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 January 15th, 2014, 09:06 AM #1 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 How is the initial conditions for PDE? 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 (t?, y?, y'?). However, for a PDE, I think that there is various way to specify the initial condition, or not? Give me examples, please!
 January 15th, 2014, 03:02 PM #2 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: How is the initial conditions for PDE? Yes it is definitely more complicate for PDE and probably there is not a general way to describe this. The main reason is that the solution of a PDE are also defined by "limit conditions" depending on this condition the solution is of course different as well as their degree of allowed initial conditions. But in general it is easy to think like this: Let u(x,t) be a function of position x and time t. If the PDE is linear and first order in time then you need to know initially only u(x,0) for any x to know your solution (even though the equation would be for instance second order in position, i.e. Schrödinger equation or any diffusion equation). If the equation is second order in time you need to know u(x,0) and $\partial u(x,0)/\partial t$ at t=0. One thing that helps to understand the initial conditions is to do at first the separation of variables, if you understand this you mainly understand the rest. However, the best to understand this is to work on practical examples.
 January 15th, 2014, 05:48 PM #3 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 Re: How is the initial conditions for PDE? I discovered 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 $u(x_0,y_0)$ $u_x(x,y)$------$u(x_0,y)$ $u_y(x,y)$------$u(x,y_0)$
 January 16th, 2014, 04:57 AM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: How is the initial conditions for PDE? Sorry, jhenrique, but that is much too limited. When you say "$u(x,y_0)$" you are assuming that the boundary condition is given along a horizontal straight line. The boundary conditions for a second order partial differential equation, with independent variables x and y, may be the value of the function, or its first derivatives, or some linear combination, $\alpha u_x+ \beta u_u$ along some boundary curve.
 January 16th, 2014, 05:56 AM #5 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 Re: How is the initial conditions for PDE? I arrived at these conclusions as follows I took a term of a general PDE $au_{xx}+2bu_{xy}+cu_{yy}+du_x+eu_y+fu+g=0$ , $u_{xy}(x,y)$ for example, and I intregated for get $u(x,y)$, so: $\int \frac{\partial^2 u}{\partial x \partial y}(x,y) dx= \int_{x_0}^{x} \frac{\partial^2 u}{\partial x \partial y}(x,y) dx + \frac{\partial u}{\partial y}(x_0,y)=\frac{\partial u}{\partial y}(x,y)$ and again: $\int \frac{\partial u}{\partial y}(x,y) dy= \int_{y_0}^{y} \frac{\partial u}{\partial y}(x,y) dy + u(x,y_0)=u(x,y)$ So, $\frac{\partial u}{\partial y}(x_0,y)$ and $u(x,y_0)$ appeared naturally in the development...
 January 16th, 2014, 01:07 PM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: How is the initial conditions for PDE? But that does not necessarily have anything to do with the boundary conditions given in the problem!
 January 16th, 2014, 01:14 PM #7 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: How is the initial conditions for PDE? But that does not necessarily have anything to do with the boundary conditions given in the problem! A typical partial differential equations problem might be "Solve $\frac{\partial^2\phi}{\partial x^2}+ \frac{\partial^2\phi}{\partial x^2}= 0$ on the region bounded by the curve $y= x^2- 4$ and the line y= 0 subject to the boundary conditions that $\phi(x,x^2- 4)= y^2= (x^2- 4)^2$ and $\phi(x, 0)= 0$.
 January 16th, 2014, 10:01 PM #8 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: How is the initial conditions for PDE? Hi HallsofIvy ! I fully agree with your answers to Jhenrique and I hope that will help him to understand. See the discussion : http://www.physicsforums.com/showthread.php?t=732954

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