Problem 8
Question
Derive the equations of the characteristics of the differential equation \(t \partial f / \partial x+x^{3} \partial f / \partial t=0\) for the quadrant \(x>0, t>0\). Hence sketch the characteristics and solve the initial value problem $$ f(x, 0)=e^{-x} $$ What is the range of influence in this problem?
Step-by-Step Solution
Verified Answer
Characteristics come from solving the ODEs, range of influence is from \(t=0\) to the characteristics in the first quadrant.
1Step 1: Understand the given differential equation
The differential equation provided is \( t \frac{\partial f}{\partial x} + x^3 \frac{\partial f}{\partial t} = 0 \). This is a first-order partial differential equation that can be solved using the method of characteristics. Characteristics are curves along which the partial differential equation becomes an ordinary differential equation (ODE).
2Step 2: Parameterize the characteristics
To find the characteristics, we first write down the system of ODEs that govern the characteristics. These are given by:\( \frac{dt}{ds} = x^3 \),\( \frac{dx}{ds} = t \),\( \frac{df}{ds} = 0 \).Here, \(s\) is a parameter along each characteristic curve.
3Step 3: Solve for the characteristic curves
Let's solve the system of differential equations:1. Solve \( \frac{dt}{ds} = x^3 \): Assuming \( x(0) = x_0 \), integrate to find the relation between \(t\) and \(s\).2. Solve \( \frac{dx}{ds} = t \): Using the solution for \(t\) from the first equation, integrate to find \(x(s)\) in terms of \(s\).3. Solve \( \frac{df}{ds} = 0 \): This implies that \(f\) is constant along each characteristic.
4Step 4: Solve the initial conditions problem
The initial condition given is \( f(x, 0) = e^{-x} \). From the third characteristic equation \( f(s) = constant \), use the initial condition where \(t=0\) to solve for \(f\). This implies that along each characteristic starting from the \(t=0\) line, \(f\) retains its initial value.
5Step 5: Sketch the characteristics
Each characteristic line is derived from the parameterized solution of the ODEs. Sketch these lines in the \(x-t\) plane starting from the initial line where \(t=0\) and extending into the region where \(x > 0, t > 0\).
6Step 6: Determine the range of influence
The range of influence for a point \((x_0, 0)\) in this context is the part of the domain \((x>0, t>0)\) that is affected by the initial condition at \((x_0, 0)\). Based on the characteristic curves, each point in this domain is influenced by its initial counterpart on the line where \(t=0\).
Key Concepts
Partial Differential EquationsCharacteristic CurvesInitial Value ProblemRange of Influence
Partial Differential Equations
Partial Differential Equations (PDEs) are equations that involve unknown functions of multiple variables and their partial derivatives. These equations are essential in capturing the behavior of various physical systems. The given problem involves a first-order PDE: \( t \frac{\partial f}{\partial x} + x^3 \frac{\partial f}{\partial t} = 0 \), where the aim is to solve it using the method of characteristics.
This particular PDE is characterized by its mixed derivatives of a function \( f \) with respect to the variables \( x \) and \( t \). This complexity often makes direct solutions challenging, necessitating an effective method like the method of characteristics that transforms the PDE into a simpler Ordinary Differential Equation (ODE) along characteristic curves.
This particular PDE is characterized by its mixed derivatives of a function \( f \) with respect to the variables \( x \) and \( t \). This complexity often makes direct solutions challenging, necessitating an effective method like the method of characteristics that transforms the PDE into a simpler Ordinary Differential Equation (ODE) along characteristic curves.
Characteristic Curves
Characteristic curves are key tools in solving PDEs, particularly first-order ones. They are the paths in the \(x-t\) plane along which the PDE transforms into an ODE, which is easier to solve. For this problem, the characteristic curves satisfy the equations \( \frac{dt}{ds} = x^3 \), \( \frac{dx}{ds} = t \), and \( \frac{df}{ds} = 0 \).
These equations describe the trajectory or path that each characteristic curve follows in the \(x-t\) plane. Solving these gives us insights into how the solution \( f \) behaves along these curves. The last equation, \( \frac{df}{ds} = 0 \), indicates that \( f \) remains constant along each of these characteristic curves, offering a powerful way to relate the initial conditions to the solution.
These equations describe the trajectory or path that each characteristic curve follows in the \(x-t\) plane. Solving these gives us insights into how the solution \( f \) behaves along these curves. The last equation, \( \frac{df}{ds} = 0 \), indicates that \( f \) remains constant along each of these characteristic curves, offering a powerful way to relate the initial conditions to the solution.
Initial Value Problem
An initial value problem in the context of partial differential equations involves finding a solution that satisfies a specified condition at a given initial time or location. Here, the initial condition given is \( f(x, 0) = e^{-x} \). This represents the state of the system at \( t = 0 \), forming the starting point for solving the PDE.
Using the method of characteristics, we apply these initial values to the equation \( \frac{df}{ds} = 0 \), confirming that \( f \) retains its initial value \( e^{-x} \) along each curve. Thus, each point on the \(t=0\) line influences only the part of the solution lying on the characteristic running through it.
Using the method of characteristics, we apply these initial values to the equation \( \frac{df}{ds} = 0 \), confirming that \( f \) retains its initial value \( e^{-x} \) along each curve. Thus, each point on the \(t=0\) line influences only the part of the solution lying on the characteristic running through it.
Range of Influence
The range of influence is a crucial concept when discussing wave propagation and PDEs. It refers to the area of influence that an initial condition exerts on the solution domain.
In our problem, a point \((x_0, 0)\) on the initial line \(t=0\) influences the solution along the characteristic curve that originates from there. Since the characteristics are derived from the solution to the differential equations, they depict the paths along which the initial values propagate through the \(x-t\) plane.
This means each point \((x_0, 0)\) dictates the behavior of \(f(x,t)\) along its characteristic curve. The range of influence for each initial point is thus crucial for constructing the overall solution profile in the domain \(x > 0, t > 0\).
In our problem, a point \((x_0, 0)\) on the initial line \(t=0\) influences the solution along the characteristic curve that originates from there. Since the characteristics are derived from the solution to the differential equations, they depict the paths along which the initial values propagate through the \(x-t\) plane.
This means each point \((x_0, 0)\) dictates the behavior of \(f(x,t)\) along its characteristic curve. The range of influence for each initial point is thus crucial for constructing the overall solution profile in the domain \(x > 0, t > 0\).
Other exercises in this chapter
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