Consider the following initial value problem (IVP), \(\rm\frac{d u}{d t} = t^2 u^{\frac{1}{5}}\), u(0) = 0. Which of the following statements are correct?
Initial Value Problem Analysis
The given initial value problem (IVP) is:
\(\frac{d u}{d t} = t^2 u^{\frac{1}{5}}\), with the initial condition \(u(0) = 0\).
The function on the right-hand side of the differential equation is \(g(t, u) = t^2 u^{\frac{1}{5}}\).
Lipschitz Condition Examination
To check if the Lipschitz condition is satisfied with respect to \(u\) in a neighbourhood of \(u=0\), we need to see if there exists a constant \(L > 0\) such that \(|g(t, u_1) - g(t, u_2)| \le L|u_1 - u_2|\) for all \(u_1, u_2\) in a neighbourhood of 0 and for \(t\) in some interval around 0.
Let's consider \(u_1 = u\) and \(u_2 = 0\), where \(u\) is close to 0. We need to check if \(|g(t, u) - g(t, 0)| \le L|u - 0|\).
\(|g(t, u) - g(t, 0)| = |t^2 u^{\frac{1}{5}} - t^2 \cdot 0^{\frac{1}{5}}| = |t^2 u^{\frac{1}{5}}| = |t^2| |u|^{\frac{1}{5}}\).
We need to see if \(|t^2| |u|^{\frac{1}{5}} \le L|u|\) for \(u \ne 0\) near 0.
This is equivalent to checking if \(\frac{|t^2| |u|^{\frac{1}{5}}}{|u|} \le L\), which simplifies to \(|t^2| |u|^{\frac{1}{5} - 1} = |t^2| |u|^{-\frac{4}{5}} \le L\).
For any fixed \(t \ne 0\), as \(u \to 0\), \(|u|^{-\frac{4}{5}} \to \infty\). Therefore, \(\frac{|t^2|}{ |u|^{\frac{4}{5}}}\) is unbounded in any neighbourhood of \(u=0\) (for \(t \ne 0\)).
Since \(\frac{\partial g}{\partial u} = \frac{\partial}{\partial u} (t^2 u^{\frac{1}{5}}) = t^2 \cdot \frac{1}{5} u^{\frac{1}{5}-1} = \frac{t^2}{5} u^{-\frac{4}{5}} = \frac{t^2}{5 u^{\frac{4}{5}}}\), which is unbounded as \(u \to 0\) for \(t \ne 0\), the function \(g(t, u)\) does not satisfy the Lipschitz condition with respect to \(u\) in any neighbourhood of \(u = 0\).
This supports the statement that the function \(g(t, u) = t^2 u^{\frac{1}{5}}\) does not satisfy the Lipschitz's condition with respect to \(u\) in any neighbourhood of \(u = 0\).
Existence and Uniqueness of Solutions
The Picard-Lindelöf theorem states that if \(g(t, u)\) is continuous and satisfies the Lipschitz condition with respect to \(u\) in a region around the initial point, then a unique solution exists. In our case, \(g(t, u) = t^2 u^{\frac{1}{5}}\) is continuous at \((0, 0)\), but it does not satisfy the Lipschitz condition near \(u=0\).
Failure of the Lipschitz condition means the uniqueness of the solution is not guaranteed by this theorem. Let's try to find solutions for the IVP by separating variables:
\(\frac{d u}{d t} = t^2 u^{\frac{1}{5}}\)
If \(u \ne 0\), we can write:
\(\frac{d u}{u^{\frac{1}{5}}} = t^2 d t\)
Integrating both sides:
\(\int u^{-\frac{1}{5}} d u = \int t^2 d t\)
\(\frac{u^{-\frac{1}{5} + 1}}{-\frac{1}{5} + 1} = \frac{t^{2+1}}{2+1} + C\)
\(\frac{u^{\frac{4}{5}}}{\frac{4}{5}} = \frac{t^3}{3} + C\)
\(\frac{5}{4} u^{\frac{4}{5}} = \frac{t^3}{3} + C\)
Applying the initial condition \(u(0) = 0\):
\(\frac{5}{4} (0)^{\frac{4}{5}} = \frac{0^3}{3} + C\)
\(0 = 0 + C \implies C = 0\)
So, \(\frac{5}{4} u^{\frac{4}{5}} = \frac{t^3}{3}\)
\(u^{\frac{4}{5}} = \frac{4}{15} t^3\)
For \(t \ge 0\), we can write \(u(t) = \left(\frac{4}{15} t^3\right)^{\frac{5}{4}}\).
