Eigenvalue Problem Analysis

The given differential equation is $((1 + x^4)y')' + \lambda y = 0$ for $x \in (0, 1)$. This is a second-order linear homogeneous differential equation. The boundary conditions are $y(0) = 0$ and $y(1) + 2y'(1) = 0$. This problem is a specific type of boundary value problem known as a Sturm-Liouville eigenvalue problem.

The standard form of a Sturm-Liouville equation is $(p(x)y')' + q(x)y + \lambda w(x)y = 0$. Comparing the given equation $((1 + x^4)y')' + \lambda y = 0$ with the standard form, we can identify the coefficients:

• $p(x) = 1 + x^4$
• $q(x) = 0$
• $w(x) = 1$

For $x \in (0, 1)$, $p(x) = 1 + x^4 > 0$ and $w(x) = 1 > 0$. The boundary conditions are of the standard type for Sturm-Liouville problems.

Energy Integral Method

To determine the nature of the eigenvalues $\lambda$, we can use the energy integral method. Multiply the differential equation by $y(x)$ and integrate over the interval $(0, 1)$:

$\int_0^1 ((1 + x^4)y')' y \, dx + \lambda \int_0^1 y^2 \, dx = 0$

Now, we integrate the first term by parts. Recall the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Let $u = y$ and $dv = ((1 + x^4)y')' \, dx$. Then $du = y' \, dx$ and $v = (1 + x^4)y'$.

$\int_0^1 ((1 + x^4)y')' y \, dx = [(1 + x^4)y' y]_0^1 - \int_0^1 (1 + x^4)y' y' \, dx$

Evaluate the boundary term $[(1 + x^4)y' y]_0^1$:

$[(1 + x^4)y' y]_0^1 = (1+1^4)y'(1)y(1) - (1+0^4)y'(0)y(0)$

$= 2y'(1)y(1) - 1 \cdot y'(0) \cdot y(0)$

Now, apply the given boundary conditions:

• At $x=0$, $y(0) = 0$. So, the term at $x=0$ is $1 \cdot y'(0) \cdot 0 = 0$.
• At $x=1$, $y(1) + 2y'(1) = 0$. This implies $y(1) = -2y'(1)$.

Substitute $y(1) = -2y'(1)$ into the term at $x=1$:

$2y'(1)y(1) = 2y'(1)(-2y'(1)) = -4(y'(1))^2$

So, the boundary term evaluates to $-4(y'(1))^2 - 0 = -4(y'(1))^2$.

The integrated equation becomes:

$-4(y'(1))^2 - \int_0^1 (1 + x^4)(y')^2 \, dx + \lambda \int_0^1 y^2 \, dx = 0$

Rearranging this equation to solve for $\lambda$:

$\lambda \int_0^1 y^2 \, dx = 4(y'(1))^2 + \int_0^1 (1 + x^4)(y')^2 \, dx$

Nature of Eigenvalues

Eigenvalues $\lambda$ correspond to non-trivial solutions $y(x)$. For a non-trivial solution, $\int_0^1 y^2 \, dx > 0$.

Let's examine the right-hand side of the equation:

$4(y'(1))^2$: This term is always non-negative, as it is a square multiplied by a positive constant.

$\int_0^1 (1 + x^4)(y')^2 \, dx$: On the interval $(0, 1)$, $1 + x^4 > 0$ and $(y')^2 \ge 0$. Therefore, the integrand $(1 + x^4)(y')^2 \ge 0$ on $(0, 1)$. The integral of a non-negative function is non-negative. So, $\int_0^1 (1 + x^4)(y')^2 \, dx \ge 0$.

Adding these two non-negative terms, the right-hand side is:

$4(y'(1))^2 + \int_0^1 (1 + x^4)(y')^2 \, dx \ge 0$

Since $\lambda \int_0^1 y^2 \, dx \ge 0$ and $\int_0^1 y^2 \, dx > 0$, it must be that $\lambda \ge 0$. This means the eigenvalues can only be non-negative (either positive or zero).

Checking for Zero Eigenvalue

We need to check if $\lambda = 0$ is a possible eigenvalue. If $\lambda = 0$, the differential equation becomes:

$((1 + x^4)y')' = 0$

Integrating once with respect to $x$:

$(1 + x^4)y' = C_1$, where $C_1$ is the integration constant.

So, $y' = \frac{C_1}{1 + x^4}$.

Integrating again:

$y(x) = C_1 \int \frac{dx}{1 + x^4} + C_2$, where $C_2$ is another integration constant.

Now, apply the boundary conditions:

Boundary condition 1: $y(0) = 0$

$0 = C_1 \int_0^0 \frac{dt}{1 + t^4} + C_2$

$0 = C_1 \cdot 0 + C_2$, so $C_2 = 0$.

The solution form is now $y(x) = C_1 \int_0^x \frac{dt}{1 + t^4}$.

Boundary condition 2: $y(1) + 2y'(1) = 0$

From $y(x)$, we have $y(1) = C_1 \int_0^1 \frac{dt}{1 + t^4}$.

From $y'(x) = \frac{C_1}{1 + x^4}$, we have $y'(1) = \frac{C_1}{1 + 1^4} = \frac{C_1}{2}$.

Substitute these into the second boundary condition:

$C_1 \int_0^1 \frac{dt}{1 + t^4} + 2 \left(\frac{C_1}{2}\right) = 0$

$C_1 \int_0^1 \frac{dt}{1 + t^4} + C_1 = 0$

$C_1 \left( \int_0^1 \frac{dt}{1 + t^4} + 1 \right) = 0$

Consider the term in the parenthesis, $\int_0^1 \frac{dt}{1 + t^4} + 1$. The integrand $\frac{1}{1 + t^4}$ is positive for $t \in [0, 1]$. Therefore, the integral $\int_0^1 \frac{dt}{1 + t^4}$ is a positive value. For example, since $1+t^4 \le 2$ for $t \in [0,1]$, $\frac{1}{1+t^4} \ge \frac{1}{2}$, so $\int_0^1 \frac{dt}{1+t^4} \ge \int_0^1 \frac{1}{2} dt = \frac{1}{2}$. Thus, $\int_0^1 \frac{dt}{1 + t^4} + 1$ is strictly greater than $1$. It is a non-zero value.

For the equation $C_1 \left( \int_0^1 \frac{dt}{1 + t^4} + 1 \right) = 0$ to hold, given that the parenthesis term is non-zero, we must have $C_1 = 0$.

If $C_1 = 0$, then $y(x) = 0 \cdot \int_0^x \frac{dt}{1 + t^4} = 0$. This is the trivial solution.

Eigenvalues are defined as values of $\lambda$ for which non-trivial solutions exist. Since $\lambda = 0$ only yields the trivial solution, $\lambda = 0$ is not an eigenvalue for this problem.

Conclusion

From the energy integral method, we found that all eigenvalues must satisfy $\lambda \ge 0$. By checking $\lambda=0$, we confirmed that it is not an eigenvalue. Therefore, all eigenvalues must be strictly positive.

Based on this analysis, the true statement is that all the eigenvalues are positive.