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Chapter 2: Problem 5

Show that the initial value problem $$ \begin{aligned} \dot{x} &=\frac{1}{2 x} \\ x(1) &=1 \end{aligned} $$ has a solution \(x(t)\) on the interval \((0, \infty)\), that \(x(t)\) is defined and continuous on \([0, \infty)\), but that \(x^{\prime}(0)\) does not exist.

Short Answer

Expert verified
\( x(t) = \sqrt{t} \). Solution is continuous on \([0, \infty)\), but \( x'(0) \) does not exist.

Step by step solution

01

- Understand the Initial Value Problem

We are given a differential equation \ \ \( \dot{x} = \frac{1}{2x} \) \ \ with an initial condition \(x(1) = 1\). Our task is to find a solution \( x(t) \), check its continuity and differentiability over the given interval.
02

- Solve the Differential Equation

Rewrite the equation as \ \ \( \frac{dx}{dt} = \frac{1}{2x} \). \ \ To solve, separate the variables: \ \ \( 2x \, dx = dt \). \ \ Integrate both sides: \ \ \( \int 2x \, dx = \int dt \). \ \ This gives us: \ \ \( x^2 = t + C \).
03

- Apply the Initial Condition

Use the initial condition \( x(1) = 1 \) to determine the constant \( C \). \ \ Substituting these values: \ \ \( 1^2 = 1 + C \). Hence, \( C = 0 \). \ \ Therefore, the solution is: \ \ \( x(t) = \sqrt{t} \).
04

- Verify the Interval

Check where \( x(t) = \sqrt{t} \) is valid. \ \ Since \( \sqrt{t} \) is defined and continuous for \( t \geq 0 \), \( x(t) \) satisfies the interval \((0, \infty) \) and is continuous on \([0, \infty)\).
05

- Examine the Derivative at t=0

To find \( x' (t) \): \ \ \( x' (t) = \frac{d}{dt} \sqrt{t} = \frac{1}{2\sqrt{t}} \). \ \ For \( t=0 \), the expression \( \frac{1}{2\sqrt{t}} \) is undefined because \( \sqrt{t} = 0 \). Therefore, \( x'(0) \) does not exist.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
A differential equation is an equation involving a function and its derivatives. They are essential in describing various physical systems and natural phenomena.
For example, consider the differential equation in the original exercise: \( \dot{x} = \frac{1}{2x} \). Here, \( \dot{x} \) represents the derivative of \( x \) with respect to time \( t \).
Solving differential equations often involves:
  • Separating variables
  • Integrating both sides
  • Applying initial conditions to find constants
These steps help find the function that satisfies both the differential equation and any given initial conditions.
Continuity
Continuity of a function means that it is smooth and unbroken over its domain.
For the solution to be continuous, it must not have any breaks, holes, or gaps.
In the original example, we solved the differential equation and found that \( x(t) = \sqrt{t} \). This function is continuous for \( t \geq 0 \) because the square root function \( \sqrt{t} \) is defined and continuous for all non-negative values of \( t \).
Therefore, the solution \( x(t) \) is continuous on the interval \([0, \infty)\).
Solving Integrals
Integration is a fundamental concept in calculus and involves finding the antiderivative or integral of a function.
To solve the given differential equation, we first separate variables to form integrable expressions: \( 2x \, dx = dt \).
Next, we integrate both sides: \ \[ \int 2x \, dx = \int dt \]
This yields: \( x^2 = t + C \).
Integrals help in solving differential equations that describe how quantities change over time or other variables.
Differentiability
Differentiability refers to whether a function has a derivative at a point or over an interval.
If a function is differentiable at every point in an interval, then it is smooth without any sharp bends or cusps.
In the exercise, the derivative of the function \( x(t) \) is given by: \( x'(t) = \frac{d}{dt} \sqrt{t} = \frac{1}{2\sqrt{t}} \).
However, at \( t = 0 \), this expression is undefined since \( \sqrt{0} \) is zero, making the denominator zero and thus, making \( x'(0) \) nonexistent.
This demonstrates that while the function is continuous over \([0, \infty)\), it is not differentiable at \( t = 0 \).

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Most popular questions from this chapter

Do problem 1 for the initial value problem $$ \begin{aligned} \dot{x} &=-x^{3} \\ x(0) &=x_{0} . \end{aligned} $$

(a) Show that the function \(f(x)=1 / x\) is not uniformly continuous on \(E=(0,1)\). Hint: \(f\) is uniformly continuous on \(E\) if for all \(\varepsilon>0\) there exists a \(\delta>0\) such that for all \(x, y \in E\) with \(|x-y|<\delta\) we have \(|f(x)-f(y)|<\varepsilon\). Thus, \(f\) is not uniformly continuous on \(E\) if there exists an \(\varepsilon>0\) such that for all \(\delta>0\) there exist \(x, y \in E\) with \(|x-y|<\delta\) such that \(|f(x)-f(y)| \geq \varepsilon\). Choose \(\varepsilon=1\) and show that for all \(\delta>0\) with \(\delta<1, x=6 / 2\) and \(y=\delta\) implies that \(x, y \in(0,1),|x-y|<\delta\) and \(|f(x)-f(y)|>1 .\) (b) Show that \(f(x)=1 / x\) does not satisfy a Lipschitz condition on \((0,1)\).

Show that the initial value problem $$ \begin{aligned} \dot{x} &=x^{3} \\ x(0) &=2 \end{aligned} $$ has a solution on an interval \((-\infty, b)\) for some \(b \in \mathbf{R}\). Sketch the solution in the \((t, x)\)-plane and note the behavior of \(x(t)\) as \(t \rightarrow b^{-}\).

Let \(V\) be a normed linear space. If \(T: V \rightarrow V\) satisfies $$ \|T(\mathbf{u})-T(\mathbf{v})\| \leq \mathrm{c}\|\mathbf{u}-\mathbf{v}\| $$ for all \(\mathbf{u}\) and \(v \in V\) with \(00\) is sufficiently small. Hint: Since \(\mathbf{f}\) is locally Lipschitz on \(E\) and \(x_{0} \in E\), there are positive constants \(\varepsilon\) and \(K_{0}\) such that the condition in Definition 2 is satisfied on \(N_{c}\left(x_{0}\right) \subset E .\) Let \(V=\left\\{\mathbf{u} \in C(I) \mid\left\|\mathbf{u}-\mathbf{x}_{0}\right\| \leq \varepsilon\right\\} .\) Then \(V\) is complete since it is a closed subset of \(C(I)\). Show that (i) for all \(\mathbf{u}, v \in\). \(V,\|T(\mathbf{u})-T(\mathbf{v})\| \leq a K_{0}\|\mathbf{u}-\mathbf{v}\|\) and that (ii) the positive constant \(a\) can be chosen sufficiently small that for \(t \in[-a, a], T \circ u(t) \in N_{e}\left(x_{0}\right)\), i.e., \(T: V \rightarrow V\).

Prove that if \(f\) is differentiable at \(x_{0}\) then there exists a \(\delta>0\) and a \(K_{0}>0\) such that for all \(x \in N_{b}\left(x_{0}\right)\) $$ \left|\mathbf{f}(\mathbf{x})-\mathbf{f}\left(\mathbf{x}_{0}\right)\right| \leq K_{0}\left|\mathbf{x}-\mathbf{x}_{0}\right| $$.
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