Folks,

Could anyone check my attached work. Mathematica seems to output u(x,y)=-3+2x for the particular solution...

Thanks

Printable View

- May 5th 2011, 11:22 AMbugatti791st Order Semi Linear PDE xu_x+yu_y=u+3
Folks,

Could anyone check my attached work. Mathematica seems to output u(x,y)=-3+2x for the particular solution...

Thanks - May 5th 2011, 01:39 PMTheEmptySet
If we parametrize x and y we have

$\displaystyle x(s) \text{ and } x(0)=1$

and

$\displaystyle y(s) \text{ and } y(0)=y_0$

Then

$\displaystyle u(x(s),y(s))$

Now if we take the derivative with respect to s we have

$\displaystyle \frac{d}{ds}u(x(s),y(s))=\frac{dx}{ds}u_x+\frac{dy }{ds}u_y=u+3$

This gives the system of equations

$\displaystyle \frac{dx}{ds}=x, \quad x(0)=1$

$\displaystyle \frac{dy}{ds}=y, \quad y(0)=y_0$

$\displaystyle \frac{du}{ds}=u+3$

Solving this gives

$\displaystyle x=e^{s} \quad y=y_0e^{s} \quad u=Ce^{s}-3$

Now if we sub out

$\displaystyle e^{s}$

we get

$\displaystyle u(x,y)=Cx-3 \implies u(1,y)=-1=C-3 \iff C=2$

$\displaystyle u(x,y)=2x-3$ - May 5th 2011, 01:49 PMJester
Here's your mistake

First let $\displaystyle e^{f} = g$ so $\displaystyle u = x g\left(\dfrac{y}{x}\right) - 3$. If $\displaystyle u(1,y) = -1$ then $\displaystyle -1 = g(y) - 3$ so $\displaystyle g(y) = 2$ so $\displaystyle u = x \cdot 2 - 3$ (your Mathematica and TheEmptySet's solution) - May 5th 2011, 11:45 PMbugatti79
Cheers Danny!

I like this alternative method, TheEmptySet. Thanks