1. **Problem Statement:** (a) Define isotherms. (b) Find the steady-state temperature distribution in a square plate of side $a=2$ with given boundary conditions. 2. **Part (a) - What are isotherms?** Isotherms are curves or lines on a surface or plane along which the temperature is constant. In other words, every point on an isotherm has the same temperature. 3. **Part (b) - Steady-state temperature distribution:** We want to find $u(x,y)$ satisfying Laplace's equation for steady-state heat conduction: $$\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ inside the square $0 < x < 2$, $0 < y < 2$. 4. **Boundary conditions:** (i) $u = 80 \sin(x)$ on the upper side $y=2$, and $u=0$ on the other three sides. (ii) $u=0$ on the vertical sides $x=0$ and $x=2$, and the horizontal sides are perfectly insulated, meaning no heat flux across them: $$\frac{\partial u}{\partial y} = 0 \text{ at } y=0 \text{ and } y=2$$ (iii) Specify boundary conditions so the steady-state solution is not identically zero. 5. **Solving (i):** Use separation of variables: assume $u(x,y) = X(x)Y(y)$. Laplace's equation becomes: $$\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda$$ For $X(x)$ with $X(0)=0$, $X(2)=0$, eigenvalues are: $$\lambda_n = \left(\frac{n\pi}{2}\right)^2, \quad X_n(x) = \sin\left(\frac{n\pi x}{2}\right)$$ For $Y(y)$: $$Y_n'' - \lambda_n Y_n = 0 \implies Y_n(y) = A_n \sinh\left(\frac{n\pi y}{2}\right) + B_n \cosh\left(\frac{n\pi y}{2}\right)$$ Since $u=0$ at $y=0$, $Y_n(0)=0 \Rightarrow B_n=0$. 6. **Apply upper boundary condition $y=2$:** $$u(x,2) = \sum_{n=1}^\infty A_n \sin\left(\frac{n\pi x}{2}\right) \sinh\left(n\pi\right) = 80 \sin(x)$$ Express $80 \sin(x)$ as Fourier sine series in terms of $\sin\left(\frac{n\pi x}{2}\right)$. Note $\sin(x)$ is not exactly $\sin\left(\frac{n\pi x}{2}\right)$, but for $n=2$, $\sin\left(\frac{2\pi x}{2}\right) = \sin(\pi x)$, which differs from $\sin(x)$. So approximate or use orthogonality to find coefficients: $$A_n = \frac{2}{2 \sinh(n\pi)} \int_0^2 80 \sin(x) \sin\left(\frac{n\pi x}{2}\right) dx$$ 7. **Part (ii) - insulated horizontal sides:** Boundary conditions: $$u=0 \text{ at } x=0, x=2$$ $$\frac{\partial u}{\partial y} = 0 \text{ at } y=0, y=2$$ Separation of variables gives: $$X_n(x) = \sin\left(\frac{n\pi x}{2}\right)$$ For $Y(y)$: $$Y_n'' - \lambda_n Y_n = 0$$ With Neumann BCs: $$Y_n'(0) = 0, \quad Y_n'(2) = 0$$ General solution: $$Y_n(y) = C_n \cosh\left(\frac{n\pi y}{2}\right) + D_n \sinh\left(\frac{n\pi y}{2}\right)$$ Apply $Y_n'(0)=0 \Rightarrow D_n=0$. Apply $Y_n'(2)=0$: $$C_n \frac{n\pi}{2} \sinh(n\pi) = 0 \Rightarrow C_n=0 \text{ unless } n=0$$ For $n=0$, $X_0(x)$ is constant, so solution is constant in $x$ and $y$. 8. **Part (iii) - Nonzero steady-state solution:** To have a nonzero solution, boundary conditions must not all be zero or homogeneous. For example, set: $$u=0 \text{ on } x=0, x=2$$ $$u=0 \text{ on } y=0$$ $$u= f(x) \neq 0 \text{ on } y=2$$ where $f(x)$ is a nonzero function, e.g., $f(x) = 100 \sin\left(\frac{\pi x}{2}\right)$. This ensures a nontrivial steady-state temperature distribution. **Final answers:** (a) Isotherms are lines of constant temperature. (b)(i) The solution is: $$u(x,y) = \sum_{n=1}^\infty A_n \sin\left(\frac{n\pi x}{2}\right) \sinh\left(\frac{n\pi y}{2}\right)$$ with $$A_n = \frac{2}{2 \sinh(n\pi)} \int_0^2 80 \sin(x) \sin\left(\frac{n\pi x}{2}\right) dx$$ (b)(ii) With insulated horizontal sides, the solution involves cosine hyperbolic terms and zero derivative BCs. (b)(iii) Nonzero BCs on at least one boundary ensure nontrivial solutions.