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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.