For X(x), the general solution is X(x) = C_1cos(kx) + C_2sin(kx). Applying X(0) = 0 gives C_1 = 0, and X(a) = 0 gives k = (nπ)/(a) for n = 1, 2, 3,. For Y(y), Y''(y) - k² Y(y) = 0 gives Y(y) = C_3 (ky) + C_4 (ky). The condition u(x, 0) = 0 leads to Y(0) = 0, so C_3 = 0.

For X(x), the general solution is X(x) = C_1cos(kx) + C_2sin(kx). Applying X(0) = 0 gives C_1 = 0, and X(a) = 0 gives k = (nπ)/(a) for n = 1, 2, 3,. For Y(y), Y''(y) - k² Y(y) = 0 gives Y(y) = C_3 (ky) + C_4 (ky). The condition u(x, 0) = 0 leads to Y(0) = 0, so C_3 = 0.

Setting y = a, we have Σ_n=1^∞ A_n (nπ) sin((nπ x)/(a)) = 40. Using the orthogonality of sine functions, the integral evaluates to (80)/(nπ)(1 - cos(nπ)). For even n, this equals 0; for odd n = 2m - 1, it equals (160)/(nπ). Thus, A_n = (160)/(nπ (nπ)) for odd n.

Setting y = a, we have Σ_n=1^∞ A_n (nπ) sin((nπ x)/(a)) = 40. Using the orthogonality of sine functions, the integral evaluates to (80)/(nπ)(1 - cos(nπ)). For even n, this equals 0; for odd n = 2m - 1, it equals (160)/(nπ). Thus, A_n = (160)/(nπ (nπ)) for odd n.

Find the steady state temperature u(x, y) in a square copper plate with side $a = 30\text{ cm}$ the faces being perfectly insulated, the upper side is kept at 40 °C and other sides are kept at 0 °C.

Answer: $u(x, y) = \frac{160}{\pi} \sum_{m=1}^{\infty} \frac{\sin\left(\frac{(2m-1)\pi x}{30}\right) \sinh\left(\frac{(2m-1)\pi y}{30}\right)}{(2m-1)\sinh((2m-1)\pi)}$

Step-by-step solution

Step 1: Set up Laplace's equation and boundary conditions

Steady-state heat conduction in two dimensions without heat generation is governed by Laplace's equation. The side of the square plate is $a = 30\text{ cm}$ . The boundary conditions are given as: $u(0, y) = 0$ , $u(a, y) = 0$ , $u(x, 0) = 0$ , and $u(x, a) = 40^\circ\text{C}$ .

Step 2: Apply separation of variables

Assuming a product solution $u(x, y) = X(x)Y(y)$ , we separate variables into two ordinary differential equations. Because $u(0, y) = 0$ and $u(a, y) = 0$ , $X(x)$ must satisfy $X(0) = 0$ and $X(a) = 0$ , requiring a negative separation constant $-k^2$ with $k = \frac{n\pi}{a}$ .

Step 3: Solve for X(x) and Y(y)

For $X(x)$ , the general solution is $X(x) = C_1\cos(kx) + C_2\sin(kx)$ . Applying $X(0) = 0$ gives $C_1 = 0$ , and $X(a) = 0$ gives $k = \frac{n\pi}{a}$ for $n = 1, 2, 3, \dots$ . For $Y(y)$ , $Y''(y) - k^2 Y(y) = 0$ gives $Y(y) = C_3\cosh(ky) + C_4\sinh(ky)$ . The condition $u(x, 0) = 0$ leads to $Y(0) = 0$ , so $C_3 = 0$ .

Step 4: Form general Fourier series solution

By the principle of superposition, the general solution is the sum of all fundamental solutions. To determine the coefficients $A_n$ , we apply the remaining non-homogeneous boundary condition at the top edge $y = a$ , where $u(x, a) = 40$ .

Step 5: Evaluate Fourier coefficients An

Setting $y = a$ , we have $\sum_{n=1}^\infty A_n \sinh(n\pi) \sin\left(\frac{n\pi x}{a}\right) = 40$ . Using the orthogonality of sine functions, the integral evaluates to $\frac{80}{n\pi}(1 - \cos(n\pi))$ . For even $n$ , this equals $0$ ; for odd $n = 2m - 1$ , it equals $\frac{160}{n\pi}$ . Thus, $A_n = \frac{160}{n\pi \sinh(n\pi)}$ for odd $n$ .

Step 6: Substitute $a = 30$ and write final solution

Substituting $a = 30\text{ cm}$ and $n = 2m - 1$ into our series expression yields the complete steady-state temperature distribution throughout the square plate.

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Find the steady state temperature u(x, y) in a square copper plate with side a=30 cma = 30\text{ cm}a=30 cm the faces being perfectly insulated, the upper side is kept at 40 °C and other sides are kept at 0 °C.