CONSTRUCTIONMETHOD 55 Problem 6.1 Spliy!

WEEK6. CONSTRUCTIONMETHOD 55 Problem 6.1 Spliy!
(a) Use an integrating factor, p(x), to make D y = y®® + 2x y® + (x2 + 1) y self-adjoint. (b) Show that the
resulting dierential operator, D í p(x)D y, is self-adjoint, ⇣w › D y⌘ = ⇣D w › y⌘ , by integrating by
parts twice. Assume that the Wronskian is W (w, y) í Û w y Û = wy® * w®y = A_p. [Hint: The goal of
Ûw® y®Û integrating by parts is to eliminate the derivatives of y in the integrals.]
(a) I have p0 = 1 and p1 = 2x so p®0 ë p1 and the equation is not self-adjoint. The integrating factor is
p(x) í exp0
so the integrating factor is
x du p1(u)1 = exp0 x du 2u1 = expx2 p0(u) 1
p(x) = expx2 = expx2. Ç 2 p (x) 1
Multiplying D y by this integrating factor, x
DÉxy= ex2Dxy= ex2 y®® +2xex2y® +(x2 +1)ex2 y=(ex2 y®)® +(x2 +1)ex2 y Ç2 which is self-adjoint.
(b) Starting with the self-adjoint form
⇣w›DÉxy⌘= dxw(ex2 y®)® + dxw(x2 +1)ex2 y
integratebypartswithu=wanddv=dx(ex2 y®)® sothatdu=dxw® andv=(ex2 y®) ⇣w›DÉxy⌘=⌧w(ex2 y®)* dxw®(ex2 y®)+ dxw(x2 +1)ex2 y Ç2
andagainwithu=w®ex2 anddv=dxy® sothatdu=dx(ex2w®)® andv=y ⇣w›DÉxy⌘=⌧wex2 y®*⌧w®ex2y+ dx(ex2w®)®y+ dxw(x2 +1)ex2 y
=⌧ex2(wy® *w®y)+ dx⇠(ex2w®)® +(x2 +1)ex2w⇡y Ç2
I see that the Wronskian has appeared, W = (wy® * w®y) = A_p = A_ ex2 , so the boundary term is zero,
Ç 2 leaving
⇣ w › DÉ x y ⌘ = ⇣ DÉ x w › y ⌘ ,
so the linear dierential operator0 1
DÉ íex2D =ex2 d2 +2xd +(x2+1) = d⇠ex2 d⇡+ex2(x2+1)
x x dx2 dx dx dx is self-adjoint as required.
© 2023 Chris O’Donovan
Code Help
WEEK6. CONSTRUCTIONMETHOD 56 I note that
(ex2_2y)®® =(ex2_2y® +xex2_2y)®
= ex2_2 y®® +xex2_2 y® +(ex2_2 +x2 ex2_2)y+xex2_2 y®
= ex2_2y®® + xy® + (x2 + 1)y = DÉxy = ex2_2Dxy
so I can integrate twice to find the solution to this ODE, ( ex2_2 y) = ax + b
y = (ax + b) e*x2_2
where a and b are constants.
© 2023 Chris O’Donovan
程序代写 CS代考加QQ: 749389476
WEEK 6. CONSTRUCTION METHOD
1.0 0.5 0.0
0.5 1.0 1.5 2.0
0.0 0.2 0.4
= 0.2 = 0.4 = 0.6 = 0.8
Figure 6.1: A plot of g(x, ⇠), Green’s function for section 6.2(d). This is the solution for x2g®® = . Problem 6.2 Self Adjoint and Green
In this problem you will construct Green’s function for the dierential operator
with boundary conditions
Dxy(x)=x2d2y+2xdy= d0x2dy1 dx2 dx dx dx
y(l) = y®(l) = 0.
(a) In order to find Green’s function for this problem you need to solve
Dx g(x, ⇠) = (x * ⇠). Solve the homogeneous version of this ODE for x ë ⇠ for the two cases
g(x,⇠) = Tg<(x,⇠), x < ⇠ g>(x, ⇠), x > ⇠.
For each case you should have two integration constants which are determined by the boundary conditions. Each of these four integration “constants” will be functions of ⇠. (b) Apply the four boundary conditions, the usual two given with the original problem (here y(l) = y®(l) = 0) as well as the two at x = ⇠, to find Green’s function, g(x, ⇠), for this problem. (c) Show that your Green’s function, g(x, ⇠), satisfies the boundary conditions. (d) Plot g(x, ⇠) as a function of x. Identify the point x = ⇠ on your plot.
(a) For x ë ⇠ I can solve the homogeneous ODE:
⇠ x 2 g √® ( x , ⇠ ) ⇡ ® = 0
© 2023 Chris O’Donovan

