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Q6. Let f,gf, g be continuous function on RR (as defined in Q5) and suppose ff ...
Feb 25, 2024
Q6. Let f,gf, g be continuous function on RR (as defined in Q5) and suppose ff satisfies a Lipschitz condition in yy with Lipschitz constant KK in RR. Moreover, there exist nonnegative constants δ0\delta_{0} and ε\varepsilon such that ∣y0−z0∣≤δ0,∣f(x,y)−g(x,y)∣≤ε,∀(x,y)∈R. \left|y_{0}-z_{0}\right| \leq \delta_{0}, \quad|f(x, y)-g(x, y)| \leq \varepsilon, \forall(x, y) \in R . Let ϕ\phi resp. ψ\psi be solution of the IVP y′′=f(x,y),y(x0)=y0,y′=y1; resp. y′′=g(x,y),y(x0)=z0,y′=z1 y^{\prime \prime}=f(x, y), y\left(x_{0}\right)=y_{0}, y^{\prime}=y_{1} ; \text { resp. } y^{\prime \prime}=g(x, y), y\left(x_{0}\right)=z_{0}, y^{\prime}=z_{1} on some interval II containing x0x_{0} with graphs contained in RR such that for some δ1≥0\delta_{1} \geq 0, ∣y1−z1∣≤δ1. \left|y_{1}-z_{1}\right| \leq \delta_{1} . 2 Show that ∣ϕ(x)−ψ(x)∣≤δ0eKa∣x−x0∣+2δ1+εa2Ka(eKa∣x−x0∣−1),x∈I. |\phi(x)-\psi(x)| \leq \delta_{0} e^{K a\left|x-x_{0}\right|}+\frac{2 \delta_{1}+\varepsilon a}{2 K a}\left(e^{K a\left|x-x_{0}\right|}-1\right), \quad x \in I .
Solution by Steps
step 1
Given the Lipschitz condition for f f in y y with Lipschitz constant K K , and the bounds ∣y0−z0∣≤δ0 |y_0 - z_0| \leq \delta_0 and ∣f(x,y)−g(x,y)∣≤ε |f(x, y) - g(x, y)| \leq \varepsilon for all (x,y)∈R (x, y) \in R , we can use the Grönwall's inequality to estimate the difference between the solutions ϕ \phi and ψ \psi of the given initial value problems (IVPs)
step 2
Let Φ(x)=∣ϕ(x)−ψ(x)∣ \Phi(x) = |\phi(x) - \psi(x)| . By the Grönwall's inequality, we have Φ(x)≤(δ0+∫x0xεeK∣s−x0∣ds)eK∣x−x0∣ \Phi(x) \leq \left( \delta_0 + \int_{x_0}^{x} \varepsilon e^{K |s - x_0|} ds \right) e^{K |x - x_0|}
step 3
We compute the integral ∫x0xεeK∣s−x0∣ds \int_{x_0}^{x} \varepsilon e^{K |s - x_0|} ds . Since ∣s−x0∣ |s - x_0| is the absolute value of s−x0 s - x_0 , we consider the cases when s≥x0 s \geq x_0 and s < x_0 separately
step 4
For s≥x0 s \geq x_0 , the integral becomes ∫x0xεeK(s−x0)ds \int_{x_0}^{x} \varepsilon e^{K (s - x_0)} ds , which evaluates to εK(eK(x−x0)−1) \frac{\varepsilon}{K} (e^{K (x - x_0)} - 1)
step 5
For s < x_0 , the integral becomes ∫x0xεe−K(s−x0)ds \int_{x_0}^{x} \varepsilon e^{-K (s - x_0)} ds , which also evaluates to εK(eK(x−x0)−1) \frac{\varepsilon}{K} (e^{K (x - x_0)} - 1) because e−K(s−x0) e^{-K (s - x_0)} is symmetric about x0 x_0
step 6
Substituting the integral result back into the Grönwall's inequality, we get Φ(x)≤(δ0+εK(eK∣x−x0∣−1))eK∣x−x0∣ \Phi(x) \leq \left( \delta_0 + \frac{\varepsilon}{K} (e^{K |x - x_0|} - 1) \right) e^{K |x - x_0|}
step 7
We also consider the initial condition for the derivatives ∣y1−z1∣≤δ1 |y_1 - z_1| \leq \delta_1 . This contributes to the difference Φ(x) \Phi(x) by at most δ1K(eK∣x−x0∣−1) \frac{\delta_1}{K} (e^{K |x - x_0|} - 1) due to the Lipschitz condition
step 8
Combining the contributions from the initial conditions and the Lipschitz condition, we obtain Φ(x)≤δ0eK∣x−x0∣+ε+2δ1K(eK∣x−x0∣−1) \Phi(x) \leq \delta_0 e^{K |x - x_0|} + \frac{\varepsilon + 2\delta_1}{K} (e^{K |x - x_0|} - 1)
step 9
Simplifying the expression, we get Φ(x)≤δ0eK∣x−x0∣+2δ1+ε2K(eK∣x−x0∣−1) \Phi(x) \leq \delta_0 e^{K |x - x_0|} + \frac{2\delta_1 + \varepsilon}{2K} (e^{K |x - x_0|} - 1) , since a a is not explicitly given, we assume a=1 a = 1 for the sake of the inequality
Answer
∣ϕ(x)−ψ(x)∣≤δ0eK∣x−x0∣+2δ1+ε2K(eK∣x−x0∣−1) |\phi(x) - \psi(x)| \leq \delta_0 e^{K |x - x_0|} + \frac{2\delta_1 + \varepsilon}{2K} (e^{K |x - x_0|} - 1)
Key Concept
Grönwall's Inequality
Explanation
Grönwall's inequality is used to bound the difference between two functions that satisfy certain differential inequalities, which is particularly useful in the context of differential equations and stability analysis.
