Solve QL PDE in parametric form

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February 9th 2011, 10:33 AM
bugatti79

Solve QL PDE in parametric form

$ \displaystyle (x+u)u_x+yu_y=u+y^2 $

Asked to derive the general solution in the given form which I have done and is as attached...

$ x=y_0^2(s)e^{2t}+te^t (u_0(s)-y_0^2(s))+(x_0(s)-y_0^2(s)e^t $

$ y=y_0(s)e^t $

$ u=(u_0(s)-y_0^2(s))e^t+y_0^2(s)e^{2t} $

The intial data curve is $x=x_0(s), y=y_0(s), u=u_0(s)$

For clarity the last line of attachment is as above which I know is correct...

How do I show that $\displaystyle u(x,y)=\frac{x-y^2}{1+ln(y)} +y^2$ for the special case u=x when y=1?
I believe this requires getting u in terms of x and y but I dont know how to do that with variables $x_0(s), y_0(s), u_0(s)$ in the above equations...
February 9th 2011, 01:48 PM
Danny

With the solution you have we find that

$u - y^2 = \left(u_0 - y_0^2\right)e^t$

$x - y^2 = \left(u_0-y_0^2\right) t e^t + \left(x_0 - y_0^2)e^t$

and dividing gives

$\dfrac{u - y^2}{x - y^2} = \dfrac{u_0-y_0^2}{\left(u_0-y_0^2\right)t + x_0-y_0^2}$

or

$\dfrac{u - y^2}{x - y^2} = \dfrac{1}{t + \dfrac{x_0-y_0^2}{u_0-y_0^2}}$ .

From your $y$ solution $t = \ln y - \ln y_0$ .
Thus,

$\dfrac{u - y^2}{x - y^2} = \dfrac{1}{\ln y - \ln y_0 + \dfrac{x_0-y_0^2}{u_0-y_0^2}}$ .

Imposing the initial curves gives your answer.
February 9th 2011, 02:54 PM
bugatti79

Quote:

Originally Posted by Danny

From your $y$ solution $t = \ln y - \ln y_0$ .
Thus,

$\dfrac{u - y^2}{x - y^2} = \dfrac{1}{\ln y - \ln y_0 + \dfrac{x_0-y_0^2}{u_0-y_0^2}}$ .

Imposing the initial curves gives your answer.

rearranging your expression then

$\displaystyle u=\frac{x-y^2}{lny-lny_0(s) +\frac{x_0+y_0^2(s)}{u_0+y_0^2(s)}}+y^2$

How is the 1+ln y determined in the denominator?

Do you mean replacing x_0(s) with x, y_0(s) with y and u_0(s) with u, because I do not get the intended answer.
February 9th 2011, 03:05 PM
Danny

Are you not told that on the initial curve that

$u_0 = x_0$ and $y_0 = 1$ ?
February 9th 2011, 03:07 PM
Danny

As for your second question - no, I did not intend on replacing $x_0$ with $x$ etc.
February 9th 2011, 09:57 PM
bugatti79

Quote:

Originally Posted by Danny

Are you not told that on the initial curve that

$u_0 = x_0$ and $y_0 = 1$ ?

I didnt understand that the special case of u=x and y=1 is on the intial data curve...it doesnt say that explicitly?

Ok because $u=u_0(s)=x, y=y_0(s)=1$ then

$\displaystyle u=\frac{x-y^2}{ln(y)-ln (1) +\frac{x-1}{x-1}}+y^2$

The final question is to determine the domain of definition of this solution. How is that tackled?

Thanks alot for your help..