Question

At an international "peace' conference a large number of delegates are seated around a circular table with each delegation sitting near its allies and diametrically opposite the delegation most bitterly opposed to it. The position of a delegate is denoted by \(\theta\), with \(0 \leq \theta \leq 2 \pi\). The fury \(f(\theta)\) felt by the delegate at \(\theta\) is the sum of his own natural hostility \(h(\theta)\) and the influences on him of each of the other delegates; a delegate at position \(\phi\) contributes an amount \(K(\theta-\phi) f(\phi)\). Thus $$ f(\theta)=h(\theta)+\int_{0}^{2 \pi} K(\theta-\phi) f(\phi) d \phi $$ Show that if \(K(\varphi)\) takes the form \(K(\varphi)=k_{0}+k_{1} \cos \psi\) then $$ f(\theta)=h(\theta)+p+q \cos \theta+r \sin \theta $$ and evaluate \(p, q\) and \(r\). A positive value for \(k_{1}\) implies that delegates tend to placate their opponents but upset their allies, whilst negative values imply that they calm their allies but infuriate their opponents. A walkout will occur if \(f(\theta)\) exceeds a certain threshold value for some \(\theta .\) Is this more likely to happen for positive or for negative values of \(k_{1} ?\)

Short answer

f(\theta) = h(\theta) + 2\pi A k_0 + \pi B k_1 \cos \theta + \pi C k_1 \sin \theta. Walkouts are more likely with negative \(k_1\).

Step-by-step solution

Express the integral equation

Start by rewriting the given integral equation. Given:\[ f(\theta) = h(\theta) + \int_{0}^{2 \pi} K(\theta - \phi) f(\phi) d \phi \]

Substitute the kernel function

Substitute the form of the kernel function into the integral. Given that \(K(\varphi) = k_0 + k_1 \cos \psi\), set \(\psi = \theta - \phi\). Then,\[ K(\theta - \phi) = k_0 + k_1 \cos(\theta - \phi) \]

Substitute and simplify the integral

Substitute this expression for \(K(\theta - \phi)\) into the original equation:\[ f(\theta) = h(\theta) + \int_{0}^{2 \pi} \left(k_0 + k_1 \cos(\theta - \phi)\right) f(\phi) d \phi \]Simplify the integral:\[ f(\theta) = h(\theta) + k_0 \int_{0}^{2 \pi} f(\phi) d \phi + k_1 \int_{0}^{2 \pi} \cos(\theta - \phi) f(\phi) d \phi \]

Evaluate the integral terms

Assume a solution of the form:\[ f(\theta) = A + B \cos \theta + C \sin \theta \]Evaluate the first term:\[ \int_{0}^{2 \pi} f(\phi) d \phi = \int_{0}^{2 \pi} (A + B \cos \phi + C \sin \phi) d \phi = 2 \pi A \]Next, use the orthogonality of sine and cosine to find:\[ \int_{0}^{2 \pi} \cos(\theta - \phi) f(\phi) d \phi \]

Utilize trigonometric identities

Use trigonometric identities to evaluate the second integral term. Start with:\[ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \]This gives us:\[ \int_{0}^{2 \pi} \cos(\theta - \phi) f(\phi) d \phi = \cos \theta \int_{0}^{2 \pi} \cos \phi f(\phi) d \phi + \sin \theta \int_{0}^{2 \pi} \sin \phi f(\phi) d \phi \]Evaluate each part separately.

Solve for B and C coefficients

Combine the results and evaluate the integrals:\[ \int_{0}^{2 \pi} \cos \phi f(\phi) d \phi = \pi B \int_{0}^{2 \pi} \sin \phi f(\phi) d \phi = \pi C \]Thus,\[ \int_{0}^{2 \pi} \cos(\theta - \phi) f(\phi) d \phi = \pi B \cos \theta + \pi C \sin \theta \]

Combine results and find coefficients

Combine all the terms to find the general form of \( f(\theta) \):\[ f(\theta) = h(\theta) + k_0 (2 \pi A) + k_1(\pi B \cos \theta + \pi C \sin \theta) \]By matching coefficients, identify \( p, q, r \) from the general form:\( f(\theta) = h(\theta) + p + q \cos \theta + r \sin \theta \), where\( p = 2\pi A k_0 \), \( q = \pi B k_1 \), and \( r = \pi C k_1 \)

Analyze the stability based on \( k_1 \)

To determine the impact of \(k_1\), consider the cases. Positive \(k_1\) implies that delegates tend to placate their opponents but upset their allies, while negative \(k_1\) implies the opposite. Since the influence contributes to the overall fury, analyze which scenario increases the likelihood of exceeding the threshold value for a walkout.

Key concepts

Circular Table Problems

In physics and various real-world scenarios, problems often involve objects or participants arranged in a circular formation. This exercise falls into such a category, where delegates are seated around a circular table. The key aspect is understanding that the position of each delegate is represented by an angle, \(\theta\), ranging from 0 to 2\pi. This setup leverages the circular symmetry to model interactions effectively.
For example, in our problem, the delegates' fury levels are influenced by their natural hostility and the contributions of other delegates. The fury felt by delegate \(\theta\) is given by an integral equation.

Orthogonality of Sine and Cosine

Understanding the orthogonality of sine and cosine functions is crucial for solving trigonometric integrals. Two functions, such as \(\sin\) and \(\cos\), are orthogonal if their integral over a specified interval is zero. In other words:
\[ \int_0^{2\pi} \cos(k\theta) \sin(m\theta) d\theta = 0 \] for any integers \(k\) and \(m\). This property simplifies integrals involving trigonometric functions significantly.
In the given problem, we assumed a solution of the form \(f(\theta) = A + B \cos(\theta) + C \sin(\theta)\). We then used orthogonality to isolate and solve the coefficients, making the complex integral manageable.

Trigonometric Integrals

Trigonometric integrals often appear in physics and engineering problems. These integrals involve functions like sine, cosine, and their combinations. Calculating these integrals typically involves leveraging trigonometric identities and properties.
For this problem, evaluating \(\int_0^{2\pi} \cos(\theta - \phi) f(\phi) d \phi\) is necessary. By using the identity:
\[ \cos(\theta - \phi) = \cos(\theta) \cos(\phi) + \sin(\theta) \sin(\phi) \] the integral becomes a sum of simpler integrals involving \(\cos(\phi)f(\phi)\) and \(\sin(\phi)f(\phi)\). This simplification, combined with orthogonality, helps extract the coefficients \ B \ and \ C\, aiding in solving for \(p\), \(q\), and \(r\) effectively.