# Solving PDE using Method of Characteristics

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 $$x$$

Thanks a mil

-

1
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

1
– pedja Oct 5 '11 at 14:23

## 1 Answer

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.

-
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

1
– 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

1
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

1
Solve for $s$ and substitute the solution into the other equation. – Sasha Oct 5 '11 at 15:42
show 5 more comments
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 19, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9420889616012573, "perplexity_flag": "head"}