1. **Problem statement:** (a) Define diabatic and adiabatic boundary conditions. (b) Explain how these boundary conditions affect the heat equation and its solutions. (c) Show that for a completely insulated bar with boundary conditions $u_x(0,t)=0$, $u_x(L,t)=0$, and initial condition $u(x,0)=f(x)$, the separation of variables leads to the solution $$u(x,t) = A_0 + \sum_{n=1}^\infty A_n \cos\left(\frac{n\pi x}{L}\right) e^{-\left(\frac{cn\pi}{L}\right)^2 t}$$ (d) Find the temperature distribution for $L=\pi$, $c=1$, and $f(x) = 1 - \frac{x}{\pi}$. 2. **Definitions:** - Diabatic boundary conditions allow heat exchange with the environment, meaning temperature or heat flux is specified at the boundary. - Adiabatic boundary conditions mean no heat transfer through the boundary, i.e., insulated boundaries with zero heat flux: $u_x=0$ at the boundary. 3. **Effect on heat equation:** The heat equation is $$u_t = c^2 u_{xx}$$ where $u(x,t)$ is temperature, $c$ is thermal diffusivity. - Diabatic conditions impose fixed temperature or flux, affecting the boundary values directly. - Adiabatic conditions impose zero flux, leading to Neumann boundary conditions $u_x=0$. 4. **Separation of variables for insulated bar:** Assume solution $u(x,t) = X(x)T(t)$. Substitute into heat equation: $$X(x) T'(t) = c^2 X''(x) T(t)$$ Divide both sides by $c^2 X(x) T(t)$: $$\frac{T'(t)}{c^2 T(t)} = \frac{X''(x)}{X(x)} = -\lambda$$ for some separation constant $\lambda$. 5. **Spatial part:** Solve $$X'' + \lambda X = 0$$ with Neumann BCs: $$X'(0) = 0, \quad X'(L) = 0$$ General solution: $$X(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x)$$ Apply BC at $x=0$: $$X'(0) = -A \sqrt{\lambda} \sin(0) + B \sqrt{\lambda} \cos(0) = B \sqrt{\lambda} = 0 \implies B=0$$ Apply BC at $x=L$: $$X'(L) = -A \sqrt{\lambda} \sin(\sqrt{\lambda} L) = 0$$ Nontrivial solution requires $$\sin(\sqrt{\lambda} L) = 0 \implies \sqrt{\lambda} L = n \pi, \quad n=0,1,2,\ldots$$ Thus, $$\lambda_n = \left(\frac{n \pi}{L}\right)^2$$ 6. **Temporal part:** Solve $$T' + c^2 \lambda T = 0 \implies T(t) = C e^{-c^2 \lambda t} = C e^{-\left(\frac{c n \pi}{L}\right)^2 t}$$ 7. **General solution:** Sum over all $n$: $$u(x,t) = A_0 + \sum_{n=1}^\infty A_n \cos\left(\frac{n \pi x}{L}\right) e^{-\left(\frac{c n \pi}{L}\right)^2 t}$$ where $A_0$ corresponds to $n=0$ mode (constant in space). 8. **Find coefficients $A_n$ from initial condition:** Given $$u(x,0) = f(x) = 1 - \frac{x}{\pi}$$ Use Fourier cosine series on $[0,L]$ with $L=\pi$: $$A_0 = \frac{1}{L} \int_0^L f(x) dx$$ $$A_n = \frac{2}{L} \int_0^L f(x) \cos\left(\frac{n \pi x}{L}\right) dx$$ Calculate $A_0$: $$A_0 = \frac{1}{\pi} \int_0^{\pi} \left(1 - \frac{x}{\pi}\right) dx = \frac{1}{\pi} \left[ x - \frac{x^2}{2 \pi} \right]_0^{\pi} = \frac{1}{\pi} \left( \pi - \frac{\pi^2}{2 \pi} \right) = \frac{1}{\pi} \left( \pi - \frac{\pi}{2} \right) = \frac{1}{2}$$ Calculate $A_n$ for $n \geq 1$: $$A_n = \frac{2}{\pi} \int_0^{\pi} \left(1 - \frac{x}{\pi}\right) \cos(n x) dx = \frac{2}{\pi} \left[ \int_0^{\pi} \cos(n x) dx - \frac{1}{\pi} \int_0^{\pi} x \cos(n x) dx \right]$$ First integral: $$\int_0^{\pi} \cos(n x) dx = \left[ \frac{\sin(n x)}{n} \right]_0^{\pi} = 0$$ Second integral by parts: Let $u = x$, $dv = \cos(n x) dx$; then $du = dx$, $v = \frac{\sin(n x)}{n}$. $$\int_0^{\pi} x \cos(n x) dx = \left. x \frac{\sin(n x)}{n} \right|_0^{\pi} - \int_0^{\pi} \frac{\sin(n x)}{n} dx = 0 - \frac{1}{n} \int_0^{\pi} \sin(n x) dx$$ Calculate: $$\int_0^{\pi} \sin(n x) dx = \left[ -\frac{\cos(n x)}{n} \right]_0^{\pi} = -\frac{\cos(n \pi) - 1}{n} = -\frac{(-1)^n - 1}{n}$$ So, $$\int_0^{\pi} x \cos(n x) dx = - \frac{1}{n} \left(-\frac{(-1)^n - 1}{n} \right) = \frac{(-1)^n - 1}{n^2}$$ Therefore, $$A_n = \frac{2}{\pi} \left(0 - \frac{1}{\pi} \cdot \frac{(-1)^n - 1}{n^2} \right) = - \frac{2}{\pi^2} \frac{(-1)^n - 1}{n^2}$$ Note that $(-1)^n - 1$ is zero for even $n$ and $-2$ for odd $n$: - For even $n$, $A_n=0$. - For odd $n$, $A_n = - \frac{2}{\pi^2} \frac{-2}{n^2} = \frac{4}{\pi^2 n^2}$. 9. **Final temperature distribution:** $$u(x,t) = \frac{1}{2} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{4}{\pi^2 n^2} \cos(n x) e^{-n^2 t}$$ This series represents the temperature distribution for the insulated bar with given initial condition.