# How is the initial conditions for PDE?

1. Jan 15, 2014

### Jhenrique

Given a PDE of order 1 and another of order 2, you could show me what is, or which are, all possible initial conditions? For an ODE of order 2, for example, the initial condition is simple, is (t₀, y₀, y'₀). However, for a PDE, I think that there is various way to specify the initial condition, or not? Give me examples, please!

2. Jan 15, 2014

### JJacquelin

3. Jan 15, 2014

### Jhenrique

Yeah but, I have seen many examples that left me confused. For a PDE of order 2, I have seen initinal conditions like:

$(u_x(x, y_0) = u_{x0}(x),\;u(x_0, y)=u_0(y))$;
$(u_x(x_0, y) = u_{x0}(y),\;u(x_0, y)=u_0(y))$;
$(u_y(x, y_0) = u_{y0}(x),\;u(x, y_0)=u_0(x))$

And others more, so, there is various way to specify the initial condition?

4. Jan 15, 2014

### JJacquelin

Fortunately there are many kind of boundary conditions, much more than the three above. Many different problems involve PDEs, especially in Physics, where the physical modelisation determines the boundary conditions.

5. Jan 15, 2014

### Jhenrique

I think no... actually, there is really many kind of boundary conditions, however, my question is wrt initial conditions and I found that is possible to analyze the necessary initial conditions for a PDE, term by term, ie...

given --------- the necessary initial conditions is

$u_{xx}(x,y)$------$u_x(x_0,y)$ and $u(x_0,y)$
$u_{yy}(x,y)$------$u_y(x,y_0)$ and $u(x,y_0)$
$u_{xy}(x,y)$------or $u_x(x,y_0)$ and $u(x_0,y)$ or $u_y(x_0,y)$ and $u(x,y_0)$ or $u(x_0,y_0)$

$u_x(x,y)$------$u(x_0,y)$
$u_y(x,y)$------$u(x,y_0)$

6. Jan 16, 2014

### JJacquelin

Why do you use the inadapted term "initial conditions" instead of the correct term "boundary conditions" ?
For example, you write u(x0,y), which means that a given function u0(y) is required on the line x=x0. A line where a condition is given is a bound and the condition is called a "boundary condition". What do you mean with "initial" ? Is it because one of the variable is the time ?
All the conditions that you show in your preceeding post are typical boundary conditions, nothing else. You are confused because you use the word "initial" instead of "boundary".
See the definition of "initial condition" : http://mathworld.wolfram.com/InitialConditions.html

Last edited: Jan 16, 2014
7. Jan 16, 2014

### Jhenrique

I understand "initial conditions" as the arbitrary constants or arbitrary functions that is inserted in the ODE or PDE. Like me that initial condition is different of boundary condition...

8. Jan 16, 2014

### JJacquelin

Why asking on a forum if you don't take account of the advices ?
In ODE, there is only one variable with symbole x, or t, or what symbol you want. So, it doesn't matter to call a condition "initial" or "boundary" (Bound reduced to a point in this case).
In PDE, they are several variables, so they cannot all be time. Of course you can loosely use "initial" instead of "boundary" if this doesn't induce confusion in your understanding of what the boundaries are for PDE.
I was insistent on the right vocabulary because you wrote a list of examples of conditions which, I am sorry to say, is non-sens. There is no relationship or correspondance, between a partial derivative, such as uxx(x,y) for example and a condition such as ux(x0,y)=u0(y) for example. A condition such as ux(x0,y)=u0(y) for example, could be specified to any PDE whatever are the partial derivatives in the PDE.

9. Jan 16, 2014

### Jhenrique

I saw in this forum some discussions/topics about the difference between initial conditions and boundary conditions for PDE and I understood that those concepts are different same for PDEs, I think...

Given:
$\frac{\partial^2 u}{\partial x^2}(x,y)$

We integrate wrt to x to get $u_x(x,y)$
$\int \frac{\partial^2 u}{\partial x^2}(x,y)dx=\int_{x_0}^{x}\frac{\partial^2 u}{\partial x^2}(x,y)dx+\frac{\partial u}{\partial x}(x_0,y)=\frac{\partial u}{\partial x}(x,y)$

Necessarily $u_x(x_0,y)$ appeared as a arbitrary function.

10. Jan 16, 2014

### JJacquelin

Do not confuse a defined integral with a PDE !

11. Jan 16, 2014

### Jhenrique

OK! Trankyou for all your coments!