IGNOU MMT-007 Solved Assignment 2024 for M.Sc. MACS

IGNOU MMT-007 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-007 Assignment Question Paper 2024

mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

  1. a) Show that f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y)=x yf(x,y)=xy
    i) satisfies a Lipschitz condition on any rectangle a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and c ≤ y ≤ d c ≤ y ≤ d c <= y <= dc \leq y \leq dc≤y≤d;
    ii) satisfies a Lipschitz condition on any strip a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and − ∞ < y < ∞ − ∞ < y < ∞ -oo < y < oo-\infty<y<\infty−∞<y<∞;
    iii) does not satisfy a Lipschitz condition on the entire plane.
    b) Use Frobenious method to find the series solution about x = 0 x = 0 x=0x=0x=0 of the equation
x ( 1 − x ) d 2 y d x 2 − ( 1 + 3 x ) d y d x − y = 0 . x ( 1 − x ) d 2 y d x 2 − ( 1 + 3 x ) d y d x − y = 0 . x(1-x)(d^(2)y)/(dx^(2))-(1+3x)(dy)/(dx)-y=0.x(1-x) \frac{d^2 y}{d x^2}-(1+3 x) \frac{d y}{d x}-y=0 .x(1−x)d2ydx2−(1+3x)dydx−y=0.
  1. a) For the following differential equation locate and classify its singular points on the x x xxx-axis
    i) x 3 ( x − 1 ) y ′ ′ − 2 ( x − 1 ) y ′ + 3 x y = 0 x 3 ( x − 1 ) y ′ ′ − 2 ( x − 1 ) y ′ + 3 x y = 0 x^(3)(x-1)y^(”)-2(x-1)y^(‘)+3xy=0x^3(x-1) y^{\prime \prime}-2(x-1) y^{\prime}+3 x y=0x3(x−1)y′′−2(x−1)y′+3xy=0
    ii) ( 3 x + 1 ) x y ′ ′ − ( x + 1 ) y ′ + 2 y = 0 ( 3 x + 1 ) x y ′ ′ − ( x + 1 ) y ′ + 2 y = 0 (3x+1)xy^(”)-(x+1)y^(‘)+2y=0(3 x+1) x y^{\prime \prime}-(x+1) y^{\prime}+2 y=0(3x+1)xy′′−(x+1)y′+2y=0
    b) Show that L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n 2 L n − 1 ( x ) L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n 2 L n − 1 ( x ) L_(n+1)(x)=(2n+1-x)L_(n)(x)-n^(2)L_(n-1)(x)\mathrm{L}_{\mathrm{n}+1}(\mathrm{x})=(2 \mathrm{n}+1-\mathrm{x}) \mathrm{L}_{\mathrm{n}}(\mathrm{x})-\mathrm{n}^2 \mathrm{~L}_{\mathrm{n}-1}(\mathrm{x})Ln+1(x)=(2n+1−x)Ln(x)−n2 Ln−1(x).
    c) Show that ∫ − 1 1 x 2 P n − 1 ( x ) P n + 1 ( x ) d x = 2 n ( n + 1 ) ( 2 n − 1 ) ( 2 n + 1 ) ( 2 n + 3 ) ∫ − 1 1   x 2 P n − 1 ( x ) P n + 1 ( x ) d x = 2 n ( n + 1 ) ( 2 n − 1 ) ( 2 n + 1 ) ( 2 n + 3 ) int_(-1)^(1)x^(2)P_(n-1)(x)P_(n+1)(x)dx=(2n(n+1))/((2n-1)(2n+1)(2n+3))\int_{-1}^1 x^2 P_{n-1}(x) P_{n+1}(x) d x=\frac{2 n(n+1)}{(2 n-1)(2 n+1)(2 n+3)}∫−11x2Pn−1(x)Pn+1(x)dx=2n(n+1)(2n−1)(2n+1)(2n+3).
    d) Construct Green’s function for the differential equation
x y ′ ′ + y ′ = 0 , 0 < x < ℓ x y ′ ′ + y ′ = 0 , 0 < x < ℓ xy^(”)+y^(‘)=0,quad0 < x < ℓx y^{\prime \prime}+y^{\prime}=0, \quad 0<x<\ellxy′′+y′=0,0<x<ℓ
under the conditions that y ( 0 ) y ( 0 ) y(0)\mathrm{y}(0)y(0) is bounded and y ( ℓ ) = 0 y ( ℓ ) = 0 y(ℓ)=0\mathrm{y}(\ell)=0y(ℓ)=0.
