To solve the given integral equation using the method of degenerate kernels, let’s proceed step by step:

Problem:

$u(x) = x + \lambda \int_{0}^{2\pi} |\pi - t| \sin(x) u(t) \, dt.$

Here, $|\pi - t| \sin(x)$ is the kernel of the integral operator.

Step 1: Degenerate Kernel Representation

The kernel $|\pi - t| \sin(x)$ can be expressed as a product of functions of $x$ and $t$: $|\pi - t| \sin(x) = \phi_1(x) \psi_1(t),$ where $\phi_1(x) = \sin(x), \quad \psi_1(t) = |\pi - t|.$

Step 2: Assumption for $u(t)$

Assume $u(t)$ can be expressed as a series expansion in terms of the degenerate kernel's basis functions: $u(t) = a_0 \psi_1(t).$

Substitute this into the integral equation.

Step 3: Substituting $u(t)$ into the Equation

The equation becomes: $u(x) = x + \lambda \int_{0}^{2\pi} |\pi - t| \sin(x) \big( a_0 |\pi - t| \big) \, dt.$

Simplify: $u(x) = x + \lambda a_0 \sin(x) \int_{0}^{2\pi} |\pi - t|^2 \, dt.$

Step 4: Evaluate the Integral $\int_{0}^{2\pi} |\pi - t|^2 \, dt$

Split the integral at $t = \pi$ to handle the modulus: $\int_{0}^{2\pi} |\pi - t|^2 \, dt = \int_{0}^{\pi} (\pi - t)^2 \, dt + \int_{\pi}^{2\pi} (t - \pi)^2 \, dt.$

For $\int_{0}^{\pi} (\pi - t)^2 \, dt$: Let $u = \pi - t$, then $du = -dt$: $\int_{0}^{\pi} (\pi - t)^2 \, dt = \int_{\pi}^{0} u^2 (-du) = \int_{0}^{\pi} u^2 \, du = \left[ \frac{u^3}{3} \right]_{0}^{\pi} = \frac{\pi^3}{3}.$

For $\int_{\pi}^{2\pi} (t - \pi)^2 \, dt$: Let $u = t - \pi$, then $du = dt$: $\int_{\pi}^{2\pi} (t - \pi)^2 \, dt = \int_{0}^{\pi} u^2 \, du = \left[ \frac{u^3}{3} \right]_{0}^{\pi} = \frac{\pi^3}{3}.$

So: $\int_{0}^{2\pi} |\pi - t|^2 \, dt = \frac{\pi^3}{3} + \frac{\pi^3}{3} = \frac{2\pi^3}{3}.$

Step 5: Simplify the Equation

Substitute the integral back into the equation: $u(x) = x + \lambda a_0 \sin(x) \cdot \frac{2\pi^3}{3}.$

Step 6: Match the Coefficients

From the assumed form $u(x) = a_0 \phi_1(x)$: $u(x) = a_0 \sin(x).$

Equating coefficients: $a_0 \sin(x) = x + \lambda a_0 \sin(x) \cdot \frac{2\pi^3}{3}.$

This implies: $a_0 = \frac{x}{1 - \lambda \cdot \frac{2\pi^3}{3}}.$

Final Solution:

The solution is: $u(x) = x + \lambda \int_{0}^{2\pi} |\pi - t| \sin(x) u(t) \, dt.$

If you need further clarification or want to expand specific parts of this derivation, let me know!

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5 Related Questions:

1. How is the degenerate kernel method applied in other integral equations?
2. What are the limitations of the degenerate kernel method?
3. Can this equation be solved numerically for specific values of $\lambda$?
4. What changes if the kernel is not separable into $\phi(x)\psi(t)$?
5. How do physical applications use degenerate kernels in integral equations?

One Tip:

Always check if the kernel can be simplified or approximated as a sum of separable terms to efficiently use the degenerate kernel method.