ODE Blow Up in Finite Time

We are given two initial value ordinary differential equations (ODEs) and asked to determine if their solutions blow up in finite time based on two assertions.

Analyzing ODE (A)

The first initial value problem is given by:

\(\frac{dx}{dt}=x^3, x(0)=1\)

This is a separable ODE. We can rewrite it as:

\(\frac{dx}{x^3} = dt\)

Integrate both sides:

\(\int \frac{dx}{x^3} = \int dt\)

This gives:

\(\int x^{-3} dx = t + C_1\)

\(-\frac{1}{2}x^{-2} = t + C_1\)

\(-\frac{1}{2x^2} = t + C_1\)

Now, use the initial condition \(x(0)=1\) to find the constant \(C_1\):

\(-\frac{1}{2(1)^2} = 0 + C_1\)

\(-\frac{1}{2} = C_1\)

Substitute \(C_1\) back into the solution equation:

\(-\frac{1}{2x^2} = t - \frac{1}{2}\)

We can rearrange this equation to solve for \(x\) in terms of \(t\):

\(\frac{1}{2x^2} = \frac{1}{2} - t\)

\(\frac{1}{x^2} = 2 \left(\frac{1}{2} - t\right) = 1 - 2t\)

\(x^2 = \frac{1}{1 - 2t}\)

\(x(t) = \pm \frac{1}{\sqrt{1 - 2t}}\)

Since the initial condition is \(x(0)=1\) (a positive value), we take the positive root:

\(x(t) = \frac{1}{\sqrt{1 - 2t}}\)

A solution blows up in finite time if \(x(t)\) approaches infinity as \(t\) approaches some finite value. In this solution, \(x(t)\) becomes undefined (goes to infinity) when the denominator is zero:

\(1 - 2t = 0\)

\(2t = 1\)

\(t = \frac{1}{2}\)

Since the solution goes to infinity at the finite time \(t = 1/2\), the solution to (A) blows up in finite time.

Therefore, assertion I is true.

Analyzing ODE (B)

The second initial value problem is given by:

\(\frac{dx}{dt}=x\sin x^2, x(0)=2.\)

This is an autonomous ODE of the form \(\frac{dx}{dt} = f(x)\) where \(f(x) = x \sin x^2\). For an autonomous ODE, the solution \(x(t)\) with initial condition \(x(t_0) = x_0\) blows up in finite time if the integral \(\int_{x_0}^{\infty} \frac{dx}{f(x)}\) converges for growth towards \(+\infty\), or \(\int_{-\infty}^{x_0} \frac{dx}{f(x)}\) converges for blow up towards \(-\infty\).

Let's look at the function \(f(x) = x \sin x^2\). The zeros of \(f(x)\) for \(x > 0\) occur when \(\sin x^2 = 0\), which means \(x^2 = n\pi\) for some positive integer \(n\). The positive zeros are \(x = \sqrt{n\pi}\) for \(n=1, 2, 3, \dots\).

• \(x = \sqrt{\pi} \approx \sqrt{3.14} \approx 1.77\)
• \(x = \sqrt{2\pi} \approx \sqrt{6.28} \approx 2.50\)
• \(x = \sqrt{3\pi} \approx \sqrt{9.42} \approx 3.07\)
• and so on...

The initial condition is \(x(0) = 2\). This value lies between the zeros \(\sqrt{\pi} \approx 1.77\) and \(\sqrt{2\pi} \approx 2.50\).

In the interval \((\sqrt{\pi}, \sqrt{2\pi})\), the value of \(x^2\) is between \(\pi\) and \(2\pi\). For angles between \(\pi\) and \(2\pi\), the sine function is negative. Thus, for \(x \in (\sqrt{\pi}, \sqrt{2\pi})\), \(\sin x^2 < 0\). Since \(x\) is positive, \(f(x) = x \sin x^2 < 0\).

Because the initial value \(x(0)=2\) is in the interval where \(f(x) < 0\), the solution \(x(t)\) will initially decrease. It will move towards the nearest zero of \(f(x)\) to its left, which is \(\sqrt{\pi}\).

As \(x\) approaches \(\sqrt{\pi}\) from the right, \(f(x) = x \sin x^2\) approaches \( \sqrt{\pi} \sin (\sqrt{\pi})^2 = \sqrt{\pi} \sin \pi = 0\). The rate of change \(\frac{dx}{dt}\) approaches zero as \(x\) approaches \(\sqrt{\pi}\).

Since the solution is bounded above by \(\sqrt{2\pi}\) and below by \(\sqrt{\pi}\) (it cannot cross the points where \(f(x)=0\) in finite time unless starting there), the solution \(x(t)\) for \(x(0)=2\) will remain between \(\sqrt{\pi}\) and \(\sqrt{2\pi}\) for all \(t > 0\) for which it exists. Specifically, it will decrease towards \(\sqrt{\pi}\) as \(t \to \infty\).

A solution that remains bounded cannot blow up to infinity. Therefore, the solution to (B) does not blow up in finite time.

Therefore, assertion II is false.

Conclusion

Assertion I is true because the solution to ODE (A) is \(x(t) = \frac{1}{\sqrt{1 - 2t}}\), which goes to infinity as \(t \to 1/2\).

Assertion II is false because the solution to ODE (B) with \(x(0)=2\) is bounded between the zeros \(\sqrt{\pi}\) and \(\sqrt{2\pi}\) of the function \(f(x)=x\sin x^2\), and thus cannot go to infinity in finite time.

Based on this analysis, statement (I) is true but statement (II) is false.