Understanding the Boundary Value Problem

The given problem is a second-order linear boundary value problem defined by the differential equation:

$$ \ddot{u}+2\dot{u}+a(t)u=\lambda u $$

with boundary conditions:

$$ \dot{u}(0)=0 \quad \text{and} \quad \dot{u}(1)=0 $$

Here, \(u\) is a positive eigenfunction, \(\lambda\) is the corresponding eigenvalue, and \(a(t)\) is a continuous function such that \(a(t) > 1\) for \(t \in [0,1]\).

Analyzing the Integral Identity

Let's examine the third statement, which proposes an integral identity:

$$ \int_0^1 {{{\left( \dot{u} \right)}^2}dt = 2\int_0^1 {u\,\dot{u}\,dt} + \int_0^1 {\left( {a\left( t \right) - \lambda } \right){u^2}dt} } $$

We can start from the given differential equation:

$$ \ddot{u} = (\lambda - a(t))u - 2\dot{u} $$

Multiply the equation by \(u\) and integrate from 0 to 1:

$$ \int_0^1 \ddot{u} u \, dt = \int_0^1 (\lambda - a(t)) u^2 \, dt - 2\int_0^1 u \dot{u} \, dt $$

Now, let's use integration by parts on the left side. Recall the integration by parts formula: \(\int f \, dg = fg - \int g \, df\). Let \(f = u\) and \(dg = \ddot{u} \, dt\). Then \(df = \dot{u} \, dt\) and \(g = \dot{u}\).

$$ \int_0^1 \ddot{u} u \, dt = \left[ \dot{u} u \right]_0^1 - \int_0^1 \dot{u} \dot{u} \, dt $$

Applying the boundary conditions \(\dot{u}(0)=0\) and \(\dot{u}(1)=0\):

$$ \left[ \dot{u} u \right]_0^1 = \dot{u}(1)u(1) - \dot{u}(0)u(0) = (0)u(1) - (0)u(0) = 0 $$

So, the left side simplifies to:

$$ \int_0^1 \ddot{u} u \, dt = - \int_0^1 (\dot{u})^2 \, dt $$

Substitute this back into the integrated differential equation:

$$ - \int_0^1 (\dot{u})^2 \, dt = \int_0^1 (\lambda - a(t)) u^2 \, dt - 2\int_0^1 u \dot{u} \, dt $$

Rearranging the terms to match the proposed identity:

$$ \int_0^1 (\dot{u})^2 \, dt = - \int_0^1 (\lambda - a(t)) u^2 \, dt + 2\int_0^1 u \dot{u} \, dt $$

$$ \int_0^1 (\dot{u})^2 \, dt = \int_0^1 (a(t) - \lambda) u^2 \, dt + 2\int_0^1 u \dot{u} \, dt $$

This matches the third statement exactly. Therefore, the integral identity is true based on the differential equation and boundary conditions.

Analyzing the Eigenvalue \(\lambda\)

Let's analyze the possible values of the eigenvalue \(\lambda\). Multiply the original differential equation by \(u\) and integrate from 0 to 1:

$$ \int_0^1 (\ddot{u}+2\dot{u}+a(t)u)u \, dt = \int_0^1 \lambda u^2 \, dt $$

$$ \int_0^1 \ddot{u}u \, dt + 2\int_0^1 \dot{u}u \, dt + \int_0^1 a(t)u^2 \, dt = \lambda \int_0^1 u^2 \, dt $$

We already know from the previous calculation that \(\int_0^1 \ddot{u}u \, dt = - \int_0^1 (\dot{u})^2 \, dt\) using integration by parts and the boundary conditions.

The term \(2\int_0^1 \dot{u}u \, dt\) can also be integrated by parts. Let \(f = u\) and \(dg = \dot{u} \, dt\). Then \(df = \dot{u} \, dt\) and \(g = u\). This doesn't seem helpful here. Let's re-examine the equation obtained after multiplying by \(u\) and integrating:

$$ \int_0^1 \ddot{u}u \, dt + 2\int_0^1 (\dot{u})^2 \, dt + \int_0^1 a(t)u^2 \, dt = \lambda \int_0^1 u^2 \, dt $$

Wait, the term \(2\dot{u}u\) cannot be directly replaced by \(2(\dot{u})^2\). Let's go back to the equation after integrating the differential equation multiplied by \(u\):

$$ \int_0^1 \ddot{u}u \, dt + \int_0^1 2\dot{u}u \, dt + \int_0^1 a(t)u^2 \, dt = \lambda \int_0^1 u^2 \, dt $$

Using \(\int_0^1 \ddot{u}u \, dt = - \int_0^1 (\dot{u})^2 \, dt\):

$$ - \int_0^1 (\dot{u})^2 \, dt + 2\int_0^1 \dot{u}u \, dt + \int_0^1 a(t)u^2 \, dt = \lambda \int_0^1 u^2 \, dt $$

This doesn't directly lead to the value of \(\lambda\).

Let's consider the equation again: \(\ddot{u}+2\dot{u}+(a(t)-\lambda)u=0\). This is a Sturm-Liouville type problem, but not in the standard form. However, we can use properties of eigenvalues and eigenfunctions.

Let's use the equation derived from multiplying by \(u\) and integrating, and the integration by parts result for \(\int \ddot{u}u\):

$$ - \int_0^1 (\dot{u})^2 \, dt + 2\int_0^1 \dot{u}u \, dt + \int_0^1 a(t)u^2 \, dt = \lambda \int_0^1 u^2 \, dt $$

We know that \(\int_0^1 2\dot{u}u \, dt = \int_0^1 \frac{d}{dt}(u^2) \, dt = [u^2]_0^1 = u(1)^2 - u(0)^2\). This doesn't seem helpful without knowing the values of \(u(0)\) and \(u(1)\).

