1. Problem 1: Find the displacement $y(x,t)$ of a string fixed at $x=0$ and $x=1$, initially at rest and given initial velocity $\lambda x(1-x)$. - The wave equation for displacement $y(x,t)$ with fixed ends is: $$\frac{\partial^2 y}{\partial t^2} = c^2 \frac{\partial^2 y}{\partial x^2}$$ with boundary conditions $y(0,t) = y(1,t) = 0$. - Initial displacement is zero: $y(x,0) = 0$, initial velocity is $\frac{\partial y}{\partial t}(x,0) = \lambda x(1-x)$. - Using separation of variables, the eigenfunctions are $\sin(n \pi x)$ with eigenfrequencies $\omega_n = n \pi c$. - The solution respecting zero initial displacement and given initial velocity is: $$y(x,t) = \sum_{n=1}^\infty B_n \sin(n\pi x) \sin(n \pi c t)$$ - The coefficients $B_n$ are found using the initial velocity condition: $$\lambda x(1-x) = \left.\frac{\partial y}{\partial t}(x,t)\right|_{t=0} = \sum_{n=1}^\infty B_n n \pi c \sin(n \pi x)$$ - Multiply both sides by $\sin(m \pi x)$ and integrate from 0 to 1: $$\int_0^1 \lambda x(1-x) \sin(m\pi x) dx = B_m n \pi c \frac{1}{2}$$ - Compute integral: $$I_m = \int_0^1 x(1-x) \sin(m \pi x) dx = \int_0^1 (x - x^2) \sin(m \pi x) dx$$ - Using integration by parts or lookup, $$I_m = \frac{2((-1)^{m+1})}{(m \pi)^3}$$ - Thus, $$B_m = \frac{2 \lambda}{c n \pi} I_m = \frac{4 \lambda ((-1)^{m+1})}{c (m \pi)^4}$$ - Final solution: $$y(x,t) = \sum_{n=1}^\infty \frac{4 \lambda (-1)^{n+1}}{c (n \pi)^4} \sin(n \pi x) \sin(n \pi c t)$$ 2. Problem 2: Temperature distribution $u(x,t)$ in a rod of length 1 cm initially steady at ends $A$ and $B$ with temperatures decreased suddenly to zero. - Initial condition: steady state temperature with boundary $u(0)=30$, $u(1)=70$. - At $t=0$, temperature at both ends set to zero instantly, the problem is now finding $u(x,t)$ with zero boundary conditions and initial temperature as prior steady state. - The steady state temperature distribution is linear: $$u(x,0) = 30 + 40 x$$ - The heat equation: $$\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}$$ - Boundary conditions: $u(0,t) = u(1,t) = 0$ after sudden change. - Using separation of variables and Fourier sine series, expand the initial condition: $$30 + 40x = \sum_{n=1}^\infty A_n \sin(n \pi x)$$ - Calculate $A_n$: $$A_n = 2 \int_0^1 (30 + 40x) \sin(n \pi x) dx = \frac{80}{n \pi} \left((-1)^{n+1} - 1\right)$$ - The solution is: $$u(x,t) = \sum_{n=1}^\infty A_n e^{-k n^2 \pi^2 t} \sin(n \pi x)$$ 3. Problem 3: Fourier transform of $$f(x) = \begin{cases} a^2 - x^2, & |x| < a \\ 0, & |x| > a > 0 \end{cases}$$ - Fourier transform: $$\hat{f}(s) = \int_{-a}^a (a^2 - x^2) e^{-i s x} dx$$ - Since $f(x)$ is even, $$\hat{f}(s) = 2 \int_0^a (a^2 - x^2) \cos(s x) dx$$ - Evaluate: $$= 2 \left[ a^2 \frac{\sin(a s)}{s} - 2 \frac{\sin(a s)}{s^3} + 2a \frac{\cos(a s)}{s^2} \right]$$ 4. Using the Fourier transform and known integrals to compute: (i) $$\int_0^\infty \frac{\sin s - s \cos s}{s^3} \cos\left(\frac{s}{2}\right) ds = \frac{\pi}{8}$$ (ii) $$\int_0^\infty \frac{\sin t - t \cos t}{t^3} dt = \frac{\pi}{4}$$ --- Final answers: - Problem 1 displacement: $$y(x,t) = \sum_{n=1}^\infty \frac{4 \lambda (-1)^{n+1}}{c (n \pi)^4} \sin(n \pi x) \sin(n \pi c t)$$ - Problem 2 temperature: $$u(x,t) = \sum_{n=1}^\infty \frac{80}{n \pi} \left((-1)^{n+1} - 1\right) e^{-k n^2 \pi^2 t} \sin(n \pi x)$$ - Problem 3 Fourier transform: $$\hat{f}(s) = 2 a^2 \frac{\sin(a s)}{s} - 4 \frac{\sin(a s)}{s^3} + 4 a \frac{\cos(a s)}{s^2}$$ - Integrals: (i) $$\int_0^\infty \frac{\sin s - s \cos s}{s^3} \cos\left(\frac{s}{2}\right) ds = \frac{\pi}{8}$$ (ii) $$\int_0^\infty \frac{\sin t - t \cos t}{t^3} dt = \frac{\pi}{4}$$