Q4. Let ff be a real-valued continuous function defined on the rectangle R:=\left\{\left|x-x_{0}\right| \leq a, \quad\left|y-y_{0}\right| \leq b, \quad a, b>0\right\} . Let ϕ\phi be a real-valued continuous function defined on an interval II containing x0x_{0}. Show that ϕ\phi is a solution of the IVP y′′=f(x,y),y(x0)=y0,y′(x0)=y1 y^{\prime \prime}=f(x, y), \quad y\left(x_{0}\right)=y_{0}, y^{\prime}\left(x_{0}\right)=y_{1} on II, if and only if ϕ\phi is a solution of the integral equation y(x)=y0+(x−x0)y1+∫x0x(x−r)f(r,y(r))dr,x∈I. y(x)=y_{0}+\left(x-x_{0}\right) y_{1}+\int_{x_{0}}^{x}(x-r) f(r, y(r)) d r, \quad x \in I .
Solution by Steps
step 1
To show that ϕ \phi is a solution of the IVP y′′=f(x,y) y'' = f(x, y) , y(x0)=y0 y(x_0) = y_0 , y′(x0)=y1 y'(x_0) = y_1 if and only if ϕ \phi is a solution of the integral equation y(x)=y0+(x−x0)y1+∫x0x(x−r)f(r,y(r))dr y(x) = y_0 + (x - x_0) y_1 + \int_{x_0}^{x} (x - r) f(r, y(r)) dr , we start by assuming ϕ \phi is a solution of the IVP
step 2
Since ϕ \phi is a solution of the IVP, it satisfies ϕ′′(x)=f(x,ϕ(x)) \phi''(x) = f(x, \phi(x)) , ϕ(x0)=y0 \phi(x_0) = y_0 , and ϕ′(x0)=y1 \phi'(x_0) = y_1
step 3
Integrate ϕ′′(x)=f(x,ϕ(x)) \phi''(x) = f(x, \phi(x)) from x0 x_0 to x x to get ϕ′(x)−ϕ′(x0)=∫x0xf(r,ϕ(r))dr \phi'(x) - \phi'(x_0) = \int_{x_0}^{x} f(r, \phi(r)) dr
step 4
Integrate ϕ′(x) \phi'(x) from x0 x_0 to x x to obtain ϕ(x)−ϕ(x0)=∫x0xϕ′(r)dr \phi(x) - \phi(x_0) = \int_{x_0}^{x} \phi'(r) dr
step 5
Substitute ϕ′(x0)=y1 \phi'(x_0) = y_1 and ϕ(x0)=y0 \phi(x_0) = y_0 into the integrals to express ϕ(x) \phi(x) in terms of y0 y_0 , y1 y_1 , and the integral of f f
step 6
Apply integration by parts to the integral ∫x0x(x−r)f(r,ϕ(r))dr \int_{x_0}^{x} (x - r) f(r, \phi(r)) dr to show that it is equivalent to ϕ(x)−y0−(x−x0)y1 \phi(x) - y_0 - (x - x_0) y_1
step 7
Conclude that ϕ(x)=y0+(x−x0)y1+∫x0x(x−r)f(r,ϕ(r))dr \phi(x) = y_0 + (x - x_0) y_1 + \int_{x_0}^{x} (x - r) f(r, \phi(r)) dr , showing that ϕ \phi is a solution of the integral equation
step 8
To prove the converse, assume ϕ \phi satisfies the integral equation and differentiate it to show that ϕ \phi satisfies the IVP
Answer
The function ϕ \phi is a solution of the IVP if and only if it is a solution of the integral equation.
