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Consider: $$x^2p+y^2q=(x+y)z$$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Thus by Lagrange's Method

$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$

$$\Rightarrow \frac{dx}{x^2}=\frac{dy}{y^2} \Rightarrow \frac{1}{x}-\frac{1}{y}=k$$

$$\Rightarrow \frac{dx}{x^2}=\frac{dz}{(x+y)z} \Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$

$$\Rightarrow \phi (\frac{1}{x}-\frac{1}{y},x+y+x\ln(z))=c$$

Or if we explore further $$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$

So can we even write $f(\frac{1}{x}-\frac{1}{y},xyz)=k$. So which one is right?

Soham

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    There are no derivatives with respect to $z$ in the equation you wrote, so you are no quite using Lagrange's method correctly.2012-08-27
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    On the other hand, I cannot see how you got from the equation you wrote at the beginning by «exploring further», the equation you wrote in the last paragraph...2012-08-27
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    @MarianoSuárez-Alvarez As I understood this is the method of multipliers where $l,m,n$ are chosen such that denom of $ldx+mdy+ndz$ is zero2012-08-27
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    But there is no $dz$ in the equation...2012-08-27
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    @MarianoSuárez-Alvarez The book I am following seem to go about Lagrange's Method like this. Do let me know if you can refer me a better book regarding the same2012-08-27
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    Whatever textbook, I am *sure* that it never writes things like $\frac{\text{something}}{0}$.... :-)2012-08-27
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    @MarianoSuárez-Alvarez I couldnt find a better link but I think this has what I mean http://www.sxccal.edu/ug_math/lecture_notes/JANUARY%202012%20-%20T%20BANERJEE/DIFFERENTIAL_EQUATION,_SEM_VI,1.docx Scroll down to Page 5 Theorem 2.12012-08-27
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    @MarianoSuárez-Alvarez hehe... welcome to the club... isnt it so very unsightly looking ? Refer this thread: http://math.stackexchange.com/questions/187201/a-question-on-lagranges-method-for-solving-partial-differential-equation2012-08-27
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    @MarianoSuárez-Alvarez As of dz it follows from the auxillary equation. More on the link given2012-08-27
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    I cannot make sense of your question, nor open the Office file you linked to. I suspect your equation would make a little bit more of sense if $u$ and $z$ were the same... I'll leave someone else to take care of this: you do not seem to have read what I wrote in my first two comments :-/2012-08-27
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    Well as I had been trying to convey there is an auxilary equation which comes up. If $Pp+Qq=R$ where $p,q$ are partial derivatives of z wrt x and y resp then $dx/P=dy/Q=dz/R$ and then you solve this auxillary equation. And from here dz comes up.2012-08-27
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    My last intervention here: have you remarked that in the equation you have both $z$ and $u$, and, as I observed a couple of times by now, that is probably a typo?2012-08-27
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    It seems that the method you're trying to apply is more usually called the [method of characteristics](http://en.wikipedia.org/wiki/Method_of_characteristics)?2012-08-27
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    It seems that Kappa light off when leaving the chatroom :-) (the chatroom died for me!)2012-08-28

3 Answers 3

1

Both are not right.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$

$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$

$\dfrac{dz}{dt}=(x+y)z=\left(-\dfrac{1}{t}-\dfrac{1}{t+y_0}\right)z$ , we have $z(x,y)=\dfrac{f(y_0)}{t(t+y_0)}=xy~f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$

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    If you consider the solution obtained : $\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=D\ $ then for $\phi(a,b)=f(a)\,b\ $ you get : $f\left(\frac 1x-\frac 1y\right)\cdot \frac {xy}z=D\ $ i.e. $\ z=xy\ f\left(\frac 1x-\frac 1y\right)/D$ like in eqworld's solution !2012-08-28
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    @Raymond: I also first thought that your two solutions are equivalent, but actually yours allows $z$ to be any function at all, since you could let $\phi\equiv C$. The explicit form $\phi(a,b)=f(a)b$ is only one particular case of $\phi(a,b)=C$.2012-08-28
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    @doraemonpaul: In any case, you were right about my answer; I edited it accordingly.2012-08-28
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    @joriki: note that $z$ appears only in my second parameter, this limits the possibilities... (I only wrote that my solution implied eqworld's)2012-08-28
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    @Raymond: Perhaps I misunderstand what you mean by $\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=C$ -- why doesn't this allow for arbitrary $z$ if we choose $\phi\equiv C$? I don't see how this is related to whether $z$ occurs only in the second parameter. Are you making any assumptions about $\phi$?2012-08-28
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    @joriki: Well with your reasoning every first order P.D.E. would have every solution. I suppose that the 'arbitrary' of the P.D.E. books means that we must have a $z$ somewhere on which we put a constraint getting an implicit or explicit relation with $z$. Consider $\phi(a,b)=a=C$ : you won't get a solution for $z$ in this case either.2012-08-28
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    @Raymond: OK, I see. I'd never seen solutions expressed in this implicit form before. The explicit form seems preferable to me, both because it's explicit and because it doesn't require additional assumptions about the arbitrary functions.2012-08-28
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    @Joriki: in many simple cases you may put it in explicit form and I suppose that physically these solutions are much most interesting (we can't do much with something implicit like $\phi(a,b)=\log(b)+exp(b)$ anyway...) but mathematicians may prefer to be exhaustive. Ok explicit solutions are more useful ! :-)2012-08-28
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    @RaymondManzoni Can you help me understand whats happening? Perhaps its not possible to do this here, can we meet up on chat or even email. My email is there on my profile page2012-08-28
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    @Soham: I can't see your email. We can chat if you want !2012-08-28
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    @RaymondManzoni yes lets do it. I am in the mathematics room2012-08-28
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    Do you mean that the general solution of a first order linear PDE whose the form of the arbitrary function should have both versions of single parameter and multiple parameters?2012-08-29
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    @doraemonpaul its more accepted to write the general solution not as $f_1(x)\cdot f_2(x)$ but as $\phi(f_1(x),f_2(x))$2012-08-29
2

