How can I verify the property $x(t)\delta(t)=x(0)\delta(t)$ in Mathematica? I tried with:
In[5]:= x[t_] := t
In[6]:= x[t] DiracDelta[t] == x[0] DiracDelta[t]
Out[6]= t DiracDelta[t] == 0
I expect the output True.
Thank you in advance.
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As noted by Bob Hanlon in the comments, $\delta(t)$ is not defined outside of integrals, so it doesn't make any sense to ask the question whether $x(t)\delta(t)=x(0)\delta(t)$.
However, what you can do is verify that both expressions behave the same under the integral, i.e. that $\int f(t) g(t)\mathrm{d}t$ is the same for $f(t)=x(t)\delta(t)$ and $f(t)=x(0)\delta(t)$:
In[1]:= Integrate[x[t] DiracDelta[t] f[t], t] == Integrate[x[0] DiracDelta[t] f[t], t]
Out[1]:= True
Here, you don't need to define anything for x[t], as this is true for all x[t].
Update 2
As noted in the comments, we should only consider definite integrals. This leads to the following definition of GeneralizedEqual:
GeneralizedEqual[f_, g_, t_, opts : OptionsPattern[]] :=
Integrate[h[t] f, {t, -Infinity, Infinity}, opts]
== Integrate[h[t] g, {t, -Infinity, Infinity}, opts]
With this definition, we can also prove the generalized version of the original equation (again, see comments), $\delta(t-T)x(t)"="\delta(t-T)x(T)$:
In[1]:= GeneralizedEqual[DiracDelta[t - T] f[t], DiracDelta[t - T] f[T], t]
Out[1]= ConditionalExpression[True, T \[Element] Reals]
Update
To make it a bit nicer to look at, you can introduce a generalized version of ==:
In[1]:= GeneralizedEqual[f_, g_, t_] := Integrate[h[t] f, t] == Integrate[h[t] g, t]
In[2]:= GeneralizedEqual[x[t] DiracDelta[t], x[0] DiracDelta[t], t]
Out[2]:= True
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$\begingroup$ Hi @Mathe172. I tried to prove Integrate[DiracDelta[t - T] f[t], t] == Integrate[DiracDelta[t - T] f[T], t], but the output is not True. Can you tell me I get True please? $\endgroup$ – Gennaro Arguzzi Aug 24 '17 at 9:34
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2$\begingroup$ Your definition of
GeneralizedEqualis not quite right, since the generalized functions are only defined via definite integrals. So this would be a better one:GeneralizedEqual[f_, g_, t_,opts:OptionsPattern[]] := Integrate[h[t] f, {t,-Infinity,Infinity},opts] == Integrate[h[t] g, {t,-Infinity,Infinity},opts]$\endgroup$ – Itai Seggev Aug 29 '17 at 22:50 -
$\begingroup$ @GennaroArguzzi See my comment my previous comment. If you used the GeneralizedEqual from there, you will find that
GeneralizedEqual[DiracDelta[t - T] f[t], DiracDelta[t - T] f[T], t]returnsConditionalExpression[True, T \[Element] Reals], and indeed the result is only valid if T is a real number. $\endgroup$ – Itai Seggev Aug 29 '17 at 23:00 -
$\begingroup$ @ItaiSeggev Thanks for pointing that out - I updated the answer to incorporate your improvements $\endgroup$ – Lukas Lang Aug 30 '17 at 9:13
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You get it directly with FunctionExpand:
In[1]:= FunctionExpand[DiracDelta[t]*f[t]]
Out[1]= DiracDelta[t] f[0]
In[2]:= FunctionExpand[DiracDelta[t-5]*f[t]]
Out[2]= DiracDelta[-5+t] f[5]
In[3]:= FunctionExpand[DiracDelta[t+2]*f[t]]
Out[3]= DiracDelta[2+t] f[-2]
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$\begingroup$ Welcome to Mathematica.SE! You can format inline code and code blocks by selecting the code and clicking the
{}button above the edit window. The edit window help button?is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful. ExecutingSetOptions[$FrontEnd, ExportMultipleCellsOptions -> {"IncludeCellLabels" -> False}]will keep theIn[]/Out[]labels from being pasted. That makes it easier for others to copy and test code. $\endgroup$ – Michael E2 Aug 24 '18 at 18:15
DiracDeltais defined in the context of an integral. Integrate both sides:Integrate[x[t] DiracDelta[t], {t, -Infinity, Infinity}] == Integrate[x[0] DiracDelta[t], {t, -Infinity, Infinity}]$\endgroup$ – Bob Hanlon Aug 23 '17 at 14:08