Faqts : Science : Mathematics : Calculus : Differential equation : Partial

+ Search
Add Entry AlertManage Folder Edit Entry Add page to http://del.icio.us/
Did You Find This Entry Useful?

5 of 17 people (29%) answered Yes
Recently 4 of 10 people (40%) answered Yes

Entry

Math:Calculus:Variable:2:Can you tell step to solve partial differential equation:product+separation

Aug 28th, 2009 12:31
Knud van Eeden, forum net tr, chat alarab, engatoo engatoo, Rockys rainwal, Raj Aryan, http://www.jaipurtravelguide.com/


----------------------------------------------------------------------
--- Knud van Eeden --- 13 April 2005 - 06:29 pm ----------------------
Math:Calculus:Variable:2:Can you tell step to solve partial 
differential equation:product+separation
---
To solve this type of partial linear differential equation with 2 
independent variables:
Assume as usual that U(t,x) is the product of two functions, where
The first of this functions is a function of t only.
The second of this functions is a function of x only.
The trick here is that this reduces the solving of the partial 
differential
equation to the solving of 2 ordinary differential equations, which in
general is much easier.
Or thus:
Step 1. Assume that the given partial differential equation U(t,x) is a
        product of two functions:
         U(t,x) = T(t) . X(x)
        where the first function T(t) is a function of t only, and the
        second function X(x) is a function of x only.
Step 2. Then fill this assumed expression for U(x,t) from step 1. in in
        the given partial differential equation and work this out,
        using the method of separation of the variables. That will
        split the equation in an expression in x only on one side of
        the equation, and an expression in t only in the other side of
        the equation. From this follows that there can only be equality
        if both sides are equal to a constant (or parameter here).
        Putting the left side and the right side respectively equal to
        this constant (=parameter), will lead here to 2 ordinary
        differential equations which when solved give you that unknown
        functions T(t) and X(x).
Step 3. Solve this ordinary differential equation T(t) for t only
Step 4. Solve this ordinary differential equation X(x) for x only
Step 5. Fill in the results of step 3. and step 4. in the equation in
        Step 1, and work this out.
        That gives the general solution for U.
Step 6. Solve for the arbitrary constants in the general solution for
        U, by filling in the values for the constants in the general
        solution, using the given boundary values. That gives you 2
        algebraic equation to solve. The equations that you get then
        are called the 'characteristic equation'.
        You will see that if you should solve for the constants, that
        this usually gives the trivial solution that this constants
        both must be zero.
        So instead solve for that parameter.
        You will see that will give you an infinite amount of discrete
        solutions for that parameter.
        Now what does U(t,0) mean? that means that you have fixed an x,
        and you are looking for the vertical U value as function of the
        time. So this are basically the vertical values of U at a
        chosen fixed position of x. If you take for U the movement of a
        wave in a string, at a fixed horizontal position, that is fixed
        x, you are putting a cardboard with a vertical slit in it in
        front of that changing wave, and you are looking how the height
        of that string varies while times goes on.
        Now what does U(0,x) mean? that means that you have fixed the
        time t, in other words you have frozen the current shape of the
        string (like a wet string which is frozen in a certain fixed
        shape in a cold arctic night). So in other words you are
        saying, U(t,0) must be of this and this fixed shape. Let us
        call this shape very general y(x). And you know that y(x) can
        be any function of x. Thus a frozen string, which you can bend
        in almost any 2 dimensional shape (keep the bending in one
        plane).
        Now the constants in your general solution should be chosen
        such that they change the general solution in this particular
        shape y(x) at a fixed time, say when t is zero.
        Now if you have only 1 constant in that general solution, you
        can usually not vary that general solution very much. And as
        this shape of that curve y(x) can take any 2 dimensional shape,
        this will certainly not be achievable in general with only one
        value of the constant or parameter. With 2 parameters values,
        already some more different frozen string shapes are possible
        to create, using the general solution. With 3 parameters values
        even more. But with an infinite amount of parameter values you
        are in business.
        And that is the reason that you are getting infinite sums of
        solutions for a partial differential equation, when you are
        asked to solve for the boundary values, say U(0,x). You use
        this solutions, e.g. varying the scaling of them to build up
        the wanted fixed shape y(x) at that chosen moment of time.
        For example, you say to yourself, OK, when I start my
        stopwatch, thus at t is zero, I want to have a string in the
        shape of a sine curve. If you translate this wish in a boundary
        value equation, you have thus the boundary condition: U(0,x) =
        sin(x)
        or as another example, you say to yourself, OK, when I start my
        stopwatch, thus at t is zero, I want to have a string in the
        shape of a parabola. If you translate this wish in a boundary
        value equation, you have thus the boundary condition: U(0,x) =
        x^2
        If you answer the question about how to get to this by choosing
        the correct values of the constants in your general equation,
        you are lead to an infinite sum. Similar to the way a Fourier
        serie can approach almost any function y(x).
        Now you can of course freeze the string at any chosen moment of
        time (e.g. by taking a picture of this string) In general you
        get at each moment a different curve y1(x), y2(x), y3(x), and
        so on. So you can equivalently, as for as corresponding
        demands:
        If you choose the time fixed at time t1, you should have the
        boundary condition: U(t1,x) = y1(x), in other words the
        particular frozen shape y1(x) at that moment of time t1.
        If you choose the time fixed at time t2, you should have the
        boundary condition: U(t2,x) = y2(x), in other words the
        particular frozen shape y2(x) at that moment of time t2.
        and so on.
Step 7. By filling in one of this solutions for the parameter in the
        given general solution, you get one so called eigenfunction.
        Each of this solutions for the parameter represents one
        eigenfunction.
        Remark: each of this eigenfunctions is also orthogonal (e.g. a
        sine or a cosine), so by taking a sum of them you can build
        almost any shape.
Step 8. The final solution for U(x,t), taking this given boundary
        values in account, is the sum of all this single solutions for
        the parameter, filled in in the general solution. Thus you get
        an infinite sum of solutions for the parameters, and thus an
        infinite sum of solutions for the eigenfunctions.