Green's function of an ODE boundary value problem

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Find the spectrum, the eigenvalues, the eigenfunctions, and the norm of the operator AA in L2([0,1])L_{2}(\left[0,1\right]), defined by the formula (Ax)(t)=∫01min⁡(t,s)x(s)ds(A x)(t) = \int_{0}^{1} \min (t, s) x(s) d s.

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Find the norm, the spectrum, the eigenvalues and the eigenfunctions of an operator AA in L2([a,b])L_{2}(\left[a, b\right]), defined by the formula (Ax)(t)=∫abK(t,s)x(s)ds,t∈[a,b](A x)(t) = \int_{a}^{b} K(t, s) x(s) d s, t \in \left[a, b\right], in the following cases:

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(1)

K(t,s)={t(1−s),t≤s,s(1−t),s≤t,a=0,b=1K(t, s) = \begin{cases} t(1-s), & t \leq s, \\ s(1-t), & s \leq t, \end{cases} \quad a = 0, b = 1;

(2)

K(t,s)={sin⁡tcos⁡s,t≤s,cos⁡tsin⁡s,s≤t,a=0,b=πK(t, s) = \begin{cases} \sin t \cos s, & t \leq s, \\ \cos t \sin s, & s \leq t, \end{cases} \quad a = 0, b = \pi;

(3)

K(t,s)={cos⁡tsin⁡s,t≤s,sin⁡tcos⁡s,s≤t,a=0,b=πK(t, s) = \begin{cases} \cos t \sin s, & t \leq s, \\ \sin t \cos s, & s \leq t, \end{cases} \quad a = 0, b = \pi;

(4)

K(t,s)={(t+1)s,t≤s,t(s+1),s≤t,a=0,b=1K(t, s) = \begin{cases} (t+1) s, & t \leq s, \\ t(s+1), & s \leq t, \end{cases} \quad a = 0, b = 1;

(5)

K(t,s)={sin⁡tsin⁡(1−s),t≤s,sin⁡(1−t)sin⁡s,s≤t,a=0,b=1K(t, s) = \begin{cases} \sin t \sin (1-s), & t \leq s, \\ \sin (1-t) \sin s, & s \leq t, \end{cases} \quad a = 0, b = 1;

(6)

K(t,s)={(1+t)(1−s),t≤s,(1−t)(1+s),s≤t,a=−1,b=1K(t, s) = \begin{cases} (1+t)(1-s), & t \leq s, \\ (1-t)(1+s), & s \leq t, \end{cases} \quad a = -1, b = 1.

Transform an integral equation μx(t)=∫abK(t,s)x(s)ds+y(t),t∈[a,b]\mu x(t) = \int_{a}^{b} K(t, s) x(s) d s+y(t), t \in [a, b], into a differential one and solve it:

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(1)

y(t)=t2,K(t,s)={π232t(2−s),t≤s,π232s(2−t),s<t,[a,b]=[0,2],μ=1y(t) = \frac{t}{2}, K(t, s) = \begin{cases} \frac{\pi^{2}}{32} t(2-s), & t \leq s, \\ \frac{\pi^{2}}{32} s(2-t), & s < t, \end{cases} \quad [a, b] = \left[0,2\right], \mu = 1;

(2)

y(t)=tet,K(t,s)={−1sinh⁡1sinh⁡tsinh⁡(s−1),t≤s,−1sinh⁡1sinh⁡ssinh⁡(t−1),s<t,[a,b]=[0,1],μ=1;y(t) = t e^{t}, \quad K(t, s) = \begin{cases} -\frac{1}{\sinh 1} \sinh t \sinh (s-1), & t \leq s, \\ -\frac{1}{\sinh 1} \sinh s \sinh (t-1), & s < t, \end{cases} \quad [a, b] = \left[0,1\right], \mu = 1 ;

(3)

y(t)=sin⁡3πt,K(t,s)={(s−1)t,0≤t≤s,(t−1)s,s<t≤1,[a,b]=[0,1],μ=0;y(t) = \sin^{3} \pi t, K(t, s) = \begin{cases} (s-1) t, & 0 \leq t \leq s, \\ (t-1) s, & s < t \leq 1, \end{cases} \quad [a, b] = \left[0,1\right], \quad \mu = 0 ;

(4)

y(t)=2t2−3t+52,K(t,s)=∣t−s∣+t2,[a,b]=[1,2],μ=0y(t) = 2 t^{2}-3 t+\frac{5}{2}, K(t, s) = \left|t-s\right|+t^{2},\left[a, b\right] = \left[1,2\right], \mu = 0.

