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I was wondering how to solve $$a(x-1)\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} =0 ?$$

$a$ is a constant.

(1) I am still having trouble to understand the method of characteristics recommended in comments. May I have some more explanation? Usually, which type of PDE can be solved by the method of characteristics?

(2) Can separation of variables apply here? Usually, which type of PDE can be solved by separation of variables?

Thanks and regards!!

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$f(x,y)=g((x-1)\mathrm{e}^{-ay})$. –  Did May 5 '11 at 21:14
    
Use the method of characteristics to get Didier Piau's result. –  Fabian May 5 '11 at 21:16
    
What is $a$? If it is a constant (and not a function), then you could use characteristics to obtain the result Didier has. –  Jonas Teuwen May 5 '11 at 22:12
    
@Jonas: a is constant. (1) I am still having trouble to understand the method of characteristics. Would you mind explain some more? (2) Also can separation of variables apply here? (3) Usually, which method is chosen for which type of PDE? Thanks! –  Tim May 5 '11 at 22:13
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This is a so cool place, I have the same problem. –  Heather Jul 14 '11 at 8:05

2 Answers 2

up vote 6 down vote accepted

The method of characteristics yields the equations

$$ \begin{eqnarray} \dot x &=& a(x-1)\;,\\ \dot y &=& 1\;,\\ \dot z &=& 0\;, \end{eqnarray} $$

with solutions

$$ \begin{eqnarray} x - 1 &=& c_1\mathrm e^{at}\;,\\ y &=& t + c_2\;,\\ z &=& c_3\;. \end{eqnarray} $$

So the function value $z=f(x,y)$ is constant along these curves. We can combine the solutions for $x(t)$ and $y(t)$ into an equation

$$(x-1)\mathrm e^{-ay}=c_4\;.$$

This tells us that $f$ only depends on $x$ and $y$ through $(x-1)\mathrm e^{-at}$, which is Didier's solution in the comments.

Yes, you can also solve this by separation of variables. The ansatz $f(x,y)=X(x)Y(y)$ leads to

$$\frac{X'}{X}a(x-1)=-\frac{Y'}{Y}\;.$$

Setting both sides equal to a constant $c$ gives you two ordinary differential equations, with solutions

$$ \begin{eqnarray} X(x)&=&c_1(x-1)^{\frac{c}{a}}\;,\\ Y(y)&=&c_2\mathrm e^{-cy}\;. \end{eqnarray} $$

That gives you solutions

$$ \begin{eqnarray} f(x,y)&=&c_3(x-1)^{\frac{c}{a}}\mathrm e^{-cy}\\ &=&c_3\left((x-1)\mathrm e^{-ay}\right)^{\frac{c}{a}}\;. \end{eqnarray} $$

Since you can choose $c$ freely, you can combine these into arbitrary functions of $(x-1)\mathrm e^{-ay}$.

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Really nice! Thanks! I was wondering what types of PDE can be solved by method of characteristics, by separation of variables and by other methods respectively? In other words, what are some guidelines for one to know what method to solve his PDE? –  Tim May 6 '11 at 6:07
    
@Tim: I don't know much about that. For separation of variables, the equation has to be linear (else you can't superimpose the solutions). For characteristics, it only has to be quasilinear. For separation of variables, the coefficients also have to be such that you can separate the variables, but you can see pretty directly when you substitute the product ansatz whether that's possible. –  joriki May 6 '11 at 6:19
    
Thanks! I was wondering in "since you can choose c freely, you can combine these into arbitrary functions of $(x−1)e^{-ay}$", why does $(.)^{c/a}$, when $c$ is arbitrary, become an arbitary function? –  Tim May 6 '11 at 6:27
    
@Tim: I didn't say that it becomes an arbitrary function, but that you can combine these into arbitrary functions -- you have one such solution for each $c$, and $c_3$ can depend arbitrarily on $c$, so you can get any (sufficiently well-behaved) function by letting $c_3(c/a)$ be its Laplace transform (see en.wikipedia.org/wiki/Laplace_transform). (Remember that separation of variables gives you only the separable solutions, not all solutions; you get all solutions as linear combinations of the separable ones.) –  joriki May 6 '11 at 6:34
    
@Tim: Sorry, I think that should be "by letting $c_3(-c/a)$ be its inverse Laplace transform". –  joriki May 6 '11 at 7:59

I would recommend you to read the book Partial Differential Equations: An Introduction by Strauss. Your PDE appears in the very beginning of the book.

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Thanks! Do you also know what page(s) on that book are relevant to solving my PDE? –  Tim May 6 '11 at 5:50

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