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Chapter 18: Problem 8

A function \(u(x, y)\) satisfies $$ 2 \frac{\partial u}{\partial x}+3 \frac{\partial u}{\partial y}=10 $$ and takes the value 3 on the line \(y=4 x\). Evaluate \(u(2,4)\).

Short Answer

Expert verified
u(2,4) = 3

Step by step solution

01

Set up the partial differential equation

Given the partial differential equation: \[2 \frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = 10\]This needs to be solved subject to the initial condition provided.
02

Identify the initial condition

The problem states that the function takes the value 3 on the line \(y = 4x\). This can be denoted as \(u(x, 4x) = 3\).
03

Solve the characteristic equations

To solve the PDE, we use the method of characteristics. The characteristic equations are:\[\frac{dx}{2} = \frac{dy}{3} = \frac{du}{10}\]From \(\frac{dx}{2} = \frac{dy}{3}\), we get:\[3 dx = 2 dy\]Integrate both sides:\[3x = 2y + C\]
04

Integrate to find the solution

To find \(u\), we need another characteristic equation. From \(\frac{dx}{2} = \frac{du}{10}\), we get:\[10 dx = 2 du\]Integrate both sides:\[10x = 2u + K\]Now solve for \(u\):\[u = 5x + \frac{K}{2}\]
05

Substitute the initial condition

Using the initial condition \(u(x, 4x) = 3\), substitute \(y = 4x\) and \(u = 3\) into the general solution to solve for the constant.\[3 = 5x + \frac{K}{2}\]Solve for \(K\):\[K=6-10x\]
06

Write the particular solution for \(u(x, y)\)

Substitute \(K=6-10x\) back into the general solution:\[u = 5x + 3 - 5x\]So \(u = 3\).
07

Evaluate at \((2,4)\)

The question asks for \(u(2,4)\).Since the solution is constant, \(u(2,4) = 3\).

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Key Concepts

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

characteristic equations
Understanding characteristic equations is essential when tackling partial differential equations (PDEs). In this exercise, we start with the PDE ewline2 \frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = 10ewlineThe method of characteristics helps convert this PDE into a more manageable form. The characteristic equations arise from equating the derivatives along the characteristic curves. For our specific PDE, the characteristic equations are set as follows: ewline\[\frac{dx}{2} = \frac{dy}{3} = \frac{du}{10}\]. ewline This gives us relations between x, y, and u along the characteristic curves. From ewline\[\frac{dx}{2} = \frac{dy}{3}\], ewline we rearrange and integrate to find: ewline\[3 dx = 2 dy\]. ewline Integrating, we get: ewline \[3x = 2y + C\],ewline where C is a constant. This line represents one characteristic curve. Similarly, from ewline\[\frac{dx}{2} = \frac{du}{10}\], ewline we rearrange and integrate to find: ewline \[10 dx = 2 du\]. ewline Integrating, we obtain: ewline \[10x = 2u + K\], ewline with K being another constant. Here, we have another characteristic relation, valuable for solving our PDE.
initial conditions
Initial conditions provide the necessary information to uniquely determine the solution to a differential equation. In this exercise, the initial condition is given as the value of the function on a specific line: ewline\[u(x, 4x) = 3\]. ewlineThis means that for any point (x, 4x) on that line, the function u(x, y) takes the value 3. We utilize this initial information to find the constants in our general solution derived from the characteristic equations. ewlineSubstituting y = 4x into the solution obtained from the characteristic equations allows us to solve for K: ewline\[3 = 5x + \frac{K}{2} \]. ewlineSolving for K, we get: ewline\[K = 6 - 10x\]. ewlineThis step is critical as it transforms our general solution into a specific and unique function compatible with the initial conditions provided.
method of characteristics
The method of characteristics is a powerful technique for solving certain partial differential equations. It reduces a PDE into a set of ordinary differential equations (ODEs), making it easier to find the solution. The method focuses on finding characteristic curves along which the solution can be more simply determined. ewlineFor our PDE ewline\[2 \frac{\frac{\partial u}{\partial x}} + 3 \frac{\frac{\partial u}{\partial y}} = 10\], ewlinewe first write the characteristic equations: ewline\[ \frac{dx}{2} = \frac{dy}{3} = \frac{du}{10}\]. ewlineWe solve these step-by-step:ewline\[\frac{dx}{2} = \frac{dy}{3}\rightarrow 3dx=2dy \rightarrow \text{integrate:} \rightarrow 3x = 2y + C\]. ewlineThen:ewline\[\frac{dx}{2} = \frac{du}{10}\rightarrow 10dx=2du \rightarrow \text{integrate:} \rightarrow 10x = 2u + K\]. ewlineCombining these, we derive the general solution and adjust it using the initial conditions to get:ewline\[u = 5x + 3 - 5x=3\]. ewlineFinally, the method verifies that the solution holds true at specific points, such as ewline(2,4)ewlineSince the solution is consistent, ewline\[u(2,4) = 3\].

