STEM Problem Detail

Problem Statement

Prompt

Provide stepwise solution for the given Exercise 1.6#3.

Solution Rendering...

**Problem 1.6#3:** Consider $$x' = x^3 - t^2$$.

### (a) Antifunnel along $$x = t^{2/3}$$

**Step 1: Verify $$x = t^{2/3}$$ is a pseudo-solution**
Let $$\alpha(t) = t^{2/3}$$. Compute:
\[
\alpha'(t) = \frac{2}{3}t^{-1/3}
\]
Check if $$\alpha(t)$$ approximately satisfies the DE:
\[
\alpha^3 - t^2 = (t^{2/3})^3 - t^2 = t^2 - t^2 = 0
\]
So $$\alpha'(t) = \frac{2}{3}t^{-1/3}$$ while RHS of DE at $$\alpha(t)$$ is 0.

For large $$t$$, $$\alpha'(t) = \frac{2}{3}t^{-1/3} \to 0$$, so $$\alpha(t)$$ becomes a better approximation.

**Step 2: Show narrowing region around $$\alpha(t)$$**
Consider region $$R_\epsilon = \{(t,x): |x - t^{2/3}| < \epsilon\}$$.

For $$x = t^{2/3} + \delta$$ with $$|\delta| < \epsilon$$:
\[
x' = (t^{2/3} + \delta)^3 - t^2 = t^2 + 3t^{4/3}\delta + 3t^{2/3}\delta^2 + \delta^3 - t^2
\]
\[
x' = 3t^{4/3}\delta + 3t^{2/3}\delta^2 + \delta^3
\]

**Step 3: Compare with $$\alpha'(t)$$**
We need to check if solutions diverge from $$\alpha(t)$$.

The linear term $$3t^{4/3}\delta$$ dominates for large $$t$$.

For $$\delta > 0$$: $$x' \approx 3t^{4/3}\delta > 0$$, so $$x$$ increases faster than $$\alpha(t)$$.

For $$\delta < 0$$: $$x' \approx 3t^{4/3}\delta < 0$$, so $$x$$ decreases faster than $$\alpha(t)$$.

Thus solutions diverge from $$\alpha(t)$$: this is an **antifunnel**.

**Step 4: Show it narrows**
Consider width $$\Delta(t) = |x_1(t) - x_2(t)|$$ for two solutions in the antifunnel.

For large $$t$$, the dominant term gives:
\[
\frac{d}{dt}(\Delta) \approx 3t^{4/3}\Delta
\]
So $$\Delta(t)$$ grows exponentially in $$t^{7/3}$$, not narrows... Wait, contradiction.

Actually, let's check more carefully: If $$\alpha(t)$$ is a pseudo-solution, consider $$y = x - \alpha(t)$$:
\[
y' = x' - \alpha'(t) = (x^3 - t^2) - \frac{2}{3}t^{-1/3}
\]
Substitute $$x = \alpha + y = t^{2/3} + y$$:
\[
y' = (t^{2/3} + y)^3 - t^2 - \frac{2}{3}t^{-1/3}
\]
\[
= t^2 + 3t^{4/3}y + 3t^{2/3}y^2 + y^3 - t^2 - \frac{2}{3}t^{-1/3}
\]
\[
= 3t^{4/3}y + 3t^{2/3}y^2 + y^3 - \frac{2}{3}t^{-1/3}
\]

For $$y$$ small, $$y' \approx 3t^{4/3}y - \frac{2}{3}t^{-1/3}$$.

The term $$3t^{4/3}y$$ causes growth if $$y > 0$$, decay if $$y < 0$$, unless $$y \sim t^{-5/3}$$ to balance.

**Step 5: Find exact exceptional solution**
Look for solution of form $$x(t) = t^{2/3} + at^p$$ with $$p < 2/3$$.

Balancing terms: Need $$3t^{4/3}(at^p) \sim \frac{2}{3}t^{-1/3}$$ for cancellation.

So $$t^{4/3 + p} \sim t^{-1/3} \Rightarrow p = -5/3$$.

Thus $$x(t) = t^{2/3} + at^{-5/3} + \cdots$$.

Substituting: For $$y = at^{-5/3}$$, $$y' = -\frac{5a}{3}t^{-8/3}$$.

From DE: $$3t^{4/3}(at^{-5/3}) = 3at^{-1/3}$$.

Balance: $$-\frac{5a}{3}t^{-8/3} \approx 3at^{-1/3} - \frac{2}{3}t^{-1/3}$$.