WEEK6. CONSTRUCTIONMETHOD
x2g√® (x, ⇠) = ↵√(⇠) g®(x,⇠)= 1↵ (⇠)
√ x2√ g√(x,⇠)=*x1↵√(⇠)+√(⇠), Ç2
and I have four integration “constants” that are functions of ⇠ (two each of ↵√ and √) as required. (b) I have x > ⇠ for both boundary conditions,
g > ( l , ⇠ ) = 0 = * l1 ↵ > ( ⇠ ) + > ( ⇠ ) g®(l,⇠)=0= 1↵ (⇠)
> l2> Ÿ↵>(⇠)=>(⇠)=0, Ç2
and g>(x, ⇠) = 0.
At x = ⇠ I have (with p(x) = x2)
⇠ ⇠ ⇠: 0 1
g⇠(x,x)*g (x,x)= ↵ (x)* (x)=0
® ⇠ ⇠:⇠ 0 ® 1 * 1 * 1
g⇠(x,x)*g (x,x)=* ↵ (x)= = ⇠< > < x2 px2 Ÿ ↵<(x)=1Ç1 & < ( x ) = x1 , Ç 1 and I know the four unknown functions. I have g>(x, ⇠) = 0 and so
(c) Atx=lIhave
=0 Ç1 so the boundary conditions are satisfied.
g < ( x , ⇠ ) = * x1 + 1⇠ g ( x , ⇠ ) = 0 1⇠ * x1 1 H ( ⇠ * x ) . Ç 2 0111 ⇠⇠:⇠0,0<⇠t. >
should have four unknown functions of t®.] (b) Next, apply the boundary conditions, y(0) = yÜ(0) = 0.
[Hint: You should have two conditions for the four unknown functions of t®.] (c) Next, apply the conditions at t = t®,
where the self-adjoint ODE is
g<(t, t) * g>(t, t) = 0
gÜ < ( t , t ) * gÜ > ( t , t ) = * 1 _ p ( t ) ,
Dty= d0pdy1+qy, dt dt
in which p, q, and y are functions of t. [Hint: You should have two further conditions for the four unknown functions of t®. You should now be able to find these four functions.] (d) What is Green’s function, g(t, t®), for
this problem? (e) Find y(t) if
f(t)=f H(t)*H(t*b) =Tf ̋_b 0>
(c) Now for the t = t® “boundary” conditions,
g<(t,t®) = 0. g>t=t® * g<t=t® = 0 I have p(t) = 1 so g  ® * g ✓  ® = * 1 dd0 dt > t=t dt< t=t *a↵>(t®) e+at® + a>(t®) e*at® = *1
↵ (t®)e+at® * (t®)e*at® =1_a Ç2
↵(t®)= 1e*at® > 2a
= 1e+a(t*t®)*e*a(t*t®)H(t*t®)
adding and subtracting I get (d) I have
(t®)=*1e+at® Ç2
gƒ(t, t®) = ↵ƒ(t®)e+at + ƒ(t®)e*at
Ÿg(t,t®)=⇠1e*at®e+at* 1e+at®e*at⇡H(t*t®) 2a 2a
= 1 sinh a(t * t®) H(t * t®) Ç 2 a
© 2023 Chris O’Donovan

WEEK6. CONSTRUCTIONMETHOD
(e) The answer is given by the convolution
ÿ®®® y(t)=(f