1
$\begingroup$

Trying to teach myself PDEs, and I'm stumped on what should probably be a very simple exercise:

Solve the equation $3u_{y}+u_{xy}=0$. And I am given the hint to let $v=u_{y}$ (it's a problem from Strauss' intro book).

Now, when you make the suggested substitution, $3u_{y}+u_{xy}=0$ becomes $3v+v_{x}=0$. The only problem is, I don't know how to solve this kind of PDE; the only types of PDE the book has really talked about at this point are ones of the form $au_{x}+bu_{y}=0$, where $a$ and $b$ are constants, and ones of the form $u_{x}+yu_{y}=0$.

What I've tried to do, therefore, is rewrite $v_{x}$ as $\frac{dv}{dx}$, subtract $3v$ from both sides, and turn it into a type of separable ODE type thing. Then, if I do that, and after substituting $v=u_{y}$ back in, I wind up getting that $u_{y}=\exp{(-3x)}\exp{(f(y))}$ Then, I suppose I'd have to integrate both sides with respect to $y$ to get the solution, but I'm worried it will take some messy integration by parts that never stops, and so I think that this method couldn't possibly be right.

Could somebody tell me the RIGHT way to do this problem? Thanks!! :)

$\endgroup$
  • $\begingroup$ Since you eliminated $y$ for $3v + v_x = 0$, you can treat it as $y_x + 3y = 0$. Use method of constant coefficients. I will post the answer as long as there are no users who post the best answers. $\endgroup$ – NasuSama Feb 9 '14 at 20:54
  • $\begingroup$ You mean the method of undetermined coefficients? $\endgroup$ – ALannister Feb 9 '14 at 20:56
  • $\begingroup$ Method of constant coefficients. That is: Set $y = e^{rx}$ where $r$ is any constant. Evaluate the ODE by substitution and algebra. $\endgroup$ – NasuSama Feb 9 '14 at 20:56
  • $\begingroup$ Gotcha. Will try. $\endgroup$ – ALannister Feb 9 '14 at 20:57
  • $\begingroup$ Yeah, it's still not working. Anyway, isn't that method only for inhomogeneous equations? Doesn't help me. If I try to get just the homogeneous part, I end up with $C\exp{(-3x)}$, which is exactly the same thing I got before, except that my constant was a constant function, $\exp{(f(y)}$, call it $g(y)$. So, I have that $u_{y}=g(y)\exp{(-3x)}$ - how do I go from that to give me an expression for $u(x,y)$? $\endgroup$ – ALannister Feb 9 '14 at 21:27
1
$\begingroup$

We are given that

$$3u_y + u_{xy} = 0$$

As the hints suggest, we use the substitution: $v = u_y$. Then, as you said before, the new ODE is

$$3v + v_x = 0$$

By the method of constant coefficients, setting $v = e^{rx}$, we have...

$$\begin{aligned} re^{rx} + 3e^{rx} &= 0\\ r + 3 &= 0\\ r &= -3 \end{aligned}$$

So we have

$$v(x) = c_1e^{-3x}$$

where $c_1$ is any constant. But since $v = u_y$,

$$u_y(x,y) = c_1e^{-3x}$$

which implies that a solution is

$$u(x,y) = c_1ye^{-3x}$$

$\endgroup$
0
$\begingroup$

Note. $$ 3u_y+u_{xy}=0 \quad\Longleftrightarrow\quad \frac{\partial}{\partial y}(3u+u_x)=0 \quad\Longleftrightarrow\quad 3u(x,y)+u_x(x,y)=f(x), $$ for some function $f=f(x)$. Next $$ \mathrm{e}^{3x}\big(3u(x,y)+u_x(x,y)\big)=\mathrm{e}^{3x}f(x) \quad\Longleftrightarrow\quad \big(\mathrm{e}^{3x}u(x,y)\big)_x =\mathrm{e}^{3x}f(x), $$ which is equivalent to $$ \mathrm{e}^{3x}u(x,y) =u(0,y)+\int_0^x \mathrm{e}^{3t}f(t)\,dt, $$ or $$ u(x,y) =\mathrm{e}^{-3x}\,u(0,y)+\int_0^x \mathrm{e}^{-3(x-t)}f(t)\,dt, $$ for a function $f$ which can be determined from the initial data.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.