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18 years, 28 days

MaplePrimes Activity

These are Posts that have been published by RedFox

Hi everybody! a few weeks ago I asked in this forum a question: First I have to say that is really nice how easliy it solves the problem. Now, I really need to see values in the solution (I mean, by given x and t). In addition, I want to create graphs of the temperture as a function of t (I mean to state x, and then to check the temperture as a function of t in this x). Only I know is to do graphs of the temp' as a function of x) How can I do all of those things? Thanks A LOT !!! RedFox (:
Hi everyone! I have a really important question about maple: while using maple to solve my problem (PDE, a few threads before), I did pay attention to the fact the maple is giving different values for 2 equal expressions! for example: assume we want to express sin(d) as a function (only) of tan(d). It's easy to see that 1/tan^2(d) is equal to [1/sin^2(d)]-1. therefore: 1/sin^2(d)=[tan^2(d)+1]/tan^2(d) from this we have immediately: sin(d)=tan(d)/sqrt(tan^2(d)+1) now, when I put sin(d) in my solution, I didn't get the same answer as I got with the expression to sin(d) above!!! Why is this happenning? and what can I do to get ovet it? (I really need to use the expression above instead of sin(d))
Hi everybody! I have to solve the following PDE (Heat equation), analytically or "numerically" (using a numeric method). My problem is mainly with the boundary conditions, which I don't know how to define: (1) U(x,t)'t = aU(x,t)''x (2) U(x,0) = T0 (3) -k*U'(0,t)x = q (4) -k*U'(d,t)x = h*(U(d,t)-T1) when a, TO, T1, k, q, d, and h are all constants! pay attention that the conditions (3) and (4) are on the derivative of U ! Thanks (: RedFox.
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