I am working on a green’s function program to solve PDE’s with the DiracDelta function in it. I am using it for some student’s and it work’s for ODE’s but not PDE"S. Can someone tell me why the following is happening. The function g’‘[x] is obviously a function of x and not t. If you were doing it by hand you can forget integration by parts.
When I enter this
LaplaceTransform[g’‘[x], t, s]
I get this
g’‘[x]/s
Obviously the s should not be on the bottom. Why am I getting this. Same for the first derivative. Enter
LaplaceTransform[g’[x], t, s]
I get this g’[x]/s
still has the s on the bottom. It works if g is a function of t
LaplaceTransform[g’[t], t, s]
is,
-g[0] + s LaplaceTransform[g[t], t, s]
which is correct, but now we have g[t] and not g[x]. Since I am writing a program I need the Laplace transform to know the difference.
I think t is just a dummy variable here for integral. It does not matter whether it is x or t.
LaplaceTransform[g'[t], t, s]
and
LaplaceTransform[g'[x], x, s]
give me the same result.
I understand that. That is not the problem. Let me put it a little better. When you plug in the Laplace Transform you have to do a integration by parts, to take the derivative of the Laplace transform and get the derivative off the Laplace transform. That would bring down one s variable. It should not do anything but add the Laplace transform to the variable g[x]. g’‘[t] - g’‘[x] = DiracDelta[x]*DiracDelta[t], taking the laplace transform would give G’‘[x] - s^2G = DiracDelta[x]e^-st .where G is now the Laplace transform. Notice nothing happened to the first term G’'[x] since we took the Laplace transform with respect to t. It is now LaplaceTransorm/dx^2, the second term changed since it is a function of t and the DiracDelta of t changed because it is a function of t. I cannot write it out better because I do not know what format of this editor. It is not a matter of a dummy variable. It is like taking the partial derivative of xy with respect to x, which gives y.
I need to learn to do post’s better because I cannot explain myself, until I can show the equations. The same thing would be true for a fourier transform. I can do it by hand but not with mathematica. I should not get the 1/s, if the variable is not t. It is like the partial derivative I did above where you hold y constant. You can take a laplace transform with respect to another variable and it has no effect but to leave the laplace transform. In order to do the laplace transforms you have to do a integration by parts, but if the variable is not the one you are taking the derivative of, you just leave it and do not do the integration by parts. I will have to learn the editor better. I could explain it so easy if I could just write out equations. Like I said I can do it by hand, but not mathematica, I do not know where the 1/s comes from, it should not be there. It would be like taking the derivative with respect to x of e^2y, it is zero. I cannot use this editor so I am having real problems communicating. I am really sorry.
Thank’s anyway, I really appreciate you trying to help.
Hi Russel, A nice write-up on posting can be found here How to post including how to put Mathematica code in your post.
Kind regards, David
However, if I read your post correctly, I think LaplaceTransform is working correctly, in that the Laplace transform of 1 is 1/s.
In[1]:= LaplaceTransform[g'[t], t, s]
Out[1]= -g[0] + s LaplaceTransform[g[t], t, s]
In[2]:= LaplaceTransform[g'[x], t, s]
Out[2]= Derivative[1][g][x]/s
In[3]:= LaplaceTransform[1, t, s]
Out[3]= 1/s
The function “LaplaceTransform” takes three arguments. The first argument is the expression to be transformed. The second is a symbol which indicates the independant variable of the first expression. The third is a symbol to be used in the output. If we take a look at your example,
LaplaceTransform[g''[x], t, s]
we see that the expression is g’‘[x] and the independant variable of it is “t”. The expression g’'[x] is a constant with respect to “t” and that’s why you get that result. Ignoring the subtilies of analysis for a bit, the input:
LaplaceTransform[expr, t, s]
Is basically just:
Integrate[expr Exp[-s t], {t, 0, Infinity}]
If your expression “expr” doesn’t have the variable of integration in it, then it’s basically just a constant as far as the function “LaplaceTransform” is concerned.
If you are still seeing issues, try writing it out as an integral using the Integrate command. I think seeing that would help us better understand what you’re looking to do.
Thanks I will learn how to post this. I started to realize that in order to explain this one I have a bigger problem than just mathematica. I think the person would also have to understand green’s functions. I can do them by hand but not with Mathematica and the answer to the problem is easy if you understand green’s functions. I do not know if I should go to the mathematics site or the physics site. You have f’[x]e^xs, you can get f[x] by integration by parts. It puts the derivative with respect to x on the e^sx and you get f[x] by itself. So you get -sf[x]e^xs so you now have f[x] by itself, with no derivatives on it. If you have f’[x]e^ys you cannot do an integration by parts, since the derivative of e^ys with respect to x is zero. So the laplace transform e^ys just stays there. If I can get someone to understand this is where the problem is, maybe they can help me. Mathematica should leave it alone, since you do not want to do a integration by parts, which gives 0. But it leaves (f’[x]e^ys)/s…[I will call a] and it is a problem with mathematica I do not understand. In other-words you have done nothing to the function or the exponential but mathematica puts in 1/s and nothing has been done to that part of the equation. David can you lead me in the right direction. It is like asking mathematica not to take a derivative by it gives and extra x or what ever variable.
You understood. Your third line is the one that gives the problem. It is treating any derivative except the one with respect to t as one. Now, I know what is happening. You can put in the derivative of x’‘’‘’‘’‘’‘’[x] and it still gives 1/s. I now see what the problem is. You, without my help were able to see what is happening. I know what is happening. I now think I know what to do. You got me started on the right path, thanks so much!
Thanks again,
Russell Barry
By the way the fourier transform works. It is also an exponential function.
FourierTransform[g’[x], t, w] g[x] is Also a function of x not t
Sqrt[2 [Pi]] DiracDelta[w] ][g]‘[x]
The g’[x] is still there and If you can remember the defination of the diracdelta function, this is right and still leaves the g’[x]
I was trying to get an answer similar to this one for the laplace transform. Only the dirac delta would be the laplace transform.
Thanks to both of you,
Have a good day