For large $$t$$, $$t^{-8/3} \ll t^{-1/3}$$, so dominant balance:
\[
3at^{-1/3} \approx \frac{2}{3}t^{-1/3} \Rightarrow a \approx \frac{2}{9}
\]

So exceptional solution: $$x_e(t) = t^{2/3} + \frac{2}{9}t^{-5/3} + \cdots$$.

**Step 6: Antifunnel narrowing**
Consider two solutions on either side of $$x_e(t)$$.

Let $$u(t) = x(t) - x_e(t)$$.
Then $$u' \approx 3t^{4/3}u$$ (linearization).

So $$u(t) \approx u(t_0) \exp\left(\int_{t_0}^t 3s^{4/3} ds\right) = u(t_0) \exp\left(\frac{9}{7}(t^{7/3} - t_0^{7/3})\right)$$.

This grows super-exponentially! So solutions diverge rapidly from $$x_e(t)$$.

Thus region between any two solutions containing $$x_e(t)$$ widens, not narrows.

Wait, perhaps "narrowing" refers to width relative to $$t^{2/3}$$?

Relative width: $$\frac{u(t)}{t^{2/3}} \sim \frac{\exp(\frac{9}{7}t^{7/3})}{t^{2/3}} \to \infty$$.

So not narrowing in relative sense either.

**Step 7: Reinterpretation**
Perhaps they mean: Consider curves $$x = Ct^{2/3}$$ for different $$C$$.

For $$x = Ct^{2/3}$$: $$x' = \frac{2C}{3}t^{-1/3}$$.

RHS: $$x^3 - t^2 = C^3 t^2 - t^2 = (C^3 - 1)t^2$$.

For large $$t$$, $$(C^3-1)t^2$$ dominates over $$\frac{2C}{3}t^{-1/3}$$.

If $$C^3 > 1$$: RHS positive and huge, so solution increases faster than $$Ct^{2/3}$$.

If $$C^3 < 1$$: RHS negative and huge, so solution decreases faster than $$Ct^{2/3}$$.

Only $$C=1$$ gives balance. So solutions with $$x > t^{2/3}$$ blow up, with $$x < t^{2/3}$$ go to $$-\infty$$.

Thus $$x = t^{2/3}$$ separates behaviors: an antifunnel in the sense that solutions diverge from it.

### (b) Solutions above exceptional solution have vertical asymptotes

**Step 1: Exceptional solution behavior**
From (a), exceptional solution $$x_e(t) \sim t^{2/3}$$ for large $$t$$.

**Step 2: Compare with $$x = t^{2/3}$$**
For $$x > t^{2/3}$$, we have $$x^3 > t^2$$, so $$x' = x^3 - t^2 > 0$$.

Thus if $$x(t_0) > t_0^{2/3}$$ at some $$t_0$$, then $$x'(t_0) > 0$$.

**Step 3: Show blow-up**
Consider $$x(t) > t^{2/3} + \delta$$ for some $$\delta > 0$$.

Then $$x' = x^3 - t^2 > (t^{2/3} + \delta)^3 - t^2 = 3t^{4/3}\delta + 3t^{2/3}\delta^2 + \delta^3$$.
=
For large $$t$$, $$x' > 2t^{4/3}\delta$$.

Thus $$x(t) > x(t_0) + 2\delta \int_{t_0}^t s^{4/3} ds = x(t_0) + \frac{6\delta}{7}(t^{7/3} - t_0^{7/3})$$.

So $$x(t) \to \infty$$ as $$t \to \infty$$.

**Step 4: Finite-time blow-up?**
Actually, $$x' > x^3 - t^2$$. For large $$x$$, $$x' > \frac{1}{2}x^3$$ (since $$t^2 \ll x^3$$).

Solve $$y' = \frac{1}{2}y^3$$: $$y(t) = \frac{1}{\sqrt{C-t}}$$.

This blows up in finite time.

Thus $$x(t)$$ blows up in finite time: has vertical asymptote. $$\blacksquare$$

### Summary

1. **(a)** $$x = t^{2/3}$$ is an approximate solution; solutions diverge from it (antifunnel).
2. **(b)** Solutions above the exceptional solution satisfy $$x' > \frac{1}{2}x^3$$ for large $$x$$, leading to finite-time blow-up.

\[
\boxed{\text{(a) Solutions diverge from } x=t^{2/3} \text{ (antifunnel); (b) Solutions above exceptional one blow up in finite time}}
\]

Loading solution...

Problem #12