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I need to solve the following by using the method of characteristics $$u\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1~,~u|_{x=y}=\frac{x}{2}$$

I have the following characteric equations:

$$\frac{dx}{ds}=u~;\frac{dy}{ds}=1~;\frac{du}{ds}=1$$

from the above I get

$$ x=us+x_{0} $$ $$ y=s +y_{0} $$ $$ u=s+u_{0} $$

I am now thinking I should go with the standard conditions $$y_0=0$$ and $$u(x,0)=f(x_0)$$

this now gives me: $$ x=uy+x_o $$ $$ y=s $$ $$ u=y+f(x_0) $$

Im confused because of the $$u$$ term in my equation for $$x$$

Can anyone please help.

Thanks a mil

doraemonpaul
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sarah jamal
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    You will be more likely to get a good answer if you were to accept those offered on earlier question of yours. It is easy, you just click the tick mark at the top-left of the answer post you found most helpful. – Sasha Oct 05 '11 at 12:50
  • thanks Sasha I didnt know about the whole 'accept' thing. blush hope that helps encourage helpers. ;) – sarah jamal Oct 05 '11 at 13:19
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    http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node8.html – Pedja Oct 05 '11 at 14:23

1 Answers1

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Your solution to characteristic equations is incorrect, which you can easily check by plugging your current solution back in. The source of the problem is that $u(s)$ is not constant, thus $x^\prime(s) = u(s)$ is not solved by $x(s) = u(s) s + x_0$.

It may help to note that $x^{\prime\prime}(s) = u^\prime(s) = 1$. Now that you are back into ODE with constant coefficients, finding solutions should be easy.

Sasha
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  • Thanks Sasha. Do i solve $$\frac{d^2x}{ds^2}=1$$ as: $$\frac{dx}{ds}=s+k$$ and so $$x=\frac{1}{2}s^2+ks+x_0$$ ? – sarah jamal Oct 05 '11 at 14:17
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    Yes, indeed. You could also use WolframAlpha to check your results. – Sasha Oct 05 '11 at 14:21
  • Ok, so the other characteristic equations were right? So I have $$y=s$$ and $$u=s+f(x_0)$$ and then using what I have above $$x_0=x-\frac{1}{2}y^2-ky$$ I am still confused about what to do about the condition $$u|_x=y=\frac{x}{2}$$ – sarah jamal Oct 05 '11 at 14:29
  • sorry i tried to edit the comment, but i was too late. its supposed to say $$u|_{x=y}=\frac{x}{2}$$ – sarah jamal Oct 05 '11 at 14:37
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    We have $u(s) = s + u_0$, $x(s) = \frac{s^2}{2} + k s + x_0$, and $y(x) = s + y_0$. The boundary conditions $\left. 2 u \right|_{x=y} = x$ means that for $s$, such that $x(s) =y(s)$, $2 u(s) = x(s)$. This leads to a system of equations for the undetermined coefficients. – Sasha Oct 05 '11 at 14:58
  • do i set up a kind of simultaneous equation thing to solve? so $$y(s)=x(s)$$ means $$s+y_0=\frac{1}{2}s^2+ks+x_0$$ this is my first equation. and $$2u(s)=x(s)$$ means $$2s+2u_0=\frac{1}{2}s^2+ks+x_0$$, this is my second equation. If I take eq2 - eq1 i will get $$s+u_0-y_0=0$$ I dont know if I am on the right track, and if I am, I dont know what to do from here. :( – sarah jamal Oct 05 '11 at 15:12
  • sorry, i tried to correct my last statement but i was too late again. @sasha I would have $$s+2u_0-y_0=0$$ – sarah jamal Oct 05 '11 at 15:19
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    Solve for $s$ and substitute the solution into the other equation. – Sasha Oct 05 '11 at 15:42
  • let us continue this discussion in chat – sarah jamal Oct 05 '11 at 15:53
  • its easier to put the math stuff here. But It says I should move it to chat. :| ... ok, so I put in the chat that I get $$s=y_0-2u_0$$ so where do I sub this into? I feel like Im going round in circles. – sarah jamal Oct 05 '11 at 16:02