In the space L2[0,1]L_{2}[0,1], find the solution of the following integral equations for all values λ∈C\lambda \in \mathbb {C}:

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(a)

x(t)−λ∫01min⁡(t,s)x(s)ds=0x(t)-\lambda \int_{0}^{1} \min (t, s) x(s) d s = 0;

(b)

x(t)+λ∫01max⁡(t,s)x(s)ds=0x(t)+\lambda \int_{0}^{1} \max (t, s) x(s) d s = 0.

In the space L2[0,1]L_{2}[0,1], find the solution of the integral equation for all values λ∈C\lambda \in \mathbb {C}:

x(t)+λ∫01(ts−min⁡(t,s))x(s)ds=y(t) x(t)+\lambda \int _{0}^{1}(t s-\min (t, s)) x(s) d s = y(t)

if

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(a)

y(t)=ty(t) = t;

(b)

y(t)=sin⁡2πty(t) = \sin 2 \pi t.

In the space L2[0,1]L_{2}[0,1], find the solution of the integral equation for all values λ∈C\lambda \in \mathbb {C}:

x(t)+λ∫01(12−∣t−s∣)x(s)ds=y(t) x(t)+\lambda \int _{0}^{1}\left(\frac{1}{2}-\left|t-s\right|\right) x(s) d s = y(t)

if

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(a)

y(t)=cos⁡πty(t) = \cos \pi t;

(b)

y(t)=cos⁡2πty(t) = \cos 2 \pi t.

In the space L2[0,1]L_{2}[0,1], find the solution of the integral equation for all values λ∈C\lambda \in \mathbb {C}:

x(t)+λ∫01(ts+max⁡(t,s))x(s)ds=y(t) x(t)+\lambda \int _{0}^{1}(t s+\max (t, s)) x(s) d s = y(t)

if

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(a)

y(t)=cos⁡πty(t) = \cos \pi t;

(b)

y(t)=cos⁡2πty(t) = \cos 2 \pi t.

In the space L2[0,π]L_{2}[0, \pi ], find the eigenvalues and eigenfunctions of the integral operator

(Ax)(t)=∫0πK(t,s)x(s)ds (A x)(t) = \int _{0}^{\pi } K(t, s) x(s) d s

if

K(t,s)={sin⁡tcos⁡s for 0⩽t⩽s⩽π,sin⁡scos⁡t for 0⩽s⩽t⩽π. K(t, s) = \begin{cases} \sin t \cos s & \text{ for } 0 \leqslant t \leqslant s \leqslant \pi , \\ \sin s \cos t & \text{ for } 0 \leqslant s \leqslant t \leqslant \pi . \end{cases}
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In the space L2[0,1]L_{2}[0,1], find the eigenvalues and eigenfunctions of the integral operator

(Ax)(t)=∫01K(t,s)x(s)ds (A x)(t) = \int _{0}^{1} K(t, s) x(s) d s

if

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(a)

K(t,s)=ts+max⁡(t,s)K(t, s) = t s+\max (t, s);

(b)

K(t,s)=ts+max⁡(t,s)−2min⁡(t,s)−K(t, s) = t s+\max (t, s)-2 \min (t, s)- -2.

