Cauchy problem, method of characteristics


by math2011
Tags: cauchy, characteristics, method
math2011
math2011 is offline
#1
Sep4-11, 09:32 AM
P: 10
This question is also posted at (with better formatting): http://www.mathhelpforum.com/math-he...cs-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?
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
hunt_mat
hunt_mat is offline
#2
Sep5-11, 05:22 AM
HW Helper
P: 1,585
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.
math2011
math2011 is offline
#3
Sep5-11, 09:15 AM
P: 10
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?

hunt_mat
hunt_mat is offline
#4
Sep5-11, 09:39 AM
HW Helper
P: 1,585

Cauchy problem, method of characteristics


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]
math2011
math2011 is offline
#5
Sep6-11, 06:25 AM
P: 10
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.


Register to reply

Related Discussions
PDE problem, Solve using Method of Characteristics Calculus & Beyond Homework 7
General Method of Characteristics Problem Differential Equations 4
First order pde cauchy problem by method of characteristics Differential Equations 1
First Order PDE Cauchy problem Using Method of Characteristics Calculus & Beyond Homework 1
A Difficult Method of Characteristics Problem Calculus & Beyond Homework 0