# PDEs solvable as ODEs

• May 9th 2009, 12:06 AM
Altair
PDEs solvable as ODEs
We have started PDEs in our MAth course in BE ME. We are following Kreyszig and I had problem understanding the basic concepts. It says "PDEs SOLVABLE AS ODEs : It happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable9s) can be treasted as parameters(s)"

There's an example I am confused in,

u_xy = -u_x where $u=u(x,y)$
The steps goes as,
$u_x = p$, $p_y = -1$, $\frac{p_y}{p}=-1$, $lnp=-y +c(x)$ and by integration with respect to x,
$u(x,y) = f(x)e^-y + g(y)$ where $f(x) = \int c(x) dx$

I do not understand the bold part and the natural log step how does $y$ come in ?
• May 9th 2009, 04:52 AM
Jester
Quote:

Originally Posted by Altair
We have started PDEs in our MAth course in BE ME. We are following Kreyszig and I had problem understanding the basic concepts. It says "PDEs SOLVABLE AS ODEs : It happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable9s) can be treasted as parameters(s)"

There's an example I am confused in,

u_xy = -u_x where $u=u(x,y)$
The steps goes as,
$u_x = p$, $p_y = -1$, $\frac{p_y}{p}=-1$, $lnp=-y +c(x)$ and by integration with respect to x,
$u(x,y) = f(x)e^-y + g(y)$ where $f(x) = \int c(x) dx$

I do not understand the bold part and the natural log step how does $y$ come in ?

Let's start with a simplier example. Suppose that $u = x^2 + f(y)$ where $f(y)$ is arbitary. So taking the $x$ derivative gives $u_x = 2x$. Now given $u_x = 2x$ find $u$. So $\partial u = 2x\, \partial x$ (separable, just like ODEs) and integrating gives $u = x^2 + c$. This would be correct if it was an ODE but it's a PDE so instead of a constant of integration, we should have a function of integration and with respect to the other variable. So in our case $u = x^2 + f(y)$

Now for your question, you have $\frac{p_y}{p} = - 1$ or $\frac{\partial p}{p} = - \partial y$ however we typical write $\frac{d p}{p} = - d y$ noting it's partial integration (integration wrt a certain variable)

so $\ln p = - y + c(x)$ or $p = e^{-y + c(x)}$

$
u_x = e^{-y} e^{ c(x)}
$

so $du = e^{-y} e^{ c(x)}dx \; \Rightarrow\; u = f(x) e^{-y} + g(y)$
where $f(x) = \int e^{ c(x)}dx$ (Since $c(x)$ is abitrary then $\int e^{ c(x)}dx$ is abritrary which we call $f(x)$).
• May 9th 2009, 10:55 AM
HallsofIvy
Quote:

Originally Posted by Altair
We have started PDEs in our MAth course in BE ME. We are following Kreyszig and I had problem understanding the basic concepts. It says "PDEs SOLVABLE AS ODEs : It happens if a PDE involves derivatives with respect to one variable only (or can be transformed to such a form), so that the other variable9s) can be treasted as parameters(s)"

There's an example I am confused in,

u_xy = -u_x where $u=u(x,y)$
The steps goes as,
$u_x = p$,
$p_y = -1$

That is incorrect. You meant $p_y= -p$ didn't you?

Quote:

, $\frac{p_y}{p}=-1$, $lnp=-y +c(x)$ and by integration with respect to x,
Well, first take the exponential of each side: exp(ln p)= exp(-y+ c(x))= exp(c(x))exp(-y). And since "c(x)" is an unknown function, g(x)= exp(c(x)) is again an unknown function: $p(x)= g(x)e^{-y}$.
Now, $p(x)= du/dx= g(x)e^{-y}$

Integrating with respect to x gives $u(x)= \left(\int g(x)dx\right)e^{-y}$. And since g(x) is an unknown function, $\int g(x)dx$ is some unknown function: call it "f(x)". That gives
Quote:

$u(x,y) = f(x)e^{-y} + g(y)$ where $f(x) = \int c(x) dx$
exept that f(x) is not " $\int c(x) dx$". It is $\int e^{c(x)}dx$

Quote:

I do not understand the bold part and the natural log step how does $y$ come in ?