Consider the Sturm-Liouville operator Lh=−h′′Lh = -h'' with a=0a=0, b=1b=1, and for each of the following boundary conditions find the eigenvalues {λn}\left\{ \lambda_{n}\right\}, the eigenvectors {en}\left\{ e_{n}\right\}, and the Green function g(x,y)g(x,y):

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(a)

h(0)=h(1)=0h(0) = h(1) = 0;

(b)

h′(0)=h′(1)=0h'(0) = h'(1) = 0;

(c)

h(0)=0h(0) = 0 and h′(1)=0h'(1) = 0;

(d)

h(0)=h′(0)h(0) = h'(0) and h(1)=−h′(1)h(1) = -h'(1).

Given the function

u(x)=∫01K(x,y)v(y) dy, u(x)=\int _{0}^{1}K(x,y)v(y)\, dy ,

where

K(x,y)={x(1−y),if x⩽y;y(1−x),if x⩾y, K(x,y)= \begin{cases} x(1-y) , & \text{if } x \leqslant y; \\ y(1-x) , & \text{if } x \geqslant y, \end{cases}

and v(y)v(y) are both continuous. Prove that the given function satisfies the equation

u′′(x)=−v(x)(0⩽x⩽1). u''(x)=-v(x) \quad (0 \leqslant x \leqslant 1) .
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At the point x=ξx = \xi of a homogeneous string (ρ=1\rho = 1), subject to the action of a strong tension T0T_{0} and fastened at the ends x=0x = 0 and x=1x = 1, a weight Q=1Q = 1 is suspended (Fig. 15.22).

Fig. 15.22Fig. 15.22

Prove that the equation of the sag of the string has the form

u=K(x,ξ)={x(1−ξ)⋅1T0for x⩽ξ,ξ(1−x)⋅1T0for x⩾ξ. u = K(x,\xi ) = \begin{cases} x(1-\xi ) \cdot \dfrac {1}{T_{0}} & \text{for } x \leqslant \xi , \\ \xi (1-x) \cdot \dfrac {1}{T_{0}} & \text{for } x \geqslant \xi . \end{cases}
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A heavy homogeneous string is fastened at the point x=1x = 1 and is subject at the point x=ξx = \xi to the action of a horizontal force Q=1Q = 1 (Fig. 15.23).

Fig. 15.23Fig. 15.23

Prove that the equilibrium equation of the string has the form

u=K(x,ξ)={−ln⁡ξ,for x⩽ξ,−ln⁡x,for x⩾ξ. u = K(x,\xi ) = \begin{cases} -\ln \xi , & \text{for } x \leqslant \xi , \\ -\ln x, & \text{for } x \geqslant \xi . \end{cases}
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Remark.
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Hint. The tension of the string at the point M(x)M(x) is taken equal to −x-x.

A homogeneous string of length l=1l = 1 is fastened by the end x=0x = 0 to a fixed axis and rotates about this axis with constant angular velocity. To the point x=ξx = \xi of such a string a force Q=1Q = 1 is applied (Fig. 15.24).

Fig. 15.24Fig. 15.24

Prove that the equation of relative equilibrium of the string has the form

u=K(x,ξ)={12ln⁡1+x1−xfor x⩽ξ,12ln⁡1+ξ1−ξfor x⩾ξ. u = K(x,\xi ) = \begin{cases} \dfrac {1}{2}\ln \dfrac {1+x}{1-x} & \text{for } x \leqslant \xi , \\ \dfrac {1}{2}\ln \dfrac {1+\xi }{1-\xi } & \text{for } x \geqslant \xi . \end{cases}
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Remark.
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Hint. The tension of the string at the point M(x)M(x), caused by the centrifugal force, is taken equal to

R=2∫x1x dx=1−x2. R = 2\int _{x}^{1} x \, dx = 1-x^{2}.

Prove that for a continuously distributed load, the equation of the sag of the string has the form

u=∫01K(x,ξ)p(ξ) dξ, u = \int _{0}^{1} K(x,\xi )p(\xi ) \, d\xi ,

where p(ξ)p(\xi ) denotes the magnitude of the load, computed per unit length of the string.

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Remark.
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Hint. Use the result of problem 5900 (problem 5900: K(x,ξ)=x(1−ξ)/T0K(x,\xi ) = x(1-\xi )/T_{0} for x⩽ξx \leqslant \xi, K(x,ξ)=ξ(1−x)/T0K(x,\xi ) = \xi (1-x)/T_{0} for x⩾ξx \geqslant \xi).