3. a) Show that between every successive pair of zeros of J 0 ( x ) J 0 ( x ) J_(0)(x)J_0(x)J0(x) there exists a zero of J 1 ( x ) J 1 ( x ) J_(1)(x)J_1(x)J1(x).
b) Using the transformation y = x 1 / 2 u , 2 x 3 / 2 = 3 z y = x 1 / 2 u , 2 x 3 / 2 = 3 z y=x^(1//2)u,2x^(3//2)=3zy=x^{1 / 2} u, 2 x^{3 / 2}=3 zy=x1/2u,2x3/2=3z find the solution of the equation y ′ ′ + x y = 0 y ′ ′ + x y = 0 y^(”)+x quad y=0y^{\prime \prime}+x \quad y=0y′′+xy=0 in terms of Bessel’s functions.
c) Show that ∫ 0 ∞ e − a x J 0 ( b x ) d x = 1 a 2 + b 2 , a > 0 b > 0 ∫ 0 ∞   e − a x J 0 ( b x ) d x = 1 a 2 + b 2 , a > 0 b > 0 int_(0)^(oo)e^(-ax)J_(0)(bx)dx=(1)/(sqrt(a^(2)+b^(2))),a > 0b > 0\int_0^{\infty} \mathrm{e}^{-\mathrm{ax}} \mathrm{J}_0(\mathrm{bx}) \mathrm{dx}=\frac{1}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}, \mathrm{a}>0 \mathrm{~b}>0∫0∞e−axJ0(bx)dx=1a2+b2,a>0 b>0.
4. a) Find the Laplace transform of cos ⁡ t t cos ⁡ t t (cos sqrtt)/(sqrtt)\frac{\cos \sqrt{t}}{\sqrt{t}}cos⁡tt.
b) If k m k m k_(m)\mathrm{k}_{\mathrm{m}}km and k n k n k_(n)\mathrm{k}_{\mathrm{n}}kn are distinct roots of Bessel function J p ( k b ) = 0 J p ( k b ) = 0 J_(p)(kb)=0\mathrm{J}_{\mathrm{p}}(\mathrm{kb})=0Jp(kb)=0 with p ≥ 0 , b > 0 p ≥ 0 , b > 0 p >= 0,b > 0\mathrm{p} \geq 0, \mathrm{~b}>0p≥0, b>0 then show that
∫ 0 b x J p ( k m x ) J p ( k n x ) d x = { 0 if m ≠ n b 2 2 [ J p + 1 ( k n b ) ] if m = n . ∫ 0 b   x J p k m x J p k n x d x = 0      if      m ≠ n b 2 2 J p + 1 k n b      if      m = n . int_(0)^(b)xJ_(p)(k_(m)x)J_(p)(k_(n)x)dx={[0,” if “,m!=n],[(b^(2))/(2)[J_(p+1)(k_(n)b)],” if “,m=n].:}\int_0^b x J_p\left(k_m x\right) J_p\left(k_n x\right) d x=\left\{\begin{array}{lll} 0 & \text { if } & m \neq n \\ \frac{b^2}{2}\left[J_{p+1}\left(k_n b\right)\right] & \text { if } & m=n \end{array} .\right.∫0bxJp(kmx)Jp(knx)dx={0 if m≠nb22[Jp+1(knb)] if m=n.
  1. a) Solve the following IBVP using the Laplace transform technique:
u t = u x x , 0 < x < 1 , t > 0 . u ( 0 , t ) = 1 , u ( 1 , t ) = 1 , t > 0 u ( x , 0 ) = 1 + sin ⁡ π x , 0 < x < 1 . u t = u x x , 0 < x < 1 , t > 0 . u ( 0 , t ) = 1 , u ( 1 , t ) = 1 , t > 0 u ( x , 0 ) = 1 + sin ⁡ π x , 0 < x < 1 . {:[u_(t)=u_(xx)”,”quad0 < x < 1″,”quadt > 0.],[u(0″,”t)=1″,”u(1″,”t)=1″,”quadt > 0],[u(x”,”0)=1+sin pix”,”quad0 < x < 1.]:}\begin{aligned} & \mathrm{u}_{\mathrm{t}}=\mathrm{u}_{\mathrm{xx}}, \quad 0<\mathrm{x}<1, \quad \mathrm{t}>0 . \\ & \mathrm{u}(0, \mathrm{t})=1, \mathrm{u}(1, \mathrm{t})=1, \quad \mathrm{t}>0 \\ & \mathrm{u}(\mathrm{x}, 0)=1+\sin \pi \mathrm{x}, \quad 0<\mathrm{x}<1 . \end{aligned}ut=uxx,0<x<1,t>0.u(0,t)=1,u(1,t)=1,t>0u(x,0)=1+sin⁡πx,0<x<1.
b) If the Fourier cosine transform of f ( x ) f ( x ) f(x)f(x)f(x) is α n e − a α α n e − a α alpha^(n)e^(-aalpha)\alpha^{\mathrm{n}} \mathrm{e}^{-\mathrm{a} \alpha}αne−aα, then show that
f ( x ) = 2 π n ! cos ⁡ ( n + 1 ) θ ( a 2 + x 2 ) n + 1 2 . f ( x ) = 2 π n ! cos ⁡ ( n + 1 ) θ a 2 + x 2 n + 1 2 . f(x)=(2)/(pi)(n!cos(n+1)theta)/((a^(2)+x^(2))^((n+1)/(2))).f(x)=\frac{2}{\pi} \frac{n ! \cos (n+1) \theta}{\left(a^2+x^2\right)^{\frac{n+1}{2}}} .f(x)=2πn!cos⁡(n+1)θ(a2+x2)n+12.