Let's reconsider the integral identity we verified:

$$ \int_0^1 (\dot{u})^2 \, dt = 2\int_0^1 u \dot{u} \, dt + \int_0^1 (a(t) - \lambda) u^2 \, dt $$

This identity is correct. Let's rearrange it to isolate \(\lambda\):

$$ \int_0^1 (a(t) - \lambda) u^2 \, dt = \int_0^1 (\dot{u})^2 \, dt - 2\int_0^1 u \dot{u} \, dt $$

$$ \int_0^1 a(t) u^2 \, dt - \lambda \int_0^1 u^2 \, dt = \int_0^1 (\dot{u})^2 \, dt - 2\int_0^1 u \dot{u} \, dt $$

$$ \lambda \int_0^1 u^2 \, dt = \int_0^1 a(t) u^2 \, dt - \int_0^1 (\dot{u})^2 \, dt + 2\int_0^1 u \dot{u} \, dt $$

Since \(u\) is a positive eigenfunction, \(u(t) > 0\) for \(t \in [0,1]\), and thus \(\int_0^1 u^2 \, dt > 0\).

Let's return to the energy-like integral derived earlier by multiplying the ODE by \(u\) and integrating by parts on \(\ddot{u}u\):

$$ - \int_0^1 (\dot{u})^2 \, dt + \int_0^1 2\dot{u}u \, dt + \int_0^1 a(t)u^2 \, dt = \lambda \int_0^1 u^2 \, dt $$

Consider the term \(\int_0^1 2\dot{u}u \, dt\). If we could integrate this out, it might help. However, let's multiply the original equation by \(e^{2t} u\) and integrate.

Another approach: Rewrite the equation as \(\ddot{u} + 2\dot{u} = (\lambda - a(t))u\). Let \(v = e^{2t}u\). Then \(\dot{v} = 2e^{2t}u + e^{2t}\dot{u} = e^{2t}(2u+\dot{u})\). \(\ddot{v} = 2e^{2t}(2u+\dot{u}) + e^{2t}(2\dot{u}+\ddot{u}) = e^{2t}(4u+4\dot{u}+\ddot{u})\). This doesn't simplify nicely.

Let's try multiplying the original equation by \(e^{2t}\) and integrate from 0 to 1:

$$ \int_0^1 e^{2t}\ddot{u} \, dt + \int_0^1 2e^{2t}\dot{u} \, dt + \int_0^1 e^{2t}a(t)u \, dt = \lambda \int_0^1 e^{2t}u \, dt $$

Notice that \(\frac{d}{dt}(e^{2t}\dot{u}) = 2e^{2t}\dot{u} + e^{2t}\ddot{u}\). So the first two terms are:

$$ \int_0^1 (e^{2t}\ddot{u} + 2e^{2t}\dot{u}) \, dt = \int_0^1 \frac{d}{dt}(e^{2t}\dot{u}) \, dt = [e^{2t}\dot{u}]_0^1 $$

Applying the boundary conditions \(\dot{u}(0)=0\) and \(\dot{u}(1)=0\):

$$ [e^{2t}\dot{u}]_0^1 = e^{2(1)}\dot{u}(1) - e^{2(0)}\dot{u}(0) = e^2(0) - e^0(0) = 0 $$

So the integrated equation becomes:

$$ 0 + \int_0^1 e^{2t}a(t)u \, dt = \lambda \int_0^1 e^{2t}u \, dt $$

$$ \int_0^1 e^{2t}a(t)u \, dt = \lambda \int_0^1 e^{2t}u \, dt $$

Since \(u\) is a positive eigenfunction, \(u(t) > 0\). Also \(e^{2t} > 0\) and \(a(t) > 1\). Therefore, \(e^{2t}a(t)u > e^{2t}u > 0\).

The integrals are: \(\int_0^1 e^{2t}a(t)u \, dt > \int_0^1 e^{2t}u \, dt > 0\). (Since \(u\) is not identically zero, the integrals are strictly positive).

From the equation \(\int_0^1 e^{2t}a(t)u \, dt = \lambda \int_0^1 e^{2t}u \, dt\), we can write:

$$ \lambda = \frac{\int_0^1 e^{2t}a(t)u \, dt}{\int_0^1 e^{2t}u \, dt} $$

Since \(a(t) > 1\), we have \(e^{2t}a(t)u > e^{2t}u\). Integrating this inequality from 0 to 1:

$$ \int_0^1 e^{2t}a(t)u \, dt > \int_0^1 e^{2t}u \, dt $$

Since both integrals are positive, their ratio must be greater than 1. Thus:

$$ \lambda = \frac{\int_0^1 e^{2t}a(t)u \, dt}{\int_0^1 e^{2t}u \, dt} > 1 $$

This shows that \(\lambda\) must be strictly greater than 1. Since \(\lambda > 1\), it is definitely true that \(\lambda > 0\). It also means that \(\lambda < 0\) and \(\lambda = 0\) are impossible.

Possible True Statements

Based on our analysis:

• The integral identity \( \int_0^1 {{{\left( \dot{u} \right)}^2}dt = 2\int_0^1 {u\,\dot{u}\,dt} + \int_0^1 {\left( {a\left( t \right) - \lambda } \right){u^2}dt} } \) is derived from the differential equation and boundary conditions, so it is possibly true.
• The eigenvalue \(\lambda\) must be strictly greater than 1, which implies \(\lambda > 0\). So, \(\lambda > 0\) is possibly true.
• Since \(\lambda > 1\), \(\lambda < 0\) is not possible.
• Since \(\lambda > 1\), \(\lambda = 0\) is not possible.

Therefore, the possibly true statements are the integral identity and \(\lambda > 0\).