Let's check if \(u(t) = 0\) for all \(t\) is a solution:
LHS: \(\frac{d u}{d t} = \frac{d(0)}{d t} = 0\)
RHS: \(t^2 u^{\frac{1}{5}} = t^2 (0)^{\frac{1}{5}} = t^2 \cdot 0 = 0\)
Since LHS = RHS, \(u(t) = 0\) is also a solution to the differential equation, and it satisfies the initial condition \(u(0) = 0\).
We have found two distinct solutions satisfying \(u(0) = 0\):
- \(u_1(t) = 0\) for all \(t\)
- \(u_2(t) = \left(\frac{4}{15} t^3\right)^{\frac{5}{4}}\) for \(t \ge 0\) (and potentially extended to \(t < 0\) if we consider complex values or restrict the domain, but for real solutions, this form is valid for \(t \ge 0\)). Note that \(u_2(0) = (\frac{4}{15} \cdot 0^3)^{5/4} = 0\).
Since we found more than one solution satisfying the initial condition, there exist more than one solution for the IVP.
Evaluating Statements
Based on our analysis:
- Statement 1: The function \(g(t, u) = t^2 u^{\frac{1}{5}}\) does not satisfy the Lipschitz's condition with respect to \(u\) in any neighbourhood of \(u = 0\). This is correct.
- Statement 2: There is no solution for the IVP. This is incorrect, as we found multiple solutions.
- Statement 3: There exist more than one solution for the IVP. This is correct, as we found \(u(t)=0\) and \(u(t) = (\frac{4}{15} t^3)^{5/4}\) satisfy the initial condition.
- Statement 4: The function \(g(t, u) = t^2 u^{\frac{1}{5}}\) satisfies the Lipschitz's condition... and hence there exists a unique solution... This is incorrect because the Lipschitz condition is not satisfied near \(u=0\), and there are multiple solutions.
The correct statements are that the Lipschitz condition fails near \(u=0\), and there exists more than one solution for the IVP.
Important Questions from Initial Value Problem
Initial value problem, \(\rm x \frac{d y}{d x}=y\), y(0) = 0, x > 0
Consider the eigenvalue problem
((1 + x4)y')' + λy = 0, x ∈ (0, 1),
y(0) = 0, y(1) + 2y'(1) = 0.
Then which of the following statements are true?
Consider the following two initial value ODEs
(A) \(\frac{dx}{dt}=x^3,x(0)=1;\)
(B) \(\frac{dx}{dt}=x\sin x^2,x(0)=2.\)
Related to these ODEs, we make the following assertions.
I. The solution to (A) blows up in finite time.
II. The solution to (B) blows up in finite time.
Which of the following statements is true?
Let y0 > 0, z0 > 0 and α > 1.
Consider the following two differential equations:
\(\begin{aligned} &(*)\left\{\begin{array}{l} \frac{d y}{d t}=y^\alpha \quad \text { for } t>0, \\ y(0)=y_0 \end{array}\right. \\ &(* *)\left\{\begin{array}{l} \frac{d z}{d t}=-z^\alpha \quad \text { for } t>0, \\ z(0)=z_0 \end{array}\right. \end{aligned}\)
We say that the solution to a differential equation exists globally if it exists for all t > 0.Which of the following statements is true?
Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem
ẋ = f(t, x), x(t0) = x0
for (t0, x0) ∈ ℝ2. Suppose that J(t0, x0) represents the maximal interval of existence for the initial value problem. Which of the following statements is true?
