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Transformieren Sie die Differentialgleichung $$ \begin{array}{r} u_{x x}-u_{y y}=0 \text { für }\left|x^{2}-y^{2}\right| \leq 1 \\ u(x, y)=x^{2}+y^{2} \text { für }\left|x^{2}-y^{2}\right|=1 \end{array} $$ auf Hyperbelkoordinaten. Hinweis: Hyperbelkoordinaten sind durch die Transformation $$ \mathbf{x}(\rho, \phi)=\left(\begin{array}{l} x(\rho, \phi) \\ y(r, \phi) \end{array}\right):=\left(\begin{array}{c} \rho \cosh \phi \\ \rho \sinh \phi \end{array}\right) $$ gegeben.

Short Answer

Expert verified
Transform the equation using hyperbolic coordinates and solve the resulting equation with the given boundary conditions.

Step by step solution

01

Understand the transformation

In the problem statement, you are given the transformation into hyperbolic coordinates. Here, \( x = \rho \cosh \phi \) and \( y = \rho \sinh \phi \). We will use this transformation to convert the partial differential equation from Cartesian coordinates to hyperbolic coordinates.
02

Find expressions for partial derivatives

To transform the differential equation, compute the partial derivatives of \( u \) with respect to \( x \) and \( y \). First, express \( u_{xx} \) and \( u_{yy} \) in terms of \( \rho \) and \( \phi \) using the chain rule: \( u_{x} = \frac{\partial u}{\partial \rho} \frac{\partial \rho}{\partial x} + \frac{\partial u}{\partial \phi} \frac{\partial \phi}{\partial x} \) and similarly for the other derivatives.
03

Apply the chain rule

Using the given transformation, express \( \frac{\partial \rho}{\partial x} \), \( \frac{\partial \phi}{\partial x} \), \( \frac{\partial y}{\partial \rho} \), and their equivalents. These will involve the derivatives of hyperbolic functions. Calculate \( \frac{\partial \rho}{\partial x} = \cosh \phi \), \( \frac{\partial \phi}{\partial x} = -\frac{\sinh \phi}{\rho \sinh^2 \phi - \rho \cosh^2 \phi} \), and similar expressions for the derivatives with respect to \( y \).
04

Transform the differential equation

Using the derivatives from Step 3, substitute into the original partial differential equation \( u_{xx} - u_{yy} = 0 \). This yields a transformed equation in terms of \( \rho \) and \( \phi \). Simplify the equation as needed.
05

Use boundary conditions

The boundary condition is \( u(x, y) = x^2 + y^2 \) for \( |x^2 - y^2| = 1 \). Using the transformation, this becomes \( u(\rho, \phi) = \rho^2 \) for \( \rho^2 = 1 \). Apply this to the transformed equation to find any remaining constants or specific form for \( u(\rho, \phi) \).
06

Solve the transformed equation

After properly setting up the transformed equation using \( \rho \) and \( \phi \), solve for \( u(\rho, \phi) \). Depending on the structure of the equation, it may be a straightforward solution or require separation of variables or another integration technique.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Differential Equations
Partial Differential Equations (PDEs) involve functions of several variables and their partial derivatives. In this context, we encounter the PDE \( u_{xx} - u_{yy} = 0 \) which needs to be transformed. The variables \( x \) and \( y \) denote Cartesian coordinates. The task is to convert this into hyperbolic coordinates, which involve different variables—\( \rho \) and \( \phi \).

PDEs are fundamental in describing various phenomena in physics and engineering, such as heat conduction, wave propagation, and quantum mechanics. The equation here is a classic form known as the Laplace equation, often used for problems involving steady-state heat distribution or potential flow, which remains simplified by subtracting one second derivative from another.

Transforming PDEs into different coordinate systems can simplify the problem and make it more approachable, as can be seen with the use of hyperbolic coordinates. This helps in realigning the problem to more natural coordinates that might better respect the symmetry or constraints of the scenario being described.
Chain Rule
The chain rule is an essential tool in calculus for finding the derivative of composite functions. In this exercise, it aids us in converting partial derivatives from Cartesian to hyperbolic coordinates.

