Problem 14
Question
Show that the "backwards" heat equation $$ \frac{\partial u}{\partial t}=-k \frac{\partial^{2} u}{\partial x^{2}} $$ suhject to \(u(0, t)=u(L, t)=0\) and \(u(x, 0)=f(x)\), is not well posed. [Hint: Show that if the data are changed an arbitrarily small amount, for example $$ f(x) \longrightarrow f(x)+\frac{1}{n} \sin \frac{n \pi x}{L} $$ for large \(n\), then the solution \(u(x, t)\) changes by a large amount.]
Step-by-Step Solution
Verified Answer
Because the solution does not have continuous dependence on the initial data, the backwards heat equation is not a well-posed problem.
1Step 1: Analyzing and Setting up the Equation
We start by considering the backwards heat equation which is given as \[ \frac{\partial u}{\partial t}=-k \frac{\partial^{2} u}{\partial x^{2}} \] Subject to \[ u(0, t)=u(L, t)=0 \] and \[ u(x, 0)=f(x) \] Now, proceed by making an arbitrarily small change to the initial condition \(f(x)\) by introducing a tweak as given: \[ f(x) \rightarrow f(x)+\frac{1}{n} \sin \frac{n \pi x}{L} \] for large n
2Step 2: Checking the Unbounded Variation
Points of interest are the solutions to the backwards heat equation with this new initialization. Now solving the equation with \( f(x)=\frac{1}{n} \sin \frac{n \pi x}{L} \), while keeping in mind the boundary conditions, we get the solution: \[ u(x, t) = e^{k(n \pi /L)^2 t} \sin (n \pi x/L) \] We can observe that as \( t \rightarrow 0^+ \), this value increases without bound for fixed \( x \), which clearly contradicts the concept of continuous dependence on the initial data.
3Step 3: Drawing Conclusion
Given that a minor modification in function \( f(x) \) at \( t=0 \) led to an unbounded alteration in solution \( u(x, t) \) as \( t \rightarrow 0^+ \), we can conclude that the backwards heat equation is not well-posed, because it does not meet the criteria of the solution depending continuously on the data.
Key Concepts
Well-posednessPartial Differential EquationsInitial Boundary Value Problem
Well-posedness
Understanding whether a mathematical problem is well-posed is crucial. The concept of well-posedness comes from the French mathematician Jacques Hadamard. This idea is primarily used in the context of differential equations. A problem is deemed well-posed if it satisfies three key criteria:
This illustrates a lack of continuous dependency, leading us to conclude that the backwards heat equation is not well-posed. Problems that are ill-posed can be difficult to solve and may exhibit large variations in outcome due to small changes in input.
- A solution exists.
- The solution is unique.
- The solution's behavior changes continuously with the initial conditions.
This illustrates a lack of continuous dependency, leading us to conclude that the backwards heat equation is not well-posed. Problems that are ill-posed can be difficult to solve and may exhibit large variations in outcome due to small changes in input.
Partial Differential Equations
Partial differential equations, or PDEs, are fundamental in modeling a variety of phenomena in science and engineering. These equations contain unknown multivariable functions and their partial derivatives. The power of PDEs lies in their ability to describe systems with varying conditions over continuous variables such as space and time.
In general, PDEs can be classified based on their characteristics such as linearity, order, and homogeneity. A common form of PDEs is the heat equation, which models the distribution of heat (or variation in temperature) in a given region over time.
In general, PDEs can be classified based on their characteristics such as linearity, order, and homogeneity. A common form of PDEs is the heat equation, which models the distribution of heat (or variation in temperature) in a given region over time.
- Order: The order of a PDE is determined by the highest derivative order it contains.
- Linearity: A linear PDE implies all terms are of the first degree in the function and its derivatives.
Initial Boundary Value Problem
An initial boundary value problem (IBVP) in the context of PDEs combines initial conditions with boundary conditions to specify a unique solution. Initial conditions define the state of a system at the beginning of the problem, while boundary conditions specify the behavior of the system at the given spatial limits.
For example, in a heat equation problem, these conditions can include the initial temperature distribution (initial condition) and the temperatures at the ends of a rod (boundary conditions).
For example, in a heat equation problem, these conditions can include the initial temperature distribution (initial condition) and the temperatures at the ends of a rod (boundary conditions).
- Initial Condition: Specifies the starting point or state of the system.
- Boundary Condition: Establishes constraints at the spatial boundaries of the problem.
- Initial condition: \( u(x, 0) = f(x) \)
- Boundary conditions: \( u(0, t) = u(L, t) = 0 \)
Other exercises in this chapter
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