You can't integrate $\mathrm dz/z$ to $\log z$ and treat $x$ and $y$ in the denominator as constant. Likewise, Raymond's answer treats $y$ as constant in integrating $y/x^2$. In the method of characteristics, since we're finding curves, there's only one independent variable at a time, and the others have to be expressed in terms of it when we want to integrate.

Thus, Raymond's answer is correct up to

$$ \frac{(x+y)\,\mathrm dx}{x^2}=\frac{\mathrm dz}z\;, $$

but then we have to express $y$ in terms of $x$,

$$ y=\left(\frac1x-k\right)^{-1}\;, $$

to obtain

$$ \left(\frac1x+\frac1x\frac1{1-kx}\right)\mathrm dx=\frac1z\mathrm dz\;. $$

Integrating this yields

$$ z=c\frac{x^2}{1-kx}\;, $$

and then substituting

$$ k=\frac1x-\frac1y $$

yields

$$ z=cxy\;, $$

which is readily confirmed to solve the given equation.

[Edit:]

As doraemonpaul rightly pointed out, that's not the general solution, since $c$ can be chosen independently for each value of $k$, so it should be

$$ z=c(k)xy=c\left(\frac1x-\frac1y\right)xy\;. $$

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    Why cant we integrate it straight away keeping y,or z as constant? After all when we differentiate we do it in the same way..2012-08-27
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    @Soham: your second method is right you just wrote $xyz=c$ instead of $\frac {xy}z=c$ (see the minus sign). You had nearly Joriki's answer (+1 of course)...2012-08-27
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    @Soham: Please elaborate; I'm not sure what form of differentiation you're referring to. Are you aware that the method of characteristics determines characteristic curves?2012-08-27
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    @joriki I meant partial differentiation of say $xyz$ wrt $x$ yields $yz$ effectively implying $yz$ is a constant. So why cant we integrate (partially wrt x) as like this: xyz + f(y,z) Should be consistent isnt it? So in the similar way why cant we integrate the LHS in the first eqn which you wrote keeping y as a constant.2012-08-27
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    @joriki No I was not aware of characteristic curves. Thanks will check it out2012-08-27
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    damn I am feeling like an idiot asking this question but can you elaborate how $y/x^2$ turned out to be $1/x(x-k)$. I think it will be $1/x(1-xk)$2012-08-27
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    @RaymondManzoni You know where I made the mistake till now? And why I didnt get $xy/z=c$ ?? I was integrating each differential thinking its coefficient as constant eg $yzdx$ was being integrated as xyz where as I should have separated the variables... Hmm... very interesting thing, dont know from where I picked this conceptual error2012-08-27
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    @Soham: It's easy to get confused about partial derivatives because our notation for them is so bad. Things would be much easier if we always marked which quantities are being considered as functions of which quantities, which quantities are being varied and which are being held fixed. In the present case, you're mixing up two different things. The partial derivatives of $z$ with respect to $x$ and $y$ are taken with $z$ considered as a function of $x$ and $y$ -- so you can imagine a surface above the $x$-$y$ plane given by $z=z(x,y)$, and you can vary $x$ or $y$ and keep the other constant...2012-08-27
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    ... whereas the *total* differentials $\mathrm dx$, $\mathrm dy$, $\mathrm dz$ are in a context where we're trying to determine characteristic *curves*, not a surface (see the Wikipedia article I linked to under the question). Along a curve, there's only a single independent variable; there are no partial derivatives; $\mathrm dx$, $\mathrm dy$, $\mathrm dz$ are all simply the change in $x$, $y$ and $z$, respectively, along the curve. Also there is no such thing as partial integration (though of course there is integration by parts, which is something completely different :-).2012-08-27
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    So when you're integrating $f(x)\mathrm dx$ or $f(y)\mathrm dy$ or $f(z)\mathrm dz$, you're integrating along one of the characteristic curves, and if you have one of the other variables still in there, you need to express it in terms of the variable with respect to which you're integrating, since it changes along the curve, so treating it as constant yields a wrong result. These things are very easy to get wrong if you try to just formally manipulate the symbols, but not so difficult to get right if you conceptually keep track of what they stand for in each case.2012-08-27
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    @joriki Thanks a lot, you actually cleared quite a thing. Now when I think about it it makes perfect sense but the insight eluded me.2012-08-27
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    @Soham: You're welcome.2012-08-27
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    @Solam: Sorry, I'd forgotten to respond to your question about $y/x^2$. You're quite right; I don't know why I had $x-k$ there; I guess I must have made a mistake copying from my paper calculations, since the end result was correct.2012-08-28
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    @joriki What did doraemonpaul comment and where?2012-08-28
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    @Soham: Not a comment, an answer. I can't tell you where because the order in which the answers are displayed depends on your selections, but if you look through all the answers to the question, you should see three of them and one is by doraemonpaul.2012-08-29
1