  1. a) Find the displacement u ( x , t ) u ( x , t ) u(x,t)u(x, t)u(x,t) of an infinite string using the method of Fourier transform given that the string is initially at rest and that the initial displacement is f ( x ) , − ∞ < x < ∞ f ( x ) , − ∞ < x < ∞ f(x),-oo < x < oo\mathrm{f}(\mathrm{x}),-\infty<\mathrm{x}<\inftyf(x),−∞<x<∞.
    b) Using Fourier integral representation show that
∫ 0 ∞ cos ⁡ ( α x ) + α sin ⁡ ( α x ) 1 + α 2 d α = { 0 if x < 0 π / 2 if x = 0 . π e − x if x > 0 ∫ 0 ∞   cos ⁡ ( α x ) + α sin ⁡ ( α x ) 1 + α 2 d α = 0 if x < 0 π / 2 if x = 0 . π e − x if x > 0 int_(0)^(oo)(cos(alphax)+alpha sin(alphax))/(1+alpha^(2))dalpha={[0,” if “,x < 0],[pi//2,” if “,x=0.],[pie^(-x),” if “,x > 0]:}\int_0^{\infty} \frac{\cos (\alpha \mathrm{x})+\alpha \sin (\alpha \mathrm{x})}{1+\alpha^2} \mathrm{~d} \alpha=\left\{\begin{array}{ccc} 0 & \text { if } & \mathrm{x}<0 \\ \pi / 2 & \text { if } & \mathrm{x}=0 . \\ \pi \mathrm{e}^{-\mathrm{x}} & \text { if } & \mathrm{x}>0 \end{array}\right.∫0∞cos⁡(αx)+αsin⁡(αx)1+α2 dα={0 if x<0π/2 if x=0.πe−x if x>0
  1. a) Using Runge-Kutta second order method with
    (i) h = 0.1 h = 0.1 h=0.1\mathrm{h}=0.1h=0.1, (ii) h = 0.2 h = 0.2 h=0.2\mathrm{h}=0.2h=0.2, solve the initial value problem
y ′ = y 2 sin ⁡ x , y ( 0 ) = 1 . y ′ = y 2 sin ⁡ x , y ( 0 ) = 1 . y^(‘)=y^(2)sin x,quady(0)=1.\mathrm{y}^{\prime}=\mathrm{y}^2 \sin \mathrm{x}, \quad \mathrm{y}(0)=1 .y′=y2sin⁡x,y(0)=1.
Upto x = 0.4 x = 0.4 x=0.4\mathrm{x}=0.4x=0.4. If the exact solution is y = sec ⁡ x y = sec ⁡ x y=sec x\mathrm{y}=\sec \mathrm{x}y=sec⁡x, obtain the error.
b) Solve the heat conduction equation u t = u x x u t = u x x u_(t)=u_(xx)\mathrm{u}_{\mathrm{t}}=\mathrm{u}_{\mathrm{xx}}ut=uxx in the region R ( 0 ≤ x ≤ 1 , t > 0 ) R ( 0 ≤ x ≤ 1 , t > 0 ) R(0 <= x <= 1,t > 0)\mathrm{R}(0 \leq \mathrm{x} \leq 1, \mathrm{t}>0)R(0≤x≤1,t>0) with the initial and boundary conditions u ( x , 0 ) = 0 , u ( 0 , t ) = 0 , u ( 1 , t ) = t u ( x , 0 ) = 0 , u ( 0 , t ) = 0 , u ( 1 , t ) = t u(x,0)=0,u(0,t)=0,u(1,t)=t\mathrm{u}(\mathrm{x}, 0)=0, \mathrm{u}(0, \mathrm{t})=0, \mathrm{u}(1, \mathrm{t})=\mathrm{t}u(x,0)=0,u(0,t)=0,u(1,t)=t using Crank-Nicolson method with h = 0.25 h = 0.25 h=0.25\mathrm{h}=0.25h=0.25 and λ = 1 λ = 1 lambda=1\lambda=1λ=1 upto two time steps.
8. a) Using second order finite Difference method, solve the boundary value problem
y ′ ′ + 5 y ′ + 4 y = 1 , y ( 0 ) = 0 , y ( 1 ) = 0 , h = 1 / 4 . y ′ ′ + 5 y ′ + 4 y = 1 , y ( 0 ) = 0 , y ( 1 ) = 0 , h = 1 / 4 . y^(”)+5y^(‘)+4y=1,y(0)=0,y(1)=0,h=1//4″. “\mathrm{y}^{\prime \prime}+5 \mathrm{y}^{\prime}+4 \mathrm{y}=1, \mathrm{y}(0)=0, \mathrm{y}(1)=0, \mathrm{~h}=1 / 4 \text {. }y′′+5y′+4y=1,y(0)=0,y(1)=0, h=1/4.
