Author 
Comment/Response 
x3

10/10/99 6:52pm
>Who knows how to solve the dif. Eq. in Mathematica 3. >Here is the Eq. a Uxx == Ut
>where U(x,t), Uxx is the second derivade of the function in x and Ut is the first derivade in t.
>
>The solution is: Exp[at][A Sin[ax]+B Cos[ax]]
>
>Please mail me. Thanks,.
While Mr. Hinton's comment to you has some merit, you are probably now somewhat confused because you know that the solution you quoted was found without having to specify any boundary conditions! Mr. Hinton's comment should be modified to say that ''there are no general solutions to the diffusion equation'', but many particular solutions exist, and they can certainly be found without specifying any boundary conditions. For example, if you assume that the diffusion equation is separable, that is, a solution can be expressed as a product of two functions: one purely in terms of the time, and the other purely in terms of the spatiaal coordinates, then particular solutions can certainly be retrieved without specifying any boundary conditions. The solution that you quoted was one of these. If you say that the soultion U then is such a product, then write the solution as U=X(x)T(t) where X is a function of x and T is a function of t. Put this solution into the diffucion equation and take the derivatives, and you will get a new differential equation 1/X Xxx = 1/T Tt, using your terminology. Now each side is separated and the left is a function of x only and the right is a function of t only. For this to be possible, then each side must be equal to some constant parameter, which let us call A. Therefore you now hav two differential equations, one in terms of x and one in terms of t XxxAx=0 and TtAT=0. Now you must solve each of these ordinary differential equations to get T and X and then you multiply them together to get your final answer U. Now where you have problems with Mathematica is that it can't solve these equations for all possible values of A. A is an arbitrary constant, and can be zero, positive real, negative real, imaginary positive, etc. You might try it on Mathematica to see the problems that arise.
For A= some number B(squared), positive real, you should get four solutions: Exp[Bx]*Exp[Sqr[B]*t] for one, and Exp[Bx]*Exp[Sqr[B]*t] for a second, Cosh[Bx]*Exp[Sqr[B]*t] for a third, and finally Sinh[Bx]*Exp[Sqr[B]*t]
For A = some number B(squared, real and negative, you get two solutions Cos[Bx]*Exp[Sqr[B]*t] and Sin[Bx]*Exp[Sqr[B]*t] . This is the soultion that you quoted in your question (since the diffusion equation is linear, you are permitted to add individual solutions together after multiplying by arbitrary constants, as you did).
For A = 0, you get another (trivial) solution, U=C+Dx, C and D arbitrary constants.
For A = iA>0 (positive and imaginary, i=Sqrt[1], you get two more solutions which are more complicated and which I won't quote here: anyway if you are going to use Mathematica, it won't give them to you anyway.
I am sorry to write so much, but I thought that it was important that you not be led to far astray by the incomplete previous comments, and hope that you see the necessity in studying mathematics at a classical level in order to understand the principles underlying most physical problems. Mathematica is a wonderful tool, but............
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