Prove that the problem of small vibrations of a fastened string can be reduced to solving the homogeneous Fredholm equation

y(x)=λ∫01K(x,ξ)y(ξ) dξ, y(x) = \lambda \int _{0}^{1} K(x,\xi )y(\xi ) \, d\xi ,

where the symmetric kernel K(x,ξ)K(x,\xi ) has the form given in problem 5901 (problem 5901: K(x,ξ)=−ln⁡ξK(x,\xi ) = -\ln \xi for x⩽ξx \leqslant \xi, K(x,ξ)=−ln⁡xK(x,\xi ) = -\ln x for x⩾ξx \geqslant \xi).

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On a homogeneous string, fastened at the ends, a continuously distributed force acts

F(ξ,t)=p0(ξ)sin⁡ωt. F(\xi ,t) = p_{0}(\xi )\sin \omega t.

Prove that the problem of forced vibrations of such a string reduces to solving the inhomogeneous Fredholm equation

y(x)=f(x)+λ∫01K(x,ξ)y(ξ) dξ. y(x) = f(x) + \lambda \int _{0}^{1} K(x,\xi )y(\xi ) \, d\xi .
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Reduce to an integral equation the problem of finding the solutions of the equation

(xy′)′+λxy=0 (xy')'+\lambda xy = 0

under the boundary conditions: y(0)y(0) bounded, y(1)=0y(1) = 0.

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Reduce to an integral equation the problem of finding the solutions of the equation

(xy′)′+(λx−n2x)y=0 (xy')'+\left(\lambda x-\frac{n^{2}}{x}\right)y = 0

under the boundary conditions: y(0)=0y(0) = 0, y(1)=0y(1) = 0.

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Construct the integral equation whose fundamental functions are the Legendre functions of order mm:

((1−x2)y′)′+(n(n+1)−m21−x2)y=0; \left((1-x^{2})y'\right)'+\left(n(n+1)-\frac{m^{2}}{1-x^{2}}\right)y = 0;

boundary condition: yy bounded at x=−1x = -1 and x=+1x = +1.

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Solve problem 5943 for the case m=0m = 0 of the Legendre polynomials.

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Remark.
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Hint. Consider the equation

((1−x2)y′)′+n(n+1)y=12. \left((1-x^{2})y'\right)'+n(n+1)y = \frac{1}{2}.

Construct the integral equation whose fundamental functions are the Hermite polynomials:

y′′+(1−x2)y+λy=0, y''+(1-x^{2})y+\lambda y = 0,

y=0y = 0 at x=±∞x = \pm \infty.

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Remark.
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Hint. Set λ=−2+μ\lambda = -2+\mu.

Construct the integral equation whose fundamental functions are the Laguerre polynomials:

xy′′+(1−x)y′+λy=0; xy''+(1-x)y'+\lambda y = 0;

yy bounded at x=0x = 0 and x=+∞x = +\infty.

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Remark.
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Hint. Set λ=−1+μ\lambda = -1+\mu.

Having found the corresponding Green's function, replace the equation

y′′+k2y=f(x),y(0)=y(1)=0 y''+k^{2}y = f(x), \quad y(0) = y(1) = 0

by an integral one.

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Having found the corresponding Green's function, replace the equation

y′′+k2y=f(x),y′(0)=hy(0),y′(1)=−Hy(1) y''+k^{2}y = f(x), \quad y'(0) = hy(0), \quad y'(1) = -Hy(1)

by an integral equation.

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Having found the corresponding Green's function G(ξ,x)G(\xi ,x), replace the equation

y′′+k2y=f(x),y(0)=0,y′(1)=−hy(1) y''+k^{2}y = f(x), \quad y(0) = 0, \quad y'(1) = -hy(1)

by an integral one.