b) Solve the wave equation u t t = u x x u t t = u x x u_(tt)=u_(xx)\mathrm{u}_{\mathrm{tt}}=\mathrm{u}_{\mathrm{xx}}utt=uxx with the initial and boundary conditions
u ( x , 0 ) = 0 , u t ( x , 0 ) = 0 , u ( 0 , t ) = 0 , u ( 1 , t ) = 100 sin ⁡ π t . u ( x , 0 ) = 0 , u t ( x , 0 ) = 0 , u ( 0 , t ) = 0 , u ( 1 , t ) = 100 sin ⁡ π t . u(x,0)=0,u_(t)(x,0)=0,u(0,t)=0,u(1,t)=100 sin pit”. “\mathrm{u}(\mathrm{x}, 0)=0, \mathrm{u}_{\mathrm{t}}(\mathrm{x}, 0)=0, \mathrm{u}(0, \mathrm{t})=0, \mathrm{u}(1, \mathrm{t})=100 \sin \pi \mathrm{t} \text {. }u(x,0)=0,ut(x,0)=0,u(0,t)=0,u(1,t)=100sin⁡πt.
with h = k = 0.25 h = k = 0.25 h=k=0.25\mathrm{h}=\mathrm{k}=0.25h=k=0.25, using the explicit method upto four time levels.
9. a) Find an approximate value of y ( 1.0 ) y ( 1.0 ) y(1.0)y(1.0)y(1.0) for the initial value problem
y ′ = x 3 − y 3 , y ( 0 ) = 1 y ′ = x 3 − y 3 , y ( 0 ) = 1 y^(‘)=x^(3)-y^(3),y(0)=1\mathrm{y}^{\prime}=\mathrm{x}^3-\mathrm{y}^3, \mathrm{y}(0)=1y′=x3−y3,y(0)=1
using the multiple method
y n + 1 = y n + h 3 [ 7 f n − 2 f n − 1 + f n − 2 ] y n + 1 = y n + h 3 7 f n − 2 f n − 1 + f n − 2 y_(n+1)=y_(n)+(h)/(3)[7f_(n)-2f_(n-1)+f_(n-2)]y_{n+1}=y_n+\frac{h}{3}\left[7 f_n-2 f_{n-1}+f_{n-2}\right]yn+1=yn+h3[7fn−2fn−1+fn−2]
with step length h = 0.2 h = 0.2 h=0.2h=0.2h=0.2. Calculate the starting values using Runge-Kutta second order method with the same h h h\mathrm{h}h.
b) Using standard five point formula, solve the Laplace equation ∇ 2 u = 0 ∇ 2 u = 0 grad^(2)u=0\nabla^2 u=0∇2u=0 in R R RRR where R R RRR is the square 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 0 <= x <= 1,0 <= y <= 10 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 10≤x≤1,0≤y≤1 subject to the boundary conditions u ( x , y ) = x 2 − y 2 u ( x , y ) = x 2 − y 2 u(x,y)=x^(2)-y^(2)\mathrm{u}(\mathrm{x}, \mathrm{y})=\mathrm{x}^2-\mathrm{y}^2u(x,y)=x2−y2 on x = 0 , y = 0 , y = 1 x = 0 , y = 0 , y = 1 x=0,y=0,y=1\mathrm{x}=0, \mathrm{y}=0, \mathrm{y}=1x=0,y=0,y=1 and 3 u + 2 ∂ u ∂ x = x 2 + y 2 3 u + 2 ∂ u ∂ x = x 2 + y 2 3u+2(delu)/(delx)=x^(2)+y^(2)3 \mathrm{u}+2 \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\mathrm{x}^2+\mathrm{y}^23u+2∂u∂x=x2+y2 on x = 1 x = 1 x=1\mathrm{x}=1x=1. Assume h = k = 1 / 2 h = k = 1 / 2 h=k=1//2\mathrm{h}=\mathrm{k}=1 / 2h=k=1/2.
10. a) Find an approximate value of y ( 1.0 ) y ( 1.0 ) y(1.0)y(1.0)y(1.0) for the initial value problem
y ′ = x − 2 y , y ( 0 ) = 1 y ′ = x − 2 y , y ( 0 ) = 1 y^(‘)=x-2y,quady(0)=1\mathrm{y}^{\prime}=\mathrm{x}-2 \mathrm{y}, \quad \mathrm{y}(0)=1y′=x−2y,y(0)=1
using Milne-Simpson’s method
y n + 1 = y n − 1 + h 3 [ f n + 1 + 4 f n + f n − 1 ] y n + 1 = y n − 1 + h 3 f n + 1 + 4 f n + f n − 1 y_(n+1)=y_(n-1)+(h)/(3)[f_(n+1)+4f_(n)+f_(n-1)]y_{n+1}=y_{n-1}+\frac{h}{3}\left[f_{n+1}+4 f_n+f_{n-1}\right]yn+1=yn−1+h3[fn+1+4fn+fn−1]
with the step length h = 0.2 h = 0.2 h=0.2h=0.2h=0.2. Calculate the starting value using Runge-Kutta fourth order method with the same h h hhh.
b) Using fourth order Taylor series method with h = 0.2 h = 0.2 h=0.2h=0.2h=0.2, solve the initial value problem
y ′ = x + cos ⁡ y , y ( 0 ) = 0 y ′ = x + cos ⁡ y , y ( 0 ) = 0 y^(‘)=x+cos y,y(0)=0\mathrm{y}^{\prime}=\mathrm{x}+\cos \mathrm{y}, \mathrm{y}(0)=0y′=x+cos⁡y,y(0)=0
upto x = 1 x = 1 x=1\mathrm{x}=1x=1.