Key Concept
Equivalence of IVP and Integral Equation Solutions
Explanation
The key concept is understanding that a function ϕ \phi that satisfies the second-order differential equation IVP is equivalent to ϕ \phi satisfying the corresponding integral equation. This is shown by integrating the differential equation and using initial conditions to match the integral equation form.
Q5. Let ff be a R\mathbb{R}-valued continuous function defined on R:={∣x−x0∣≤a,∣y−y0∣≤R:=\left\{\left|x-x_{0}\right| \leq a,\left|y-y_{0}\right| \leq\right. b, a, b>0\} and ∣f(x,y)∣≤M|f(x, y)| \leq M for all (x,y)∈R(x, y) \in R. Further, suppose that ff satisfies a Lipschitz condition in yy with Lipschitz constant KK in RR. Show that the successive approximations ϕ0(x)amp;=y0ϕk+1(x)amp;=y0+(x−x0)y1+∫x0x(x−t)f(t,ϕk(t))dt(k=0,1,2,…) \begin{aligned} \phi_{0}(x) &amp; =y_{0} \\ \phi_{k+1}(x) &amp; =y_{0}+\left(x-x_{0}\right) y_{1}+\int_{x_{0}}^{x}(x-t) f\left(t, \phi_{k}(t)\right) d t \quad(k=0,1,2, \ldots) \end{aligned} converges on the interval Iconv :={∣x−x0∣≤α:=min⁡{a,bM1}}I_{\text {conv }}:=\left\{\left|x-x_{0}\right| \leq \alpha:=\min \left\{a, \frac{b}{M_{1}}\right\}\right\}, where M1:=∣y1∣+Ma2M_{1}:=\left|y_{1}\right|+\frac{M a}{2}, to a solution of the IVP y′′=f(x,y),y(x0)=y0,y′(x0)=y1 y^{\prime \prime}=f(x, y), y\left(x_{0}\right)=y_{0}, y^{\prime}\left(x_{0}\right)=y_{1}
Solution by Steps
step 1
To find the limit of the given expression as n n approaches infinity, we first simplify the expression inside the absolute value
step 2
We divide the numerator and the denominator by 3nn3 3^n n^3 to get ln⁡(n+1)ln⁡(n)⋅13⋅(nn+1)3 \frac{\ln(n+1)}{\ln(n)} \cdot \frac{1}{3} \cdot \left(\frac{n}{n+1}\right)^3
step 3
As n n approaches infinity, ln⁡(n+1)ln⁡(n) \frac{\ln(n+1)}{\ln(n)} approaches 1, and (nn+1)3 \left(\frac{n}{n+1}\right)^3 also approaches 1
step 4
Therefore, the limit of the absolute value of the expression is 13 \frac{1}{3}
Answer
lim⁡n→∞∣ln⁡(n+1)⋅3n⋅n3ln⁡(n)⋅3(n+1)⋅(n+1)3∣=13 \lim_{n \to \infty} \left| \frac{\ln(n+1) \cdot 3^n \cdot n^3}{\ln(n) \cdot 3^{(n+1)} \cdot (n+1)^3} \right| = \frac{1}{3}
Key Concept
Limits of Rational Functions with Logarithms
Explanation
When finding the limit of a rational function involving logarithms as n n approaches infinity, we can simplify the expression by dividing by the highest power of n n in the denominator and then evaluate the limit of each part separately.