I think that your work using the characteristics method was right at the start including :

$$\frac{dx}{x^2}=\frac{dy}{y^2} \Rightarrow \frac{1}{x}-\frac{1}{y}=k$$

But the implication here was wrong :

$$\frac{dx}{x^2}=\frac{dz}{(x+y)z} \Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$

because the $x$ appears at the denominator of $dz$ as well as the numerator of $dx$ so that I think this should be : $$\frac{dx}{x^2}=\frac{dz}{(x+y)z} \implies \frac{(x+y)\,dx}{x^2}=\frac{dz}z$$ with Joriki's correction and using $\ y=\dfrac 1{\frac 1x-k}=\dfrac x{1-kx}$ we get : $$\left(\frac 1x+\frac 1x+\frac k{1-kx}\right)dx=\frac{dz}z\implies 2\ln(x)-\ln(1-kx)-\ln(z)=C_0$$ $$\implies \ln\left(\frac {x^2}{1-kx}\right)-\ln(z)=\ln(xy)-\ln(z)=C_0$$ (we replaced by $y$ again to remove the $k$ constant) $$\implies \phi\left(\frac{1}{x}-\frac{1}{y},\ \frac {xy}z\right)=C$$ (you had another error here : forgetting the denominator of the second parameter)

After that you wrote :

EDITED:
Or if we explore further $$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$

If you divide $\ yzdx+xzdy-xydz=0$ by $xyz\ $ you get : $\dfrac {dx}x +\dfrac {dy}y -\dfrac {dz}z = 0\implies\dfrac {xy}z=C_1$

So that the other way to write the solution doesn't differ from the first one : $$\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=C$$

Because of the arbitrary character of $\phi$ other equivalent parameters could have been obtained/chosen. Like replacing one of the parameters by the product $\dfrac{y-x}z$ or something more elaborate.

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    Edited... and Ray, thanks for showing interest in the question. Much appreciated. Thanks2012-08-27
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    It seems you ended up making a similar mistake as the OP -- you treat $y$ as constant in integrating with respect to $x$ -- see my answer.2012-08-27
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    @Joriki: Oops you are right. Thanks for the correction !2012-08-27
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    @Joriki: Sorry to bother you again with this but do you see something wrong in my (updated) answer (the downvote is of course unexplained !) ?2012-08-28
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    The only problems I see are that it should be $+k/(1-kx)$ (but the minus sign for the logarithm after integration is correct), and that you fudged over the point that the integral of $1/x$ is $\log|x|$, not $\log x$ (but I sort of did that, too). (I didn't downvote.)2012-08-28
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    @joriki: It seem we got both downvoted 4 hours earlier... Anyway thanks for the verification and all your excellent explications in this thread and elsewhere !2012-08-28
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    Thanks :-) I hadn't noticed before, but doraemonpaul posted an answer around that time saying that ours were wrong. I think he was right about mine but not about yours :-) (See my comment under his answer.)2012-08-28
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    @joriki: I noticed the same thing and posted another one before seeing your two comments. Cheers,2012-08-28
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    Sorry about the comment confusion; I deleted some because I changed my mind twice :-)2012-08-28
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    @joriki: I needed longer yesterday evening but got my mind fixed on this ! We all learn best from our errors. Cheers !2012-08-28
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    @RaymondManzoni I am sorry my internet is acting up I am not able to connect to the chat server. Lets call it a day today, I will ask you the cauchy problem tomorrow.2012-08-28
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    Thanks a lot for the time and effort2012-08-28
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    @Soham: It's not you, its the chatserver I think (I had the same problem). Ok to see the Cauchy problem another day (it's better to ask a question here...). Good night (I suppose it's late for you) ! Glad it helped a little,2012-08-28