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: $$u(x,t)=2e^{-\frac{{\pi }^2}{8}t}\sin \left(\frac{\pi x}{2}\right)-e^{-\frac{4{\pi }^2}{8}t}\sin (\pi x)+4e^{-\frac{16{\pi }^2}{8}t}\sin (2\pi x).$$

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: $$X^{″}(x)T(t)=4X(x)T^{\prime }(t).$$ Now, we'll divide both sides by $X(x)T(t)$, which yields: $$\frac{X^{″}(x)}{X(x)}=4\frac{T^{\prime }(t)}{T(t)}.$$ 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 $-{\lambda }^2$: $$\frac{X^{″}(x)}{X(x)}=4\frac{T^{\prime }(t)}{T(t)}=-{\lambda }^2.$$ #Step 2: Solve for X(x)#

We now have two ordinary differential equations (ODE): $$X^{″}(x)+{\lambda }^{2}X(x)=0.$$ 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: $$u(0,t)=0,u(2,t)=0.$$ 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): $$4T^{\prime }(t)+{\lambda }^{2}T(t)=0.$$ Integrating, we get: $$T^{\prime }(t)=-\frac{{\lambda }^2}{4}T(t).$$ 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: $$u(x,t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{B}_n{C}_n{e}^{-\frac{n^2{\pi }^2}{8}t}\sin (n\pi x/2).$$ 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: $$B_1{C}_1=2,B_2{C}_2=-1,B_4{C}_4=4.$$ with all other $B_n{C}_n=0$. #Step 6: Final Solution#

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

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical equations that involve functions and their partial derivatives with respect to various variables. They are crucial in modeling and analyzing dynamic systems found in physics and engineering. A PDE can describe phenomena such as heat conduction, wave propagation, and fluid dynamics. In this exercise, we explore a specific example of a PDE known as the heat equation, which models the distribution of heat in a given domain over time. The heat equation is expressed as: $$u_{xx}=4u_t$$ Here, $u_{xx}$ represents the second spatial derivative (concerning the variable $x$), and $u_t$ is the first time derivative. Solving PDEs like this involves finding functions that satisfy the equation under the given conditions, known as boundary and initial conditions. These conditions provide necessary information about the behavior of the function at specific points, helping to determine a unique solution.

Boundary Conditions

Boundary conditions are an essential component of solving PDEs, as they specify the behavior of a function at the boundaries of the domain. In the context of the heat equation, boundary conditions are necessary to ensure a well-defined and unique solution over time. In this problem, the boundary conditions are given by:

• $u(0,t)=0$
• $u(2,t)=0$

These conditions describe how the system behaves at the edges $x=0$ and $x=2$, indicating that the temperature remains zero at these points for all times $t>0$. By applying these conditions to the general solution $X(x)$, we ensure that any potential solutions satisfy these boundary constraints, thus aiding in the selection of appropriate functions. Furthermore, boundary conditions help in determining constants in the general form of the solution, allowing for the extraction of a specific solution that fits the entire problem.

Fourier Series Expansion

The Fourier series expansion is a powerful mathematical tool used to express a function as an infinite sum of sines and cosines. This technique is particularly useful in solving PDEs, as it allows us to handle complex boundary and initial conditions by decomposing them into simpler, well-understood trigonometric forms. In this exercise, the initial condition is given by the function: $$u(x,0)=2\sin \left(\frac{\pi x}{2}\right)-\sin (\pi x)+4\sin (2\pi x)$$ To match this condition with our solution, we use a Fourier sine series expansion. This involves writing the solution as a sum of terms of the form $B_n{C}_n{e}^{-\frac{n^2{\pi }^2}{8}t}\sin \left(\frac{n\pi x}{2}\right)$. By comparing this series with the provided initial condition, we determine the coefficients $B_n{C}_n$ for each relevant $n$. Thus, Fourier series expansion helps us solve the PDE by matching the infinite series solution with the initial data provided.

Separation of Variables

The separation of variables is a method used to transform a PDE into simpler ordinary differential equations (ODEs), which are easier to solve. This approach assumes that the solution can be expressed as a product of functions, each dependent on a single variable. In this problem, we assume that the solution $u(x,t)$ can be written as $X(x)T(t)$. Substituting this expression into the original PDE, we separate the variables to obtain: $$\frac{X^{″}(x)}{X(x)}=4\frac{T^{\prime }(t)}{T(t)}=-{\lambda }^2$$ This technique reduces the PDE into two ODEs: one for $X(x)$ and another for $T(t)$. Once separated, these ODEs can be individually solved. $X(x)$ satisfies the equation $X^{″}(x)+{\lambda }^{2}X(x)=0$, whose solutions involve trigonometric functions. Meanwhile, $T(t)$ solves the equation $4T^{\prime }(t)+{\lambda }^{2}T(t)=0$, with solutions taking an exponential form. The separation of variables simplifies tackling complex PDEs by breaking them down into manageable parts.