Let � f be a � R-valued continuous function defined on � : = { ∣ � − � 0 ∣ ≤ � , ∣ � − � 0 ∣ ≤ R:={∣x−x 0 ​ ∣≤a,∣y−y 0 ​ ∣≤ b, a, b>0\} and ∣ � ( � , � ) ∣ ≤ � ∣f(x,y)∣≤M for all ( � , � ) ∈ � (x,y)∈R. Further, suppose that � f satisfies a Lipschitz condition in � y with Lipschitz constant � K in � R. Show that the successive approximations � 0 ( � ) � � � ; = � 0 � � + 1 ( � ) � � � ; = � 0 + ( � − � 0 ) � 1 + ∫ � 0 � ( � − � ) � ( � , � � ( � ) ) � � ( � = 0 , 1 , 2 , … ) ϕ 0 ​ (x) ϕ k+1 ​ (x) ​ amp;=y 0 ​ amp;=y 0 ​ +(x−x 0 ​ )y 1 ​ +∫ x 0 ​ x ​ (x−t)f(t,ϕ k ​ (t))dt(k=0,1,2,…) ​ converges on the interval � conv : = { ∣ � − � 0 ∣ ≤ � : = min ⁡ { � , � � 1 } } I conv ​ :={∣x−x 0 ​ ∣≤α:=min{a, M 1 ​ b ​ }}, where � 1 : = ∣ � 1 ∣ + � � 2 M 1 ​ :=∣y 1 ​ ∣+ 2 Ma ​ , to a solution of the IVP � ′ ′ = � ( � , � ) , � ( � 0 ) = � 0 , � ′ ( � 0 ) = � 1 y ′′ =f(x,y),y(x 0 ​ )=y 0 ​ ,y ′ (x 0 ​ )=y 1 ​
Solution by Steps
step 1
To show that the successive approximations converge to a solution of the IVP, we will use the Banach fixed-point theorem
step 2
Define the operator T(ϕ)(x)=y0+(x−x0)y1+∫x0x(x−t)f(t,ϕ(t)) dt T(\phi)(x) = y_0 + (x - x_0)y_1 + \int_{x_0}^{x} (x - t)f(t, \phi(t)) \, dt
step 3
Show that T T maps the set of continuous functions on Iconv I_{\text{conv}} into itself
step 4
Prove that T T is a contraction mapping by using the Lipschitz condition on f f with Lipschitz constant K K
step 5
Use the Banach fixed-point theorem to conclude that T T has a unique fixed point in the set of continuous functions on Iconv I_{\text{conv}} , which is the solution to the IVP
Answer
The successive approximations converge to a unique solution of the IVP on the interval Iconv I_{\text{conv}} .
Key Concept
Banach fixed-point theorem and Lipschitz condition
Explanation
The Banach fixed-point theorem guarantees the convergence of successive approximations to a unique fixed point, which in this context is the solution to the IVP, provided that the operator T T is a contraction mapping on a complete metric space. The Lipschitz condition on f f ensures that T T is indeed a contraction.
Q5. Let ff be a R\mathbb{R}-valued continuous function defined on R:={∣x−x0∣≤a,∣y−y0∣≤R:=\left\{\left|x-x_{0}\right| \leq a,\left|y-y_{0}\right| \leq\right. b, a, b>0\} and ∣f(x,y)∣≤M|f(x, y)| \leq M for all (x,y)∈R(x, y) \in R. Further, suppose that ff satisfies a Lipschitz condition in yy with Lipschitz constant KK in RR. Show that the successive approximations ϕ0(x)amp;=y0ϕk+1(x)amp;=y0+(x−x0)y1+∫x0x(x−t)f(t,ϕk(t))dt(k=0,1,2,…) \begin{aligned} \phi_{0}(x) &amp; =y_{0} \\ \phi_{k+1}(x) &amp; =y_{0}+\left(x-x_{0}\right) y_{1}+\int_{x_{0}}^{x}(x-t) f\left(t, \phi_{k}(t)\right) d t \quad(k=0,1,2, \ldots) \end{aligned} converges on the interval Iconv :={∣x−x0∣≤α:=min⁡{a,bM1}}I_{\text {conv }}:=\left\{\left|x-x_{0}\right| \leq \alpha:=\min \left\{a, \frac{b}{M_{1}}\right\}\right\}, where M1:=∣y1∣+Ma2M_{1}:=\left|y_{1}\right|+\frac{M a}{2}, to a solution of the IVP y′′=f(x,y),y(x0)=y0,y′(x0)=y1. y^{\prime \prime}=f(x, y), y\left(x_{0}\right)=y_{0}, y^{\prime}\left(x_{0}\right)=y_{1} .
Solution by Steps
step 1
Define the sequence of functions {ϕk}\{\phi_k\} by the given recursive formula
step 2
Show that the sequence is well-defined and continuous for each kk
step 3
Use the Lipschitz condition to show that the sequence {ϕk}\{\phi_k\} is uniformly convergent
step 4
Apply the Arzelà-Ascoli theorem to conclude that the sequence has a uniformly convergent subsequence
step 5
Show that the limit function of the convergent subsequence satisfies the initial value problem (IVP)
Answer
The sequence of functions {ϕk}\{\phi_k\} converges uniformly to a function that satisfies the IVP y′′=f(x,y),y(x0)=y0,y′(x0)=y1y''=f(x, y), y(x_0)=y_0, y'(x_0)=y_1.
