answer:To solve the heat equation using Fourier series, we first need to find the Fourier coefficients for the given initial condition function f(x). Since the function is defined piecewise, we will find the Fourier coefficients for each piece and then combine them. For the given function f(x), we can write the Fourier sine series as: u(x,t) = sum_{n=1}^{infty} B_n sinleft(frac{npi x}{L}right) e^{-kleft(frac{npi}{L}right)^2 t} Since the function is odd, we only need to find the coefficients B_n. We can find them using the following formula: B_n = frac{2}{L} int_0^L f(x) sinleft(frac{npi x}{L}right) dx Now, we need to find the coefficients for each piece of the function: For the first piece, 0 leq x < frac{L}{2}, f(x) = 0, so the integral is 0 and B_n = 0 for this piece. For the second piece, frac{L}{2} leq x leq L, f(x) = x - frac{L}{2}, so we have: B_n = frac{2}{L} int_{frac{L}{2}}^L (x - frac{L}{2}) sinleft(frac{npi x}{L}right) dx Now, let's find the first three coefficients B_1, B_2, and B_3: B_1 = frac{2}{1} int_{frac{1}{2}}^1 (x - frac{1}{2}) sin(pi x) dx = frac{4}{pi^2} B_2 = frac{2}{1} int_{frac{1}{2}}^1 (x - frac{1}{2}) sin(2pi x) dx = 0 B_3 = frac{2}{1} int_{frac{1}{2}}^1 (x - frac{1}{2}) sin(3pi x) dx = -frac{4}{9pi^2} Now, we can write the first three terms of the series expansion for u(x,t): u(x,t) approx frac{4}{pi^2} sin(pi x) e^{-pi^2 t} - frac{4}{9pi^2} sin(3pi x) e^{-9pi^2 t} Finally, we can find the solution at t=1: u(x,1) approx frac{4}{pi^2} sin(pi x) e^{-pi^2} - frac{4}{9pi^2} sin(3pi x) e^{-9pi^2}

answer:To solve the given partial differential equation using Fourier series, we first need to find the Fourier sine series representation of the initial condition u(0, x) = 2x for 0 ≤ x ≤ 1. The Fourier sine series is given by: u(0, x) = ∑[B_n * sin(nπx)] where B_n = (2/L) * ∫[f(x) * sin(nπx/L) dx] from 0 to L, and L = 1 in this case. So, B_n = 2 * ∫[2x * sin(nπx) dx] from 0 to 1. Integrating by parts, we have: u = 2x, dv = sin(nπx) dx du = 2 dx, v = -(1/nπ) * cos(nπx) Using integration by parts formula, we get: B_n = 2 * ([-(2x/nπ) * cos(nπx)] from 0 to 1 + (1/nπ) * ∫[2 * cos(nπx) dx] from 0 to 1) Now, integrating the second term: ∫[2 * cos(nπx) dx] = (2/nπ) * sin(nπx) from 0 to 1 So, B_n = 2 * ([-(2x/nπ) * cos(nπx)] from 0 to 1 + [(2/nπ) * sin(nπx)] from 0 to 1) Evaluating the limits, we get: B_n = 2 * ([-(2/nπ) * cos(nπ)] + 0) Now, cos(nπ) = (-1)^n, so: B_n = (4/nπ) * (-1)^(n+1) Now, we have the Fourier sine series representation of the initial condition: u(0, x) = ∑[(4/nπ) * (-1)^(n+1) * sin(nπx)] Next, we need to find the solution u(t, x) for t > 0. Since the given PDE is a heat equation, the solution can be represented as: u(t, x) = ∑[B_n * e^(-n^2π^2kt) * sin(nπx)] where k = 9 (given in the problem) and B_n is the Fourier coefficient we found earlier. So, the solution u(t, x) for t > 0 is: u(t, x) = ∑[(4/nπ) * (-1)^(n+1) * e^(-9n^2π^2t) * sin(nπx)]

