Question

Find the solution of the heat conduction problem $$ \begin{aligned} u_{x x} &=4 u_{t}, \quad 00 \\ u(0, t) &=0, \quad u(2, t)=0, \quad t>0 \\ u(x, 0) &=2 \sin (\pi x / 2)-\sin \pi x+4 \sin 2 \pi x, \quad 0 \leq x \leq 2 \end{aligned} $$

Short answer

Question: Determine the temperature distribution u(x, t) for the given heat conduction problem described by the partial differential equation with boundary and initial conditions: PDE: \(u_{xx} = 4u_t\), \(0 < x < 2\), \(t > 0\) Boundary Conditions: \(u(0, t) = 0\), \(u(2, t) = 0\) Initial Condition: \(u(x, 0) = 2 \sin(\pi x/2) - \sin(\pi x) + 4\sin(2\pi x)\) Answer: The temperature distribution u(x, t) for the given heat conduction problem is: $$ u(x,t) = 2 e^{-\frac{\pi^2}{8}t} \sin\left(\frac{\pi x}{2}\right) - e^{-\frac{4\pi^2}{8}t} \sin(\pi x) + 4 e^{-\frac{16\pi^2}{8}t} \sin(2\pi x). $$

Step-by-step solution

We assume a solution of the form \(u(x, t) = X(x)T(t)\). Plugging this into the PDE, we get: $$ X''(x)T(t) = 4X(x)T'(t). $$ Now, we'll divide both sides by \(X(x)T(t)\), which yields: $$ \frac{X''(x)}{X(x)} = 4\frac{T'(t)}{T(t)}. $$ Since the left side is a function of \(x\) only and the right side is a function of \(t\) only, they must be equal to a constant. Let's call it \(-\lambda^2\): $$ \frac{X''(x)}{X(x)} = 4\frac{T'(t)}{T(t)}= -\lambda^2. $$ #Step 2: Solve for X(x)#

We now have two ordinary differential equations (ODE): $$ X''(x) + \lambda^2X(x) = 0. $$ This is a second-order linear ODE, and we know that its general solution is: $$ X(x) = A\cos(\lambda x) + B\sin(\lambda x), $$ where \(A\) and \(B\) are constants. #Step 3: Apply Boundary Conditions for X(x)#

We are given two boundary conditions for the PDE: $$ u(0, t) = 0, \quad u(2, t) = 0. $$ Plugging these into our assumed solution and solving for X(x) yields: $$ X(0) = A\cos(0) + B\sin(0) = 0 \Rightarrow A = 0, $$ and $$ X(2) = B\sin(2\lambda) = 0. $$ As we don't want the trivial solution of \(X(x) = 0\), we need \(\sin(2\lambda) = 0\), which implies \(\lambda = n\pi/2\) for \(n \in \mathbb{Z}\). Our final solution for X(x) is then: $$ X_n(x) = B_n\sin(n\pi x/2), $$ where \(n \in \mathbb{Z}\) and \(B_n\) is a constant that depends on \(n\). #Step 4: Solve T(t) ODE#

Now let's solve the ODE for T(t): $$ 4T'(t) + \lambda^2 T(t) = 0. $$ Integrating, we get: $$ T'(t) = -\frac{\lambda^2}{4} T(t). $$ The solution to this ODE is an exponential: $$ T_n(t) = C_n e^{-\frac{n^2\pi^2}{8}t}, $$ where \(C_n\) is a constant that depends on \(n\). #Step 5: General Solution & Apply Initial Condition#

Now that we have solutions for both \(X_n(x)\) and \(T_n(t)\), we can form the general solution as an infinite sum: $$ u(x,t) = \sum_{n = -\infty}^{\infty} B_n C_n e^{-\frac{n^2\pi^2}{8}t} \sin(n\pi x/2). $$ Applying the initial condition \(u(x, 0) = 2 \sin(\pi x/2) - \sin(\pi x) + 4\sin(2\pi x)\), we find the coefficients \(B_n C_n\) using a Fourier sine series expansion. Comparing with our initial condition, we get: $$ B_1C_1 = 2, \quad B_2C_2 = -1, \quad B_4C_4 = 4. $$ with all other \(B_nC_n = 0\). #Step 6: Final Solution#

Plugging the values of the coefficients \(B_nC_n\) into the general solution, we finally obtain the solution to the heat conduction problem: $$ u(x,t) = 2 e^{-\frac{\pi^2}{8}t} \sin\left(\frac{\pi x}{2}\right) - e^{-\frac{4\pi^2}{8}t} \sin(\pi x) + 4 e^{-\frac{16\pi^2}{8}t} \sin(2\pi x). $$