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Let K(x;y)={x(1−y), if x⩽y,y(1−x), if x>y,K(x ; y) = \begin{cases} x(1-y), & \text{ if } x \leqslant y, \\ y(1-x), & \text{ if } x > y, \end{cases} and let φ(y)\varphi (y) be a function continuous on the segment [0;1][0 ; 1]. Show that the function

u(x)=∫01K(x;y)φ(y)dy u(x) = \int _{0}^{1} K(x ; y) \varphi (y) d y

satisfies on the segment [0;1][0 ; 1] the equation u′′(x)=−φ(x)u^{\prime \prime }(x) = -\varphi (x).

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Consider the operator K:L2([0,1])→L2([0,1])K: L^{2}([0,1]) \rightarrow L^{2}([0,1]) defined by

(Kf)(t)=∫01k(t,s)f(s)ds (K f)(t) = \int _{0}^{1} k(t, s) f(s) d s

where k(s,t)=min⁡{t,s:t,s∈[0,1]}k(s, t) = \min \left\{ t, s: t, s \in [0,1]\right\}.

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(a)

Prove that KK is a compact self-adjoint operator.

(b)

Find the spectrum σ(K)\sigma (K) and the norm ∥K∥\left\| K\right\|.

Problem 3.4.16 from Rudin W. "Functional Analysis" [Rud91]More problems in Chapter 3.4 "Duality in Banach Spaces"

Define

K(s,t)={(1−s)t if 0≤t≤s(1−t)s if s≤t≤1 K(s, t) = \begin{cases} (1-s) t & \text{ if } 0 \leq t \leq s \\ (1-t) s & \text{ if } s \leq t \leq 1\end{cases}

and define T∈B(L2(0,1))T \in \mathscr {B}\left(L^{2}(0,1)\right) by

(Tf)(s)=∫01K(s,t)f(t)dt(0≤s≤1) (T f)(s) = \int _{0}^{1} K(s, t) f(t) d t \quad (0 \leq s \leq 1)
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(a)

Show that the eigenvalues of TT are (nπ)−2,n=1,2,3,…(n \pi )^{-2}, n = 1,2,3, \ldots, that the corresponding eigenfunctions are sin⁡nπx\sin n \pi x, and that each eigenspace is onedimensional. Hint: If λ≠0\lambda \neq 0, the equation Tf=λfT f = \lambda f implies that ff is infinitely differentiable, that λf′′+f=0\lambda f^{\prime \prime }+f = 0, and that f(0)=f(1)=0f(0) = f(1) = 0. The case λ=0\lambda = 0 can be treated separately.

(b)

Show that the above eigenfunctions form an orthogonal basis for L2(0,1)L^{2}(0,1).

(c)

Suppose g(t)=∑cnsin⁡nπtg(t) = \sum c_{n} \sin n \pi t. Discuss the equation Tf−λf=gT f-\lambda f = g.

(d)

Show that TT is also a compact operator on CC, the space of all continuous functions on [0,1]. Hint: If {fi}\left\{ f_{i}\right\} is uniformly bounded, then {Tfi}\left\{ T f_{i}\right\} is equicontinuous.

Consider the following three boundary value problems. The differential equation is

f′′−f=g f^{\prime \prime }-f = g

where g∈L2([0,1])g \in L^{2}([0,1]) is given. The choices of boundary conditions are

(i) f(0)=f(1)=0f(0) = f(1) = 0.

(ii) f′(0)=f′(1)=0f^{\prime }(0) = f^{\prime }(1) = 0.

(iii) f(0)=f(1)f(0) = f(1) and f′(0)=f′(1)f^{\prime }(0) = f^{\prime }(1).

Show that each of these problems has a unique solution ff such that f′f^{\prime } is absolutely continuous and f′′∈L2([0,1])f^{\prime \prime } \in L^{2}([0,1]). Hint: Combine Example 13.4 with Theorem 13.13.

Do this also by solving the problems explicitly.

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Show that if

∫Gab(x,y)m1(dy)=∫Gab(x,y)m2(dy) \int G_{a b}(x, y) m_{1}(d y)=\int G_{a b}(x, y) m_{2}(d y)

for all x,ax, a, and bb, then m1=m2m_{1}=m_{2}.

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