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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:


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

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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 5 '11 at 12:50
thanks Sasha I didnt know about the whole 'accept' thing. blush hope that helps encourage helpers. ;) – sarah jamal Oct 5 '11 at 13:19
up vote 1 down vote accepted

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.

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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 5 '11 at 14:17
Yes, indeed. You could also use WolframAlpha to check your results. – Sasha Oct 5 '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 5 '11 at 14:29
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 5 '11 at 14:58
Solve for $s$ and substitute the solution into the other equation. – Sasha Oct 5 '11 at 15:42

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