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APM346-2012 => APM346 Math => Home Assignment 1 => Topic started by: Djirar on September 22, 2012, 01:29:40 PM

Title: Problem 3
Post by: Djirar on September 22, 2012, 01:29:40 PM
Hello,

For this question I managed to find a solution that satisfies the conditions, but that only depends on X and a constant.

In general, if I find a solution that satisfies all conditions is this solution correct regardless of the method used to find it?
Title: Re: Problem 3
Post by: Victor Ivrii on September 22, 2012, 01:45:48 PM
Hello,

For this question I managed to find a solution that satisfies the conditions, but that only depends on X and a constant.

In general, if I find a solution that satisfies all conditions is this solution correct regardless of the method used to find it?

You need to find all solutions and justify that there are no other solutions. The method of characteristics we studied ensures this if correctly applied. Your home-brewed method may not.
Title: Re: Problem 3
Post by: Djirar on September 22, 2012, 01:52:35 PM
I did use the method of characteristics, but I parametrized x and y in terms of s and integrated with respect to s . The thing is my solution doesn't depend on Y, is this ok ?
Title: Re: Problem 3
Post by: Victor Ivrii on September 22, 2012, 02:33:13 PM
I did use the method of characteristics, but I parametrized x and y in terms of s and integrated with respect to s . The thing is my solution doesn't depend on Y, is this ok ?

I have no idea what is $s$. TA who will check your paper probably has no idea either. You need to return to the original coordinates $x,y$.
Title: Re: Problem 3
Post by: Djirar on September 22, 2012, 03:02:04 PM
I got it, thank you for your help.
Title: Re: Problem 3
Post by: Aida Razi on September 24, 2012, 09:00:44 PM
Solution is attached!
Title: Re: Problem 3
Post by: James McVittie on September 25, 2012, 07:29:46 AM
Find the solution:

$u_{x}+3u_{y}=xy$

$u_{x=0}=0$

By examining Integral Lines:

$\frac{1}{dx}=\frac{3}{dy}=\frac{xy}{du}$ <very ugly form. Never use it $\color{blue} {\frac{dx}{1}=\frac{dy}{3}=\frac{du}{xy}}$

Then from the first equality:

$3x=y+C$ where C is some constant.

$y=3x-C$

Then again from the Integral Lines:

$dx(xy)=du$

$dx(x(3x-C))=du$

$u=x^{3}-\frac{C}{2}x^{2}+C_{1}$

Then by the initial condition:

$u(x=0)=0$

$C_{1}=0$

Therefore,$C=3x-y$

Therefore,

$u(x,y)=x^{3}-\frac{1}{2}x^{2}C$

$u(x,y)=x^{3}-\frac{1}{2}x^{2}(3x-y)$

$u(x,y)=-\frac{1}{2}x^{3}+\frac{1}{2}x^{2}y$

Check:

$u_{x}(x,y)=-\frac{3}{2}x^{2}+xy$

$u_{y}(x,y)=\frac{1}{2}x^{2}$

$u_{x}+3u_{y}=xy$
Title: Re: Problem 3
Post by: Victor Ivrii on September 25, 2012, 10:09:56 AM
Now it is correct.