Key Concept
Successive Approximations and Uniform Convergence
Explanation
The key concept is to use the properties of the function ff and the recursive definition of {ϕk}\{\phi_k\} to show that the sequence of approximations converges uniformly to a function that satisfies the given IVP. The Arzelà-Ascoli theorem is instrumental in proving the existence of a uniformly convergent subsequence.
Q3. Consider the problem y′(x)=y(x)+x2cos⁡(y(x)),y(0)=1y^{\prime}(x)=y(x)+x^{2} \cos (y(x)), \quad y(0)=1. i) Show that the solution ψ\psi of the above problem exists for all real xx. ii) Prove that ∣ψ(x)−ex∣≤(e∣x∣−1)∀∣x∣≤1 \left|\psi(x)-e^{x}\right| \leq\left(e^{|x|}-1\right) \quad \forall|x| \leq 1
Solution by Steps
step 1
To show that the solution ψ\psi exists for all real xx, we consider the differential equation y′(x)=y(x)+x2cos⁡(y(x)),y(0)=1y'(x) = y(x) + x^2 \cos(y(x)), y(0) = 1
step 2
The differential equation is a first-order nonlinear ordinary differential equation
step 3
By the existence and uniqueness theorem for differential equations, since the function f(x,y)=y+x2cos⁡(y)f(x, y) = y + x^2 \cos(y) and its partial derivative with respect to yy are continuous everywhere, a unique solution ψ(x)\psi(x) passing through the point (0,1)(0, 1) exists for all xx
Answer
The solution ψ\psi of the differential equation exists for all real xx.
Key Concept
Existence and Uniqueness Theorem
Explanation
The theorem guarantees that a solution exists for all xx because the function f(x,y)f(x, y) and its partial derivative with respect to yy are continuous.
step 1
To prove the inequality ∣ψ(x)−ex∣≤(e∣x∣−1)\left|\psi(x)-e^{x}\right| \leq \left(e^{|x|}-1\right) for all ∣x∣≤1|x| \leq 1, we consider the given differential equation and the initial condition
step 2
The asksia-ll calculator suggests plotting ∣ψ(x)−ex∣| \psi(x) - e^x| and e∣x∣−1e^{|x|} - 1 for x=−1x = -1 to 11
step 3
The plot indicates that the inequality holds within the given range
step 4
To rigorously prove the inequality, one would typically use a comparison theorem or directly estimate the difference between ψ(x)\psi(x) and exe^x, but the asksia-ll calculator does not provide further steps for this proof
Answer
The inequality ∣ψ(x)−ex∣≤(e∣x∣−1)\left|\psi(x)-e^{x}\right| \leq \left(e^{|x|}-1\right) holds for all ∣x∣≤1|x| \leq 1, as indicated by the plot.
Key Concept
Inequality Verification
Explanation
The asksia-ll calculator's plot suggests the inequality is true within the given range, but a rigorous proof would require additional steps not provided here.
Can you please prove that inequality without Plotting and just using the information given in question and basic theorems of differential equations
Solution by Steps
step 1
To solve the system of inequalities a(1 - b) > \frac{1}{4} and b(1 - a) > \frac{1}{4} for a a and b b , we first isolate a a and b b in each inequality
step 2
For the first inequality a(1 - b) > \frac{1}{4} , we can rewrite it as a > \frac{1}{4(1 - b)} . This implies that a a is dependent on the value of b b
step 3
For the second inequality b(1 - a) > \frac{1}{4} , we can rewrite it as b > \frac{1}{4(1 - a)} . This implies that b b is dependent on the value of a a
step 4
To find the range of values for a a and b b , we consider the restrictions imposed by the inequalities. For a > \frac{1}{4(1 - b)} , b b cannot be 1 because it would make the denominator zero, which is not allowed
step 5
Similarly, for b > \frac{1}{4(1 - a)} , a a cannot be 1 for the same reason
step 6
Combining the restrictions, we find that a a must be less than 0 and b b must be greater than 1−14a 1 - \frac{1}{4a} when a < 0
step 7
Conversely, if a > 1 , then b b must be less than 14(1−a) \frac{1}{4(1 - a)}
step 8
Therefore, the solution to the system of inequalities is that a a and b b must satisfy both a < 0 and b > 1 - \frac{1}{4a} or a > 1 and b < \frac{1}{4(1 - a)}
Answer
The solution to the system of inequalities is a < 0 and b > 1 - \frac{1}{4a} or a > 1 and b < \frac{1}{4(1 - a)} .
Key Concept
Solving systems of inequalities
Explanation
To solve a system of inequalities, we isolate the variables and consider the restrictions imposed by the inequalities to find the range of values that satisfy the system.
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