
#1
Sep411, 09:32 AM

P: 10

This question is also posted at (with better formatting): http://www.mathhelpforum.com/mathhe...cs187192.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? 



#2
Sep511, 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. 



#3
Sep511, 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? 



#4
Sep511, 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] 



#5
Sep611, 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. 


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