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Chapter 2: Problem 5

Show that \(y=0\) and \(y=t|t|\) both satisfy the initial value problem \(d y / d t=2 \sqrt{|y|}, y(0)=\) 0 . Does this contradict the Existence and Uniqueness Theorem?

Short Answer

Expert verified
Both solutions satisfy the initial value problem. The lack of uniqueness does not contradict the theorem, as its conditions are not met.

Step by step solution

01

Verify First Solution

Check if the first solution, which is given by the equation \(y = 0\), satisfies the initial value problem. \(\frac{dy}{dt} = 0\) because \(y\) is constant. Substitute into the right-hand side of the differential equation \(\frac{dy}{dt} = 2 \sqrt{|y|}\): \[ 0 = 2 \sqrt{|0|} = 0 \]Since both sides of the equation are equal, \(y = 0\) is a solution.
02

Verify Second Solution

Check if the second solution, given by \(y = t|t|\) satisfies the initial value problem. First, compute the derivative of \(y = t|t|\) with respect to \(t\):\[\frac{dy}{dt} = \frac{d}{dt}(t|t|) = \begin{cases} 2t & \text{if } t > 0 \ 0 & \text{if } t = 0 \ -2t & \text{if } t < 0 \end{cases}\]Next, substitute \(y = t|t|\) and its derivative into the differential equation \(\frac{dy}{dt} = 2 \sqrt{|y|}\):For \(t > 0\): \[2t = 2 \sqrt{|t^2|} = 2 \sqrt{t^2} = 2t\]For \(t < 0\): \[-2t = 2 \sqrt{|t^2|} = 2 \sqrt{t^2} = -2t\]For \(t = 0\): \[0 = 2 \sqrt{0} = 0\]In all cases, the differential equation is satisfied, so \(y = t|t|\) is also a solution.
03

Initial Condition Check

Check whether both solutions satisfy the initial condition \(y(0) = 0\). Clearly, for \(y = 0\), we have \(y(0) = 0\). Similarly, for \(y = t|t|\), at \(t = 0\), \(y(0) = 0|0| = 0\). Therefore, both solutions satisfy the initial condition.
04

Analyze Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem states that for the initial value problem \(\frac{dy}{dt} = f(t, y) \), \(y(t_0) = y_0\), if \(f(t, y)\) and \(\frac{\partial f}{\partial y}\) are continuous near the point \((t_0, y_0)\), then there exists a unique solution. The function \(f(t, y) = 2 \sqrt{|y|}\) is continuous in \(y\), but \(\frac{\partial f}{\partial y} = \frac{1}{\sqrt{|y|}}\) is not continuous at \(y = 0\). Hence, the conditions for the Existence and Uniqueness Theorem are not satisfied, which means the existence of multiple solutions doesn鈥檛 contradict the theorem.

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Key Concepts

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

Existence and Uniqueness Theorem
The Existence and Uniqueness Theorem is a key principle in differential equations that helps us understand whether a given initial value problem has a single solution or possibly multiple solutions. The theorem states that for a first-order differential equation of the form \( \frac{dy}{dt} = f(t, y) \) with an initial condition \( y(t_0) = y_0 \), there will be a unique solution if two criteria are met:
  • The function \( f(t, y) \) is continuous near the point \((t_0, y_0)\).
  • The partial derivative \( \frac{\foot\frac{\footfraction}{foot\beginb}^{foot t_{\foot

In this specific problem, we look at the differential equation \( \frac{dy}{dt} = 2 \foot\beginsqrt{\footabs{\foot y_{\foot}\foot\frac}} \), with \( footy_{\foot}(0) = foot0_{\foot} \). While \( \foot f_{\foot}\foot (t, y) = 2 \foot\beginsqrt{\foot\footabs{\foot y }\foot\}) \foot is continuous in \{foot y \}) \), the partial derivative \foot\foot \foot\frac\{footpartial \foot\foot f_{\foot\frac{}}\foot}{footpartial foot \})\ ,foot foot at \foot,\foot {abs}={foot \foot y_{\foot0 \foot}.abs}\foot foot \footnot continuous y=foot \foot. abs foot\foot at \foot.}abs
Continuous Functions
A continuous function is one that does not have any abrupt changes or jumps. Intuitively, if you can draw a function without lifting your pen from the paper, the function is continuous.
When we deal with differential equations, continuity plays a significant role in ensuring the reliability of solutions. The function \( f(t, y) \) must be continuous near the point where we start solving the equation for the Existence and Uniqueness Theorem to hold.
In the given problem, \( f(t, y) = 2 \foot\beginsqrt{\foot abs \foot{ y_{\foot}}} \) is indeed continuous in \( y_{\foot \). This means that for any small interval around any point \( (t, 0) \), there are no discontinuities which ensures that values of \( f(t, y)\) and small changes form a smooth curve consistent in analysis .
However essential function may continuous not always imply an unique solution exists, as evident from problem, wherein, despite continuity multiple satisfied 'condition'.
Derivative Calculation
Understanding how to calculate derivatives is crucial for solving differential equations. The derivative of a function measures how that function changes as the input changes. In other words, it tells us the rate at which one quantity changes with respect to another.
In this problem, we start by finding the derivatives of the two solutions that we suspected might work. For the first potential solution \( y = 0 \), the derivative is straightforward since it remains zero for all values of \( t \). Thus we can say:

\footfrac{dy}{dt}= 0
Next derivative potential solution \( y = t |t| \) . Break down into cases whether greater than equal zero other : :
\footfrac{d}{dt}(t |t|) =
\begin{ cases \foot2 t & , text greater than zero equal 't'

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Most popular questions from this chapter

\(2 \ln t d t-\ln \left(4 y^{2}\right) d y=0\)

\(t d y-\left(y+\sqrt{t^{2}+y^{2}}\right) d t=0, y(1)=0\)

(a) Show that if \(y=y_{1}(t)\) is a solution of \(d y / d t+p(t) y=q(t)\) then \(y(t)=y_{1}(t)\) is a solution of \(d^{2} y / d t^{2}+p^{\prime}(t) y+p(t) y^{\prime}=\) \(q^{\prime}(t)\). (b) Solve \(d y / d t-\frac{1}{t} y=\ln t, t>0\). (c) Use the results obtained in (a) and (b) to solve $$ \frac{d^{2} y}{d t^{2}}-\frac{1}{t} \frac{d y}{d t}+\frac{1}{t^{2}} y=\frac{1}{t}, \quad t>0 . $$

Find a general solution of the equation $$ \left(2 t-y^{2} \sin t y\right) d t+(\cos t y-t y \sin t y) d y=0 . $$ Graph various solutions on the rectangle \([0,3 \pi] \times[0,3 \pi]\).

\(\left(5 t y+4 y^{2}+1\right) d t+\left(t^{2}+2 t y\right) d y=0\)
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