\(Sin^2\left(\theta \:\right)+Cos^2\left(\theta \right)=1\)

MMT-007 Sample Solution 2024

mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

  1. a) Show that f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y)=x yf(x,y)=xy
    i) satisfies a Lipschitz condition on any rectangle a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and c ≤ y ≤ d c ≤ y ≤ d c <= y <= dc \leq y \leq dc≤y≤d;
ii) satisfies a Lipschitz condition on any strip a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and − ∞ < y < ∞ − ∞ < y < ∞ -oo < y < oo-\infty<y<\infty−∞<y<∞;
iii) does not satisfy a Lipschitz condition on the entire plane.
Answer:
To show that the function f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y) = xyf(x,y)=xy satisfies a Lipschitz condition on certain domains, we need to find a constant L L LLL such that for all points ( x 1 , y 1 ) ( x 1 , y 1 ) (x_(1),y_(1))(x_1, y_1)(x1,y1) and ( x 2 , y 2 ) ( x 2 , y 2 ) (x_(2),y_(2))(x_2, y_2)(x2,y2) in the domain,
| f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | ≤ L ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 | f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | ≤ L ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 |f(x_(1),y_(1))-f(x_(2),y_(2))| <= Lsqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))|f(x_1, y_1) – f(x_2, y_2)| \leq L\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}|f(x1,y1)−f(x2,y2)|≤L(x1−x2)2+(y1−y2)2
i) On a rectangle a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and c ≤ y ≤ d c ≤ y ≤ d c <= y <= dc \leq y \leq dc≤y≤d:
Consider any two points ( x 1 , y 1 ) ( x 1 , y 1 ) (x_(1),y_(1))(x_1, y_1)(x1,y1) and ( x 2 , y 2 ) ( x 2 , y 2 ) (x_(2),y_(2))(x_2, y_2)(x2,y2) in the rectangle. Then,
| f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | = | x 1 y 1 − x 2 y 2 | = | x 1 y 1 − x 1 y 2 + x 1 y 2 − x 2 y 2 | ≤ | x 1 | | y 1 − y 2 | + | y 2 | | x 1 − x 2 | ≤ max { | x 1 | , | y 2 | } ( | x 1 − x 2 | + | y 1 − y 2 | ) | f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | = | x 1 y 1 − x 2 y 2 | = | x 1 y 1 − x 1 y 2 + x 1 y 2 − x 2 y 2 | ≤ | x 1 | | y 1 − y 2 | + | y 2 | | x 1 − x 2 | ≤ max { | x 1 | , | y 2 | } ( | x 1 − x 2 | + | y 1 − y 2 | ) {:[|f(x_(1)”,”y_(1))-f(x_(2)”,”y_(2))|=|x_(1)y_(1)-x_(2)y_(2)|],[=|x_(1)y_(1)-x_(1)y_(2)+x_(1)y_(2)-x_(2)y_(2)|],[ <= |x_(1)||y_(1)-y_(2)|+|y_(2)||x_(1)-x_(2)|],[ <= max{|x_(1)|”,”|y_(2)|}(|x_(1)-x_(2)|+|y_(1)-y_(2)|)]:}\begin{align*} |f(x_1, y_1) – f(x_2, y_2)| &= |x_1y_1 – x_2y_2| \\ &= |x_1y_1 – x_1y_2 + x_1y_2 – x_2y_2| \\ &\leq |x_1||y_1 – y_2| + |y_2||x_1 – x_2| \\ &\leq \max\{|x_1|, |y_2|\}(|x_1 – x_2| + |y_1 – y_2|) \end{align*}|f(x1,y1)−f(x2,y2)|=|x1y1−x2y2|=|x1y1−x1y2+x1y2−x2y2|≤|x1||y1−y2|+|y2||x1−x2|≤max{|x1|,|y2|}(|x1−x2|+|y1−y2|)
Since a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and c ≤ y ≤ d c ≤ y ≤ d c <= y <= dc \leq y \leq dc≤y≤d, we can take L = max { | a | , | b | , | c | , | d | } L = max { | a | , | b | , | c | , | d | } L=max{|a|,|b|,|c|,|d|}L = \max\{|a|, |b|, |c|, |d|\}L=max{|a|,|b|,|c|,|d|}. Then,
| f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | ≤ L ( | x 1 − x 2 | + | y 1 − y 2 | ) ≤ L 2 ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 | f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | ≤ L ( | x 1 − x 2 | + | y 1 − y 2 | ) ≤ L 2 ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 |f(x_(1),y_(1))-f(x_(2),y_(2))| <= L(|x_(1)-x_(2)|+|y_(1)-y_(2)|) <= Lsqrt2sqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))|f(x_1, y_1) – f(x_2, y_2)| \leq L(|x_1 – x_2| + |y_1 – y_2|) \leq L\sqrt{2}\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2}|f(x1,y1)−f(x2,y2)|≤L(|x1−x2|+|y1−y2|)≤L2(x1−x2)2+(y1−y2)2
So, f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y) = xyf(x,y)=xy satisfies a Lipschitz condition on any rectangle a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and c ≤ y ≤ d c ≤ y ≤ d c <= y <= dc \leq y \leq dc≤y≤d with Lipschitz constant L 2 L 2 Lsqrt2L\sqrt{2}L2.
