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Hw1 p3

Solution 2

I got tired of writing so much text with hand and figured that typing would be faster. First solution was trivial solution, written by hand.

Assuming \(y \neq 0\), the variables can be separated: $$ \frac{dy}{dt} = ty^a \Rightarrow y^{-a} dy = t \, dt $$ $$ \int y^{-a} dy = \int t \, dt $$ $$ \frac{y^{-a+1}}{-a+1} = \frac{t^2}{2} + C $$ Since \(a < 1\), there is \(1-a > 0\): $$ \frac{y^{1-a}}{1-a} = \frac{t^2}{2} + C $$ Appply \(y(0)=0\) $$ \frac{0^{1-a}}{1-a} = \frac{0^2}{2} + C \Rightarrow 0=C $$ $$ \frac{y^{1-a}}{1-a} = \frac{t^2}{2} $$ $$ y^{1-a} = (1-a)\frac{t^2}{2} $$ $$ y_2(t) = \left( \frac{1-a}{2}t^2 \right)^{\frac{1}{1-a}} $$

initial condition: \(y_2(0) = \left( \frac{1-a}{2} \cdot 0^2 \right)^{\frac{1}{1-a}} = 0\) is satisfied. This solution \(y_2(t)\) is non-trivial for \(t \neq 0\).

Since there are 2 distinct solutions the solution to the IVP is not unique.

Explanation

Use Existence-Uniqueness theorem.

Check the conditions for the problem.

\(f(t,y) = ty^a\) \((t_0, y_0) = (0,0)\)

  1. Continuity: The function \(f(t,y) = ty^a\) is continuous at \((0,0)\) for \(a>0\). If \(a \le 0\), it is not continuous at \(y=0\). The problem is stated for \(a < 1\).

  2. Lipschitz Condition:

$$ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(ty^a) = aty^{a-1} = \frac{at}{y^{1-a}} $$ The theorem needs this partial derivative to be bounded in a rectangle \(R\) around the initial point \((0,0)\). Because thhe problem says \(a < 1\), the exponent \(1-a\) is positive. Consider \(\frac{\partial f}{\partial y}\) as \(y \to 0\): $$ \lim_{y\to 0} \left| \frac{\partial f}{\partial y} \right| = \lim_{y\to 0} \left| \frac{at}{y^{1-a}} \right| = \infty \quad (\text{for any } t \neq 0) $$ The partial derivative \(\frac{\partial f}{\partial y}\) is unbounded in any rectangle containing the line \(y=0\). So the function \(f(t,y)\) is not Lipschitz continuous with respect to \(y\) in any neighborhood of the initial point \((0,0)\).

The hypotheses of the Existence-Uniqueness Theorem are not satisfied. The Lipschitz condition fails at the initial point. The theorem does not guarantee a unique solution. The existence of two distinct solutions does not violate the theorem because the theorem's conditions were never met.