partial differential equations

**dlbsd** · April 8th 2009, 07:32 PM

Hi, I just started studying PDE's so sorry for the newbie questions

I have two questions

1. Solve
$au_x + bu_y + cu = 0$

2. Solve
$u_x + u_y = 1$

Here is what I have so far for #1:
Using change of variables
$x' = ax + by$
$y' = bx - ay$

$u_x = au_{x'} + bu_{y'}$
$u_y = bu_{x'} - au_{y'}$

$a(au_{x'} + bu_{y'}) + b(bu_{x'} - au_{y'}) + cu = 0$
$(a^2 + b^2)u_{x'} + cu = 0$

This is where I get stuck. How do i solve for u in this case?

For #2, I'm not sure how to start but I want to say its unsolvable because its inhomogeneous? (probably way wrong in saying that haha)

**Danny** · April 9th 2009, 05:32 AM

Originally Posted by dlbsd

Hi, I just started studying PDE's so sorry for the newbie questions

I have two questions

1. Solve
$au_x + bu_y + cu = 0$

2. Solve
$u_x + u_y = 1$

Here is what I have so far for #1:
Using change of variables
$x' = ax + by$
$y' = bx - ay$

$u_x = au_{x'} + bu_{y'}$
$u_y = bu_{x'} - au_{y'}$

$a(au_{x'} + bu_{y'}) + b(bu_{x'} - au_{y'}) + cu = 0$
$(a^2 + b^2)u_{x'} + cu = 0$

This is where I get stuck. How do i solve for u in this case?

For #2, I'm not sure how to start but I want to say its unsolvable because its inhomogeneous? (probably way wrong in saying that haha)

What you're trying to do is introduce a new coordinate system to tranform you pde to one that is easier (i.e. missing a derivative) you'll find in general, this will take you to trying to solve part of the original problem.

For your first problem, it's separable so

$\frac{du}{u} = - \frac{c}{a^2+b^2}\, dx'$ so $\ln u = -\frac{c}{a^2+b^2} x' + f(y')$

For your second, you can use your change of variables (a=1, b=1) to get

$2 u_{x'} = 1$ - again, separable.

Have you learned the method of characteristics yet? It's really much quicker.

**chisigma** · April 9th 2009, 05:56 AM

Just a minimum of theory...

... a linear first order PDE is written as...

$\alpha(x,y,u) \cdot u_{x} + \beta(x,y,u) \cdot u_{y} = \gamma (x,y,u)$ (1)

... and it is equivalent to the system of ordinary equations...

$\frac {dx}{\alpha(x,y,u)} = \frac {dy}{\beta(x,y,u)} =\frac {du}{\gamma(x,y,u)}$ (2)

If the solution of (2) is...

$c_{1}= \varphi_{1} (x,y,u)$

$c_{2}= \varphi_{2} (x,y,u)$ (3)

... where $c_{1}$ and $c_{2}$ are two 'arbitrary constants', the solution of 1 is...

$G \{ \varphi_{1} (x,y,u), \varphi_{2} (x,y,u) \}=0$ (5)

... where G(*,*) is arbitrary, with G(*,*) and its partial first order derivatives are continous...

Let's start with #2 because it is more simple...

$u_{x} + u_{y} = 1$ (5)

Here $\alpha=\beta=\gamma=1$ , so that the system (2) is...

$dx=dy$

$dy=du$ (6)

... and it is easily solved...

$c_{1}= x - y$

$c_{2}= y - u$ (7)

Therefore solution of (5) is...

$G \{ x - y, y - u \}=0$ (8)

And now we will 'attack' #1...

$a\cdot u_{x} + b\cdot u_{y} + c\cdot u =0$ (9)

Here is $\alpha=a,\beta=b, \gamma=-c\cdot u$ , so that (2) is...

$\frac{dx}{a}=\frac{dy}{b}$

$\frac{dy}{b}= -\frac {du}{c\cdot u}$ (10)

... the solution of which is...

$c_{1}= \frac{x}{a} - \frac{y}{b}$

$c_{2}= u^{\frac{1}{c}}\cdot e^{\frac{y}{b}}$ (11)

Solution of (9) is then...

$G \{ \frac{x}{a}-\frac{y}{b},u^{\frac{1}{c}}\cdot e^{\frac{y}{b}} \}=0$ (12)

Kind regards

$\chi$ $\sigma$

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