The following two-point boundary value problem \(\left\{\begin{aligned} y^{\prime \prime}(x)+λ y(x) &=0 \text { for } x ∈(0, \pi) \\ y(0) &=0 \\ y(\pi) &=0 \end{aligned}\right.\) has a trivial solution y = 0. It also has a non-trivial solution for
Boundary Value Problem Analysis
We are given a two-point boundary value problem:
\[ \left\{ \begin{aligned} y^{\prime \prime}(x)+\lambda y(x) &=0 & \text { for } x \in(0, \pi) \\ y(0) &=0 \\ y(\pi) &=0 \end{aligned} \right. \]
We need to find the values of the parameter \(\lambda\) for which this problem has a non-trivial solution (a solution other than \(y(x) = 0\) for all \(x\)). This type of problem is an eigenvalue problem, where the values of \(\lambda\) for which non-trivial solutions exist are called eigenvalues.
The differential equation is \(y^{\prime \prime}(x)+\lambda y(x) = 0\). The characteristic equation is \(r^2 + \lambda = 0\), which gives \(r^2 = -\lambda\).
Lambda Cases for Non-Trivial Solutions
We consider three cases based on the value of \(\lambda\):
Case 1: \(\lambda < 0\)
Let \(\lambda = -\mu^2\) where \(\mu > 0\). The characteristic equation becomes \(r^2 - \mu^2 = 0\), so \(r = \pm \mu\). The general solution is \(y(x) = c_1 e^{\mu x} + c_2 e^{-\mu x}\).
Applying the boundary conditions:
- \(y(0) = 0 \Rightarrow c_1 e^0 + c_2 e^0 = 0 \Rightarrow c_1 + c_2 = 0 \Rightarrow c_2 = -c_1\).
- So, \(y(x) = c_1 e^{\mu x} - c_1 e^{-\mu x} = c_1 (e^{\mu x} - e^{-\mu x}) = 2 c_1 \sinh(\mu x)\).
- \(y(\pi) = 0 \Rightarrow 2 c_1 \sinh(\mu \pi) = 0\). Since \(\mu > 0\) and \(\pi > 0\), \(\mu \pi > 0\), and \(\sinh(z) = 0\) only when \(z = 0\). Thus, \(\sinh(\mu \pi) \neq 0\).
- Therefore, we must have \(c_1 = 0\). Since \(c_2 = -c_1\), \(c_2 = 0\) as well.
This gives only the trivial solution \(y(x) = 0\). So, there are no non-trivial solutions when \(\lambda < 0\).
Case 2: \(\lambda = 0\)
The differential equation becomes \(y^{\prime \prime}(x) = 0\). The general solution is \(y(x) = Ax + B\).
Applying the boundary conditions:
- \(y(0) = 0 \Rightarrow A(0) + B = 0 \Rightarrow B = 0\).
- So, \(y(x) = Ax\).
- \(y(\pi) = 0 \Rightarrow A\pi = 0 \Rightarrow A = 0\).
This gives only the trivial solution \(y(x) = 0\). So, there are no non-trivial solutions when \(\lambda = 0\).
Case 3: \(\lambda > 0\)
Let \(\lambda = \omega^2\) where \(\omega > 0\). The characteristic equation becomes \(r^2 + \omega^2 = 0\), so \(r = \pm i\omega\). The general solution is \(y(x) = c_1 \cos(\omega x) + c_2 \sin(\omega x)\).
Applying the boundary conditions:
- \(y(0) = 0 \Rightarrow c_1 \cos(0) + c_2 \sin(0) = 0 \Rightarrow c_1(1) + c_2(0) = 0 \Rightarrow c_1 = 0\).
- So, \(y(x) = c_2 \sin(\omega x)\).
- \(y(\pi) = 0 \Rightarrow c_2 \sin(\omega \pi) = 0\).
For a non-trivial solution, we require \(y(x)\) not to be identically zero, which means \(c_2 \neq 0\). If \(c_2 \neq 0\), then we must have \(\sin(\omega \pi) = 0\).
The sine function is zero at integer multiples of \(\pi\). So, \(\omega \pi = n \pi\) for some integer \(n\). Since \(\omega > 0\), we require \(n\) to be a positive integer, i.e., \(n = 1, 2, 3, \ldots\). Thus, \(\omega = n\) for \(n \in \mathbb{N}\).
Substituting back \(\lambda = \omega^2\), we find that non-trivial solutions exist for \(\lambda = n^2\) where \(n \in \mathbb{N}\) (positive integers). The corresponding non-trivial solutions are \(y_n(x) = c_n \sin(nx)\) for some non-zero constant \(c_n\).
Eigenvalues and Options
The values of \(\lambda\) for which non-trivial solutions exist are the eigenvalues, which are \(\lambda = 1^2, 2^2, 3^2, \ldots\), or \(\lambda = 1, 4, 9, \ldots\). These are \(\lambda = n^2\) for \(n \in \mathbb{N}\).
Let's look at the given options:
- Option 1: no values of \(\lambda \in \mathbb{R}\). This is incorrect, as we found non-trivial solutions for \(\lambda = n^2\) with \(n \in \mathbb{N}\).
- Option 2: \(\lambda = 1\). This corresponds to the eigenvalue when \(n=1\). A non-trivial solution exists for \(\lambda = 1\). This option is correct.
- Option 3: \(\lambda = n\)2 for all \(n \in \mathbb{N}\) , \(n > 1\). This corresponds to the eigenvalues when \(n=2, 3, 4, \ldots\). Non-trivial solutions exist for these values of \(\lambda\). This option is correct.
- Option 4: \(\lambda \le 0\). As shown in Case 1 and Case 2, for \(\lambda \le 0\), only the trivial solution exists. This option is incorrect.
The boundary value problem has a non-trivial solution for \(\lambda = n^2\), where \(n\) is a positive integer (\(n \in \mathbb{N}\)). This includes \(\lambda = 1\) (when \(n=1\)) and \(\lambda = n^2\) for \(n > 1\).
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?
