Question

In each exercise, solve the initial-boundary value problem $$ \begin{array}{lcc} u_{t}=\kappa u_{\text {I? }}, & 0

Short answer

Question: Find the solution for the given initial-boundary value problem (IBVP) with nonhomogeneous boundary conditions for a heat equation \(u_t = \kappa u_{xx}\) with \(\kappa = 0.5\), boundary conditions \(u(0,t)=T_0=0\) and \(u(l,t)=T_1=100\), initial condition \(u(x,0)=25x+80\sin(\pi x)\cos(\pi x)\), and length \(l=4\). Answer: The solution for the given IBVP is \(u(x,t)=40\sin(\frac{\pi}{2}x)e^{-\frac{\pi^2}{4}t}+25x\).

Step-by-step solution

Nonhomogeneous to homogeneous boundary conditions

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 \(v(x,0)=g(x)=f(x)-G(x) = 25x+80\sin(\pi x)\cos(\pi x)-25x=80\sin(\pi x)\cos(\pi x)\).

Separation of Variables

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 \(-\omega^2\).

Solve the ODEs for X and T

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 \(n\): $$v(x,t)=\sum_{n=1}^{\infty}c_n\sin(\frac{n\pi}{l}x)e^{-\kappa\left(\frac{n\pi}{l}\right)^{2}t}$$

Find the Fourier coefficients

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: $$v(x,t)=\frac{160}{\pi}\sin(\frac{2\pi}{l}x)e^{-\kappa\left(\frac{2\pi}{l}\right)^{2}t}$$

Find the solution for u(x,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.

Key concepts

Heat Equation

The heat equation is a fundamental concept in partial differential equations that describes how heat diffuses through a given region over time. It is expressed as \( u_t = \kappa u_{xx} \), where \( u(x,t) \) represents the temperature at a point \( x \) and time \( t \), and \( \kappa \) is the thermal diffusivity constant. This equation helps understand the gradual change of temperature in a medium by modeling the flow of heat.

• Heat further moves from hot to cooler areas naturally until an equilibrium is reached.
• In our problem, the heat equation is used to examine the behavior of temperature in a rod of length \( l=4 \).
• Initial temperature distribution and boundary conditions work together to determine the evolution of temperature over time.

This equation is commonly employed in physics and engineering to solve problems regarding heat transfer in solids.

Initial-Boundary Value Problem

An initial-boundary value problem (IBVP) combines both the initial conditions and boundary conditions to solve differential equations. In our exercise, the boundary conditions were given as \( u(0, t) = T_0 \) and \( u(l, t) = T_1 \), with initial conditions defined by \( u(x, 0) = f(x) \). Here, the task is to find the temperature distribution \( u(x,t) \).
The solution to an IBVP involves satisfying the conditions simultaneously:

• Initial conditions are given to specify the state of a system at \( t=0 \).
• Boundary conditions tell us how the edges of the domain behave, with the temperatures at the ends of the rod being constant values \( T_0 \) and \( T_1 \).

This problem setup is common in real-world applications where conditions at both the start and boundary of the region affect the results.

Fourier Series

A Fourier series is a way to represent a function as a sum of sinusoidal components. In our solution, Fourier series helps to express the solution in a form that resonates with the heat distribution along the rod. We represent \( v(x, t) \) as a series of sines and cosines, which accommodates oscillatory behavior in infinite terms:\[ v(x,t) = \sum_{n=1}^{\infty} c_n \sin\left(\frac{n\pi}{l} x\right) e^{-\kappa \left( \frac{n\pi}{l} \right)^2 t} \]Here, the coefficients \( c_n \) are determined using the initial condition through orthogonality properties, ensuring only the relevant modes contribute to the solution.

• This approach simplifies complex boundary functions into manageable components.
• For periodic conditions or even arbitrary initial setups, Fourier series is versatile, enabling these transformations.

Separation of Variables

Separation of variables is a method used to solve partial differential equations by splitting them into simpler, separable ordinary differential equations (ODEs). In this approach, we hypothesize a solution for \( v(x,t) \) of the form \( v(x,t) = X(x)T(t) \). This assumption allows the heat equation to be divided into:

• \( X''(x) + \omega^2 X(x) = 0 \)
• \( T'(t) = -\kappa \omega^2 T(t) \)

Each part is solved independently, leading to solutions in terms of sine and exponential functions, attributed to boundary and initial conditions, respectively. This method is highly effective for linear systems where the solution must satisfy both spatial and temporal constraints in separate steps.