ii) On a strip a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and − ∞ < y < ∞ − ∞ < y < ∞ -oo < y < oo-\infty < y < \infty−∞<y<∞:
Using a similar argument as in part (i), we can show that f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y) = xyf(x,y)=xy satisfies a Lipschitz condition on any strip a ≤ x ≤ b a ≤ x ≤ b a <= x <= ba \leq x \leq ba≤x≤b and − ∞ < y < ∞ − ∞ < y < ∞ -oo < y < oo-\infty < y < \infty−∞<y<∞ with a Lipschitz constant L = max { | a | , | b | } L = max { | a | , | b | } L=max{|a|,|b|}L = \max\{|a|, |b|\}L=max{|a|,|b|}.
iii) On the entire plane:
To show that f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y) = xyf(x,y)=xy does not satisfy a Lipschitz condition on the entire plane, consider the points ( x 1 , y 1 ) = ( 1 , n ) ( x 1 , y 1 ) = ( 1 , n ) (x_(1),y_(1))=(1,n)(x_1, y_1) = (1, n)(x1,y1)=(1,n) and ( x 2 , y 2 ) = ( 0 , 0 ) ( x 2 , y 2 ) = ( 0 , 0 ) (x_(2),y_(2))=(0,0)(x_2, y_2) = (0, 0)(x2,y2)=(0,0) for some large n ∈ N n ∈ N n inNn \in \mathbf{N}n∈N. Then,
| f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | = | 1 ⋅ n − 0 ⋅ 0 | = n | f ( x 1 , y 1 ) − f ( x 2 , y 2 ) | = | 1 ⋅ n − 0 ⋅ 0 | = n |f(x_(1),y_(1))-f(x_(2),y_(2))|=|1*n-0*0|=n|f(x_1, y_1) – f(x_2, y_2)| = |1 \cdot n – 0 \cdot 0| = n|f(x1,y1)−f(x2,y2)|=|1⋅n−0⋅0|=n
However,
( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 = 1 2 + n 2 ≈ n ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 = 1 2 + n 2 ≈ n sqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))=sqrt(1^(2)+n^(2))~~n\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2} = \sqrt{1^2 + n^2} \approx n(x1−x2)2+(y1−y2)2=12+n2≈n
for large n n nnn. If f f fff were Lipschitz on the entire plane, there would exist a constant L L LLL such that n ≤ L ⋅ n n ≤ L ⋅ n n <= L*nn \leq L \cdot nn≤L⋅n for all n n nnn, which is impossible. Therefore, f ( x , y ) = x y f ( x , y ) = x y f(x,y)=xyf(x, y) = xyf(x,y)=xy does not satisfy a Lipschitz condition on the entire plane.
b) Use Frobenious method to find the series solution about x = 0 x = 0 x=0x=0x=0 of the equation
x ( 1 − x ) d 2 y d x 2 − ( 1 + 3 x ) d y d x − y = 0 . x ( 1 − x ) d 2 y d x 2 − ( 1 + 3 x ) d y d x − y = 0 . x(1-x)(d^(2)y)/(dx^(2))-(1+3x)(dy)/(dx)-y=0.x(1-x) \frac{d^2 y}{d x^2}-(1+3 x) \frac{d y}{d x}-y=0 .x(1−x)d2ydx2−(1+3x)dydx−y=0.
Answer:
To solve the differential equation using the Frobenius method, we assume a solution of the form:
y = ∑ n = 0 ∞ a n x n + r y = ∑ n = 0 ∞   a n x n + r y=sum_(n=0)^(oo)a_(n)x^(n+r)y = \sum_{n=0}^{\infty} a_n x^{n+r}y=∑n=0∞anxn+r
where r r rrr is a constant to be determined. We then substitute this series into the differential equation and solve for the coefficients a n a n a_(n)a_nan.
