Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 10 ppt

Thus the Green function isSince we are dealing with an Euler equation, we substitute y = xλ to find the homogeneous solutions... Since this problem has both an inhomogeneous term in the

The general solution of the differential equation is thus

y = c1cos x + c2sin x + x.Applying the two initial conditions gives us the equations

x2y00(x) − xy0(x) + y(x) = 0.Substituting y = xλ into the homogeneous differential equation yields

x2λ(λ − 1)xλ−2− xλxλ+ xλ = 0

λ2− 2λ + 1 = 0(λ − 1)2 = 0

λ = 1

The homogeneous solutions are

y1 = x, y2 = x log x.The Wronskian of the homogeneous solutions is

W [x, x log x] =

x x log x

1 1 + log x

= x + x log x − x log x

= x

Writing the inhomogeneous equation in the standard form:

The Wronskian of the homogeneous solutions is

W [cos x, sin x] =

= cos2t + sin2t = 1

We use variation of parameters to find a particular solution

yp = − cos t

Zg(t) sin t dt + sin t

Zg(t) cos t dt

The general solution can be written in the form,

y(t) =



c1−

Z t a

g(τ ) sin τ dτ

cos t +



c2+

Z t b

g(τ ) cos τ dτ

sin t

2 Since the initial conditions are given at t = 0 we choose the lower bounds of integration in the general solution

g(τ ) sin τ dτ

cos t +



c2+

Z t 0

g(τ ) cos τ dτ

sin tThe initial condition y(0) = 0 gives the constraint, c1 = 0 The derivative of y(t) is then,

y0(t) = −g(t) sin t cos t +

Z t 0

g(τ ) sin τ dτ sin t + g(t) cos t sin t +



c2+

Z t 0

g(τ ) cos τ dτ

cos t,

y0(t) =

Z t 0

g(τ ) sin τ dτ sin t +



c2+

Z t 0

g(τ ) cos τ dτ

cos t

The initial condition y0(0) = 0 gives the constraint c2 = 0 The solution subject to the initial conditions is

y =

Z t 0

g(τ )(sin t cos τ − cos t sin τ ) dτ

y =

Z t 0

g(τ ) sin(t − τ ) dτ

3 The solution of the initial value problem

y00+ y = sin(λt), y(0) = 0, y0(0) = 0,is

y =

Z t 0

sin(λτ ) sin(t − τ ) dτ

Figure 21.5: Non-resonant Forcing

For λ 6= 1, this is

y = 12

Z t 0

cos(t − τ − λτ ) − cos(t − τ + λτ )
dτ

= 12

= 12

The solution is the sum of two periodic functions of period 2π and 2π/λ This solution is plotted in Figure 21.5

on the interval t ∈ [0, 16π] for the values λ = 1/4, 7/8, 5/2

Figure 21.6: Resonant Forcing

For λ = 1, we have

y = 12

Z t 0

cos(t − 2τ ) − cos(tau)
dτ

= 12



−1

2sin(t − 2τ ) − τ cos t

t 0

u01y1+ u02y2+ u03y3 = 0

u01y01+ u02y20 + u03y30 = 0Differentiating the expression for yp,

u01y100+ u1y0001 + u02y002 + u2y2000+ u03y300+ u3y0003 + p2(u1y001 + u2y200+ u3y003) + p1(u1y01+ u2y20 + u3y03)

+ p0(u1y1+ u2y2+ u3y3) = f (x)

u01y100+ u02y002 + u03y300+ u1L[y1] + u2L[y2] + u3L[y3] = f (x)

u01y001 + u02y200+ u03y300= f (x)

With the two constraints, we have the system of equations,

j, the particular solution is

yp = y1

Z(y2y03− y0

2y3)f (x)

W (x) dx + y2

Z(y3y10 − y0

3y1)f (x)

W (x) dx + y3

Z(y1y20 − y0

Thus the Green function is

Since we are dealing with an Euler equation, we substitute y = xλ to find the homogeneous solutions

λ(λ − 1) + λ − 1 = 0(λ − 1)(λ + 1) = 0

y1 = x, y2 = 1

x

Variation of Parameters The Wronskian of the homogeneous solutions is

W (x) =

1 −1/x2

Green Function Since this problem has both an inhomogeneous term in the differential equation and neous boundary conditions, we separate it into the two problems

Thus the solution is

u(x) =

Z ∞ 0

G(x|ξ)ξ2dξ

=

Z x 0

The Green function satisfies

G000(x|ξ) + p2(x)G00(x|ξ) + p1(x)G0(x|ξ) + p0(x)G(x|ξ) = δ(x − ξ)

First note that only the G000(x|ξ) term can have a delta function singularity If a lower derivative had a delta functiontype singularity, then G000(x|ξ) would be more singular than a delta function and there would be no other term in theequation to balance that behavior Thus we see that G000(x|ξ) will have a delta function singularity; G00(x|ξ) will have

a jump discontinuity; G0(x|ξ) will be continuous at x = ξ Integrating the differential equation from ξ− to ξ+ yields

G00(ξ+|ξ) − G00(ξ−|ξ) = 1.

Thus we have the three continuity conditions:

G00(ξ+|ξ) = G00(ξ−|ξ) + 1

G0(ξ+|ξ) = G0(ξ−|ξ)G(ξ+|ξ) = G(ξ−|ξ)Solution 21.12

Variation of Parameters Consider the problem

x2y00− 2xy0+ 2y = e−x, y(1) = 0, y0(1) = 1

Previously we showed that two homogeneous solutions are

y1 = x, y2 = x2.The Wronskian of these solutions is

W (x) =

... u03< /sub>y3< /sub>00+ u3< /sub>y0003< /sub> + p2(u1y001 + u2y200+ u3< /sub>y003< /sub>)... class="text_page_counter">Trang 10< /span>

u01y1+ u02y2+ u03< /sub>y3< /small>...