Cauchy problem, method of characteristics

In summary, the Cauchy problem \displaystyle \frac{1}{2x}u_x + xu u_y + u^2 = 0, subject to \displaystyle u(x,x) = \frac{1}{x^2}, x > 0, can be solved using the characteristic equations \displaystyle x_t = \frac{1}{2x}, y_t = xu, u_t = -u^2 and the initial conditions x(0,s) = s, y(0,s) = s, u(0,s) = \frac{1}{s^2}. After solving the characteristic equations, we obtain the solution u = \frac{1}{x^2}, which is independent
  • #1
math2011
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0
This question is also posted at (with better formatting): http://www.mathhelpforum.com/math-help/f59/cauchy-problem-method-characteristics-187192.html.

Solve the following Cauchy problem
[tex]\displaystyle \frac{1}{2x}u_x + xu u_y + u^2 = 0[/tex],
subject to
[tex]\displaystyle u(x,x) = \frac{1}{x^2}, x > 0[/tex].

Attempt:

The characteristic equations are [tex]\displaystyle x_t = \frac{1}{2x}, y_t = xu, u_t = -u^2[/tex].

The initial conditions are [tex]x(0,s) = s, y(0,s) = s, u(0,s) = \frac{1}{s^2}[/tex].

The Jacobian is [tex]J = \begin{vmatrix}\frac{1}{2s} & \frac{1}{s} \\1 & 1\end{vmatrix} = - \frac{1}{2s}[/tex] and hence we expect a unique solution when [tex]s \ne \pm \infty[/tex] and [tex]s \ne 0[/tex]. (Is this correct?)

Now solve the characteristic equations.

[tex]\displaystyle \frac{dx}{dt} = \frac{1}{2x}[/tex]
[tex]2x dx = dt[/tex]
[tex]x^2 = t + f_1(s)[/tex].
Apply initial condition to get [tex]f_1(s) = s^2[/tex] and hence [tex]x = \sqrt{t + s^2}[/tex].

[tex]\displaystyle \frac{du}{dt} = -u^2[/tex]
[tex]\frac{1}{u^2} du = - dt[/tex]
[tex]u^{-1} = t + f_2(s)[/tex]
[tex]u = \frac{1}{t + f_2(s)}[/tex].
Apply initial condition to get [tex]f_2(s) = s^2[/tex] and hence [tex]\displaystyle u = \frac{1}{t + s^2} = \frac{1}{x^2}[/tex]. (Is this it for the question? Why is u independent of y? What have I done wrong?)

Substitute above x and y into characteristic equation [tex]y_t = xu[/tex] and we get [tex]\displaystyle y = \frac{1}{\sqrt{t + s^2}}[/tex]. Integrate over t and we get [tex]y = 2\sqrt{t + s^2} + f_3(s)[/tex]. Apply initial condition we get [tex]f_3(s) = -s[/tex] and [tex]y = 2 \sqrt{t + s^2} - s[/tex].

From expressions of x and y obtained above we get
[tex]t = x^2 - s^2[/tex]
[tex]\displaystyle t = \frac{1}{4}(y + s)^2 - s^2[/tex].

Therefore the characteristics is [tex](y + s)^2 = 4 x^2[/tex]. (Do I need this characteristics at all? What should I do with it?)


Is the above attempt correct?
 
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  • #2
You have for y:
[tex]
\frac{dy}{dt}=\frac{1}{\sqrt{s^{2}+t}}\Rightarrow y(t)-s=\int_{0}^{t}\frac{du}{\sqrt{s^{2}+u}}
[/tex]
This will give a different y than you have and it's where you have made a mistake I think. Once you have x and y in terms of s and t ten you should be able to get u in terms of x and y.
 
  • #3
Why does [tex]\frac{dy}{dt} = \frac{1}{\sqrt{s^2 + t}}[/tex] imply [tex]y(t) - s = \int^t_0 \frac{1}{\sqrt{s^2 + u}} du[/tex]?

How do you get the -s on the LHS of the second equation?

Why is the integration on the RHS only from 0 to t? (I just realized that s > 0, is this related to the integral?)

What is the reason for substituting t with u and integrating from 0 to t?
 
  • #4
Integrate both sides and use the initial conditions:
[tex]
\int_{0}^{t}\frac{dy}{dt}dt=\int_{0}^{t}\frac{dr}{\sqrt{s^{2}+r}}=\left[ y\right]_{0}^{t}=y(t)-y(0)=y(t)-s
[/tex]
Use a change of variables [itex]v=r+s^{2}[/itex] and the integral becomes:
[tex]\int_{0}^{t}\frac{dr}{\sqrt{s^{2}+r}}=\int_{s^{2}}^{t+s^{2}}\frac{dv}{\sqrt{v}}
[/tex]
 
  • #5
I tried this and it turns out to be the same as what I got before.
[tex]\int^t_0 \frac{1}{\sqrt{s^2 + r}} dr = \int^{s^2+t}_{s^2} \frac{1}{\sqrt{v}} dv = \left[2 v^{\frac{1}{2}} \right]^{s^2+t}_{s^2} = 2\sqrt{s^2 + t} - 2s[/tex]
[tex]y(t) = 2\sqrt{s^2 + t} - s[/tex]
Then
[tex]s = 2x - y[/tex]
[tex]t = x^2 - (2x - y)^2[/tex]
and
[tex]u = \frac{1}{t + s^2} = \frac{1}{x^2 - (2x - y)^2 + (2x - y)^2} = \frac{1}{x^2}[/tex]
Maybe this is the correct solution.
 

1. What is the Cauchy problem in the context of mathematical analysis?

The Cauchy problem is a type of initial value problem in mathematics that involves finding a solution to a partial differential equation (PDE) given certain initial conditions. It is named after the French mathematician Augustin-Louis Cauchy, who first studied this type of problem in the 19th century.

2. What is the method of characteristics?

The method of characteristics is a technique used to solve first-order linear PDEs, including the Cauchy problem. It involves finding a set of curves, called characteristics, that satisfy the PDE and then using these curves to construct a solution to the problem.

3. How does the method of characteristics work?

The method of characteristics works by first finding the characteristics of the PDE, which are curves along which the PDE can be reduced to a set of ordinary differential equations (ODEs). Then, the ODEs can be solved to obtain a general solution, which can be used to determine the solution to the Cauchy problem.

4. What types of PDEs can be solved using the method of characteristics?

The method of characteristics is most commonly used to solve first-order linear PDEs, but it can also be applied to some nonlinear PDEs. It is particularly useful for PDEs with constant coefficients.

5. What are some applications of the Cauchy problem and method of characteristics in science?

The Cauchy problem and method of characteristics have applications in various fields of science, including fluid dynamics, electromagnetism, and quantum mechanics. They can be used to model and predict the behavior of physical systems, such as the flow of fluids and the propagation of electromagnetic waves.

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