First, we calculate the derivatives:
d y d x = ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 d y d x = ∑ n = 0 ∞   ( n + r ) a n x n + r − 1 (dy)/(dx)=sum_(n=0)^(oo)(n+r)a_(n)x^(n+r-1)\frac{dy}{dx} = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1}dydx=∑n=0∞(n+r)anxn+r−1
d 2 y d x 2 = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 d 2 y d x 2 = ∑ n = 0 ∞   ( n + r ) ( n + r − 1 ) a n x n + r − 2 (d^(2)y)/(dx^(2))=sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r-2)\frac{d^2y}{dx^2} = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2}d2ydx2=∑n=0∞(n+r)(n+r−1)anxn+r−2
Substituting these into the differential equation, we get:
x ( 1 − x ) ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 − ( 1 + 3 x ) ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 − ∑ n = 0 ∞ a n x n + r = 0 x ( 1 − x ) ∑ n = 0 ∞   ( n + r ) ( n + r − 1 ) a n x n + r − 2 − ( 1 + 3 x ) ∑ n = 0 ∞   ( n + r ) a n x n + r − 1 − ∑ n = 0 ∞   a n x n + r = 0 x(1-x)sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r-2)-(1+3x)sum_(n=0)^(oo)(n+r)a_(n)x^(n+r-1)-sum_(n=0)^(oo)a_(n)x^(n+r)=0x(1-x) \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} – (1+3x) \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} – \sum_{n=0}^{\infty} a_n x^{n+r} = 0x(1−x)∑n=0∞(n+r)(n+r−1)anxn+r−2−(1+3x)∑n=0∞(n+r)anxn+r−1−∑n=0∞anxn+r=0
Expanding and combining like terms, we have:
∑ n = 0 ∞ [ ( n + r ) ( n + r − 1 ) − ( n + r ) + 1 ] a n x n + r − ∑ n = 0 ∞ [ 3 ( n + r ) + ( n + r ) ( n + r − 1 ) ] a n x n + r + 1 = 0 ∑ n = 0 ∞   [ ( n + r ) ( n + r − 1 ) − ( n + r ) + 1 ] a n x n + r − ∑ n = 0 ∞   [ 3 ( n + r ) + ( n + r ) ( n + r − 1 ) ] a n x n + r + 1 = 0 sum_(n=0)^(oo)[(n+r)(n+r-1)-(n+r)+1]a_(n)x^(n+r)-sum_(n=0)^(oo)[3(n+r)+(n+r)(n+r-1)]a_(n)x^(n+r+1)=0\sum_{n=0}^{\infty} [(n+r)(n+r-1) – (n+r) + 1] a_n x^{n+r} – \sum_{n=0}^{\infty} [3(n+r) + (n+r)(n+r-1)] a_n x^{n+r+1} = 0∑n=0∞[(n+r)(n+r−1)−(n+r)+1]anxn+r−∑n=0∞[3(n+r)+(n+r)(n+r−1)]anxn+r+1=0
Equating coefficients of x n + r x n + r x^(n+r)x^{n+r}xn+r and x n + r + 1 x n + r + 1 x^(n+r+1)x^{n+r+1}xn+r+1 to zero, we get the recurrence relations:
( n + r ) ( n + r − 2 ) a n − ( 3 ( n + r ) + ( n + r ) ( n + r − 1 ) ) a n + 1 = 0 ( n + r ) ( n + r − 2 ) a n − ( 3 ( n + r ) + ( n + r ) ( n + r − 1 ) ) a n + 1 = 0 (n+r)(n+r-2)a_(n)-(3(n+r)+(n+r)(n+r-1))a_(n+1)=0(n+r)(n+r-2) a_n – (3(n+r) + (n+r)(n+r-1)) a_{n+1} = 0(n+r)(n+r−2)an−(3(n+r)+(n+r)(n+r−1))an+1=0
Simplifying, we find:
( n + r ) ( n + r − 2 ) a n − ( n + r + 1 ) ( n + r + 3 ) a n + 1 = 0 ( n + r ) ( n + r − 2 ) a n − ( n + r + 1 ) ( n + r + 3 ) a n + 1 = 0 (n+r)(n+r-2)a_(n)-(n+r+1)(n+r+3)a_(n+1)=0(n+r)(n+r-2) a_n – (n+r+1)(n+r+3) a_{n+1} = 0(n+r)(n+r−2)an−(n+r+1)(n+r+3)an+1=0
For n = 0 n = 0 n=0n = 0n=0, we have the indicial equation:
r ( r − 2 ) = 0 r ( r − 2 ) = 0 r(r-2)=0r(r-2) = 0r(r−2)=0
which gives r = 0 r = 0 r=0r = 0r=0 or r = 2 r = 2 r=2r = 2r=2.
For r = 0 r = 0 r=0r = 0r=0, the recurrence relation becomes:
n ( n − 2 ) a n − ( n + 1 ) ( n + 3 ) a n + 1 = 0 n ( n − 2 ) a n − ( n + 1 ) ( n + 3 ) a n + 1 = 0 n(n-2)a_(n)-(n+1)(n+3)a_(n+1)=0n(n-2) a_n – (n+1)(n+3) a_{n+1} = 0n(n−2)an−(n+1)(n+3)an+1=0