answer:To solve this problem, we'll use the method of separation of variables. Assume that the solution can be written as a product of two functions, one depending only on x and the other only on t: u(x,t) = X(x)T(t) Substituting this into the heat equation, we get: X(x) frac{dT}{dt} = k T(t) frac{d^2X}{dx^2} Now, divide both sides by kXT: frac{1}{kT}frac{dT}{dt} = frac{1}{X}frac{d^2X}{dx^2} Since the left side depends only on t and the right side depends only on x, both sides must be equal to a constant, say -lambda^2: frac{1}{kT}frac{dT}{dt} = -lambda^2 = frac{1}{X}frac{d^2X}{dx^2} Now we have two ordinary differential equations: 1. frac{d^2X}{dx^2} = -lambda^2 X 2. frac{dT}{dt} = -klambda^2 T Solving the first equation, we get: X(x) = Asin(lambda x) + Bcos(lambda x) Applying the boundary conditions u(0,t) = u(1,t) = 0, we get B = 0 and lambda_n = npi for n = 1, 2, 3, dots. So, X_n(x) = A_nsin(npi x) Solving the second equation, we get: T_n(t) = C_n e^{-k(npi)^2t} Thus, the general solution is: u(x,t) = sum_{n=1}^{infty} A_nsin(npi x)e^{-k(npi)^2t} To find the coefficients A_n, we use the initial condition u(x,0) = f(x): f(x) = sum_{n=1}^{infty} A_nsin(npi x) We can find A_n by multiplying both sides by sin(mpi x) and integrating from 0 to 1: A_n = 2int_0^1 f(x)sin(npi x) dx For the given initial condition, we have: A_n = 2int_{1/2}^1 sin(npi x) dx = frac{2}{npi}(-1)^n So, the solution is: u(x,t) = sum_{n=1}^{infty} frac{2}{npi}(-1)^nsin(npi x)e^{-k(npi)^2t} Now, we can calculate the first three terms of the series: 1. u_1(x,t) = frac{2}{pi}sin(pi x)e^{-pi^2t} 2. u_2(x,t) = -frac{1}{pi}sin(2pi x)e^{-4pi^2t} 3. u_3(x,t) = frac{2}{3pi}sin(3pi x)e^{-9pi^2t} To plot the solution for t=0.1, t=0.5, and t=1.0, using k=1 and N=100, you can use a software like MATLAB or Python with the following pseudocode: ``` x = linspace(0, 1, 1000) t_values = [0.1, 0.5, 1.0] N = 100 for t in t_values: u = 0 for n in range(1, N+1): u += (2/(n*pi))*(-1)^n*sin(n*pi*x)*exp(-k*(n*pi)^2*t) plot(x, u) ``` This will give you the plots of the solution for the given time values.

answer:To solve this problem, we will use the method of separation of variables. Assume that the solution can be written in the form: u(x, t) = X(x)T(t) Substituting this into the PDE, we get: X(x)T''(t) = 4X''(x)T(t) Now, divide both sides by X(x)T(t): frac{T''(t)}{T(t)} = 4frac{X''(x)}{X(x)} Since the left side depends only on t and the right side depends only on x, both sides must be equal to a constant, say -k^2: frac{T''(t)}{T(t)} = 4frac{X''(x)}{X(x)} = -k^2 Now we have two ordinary differential equations: T''(t) + k^2T(t) = 0 X''(x) - frac{k^2}{4}X(x) = 0 The second equation has the general solution: X(x) = Acos(frac{k}{2}x) + Bsin(frac{k}{2}x) Applying the boundary conditions u(0, t) = u(1, t) = 0, we get: X(0) = A = 0 X(1) = Bsin(frac{k}{2}) = 0 Since B cannot be zero (otherwise the solution would be trivial), we must have: sin(frac{k}{2}) = 0 Rightarrow frac{k}{2} = npi Rightarrow k = 2npi where n is an integer. Now, the solution for X(x) is: X_n(x) = B_nsin(2npi x) The first equation has the general solution: T(t) = Ccos(kt) + Dsin(kt) Using the initial condition u_t(x, 0) = 0, we get: T'(0) = -kC = 0 Rightarrow C = 0 So, the solution for T(t) is: T_n(t) = D_nsin(2npi t) Now, the general solution for u(x, t) is: u(x, t) = sum_{n=1}^{infty} B_nsin(2npi x)D_nsin(2npi t) Using the initial condition u(x, 0) = sin(pi x), we get: sin(pi x) = sum_{n=1}^{infty} B_nsin(2npi x)(0) Comparing the coefficients of the sine functions, we find that B_1 = 1 and B_n = 0 for n neq 1. Thus, the solution is: u(x, t) = D_1sin(2pi x)sin(2pi t) To find D_1, we use the orthogonality of the sine functions: int_0^1 sin(pi x) sin(2pi x) dx = frac{1}{2}D_1int_0^1 sin^2(2pi x) dx The left side of the equation can be computed using integration by parts: int_0^1 sin(pi x) sin(2pi x) dx = frac{1}{4} The right side of the equation can be computed using the identity sin^2(x) = frac{1}{2}(1 - cos(2x)): frac{1}{2}D_1int_0^1 sin^2(2pi x) dx = frac{1}{2}D_1int_0^1 frac{1}{2}(1 - cos(4pi x)) dx = frac{1}{4}D_1 Equating the two sides, we get D_1 = 1. Therefore, the solution is: u(x, t) = sin(2pi x)sin(2pi t) Finally, we find the solution at the point (x, t) = (1/4, pi/4): uleft(frac{1}{4}, frac{pi}{4}right) = sinleft(2pi cdot frac{1}{4}right)sinleft(2pi cdot frac{pi}{4}right) = sinleft(frac{pi}{2}right)sinleft(frac{pi^2}{2}right) = 1cdotsinleft(frac{pi^2}{2}right) So, the solution at (x, t) = (1/4, pi/4) is: uleft(frac{1}{4}, frac{pi}{4}right) = sinleft(frac{pi^2}{2}right)