MMT-007 Solved Assignment 2024

Next Lesson

Sample Solution

MMT-007 Solved Assignment 2024 Sample Solution
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.
Verified Answer
5/5
Back to Course
Next Lesson
Scroll to Top
Scroll to Top