or
a n + 1 = n ( n − 2 ) ( n + 1 ) ( n + 3 ) a n a n + 1 = n ( n − 2 ) ( n + 1 ) ( n + 3 ) a n a_(n+1)=(n(n-2))/((n+1)(n+3))a_(n)a_{n+1} = \frac{n(n-2)}{(n+1)(n+3)} a_nan+1=n(n−2)(n+1)(n+3)an
Since a 0 a 0 a_(0)a_0a0 is arbitrary, let’s choose a 0 = 1 a 0 = 1 a_(0)=1a_0 = 1a0=1 for simplicity. Then, we can find the next few coefficients:
For n = 0 n = 0 n=0n = 0n=0:
a 1 = 0 ( 0 − 2 ) ( 0 + 1 ) ( 0 + 3 ) a 0 = 0 a 1 = 0 ( 0 − 2 ) ( 0 + 1 ) ( 0 + 3 ) a 0 = 0 a_(1)=(0(0-2))/((0+1)(0+3))a_(0)=0a_1 = \frac{0(0-2)}{(0+1)(0+3)} a_0 = 0a1=0(0−2)(0+1)(0+3)a0=0
For n = 1 n = 1 n=1n = 1n=1:
a 2 = 1 ( 1 − 2 ) ( 1 + 1 ) ( 1 + 3 ) a 1 = 0 a 2 = 1 ( 1 − 2 ) ( 1 + 1 ) ( 1 + 3 ) a 1 = 0 a_(2)=(1(1-2))/((1+1)(1+3))a_(1)=0a_2 = \frac{1(1-2)}{(1+1)(1+3)} a_1 = 0a2=1(1−2)(1+1)(1+3)a1=0
For n = 2 n = 2 n=2n = 2n=2:
a 3 = 2 ( 2 − 2 ) ( 2 + 1 ) ( 2 + 3 ) a 2 = 0 a 3 = 2 ( 2 − 2 ) ( 2 + 1 ) ( 2 + 3 ) a 2 = 0 a_(3)=(2(2-2))/((2+1)(2+3))a_(2)=0a_3 = \frac{2(2-2)}{(2+1)(2+3)} a_2 = 0a3=2(2−2)(2+1)(2+3)a2=0
For n = 3 n = 3 n=3n = 3n=3:
a 4 = 3 ( 3 − 2 ) ( 3 + 1 ) ( 3 + 3 ) a 3 = 0 a 4 = 3 ( 3 − 2 ) ( 3 + 1 ) ( 3 + 3 ) a 3 = 0 a_(4)=(3(3-2))/((3+1)(3+3))a_(3)=0a_4 = \frac{3(3-2)}{(3+1)(3+3)} a_3 = 0a4=3(3−2)(3+1)(3+3)a3=0
And so on. It appears that all coefficients after a 0 a 0 a_(0)a_0a0 are zero. Therefore, the solution for r = 0 r = 0 r=0r = 0r=0 is:
y = a 0 = 1 y = a 0 = 1 y=a_(0)=1y = a_0 = 1y=a0=1
Now, let’s consider the case r = 2 r = 2 r=2r = 2r=2:
For r = 2 r = 2 r=2r = 2r=2, the recurrence relation becomes:
( n + 2 ) ( n ) a n − ( n + 3 ) ( n + 5 ) a n + 1 = 0 ( n + 2 ) ( n ) a n − ( n + 3 ) ( n + 5 ) a n + 1 = 0 (n+2)(n)a_(n)-(n+3)(n+5)a_(n+1)=0(n+2)(n) a_n – (n+3)(n+5) a_{n+1} = 0(n+2)(n)an−(n+3)(n+5)an+1=0
or
a n + 1 = ( n + 2 ) ( n ) ( n + 3 ) ( n + 5 ) a n a n + 1 = ( n + 2 ) ( n ) ( n + 3 ) ( n + 5 ) a n a_(n+1)=((n+2)(n))/((n+3)(n+5))a_(n)a_{n+1} = \frac{(n+2)(n)}{(n+3)(n+5)} a_nan+1=(n+2)(n)(n+3)(n+5)an
Again, let’s choose a 0 = 1 a 0 = 1 a_(0)=1a_0 = 1a0=1. Then, we can find the next few coefficients:
For n = 0 n = 0 n=0n = 0n=0:
a 1 = 2 ( 0 ) ( 0 + 3 ) ( 0 + 5 ) a 0 = 0 a 1 = 2 ( 0 ) ( 0 + 3 ) ( 0 + 5 ) a 0 = 0 a_(1)=(2(0))/((0+3)(0+5))a_(0)=0a_1 = \frac{2(0)}{(0+3)(0+5)} a_0 = 0a1=2(0)(0+3)(0+5)a0=0
For n = 1 n = 1 n=1n = 1n=1:
a 2 = 3 ( 1 ) ( 1 + 3 ) ( 1 + 5 ) a 1 = 0 a 2 = 3 ( 1 ) ( 1 + 3 ) ( 1 + 5 ) a 1 = 0 a_(2)=(3(1))/((1+3)(1+5))a_(1)=0a_2 = \frac{3(1)}{(1+3)(1+5)} a_1 = 0a2=3(1)(1+3)(1+5)a1=0
And so on. It appears that all coefficients after a 0 a 0 a_(0)a_0a0 are also zero in this case. Therefore, the solution for r = 2 r = 2 r=2r = 2r=2 is:
y = a 0 x 2 = x 2 y = a 0 x 2 = x 2 y=a_(0)x^(2)=x^(2)y = a_0 x^2 = x^2y=a0x2=x2
In conclusion, the series solutions about x = 0 x = 0 x=0x=0x=0 of the equation x ( 1 − x ) d 2 y d x 2 − ( 1 + 3 x ) d y d x − y = 0 x ( 1 − x ) d 2 y d x 2 − ( 1 + 3 x ) d y d x − y = 0 x(1-x)(d^(2)y)/(dx^(2))-(1+3x)(dy)/(dx)-y=0x(1-x) \frac{d^2 y}{d x^2}-(1+3 x) \frac{d y}{d x}-y=0x(1−x)d2ydx2−(1+3x)dydx−y=0 are y = 1 y = 1 y=1y = 1y=1 and y = x 2 y = x 2 y=x^(2)y = x^2y=x2.
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