For example, to find \( u_x \) in terms of \( \rho \) and \( \phi \), we use the chain rule to express:
  • \( u_x = \frac{\partial u}{\partial \rho} \cdot \frac{\partial \rho}{\partial x} + \frac{\partial u}{\partial \phi} \cdot \frac{\partial \phi}{\partial x} \)
Similarly, apply the chain rule for other derivatives.

This approach allows us to break down complex derivatives into more manageable parts, especially when dealing with transformations involving non-linear functions, such as hyperbolic functions. The chain rule is a bridge between different frameworks of expressing physical phenomena, vital for applications requiring variable transformations.
Boundary Conditions
Boundary conditions are additional constraints that a solution must satisfy along the boundaries of the domain. In our given problem, the boundary condition is \( u(x, y) = x^2 + y^2 \) for \( |x^2 - y^2| = 1 \).

When translated into hyperbolic coordinates, this condition becomes \( u(\rho, \phi) = \rho^2 \) for \( \rho^2 = 1 \). Such conditions are crucial in finding a unique solution to a PDE. They specify what the solution must be at the edges of the spatial region under consideration.

There are different types of boundary conditions:
  • **Dirichlet boundary conditions** specify the value of a function directly.
  • **Neumann boundary conditions** specify the value of the function's derivative.
In this problem, we observe a Dirichlet condition because the specific value of \( u \) is prescribed on the region boundary. Appropriate consideration of the boundary conditions ensures the physical and mathematical relevance of the solution chosen.

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Most popular questions from this chapter

Zeigen Sie, dass für eine zweimal stetig differenzierbare Lösung \(u(x, y)\) des Randwertproblems $$ -\Delta u=f \quad \text { in } \Omega, \quad u=0 \quad \operatorname{auf} \Gamma_{d}, \quad \frac{\partial u}{\partial \mathbf{n}}=q \quad \operatorname{auf} \Gamma_{n} $$ die Integralgleichung $$ \int_{\Omega}[\nabla u \cdot \nabla h-f h] d F-\int_{\Gamma_{n}} q h d s=0 $$ gilt, wobei \(f, q\) gegebene, integrierbare Funktionen sind und \(\Gamma=\Gamma_{d} \cup \Gamma_{n}, \Gamma_{d} \cap\) \(\Gamma_{n}=\emptyset\), der Rand von \(\Omega\) ist. \(h\) sei stetig differenzierbar und auf \(\Gamma_{d}\) gleich Null. \(\frac{\partial}{\partial \mathrm{n}}\) bezeichnet die Ableitung in Richtung der äußeren Normalen auf dem Rand \(\operatorname{von} \Omega\).

Bestimmen Sie eine divergenzfreie Lösung \(\mathbf{u}(x, y)=(u(x, y), v(x, y))\), d.h. \(\operatorname{div} \mathbf{u}=0\), des Differentialgleichungssystems $$ \begin{aligned} u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y} &=-\frac{\partial p}{\partial x}+\frac{1}{R e} \Delta u \\ u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y} &=-\frac{\partial p}{\partial y}+\frac{1}{R e} \Delta v \end{aligned} $$ in einem Rechteckgebiet \(\Omega=\\{(x, y) \mid 0

Zeigen Sie, dass für ein 3-mal stetig differenzierbares divergenzfreies Geschwindigkeitsfeld \(\mathbf{u}=(u, v)\) aus den instationären STOKES- Gleichungen $$ \frac{\partial \mathbf{u}}{\partial t}=-\operatorname{grad} p+\frac{1}{R e} \Delta \mathbf{u} $$ für den Druck die Gleichung \(\Delta p=0\) folgt.

Bestimmen Sie mit Hilfe des Separationsansatzes Lösungen \(u(x, y)\) bzw. \(u(x, t)\) für die partiellen Differentialgleichungen (a) \(u_{x x}=4 u_{y}, u(0, y)=u(\pi, y)=0\), (b) \(a^{2} u_{x x}=u_{t t}, a>0\).

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