91Ó°ÊÓ

Using the ideas from the equations (17)-(20), we introduce a new function \(v(x,t)\) such that \(u(x,t)=v(x,t)+G(x)\), where \(G(x) = \frac{T_1 - T_0}{l}x\). We need to find the function \(v(x,t)\) that satisfies the homogeneous boundary conditions \(v(0,t)=0\) and \(v(l,t)=0\), and the new initial condition \(v(x,0)=g(x)=f(x)-G(x) = 25x+80\sin(\pi x)\cos(\pi x)-25x=80\sin(\pi x)\cos(\pi x)\).

Assume a solution of the form \(v(x,t)=X(x)T(t)\). Plug this into the heat equation for \(v(x,t)\): $$\frac{dX(x)}{dx}\left(\frac{dT(t)}{dt}\right)=\frac{\kappa X''(x)T(t)}{X(x)T(t)}.$$ Divide both sides by \(XT\) to obtain: $$\frac{1}{\kappa}\frac{dT}{dt} = \frac{X''}{X} = \text{constant}$$ Let this constant be \(-\omega^2\).

We now have two ordinary differential equations (ODEs) to solve: 1. For \(X(x)\): \(X''(x) + \omega^2 X(x) = 0\), with boundary conditions \(X(0)=0\) and \(X(l)=0\). 2. For \(T(t)\): \(T'(t) = -\kappa \omega^2 T(t)\). For the first ODE, we have a standard eigenvalue problem with solutions in the form of a linear combination of sines and cosines: \(X(x) = A\cos(\omega x) + B\sin(\omega x)\). With the boundary conditions, we get \(X(0)=A=0\) and \(X(l)=B\sin(\omega l)=0\). Since \(B \neq 0\), this implies \(\omega l = n\pi\) for \(n=1,2,3,...\). So, \(\omega_n = \frac{n \pi}{l}\) and \(X_n(x) = B_n\sin(\omega_n x)\). For the second ODE, we have a first-order linear ODE, which has an exponential solution: \(T_n(t) = C_n e^{-\kappa \omega_n^2 t}\), where \(C_n\) is a constant. Hence, the total solution is a summation over all possible \(n\): $$v(x,t)=\sum_{n=1}^{\infty}c_n\sin(\frac{n\pi}{l}x)e^{-\kappa\left(\frac{n\pi}{l}\right)^{2}t}$$

Using the initial condition \(v(x,0)=80\sin(\pi x)\cos(\pi x)\), we determine the coefficients \(c_n\) using orthogonality: $$c_n = \frac{2}{l}\int_0^l{80\sin(\pi x)\cos(\pi x)\sin(\frac{n\pi}{l}x)~dx}$$ Note that only \(n=2\) contributes to the sum, so \(c_2=\frac{160}{\pi}\) and \(c_n=0\) for \(n\neq2\). The solution for \(v(x,t)\) is thus: $$v(x,t)=\frac{160}{\pi}\sin(\frac{2\pi}{l}x)e^{-\kappa\left(\frac{2\pi}{l}\right)^{2}t}$$

Now we return to the function \(u(x,t)\) using the transformation from Step 1: $$u(x,t)=v(x,t)+G(x)=\frac{160}{\pi}\sin(\frac{2\pi}{l}x)e^{-\kappa\left(\frac{2\pi}{l}\right)^{2}t}+\frac{T_1-T_0}{l}x$$ Finally, substituting the given values for \(\kappa\), \(l\), \(T_0\), and \(T_1\): $$u(x,t)=40\sin(\frac{\pi}{2}x)e^{-\frac{\pi^2}{4}t}+25x$$ This is the solution to the given initial-boundary value problem.