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Most popular questions from this chapter

A sheet of material of thickness \(w\), specific heat capacity \(c\) and thermal conductivity \(k\) is isolated in a vacuum, but its two sides are exposed to fluxes of radiant heat of strengths \(J_{1}\) and \(J_{2}\). Ignoring short-term transients, show that the temperature difference between its two surfaces is steady at \(\left(J_{2}-J_{1}\right) w / 2 k\), whilst their average temperature increases at a rate \(\left(J_{2}+J_{1}\right) / c w\).

The non-relativistic Schrödinger equation (18.7) is similar to the diffusion equation in having different orders of derivatives in its various terms; this precludes solutions that are arbitrary functions of particular linear combinations of variables. However, since exponential functions do not change their forms under differentiation, solutions in the form of exponential functions of combinations of the variables may still be possible. Consider the Schrödinger equation for the case of a constant potential, i.e. for a free particle, and show that it has solutions of the form \(A \exp (l x+m y+n z+\lambda t\) ) where the only requirement is that $$ -\frac{\hbar^{2}}{2 m}\left(l^{2}+m^{2}+n^{2}\right)=i \hbar \lambda $$ In particular, identify the equation and wavefunction obtained by taking \(\lambda\) as \(-i E / \hbar\), and \(l, m\) and \(n\) as \(i p_{x} / \hbar, i p_{y} / \hbar\) and \(i p_{z} / \hbar\) respectively, where \(E\) is the energy and \(p\) the momentum of the particle; these identifications are essentially the content of the de Broglie and Einstein relationships.

An incompressible fluid of density \(\rho\) and negligible viscosity flows with velocity \(v\) along a thin straight tube, perfectly light and flexible, of cross-section \(A\) and held under tension \(T\). Assume that small transverse displacements \(u\) of the tube are governed by $$ \frac{\partial^{2} u}{\partial t^{2}}+2 v \frac{\partial^{2} u}{\partial x \partial t}+\left(v^{2}-\frac{T}{\rho A}\right) \frac{\partial^{2} u}{\partial x^{2}}=0 $$ (a) Show that the general solution consists of a superposition of two waveforms travelling with different speeds. (b) The tube initially has a small transverse displacement \(u=a \cos k x\) and is suddenly released from rest. Find its subsequent motion.

Find the most general solutions \(u(x, y)\) of the following equations consistent with the boundary conditions stated: (a) \(y \frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=0, \quad u(x, 0)=1+\sin x\) (b) \(i \frac{\partial u}{\partial x}=3 \frac{\partial u}{\partial y}, \quad u=(4+3 i) x^{2}\) on the line \(x=y\); (c) \(\sin x \sin y \frac{\partial u}{\partial x}+\cos x \cos y \frac{\partial u}{\partial y}=0, \quad u=\cos 2 y\) on \(x+y=\pi / 2\) (d) \(\frac{\partial u}{\partial x}+2 x \frac{\partial u}{\partial y}=0, \quad u=2\) on the parabola \(y=x^{2}\).

In those cases in which it is possible to do so, evaluate \(u(2,2)\), where \(u(x, y)\) is the solution of $$ 2 y \frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=2 x y\left(2 y^{2}-x^{2}\right) $$ that satisfies the (separate) boundary conditions given below. (a) \(u(x, 1)=x^{2}\) for all \(x\). (b) \(u(x, 1)=x^{2}\) for \(x \geq 0\). (c) \(u(x, 1)=x^{2}\) for \(0 \leq x \leq 3\) (d) \(u(x, 0)=x\) for \(x \geq 0\) (e) \(u(x, 0)=x\) for all \(x\). (f) \(u(1, \sqrt{10})=5\) (g) \(u(\sqrt{10}, 1)=5\).
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