Question

Consider the partial differential equation \(u_{t}(x, t)=\kappa u_{x x}(x, t)+\alpha u(x, t)\), where \(\alpha\) is a constant. (a) Suppose we introduce a new dependent variable \(w(x, t)\) by defining \(u(x, t)=\) \(e^{\delta t} w(x, t)\), where \(\delta\) is a constant. Show that if \(\delta\) is chosen properly, then \(w(x, t)\) is a solution of \(w_{t}(x, t)=\kappa w_{x x}(x, t)\). What is the value \(\delta\) ? (b) Show that \(w(x, t)=e^{-4 \pi^{2} t} \cos 2 \pi x\) is a solution of the initial-boundary value problem $$ \begin{array}{lc} w_{t}(x, t)=w_{x x}(x, t), \quad 0

Short answer

Question: Find the solution to the partial differential equation (PDE) \(u_t(x,t) = \kappa u_{xx}(x,t) + \alpha u(x,t)\), with the transformation \(u(x,t) = e^{\delta t}w(x,t)\), where \(\kappa = -4\pi^2\), \(\alpha = -4\), and given that the function \(w(x,t) = e^{-4\pi^2 t}\cos(2\pi x)\) is a solution for the simplified PDE with the specified boundary conditions. Answer: The solution to the given PDE is \(u(x,t) = e^{-(4+4\pi^2)t}\cos(2\pi x)\).

Step-by-step solution

Part (a): Find the value of \(\delta\) that simplifies the PDE involving \(w(x,t)\)

We are given the transformation \(u(x,t) = e^{\delta t}w(x,t)\). Let's compute \(u_t(x,t)\), \(u_{x}(x,t)\), and \(u_{xx}(x,t)\) in terms of \(w(x,t)\), \(w_t(x,t)\), \(w_x(x,t)\), and \(w_{xx}(x,t)\): 1. \(u_t(x,t) = \delta e^{\delta t}w(x,t) + e^{\delta t}w_t(x,t) = (\delta w(x,t) + w_t(x,t)) e^{\delta t}\) 2. \(u_x(x,t) = e^{\delta t}w_x(x,t)\) 3. \(u_{xx}(x,t) = e^{\delta t}w_{xx}(x,t)\) Now, substitute these derivatives into the given PDE, \(u_t(x,t) = \kappa u_{xx}(x,t) + \alpha u(x,t)\), and get: \((\delta w(x,t) + w_t(x,t))e^{\delta t} = \kappa (e^{\delta t}w_{xx}(x,t)) + \alpha (e^{\delta t}w(x,t))\) Divide both sides by \(e^{\delta t}\), which is always positive: \(\delta w(x,t) + w_t(x,t) = \kappa w_{xx}(x,t) + \alpha w(x,t)\) Now, we need to choose the value of \(\delta\) such that the transformed equation is \(w_t(x,t) = \kappa w_{xx}(x,t)\). To obtain this, let \(\delta = -\alpha\) so that the equation becomes: \(w_t(x,t) = \kappa w_{xx}(x,t)\) Thus, the value of \(\delta\) is \(-\alpha\).

Part (b): Verify the given function as a solution for the simplified PDE

We are given the function \(w(x,t) = e^{-4\pi^2 t}\cos(2\pi x)\) and asked to show that it is a solution of the given initial-boundary value problem: 1. Compute the first and second derivatives of w(x,t) with respect to x and t: - \(w_t(x,t) = -4\pi^2 e^{-4\pi^2 t}\cos(2\pi x)\) - \(w_x(x,t) = -2\pi e^{-4\pi^2 t}\sin(2\pi x)\) - \(w_{xx}(x,t) = -4\pi^2 e^{-4\pi^2 t}\cos(2\pi x)\) 2. Substitute the derivatives into the simplified PDE, \(w_t(x,t) = w_{xx}(x,t)\): - \(-4\pi^2 e^{-4\pi^2 t}\cos(2\pi x) = -4\pi^2 e^{-4\pi^2 t}\cos(2\pi x)\) This equation is true, so the given function satisfies the simplified PDE. 3. Check the initial and boundary conditions: - \(w_x(0,t) = -2\pi e^{-4\pi^2 t}\sin(0) = 0\) - \(w_x(1,t) = -2\pi e^{-4\pi^2 t}\sin(2\pi) = 0\) - \(w(x,0) = e^{0}\cos(2\pi x) = \cos(2\pi x)\) The given function satisfies the initial and boundary conditions, so it is indeed a solution for the initial-boundary value problem.

Part (c): Solve the original PDE using the results in Parts (a) and (b)

Based on Part (a), we know that the transformation \(u(x,t) = e^{\delta t}w(x,t)\) simplifies the PDE if \(\delta = -\alpha\). From Part (b), we have found that \(w(x,t) = e^{-4\pi^2 t}\cos(2\pi x)\) is a solution for the simplified PDE. Now, we have \(\delta = -\alpha = -4\), so we can find \(u(x,t)\): \(u(x,t) = e^{-4t}e^{-4\pi^2 t}\cos(2\pi x) = e^{-(4+4\pi^2)t}\cos(2\pi x)\) Therefore, the solution to the original PDE is \(u(x,t) = e^{-(4+4\pi^2)t}\cos(2\pi x)\).

Key concepts

Initial-Boundary Value Problem

Initial-boundary value problems (IBVPs) combine elements of both initial value problems and boundary value problems, crucial in solving partial differential equations (PDEs). They require conditions to be met both initially at time zero and along spatial boundaries.
This means when dealing with PDEs, we must provide specific information:

• Initial conditions at time zero. These dictate the state of the system at the start of interest.
• Boundary conditions along the spatial domain. These define how the system behaves at the edges of the region considered.

In the given problem, the equation involves both spatial and temporal variables. At time zero, the function equals a given cosine function, and at the boundaries, the spatial derivative vanishes. This structure sets the stage for solving the PDE under given constraints and finding a specific function that satisfies all conditions.

PDE Transformations

Transformations play a crucial role in simplifying and solving partial differential equations. They often convert a complicated PDE into a simpler form, making it easier to find solutions. In this problem, a transformation was applied where a new dependent variable, \( w(x, t) \), was introduced through the relation \( u(x, t) = e^{\delta t}w(x, t) \).

This specific transformation targets the elimination of the extra term \( \alpha u(x, t) \) in the original PDE:

\[ u_{t}(x, t)= \kappa u_{x x}(x,t)+\alpha u(x, t)\]
By selecting \(\delta = -\alpha\), the PDE simplifies notably to:
\[ w_{t}(x, t) = \kappa w_{x x}(x, t)\]
Such transformations maintain the core structure of the solution while simplifying complexities. This approach highlights the elegance and power of mathematical operations to facilitate problem-solving in differential equations.

Cosine Function Solutions

The cosine function proves particularly useful in solving PDEs linked to wave-like or oscillatory phenomena. In this exercise, the solution is illustrated using a specific cosine function:
\[w(x, t) = e^{-4\pi^2 t} \cos(2\pi x)\]
This solution addresses the IBVP by satisfying both the boundary and initial conditions. Here's why cosine functions are fitting in this context:

• Cosine functions are periodic, making them naturally fit problems involving periodic boundary conditions.
• Derivatives of cosine functions yield sines and cosines, facilitating elegant simplification of derivatives required in PDEs.
• The initial value matches the cosine function exactly when \( t = 0 \).

These properties collectively ensure that the chosen cosine function is an effective solution to the given PDE, fulfilling its required properties over time and space.