x, 0) = f (x), u(0, t) = 0, lim x u(x, t) = 0. Use odd reflection of fundamental solution for whole line S = e (x x 0) 2 /(4Dt) / 4πDt, S(x, x 0, t) = Therefore the solution u is just u(x, t) = 0 1 [ ] e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) 4πDt f (x 0 ) 4πDt [e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) ] dx 0.

The age of the earth Lord Kelvin: simple model of temperature of earth u(x, t) at depth x and time t u t = Du xx, x > 0, u(x, 0) = U 0, u(0, t) = 0. Scale chosen so u = 0 on surface; assumes initially constant temperature (U 0 ) throughout the molten earth.

The age of the earth Lord Kelvin: simple model of temperature of earth u(x, t) at depth x and time t u t = Du xx, x > 0, u(x, 0) = U 0, u(0, t) = 0. Scale chosen so u = 0 on surface; assumes initially constant temperature (U 0 ) throughout the molten earth. We found solution u(x, t) = 0 f (x 0 ) 4πDt [e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) ] dx 0.

The age of the earth Lord Kelvin: simple model of temperature of earth u(x, t) at depth x and time t u t = Du xx, x > 0, u(x, 0) = U 0, u(0, t) = 0. Scale chosen so u = 0 on surface; assumes initially constant temperature (U 0 ) throughout the molten earth. We found solution u(x, t) = 0 f (x 0 ) 4πDt [e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) ] dx 0. Temperature gradient µ at surface is therefore µ = u x (0, t) = U 0 4πDt 1 Dt 0 x 0 e x 2 0 /(4Dt) dx 0 = U 0 πdt.

The age of the earth Lord Kelvin: simple model of temperature of earth u(x, t) at depth x and time t u t = Du xx, x > 0, u(x, 0) = U 0, u(0, t) = 0. Scale chosen so u = 0 on surface; assumes initially constant temperature (U 0 ) throughout the molten earth. We found solution u(x, t) = 0 f (x 0 ) 4πDt [e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) ] dx 0. Temperature gradient µ at surface is therefore µ = u x (0, t) = U 0 4πDt 1 Dt 0 x 0 e x 2 0 /(4Dt) dx 0 = U 0 πdt. This relates the age of earth t to quantities we can estimate U 0 melting temp. of iron 10 4 C, D 10 3 m 2 /s, µ 10 2 C/m,

The age of the earth Lord Kelvin: simple model of temperature of earth u(x, t) at depth x and time t u t = Du xx, x > 0, u(x, 0) = U 0, u(0, t) = 0. Scale chosen so u = 0 on surface; assumes initially constant temperature (U 0 ) throughout the molten earth. We found solution u(x, t) = 0 f (x 0 ) 4πDt [e (x x 0) 2 /(4Dt) e (x+x 0) 2 /(4Dt) ] dx 0. Temperature gradient µ at surface is therefore µ = u x (0, t) = U 0 4πDt 1 Dt 0 x 0 e x 2 0 /(4Dt) dx 0 = U 0 πdt. This relates the age of earth t to quantities we can estimate U 0 melting temp. of iron 10 4 C, D 10 3 m 2 /s, µ 10 2 C/m, which gives t 3 10 7 years!!??

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