Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 10 pps

With this new boundary condition, we can solve the problem with the Fourier sine transform.. We take the sine transform of the partial differential equation and the initial condition.. W

Z x+τ x−τ

g(ξ) dξFinally we make the change of variables t = τ /c, u(x, t) = v(x, τ ) to obtain D’Alembert’s solution of the wave equation,

u(x, t) = 1

2(f (x − ct) + f (x + ct)) +

12c

Z x+ct x−ct

∂

∂t, v(x, τ ) = u(x, t),the problem becomes

gxx(x; ξ) − s2g(x; ξ) = δ(x − ξ), g(±∞; ξ) bounded

esx is a homogeneous solution that is bounded at x = −∞ e−sx is a homogeneous solution that is bounded at

x = +∞ The Wronskian of these solutions is

W (x) =

esx e−sx

s esx −s e−sx

−1 2sesξe−sx for x > ξ, = −

12se

Z τ

−τ

g(x − ξ) dξ

v(x, τ ) = 1

2(f (x − τ ) + f (x + τ )) +

12c

Z x+τ x−τ

Z x+ct x−ct

g(ξ) dξ

Solution 44.7

1 We take the Laplace transform of Equation 44.1

s ˆφ − φ(x, 0) = a2φˆxxˆ

φxx− s

We take the Laplace transform of the initial condition, φ(0, t) = f (t), and use that ˆφ(x, s) vanishes as x → ∞

to obtain boundary conditions for ˆφ(x, s)

ˆφ(0, s) = ˆf (s), φ(∞, s) = 0ˆThe solutions of Equation 44.3 are

exp



±

√s

a x

.The solution that satisfies the boundary conditions is

ˆφ(x, s) = ˆf (s) exp



−

√s

a x



We write this as the product of two Laplace transforms.

ˆφ(x, s) = ˆf (s)L

x2a√

Z t 0

2 Consider the case f (t) = 1

φ(x, t) = x

2a√π

Z t 0

ξ = x2a√

τ, dξ = −

x4aτ3/2

, for 0 < t < T

Consider t > T

φ(x, t) = x

2a√π

Z t t−T

φ(x, t) = −√2

π

Z x/(2a√t) x/(2a√t−T )

e−ξ2 dξ

φ(x, t) = erf

x2a√

t − T



− erf

x2a√t

utx = κuxxx.Subtracting α times the former equation from the latter yields,

utx− αut= κuxxx− ακuxx,

Thus v satisfies the same partial differential equation as u This is because the equation for u is linear and homogeneousand v is a linear combination of u and its derivatives The problem for v is,

vt= κvxx, x > 0, t > 0,v(0, t) = 0, v(x, 0) = f0(x) − αf (x)

With this new boundary condition, we can solve the problem with the Fourier sine transform We take the sine transform

of the partial differential equation and the initial condition

v(ω, 0) = Fs[f0(x) − αf (x)]

ˆ

vt(ω, t) = −κω2ˆv(ω, t)ˆ

v(ω, 0) = Fs[f0(x) − αf (x)]

Now we have a first order, ordinary differential equation for ˆv The general solution is,

ˆv(ω, t) = c e−κω2t.The solution subject to the initial condition is,

ˆv(ω, t) = Fs[f0(x) − αf (x)] e−κω2t.Now we take the inverse sine transform to find v We utilize the Fourier cosine transform pair,

κte

−x 2 /(4κt)



Recall that the Fourier sine convolution theorem is,

Fs 12π

Z ∞ 0

Z ∞ 0

e−α(ξ−x)v(ξ, t) dξ

Solution 44.9

Z ∞ 0

ω e−cω2sin(ωx) dω = − ∂

∂x

Z ∞ 0

e−cω2cos(ωx) dω

= −12

r πc

ut= uxx, x > 0, t > 0,u(0, t) = g(t), u(x, 0) = 0

We take the Fourier sine transform of the partial differential equation and the initial condition

ˆ

ut(ω, t) = −ω2u(ω, t) +ˆ ω

πg(t), u(ω, 0) = 0ˆNow we have a first order, ordinary differential equation for ˆu(ω, t)

π e

−ω 2 t

Z t 0

g(τ ) eω2τ dτ + c(ω) e−ω2t

The initial condition is satisfied for c(ω) = 0.

ˆu(ω, t) = ω

π

Z t 0

g(τ ) e−ω2(t−τ ) dτ



u(x, t) =

Z t 0

2√π(t − τ )3/2 e−x2/(4(t−τ )) dτ

u(x, t) = x

2√π

Z t 0

We take the Fourier sine transform of the partial differential equation and the boundary conditions

−ω2u(ω, y) +ˆ k

πu(0, y) + ˆuyy(ω, y) = 0ˆ

uyy(ω, y) − ω2u(ω, y) = −ˆ k

πf (y), u(ω, 0) = ˆˆ u(ω, 1) = 0

This is an inhomogeneous, ordinary differential equation that we can solve with Green functions The homogeneoussolutions are

sinh(ωy) sinh(ω(y − 1))

ω cosh(ωy) ω cosh(ω(y − 1))

... satisfies the same partial differential equation as u This is because the equation for u is linear and homogeneousand v is a linear combination of u and its derivatives The problem for v is,

vt=...

L[u(x)] = f (x) for x ∈ Ω, B[u(x)] = for x ∈ ∂Ω ( 45. 1 )For example, L[u] might be

L[u] = ut− κ∆u, or L[u] = utt− c2∆u

and B[u] might be u... transform to obtain the solution for u(x, y) To this we use the Fourier transformpair,

F

2c

x2+ c2



= e−c|ω| ,and the