91Ó°ÊÓ

Let u(x, y) = X(x)Y(y). Substituting this expression into Laplace's equation: $$\nabla^2 u(x,y) = X''Y + XY'' = 0.$$ Dividing through by \(XY\), we get: $$\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0.$$ Since these functions are independent, we can say that $$\frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = k,$$ where k is a constant.

Now, we have two ordinary differential equations to solve: 1. \(X''(x) = kX(x)\) 2. \(Y''(y) = -kY(y)\) Solving the first equation, we get: $$X''(x) - kX(x) = 0.$$ Let k = \(m^2,\) so the equation becomes: $$X''(x) - m^2X(x) = 0.$$ The general solution of this equation is: $$X(x) = c_1\: e^{mx} + c_2\:e^{-mx}.$$ From the boundary conditions, \(u(0, y) = X(0)Y(y) = 0,\) so \(X(0)=0\). Substituting x=0, we get: $$X(0) = c_1\: e^{0} + c_2\:e^{0} = c_1 + c_2 = 0 \implies c_1 = -c_2.$$ So, the equation becomes: $$X(x) = c_1(e^{mx} - e^{-mx}).$$ Now, using the boundary condition \(u(a, y) = X(a)Y(y) = 0,\) so \(X(a)=0\). Substituting x=a, we get: $$X(a) = c_1(e^{ma} - e^{-ma}) = 0,$$ which means that \(e^{ma} = e^{-ma}\) or \(m = n \pi/a\) for integer values of n. Therefore, the general solution of X(x) becomes: $$X_n(x) = c_n\: \sin\left(\frac{n\pi}{a}x\right),$$ where n = 1, 2, 3, .... Now solving the second equation, \(Y''(y) + m^2Y(y) = 0,\) we get the general solution: $$Y_n(y) = A_n\: e^{-n\pi y/a} + B_n\: e^{n\pi y/a}.$$

Using the additional boundary condition that \(Y(y)\rightarrow 0\) as \(y\rightarrow \infty\), we find out that the terms with growing exponentials are not useful, so we consider only the terms containing the declining exponentials: $$Y_n(y) = A_n\: e^{-n\pi y/a}.$$ Now, we form the solution u(x, y) by combining the solutions for X(x) and Y(y): $$u(x, y) = \sum_{n=1}^\infty A_n \: \sin\left(\frac{n\pi}{a}x\right) e^{-n\pi y/a}.$$

For this specific function, we apply the boundary condition \(u(x, 0) = f(x)\). So, let \(f(x) = x(a-x)\). We have: $$u(x, 0) = f(x) = \sum_{n=1}^\infty A_n \sin \left(\frac{n\pi}{a}x\right).$$ We can use the Fourier sine series representaion for our function f(x), which is $$f(x) = \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi}{a}x\right),$$ where $$b_n = \frac{2}{a}\int_0^a f(x) \sin\left(\frac{n\pi}{a}x\right) dx.$$ By comparing these two expressions, we can deduce that \(A_n = b_n\). To find \(b_n\), we will integrate: $$b_n = \frac{2}{a}\int_0^a x(a-x) \sin\left(\frac{n\pi}{a}x\right) dx.$$ I won't do the entire integration process here, but after integrating we get: $$b_n = -\frac{8a}{n^3\pi^3} \left[1 - \left(-1\right)^n \right].$$ Finally, plugging this value into the solution for u(x,y), we have: $$u(x,y) = \sum_{n=1}^\infty -\frac{8a}{n^3\pi^3} \left[1 - \left(-1\right)^n \right] \sin\left(\frac{n\pi}{a}x\right) e^{-n\pi y/a}.$$

We have the solution for u(x,y): $$u(x,y) = -\frac{8\cdot 5}{\pi^3} \sum_{n=1}^\infty \frac{1 - \left(-1\right)^n}{n^3} \sin\left(\frac{n\pi}{5}x\right) e^{-n\pi y/5}.$$ We need to find the smallest value of \(y_0\) for which \(u(x, y) \leq 0.1\) for all \(y \geq y_0\). Since we have a negative term multiplying the whole series, and all terms in the series are positive, the smallest value of u(x,y) will occur when n=1 and x=a/2: $$-\frac{40}{\pi^3} e^{-\pi y/5} \leq 0.1$$ Solving for y, we get: $$y \geq \frac{5}{\pi} \ln \frac{400}{\pi^3} \approx 7.672.$$ Therefore, the smallest value of \(y_0\) for which \(u(x, y) \leq 0.1\) for all \(y \geq y_0\) is approximately \(y_0 \approx 7.672\).