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
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
hunt_mat
hunt_mat is offline
#2
Sep5-11, 05:22 AM
HW Helper
P: 1,584
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: 1,584

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