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Finding a fundamental solution of a linear differential operator $f''-c^2f$, I try in 12.3.1 on Windows 10

DSolve[f''[x] - c^2*f[x] == DiracDelta[x - y], f[x], x]

{{f[x] -> E^(c x) C[1] + E^(-c x) C[2] - ( E^(-c x - c y) (-E^(2 c x) + E^(2 c y)) HeavisideTheta[x - y])/( 2 c)}}

To be sure, I also execute (following the documentation to DiracDelta, namely the "Properties" section, and Encyclopedia of Mathematics)

DSolve[f''[x] - c^2*f[x] == DiracDelta[-x + y], f[x], x]

{{f[x] -> E^(c x) C[1] + E^(-c x) C[2] + E^(-c x) (E^(2 c x) Inactive[Integrate][(E^(-c K[1]) DiracDelta[y - K[1]])/( 2 c), {K[1], 1, x}] + Inactive[Integrate][-((E^(c K[2]) DiracDelta[y - K[2]])/( 2 c)), {K[2], 1, x}])}}

Which of these different results is correct?

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    $\begingroup$ The first one is the correct one , I think. The second solution is a general solution, which Mathematica couldn't evaluate further. May be because Mathematica doesn't recognize the symmetrie of DiracDelta: DiracDela[x]== DiracDelta[-x] ->True but DiracDela[x-y]== DiracDelta[y-x] ->unevaluated $\endgroup$
    – Ulrich Neumann
    Nov 19 '21 at 13:34
  • $\begingroup$ My unsuccessful attempt is DSolve[f''[x] - c^2*f[x] == eps/Pi/(eps^2 + (x - y)^2), f[x], x] and then eps tends to zero from above. The result is E^(c x) C[1] + E^(-c x) C[2]. $\endgroup$
    – user64494
    Nov 19 '21 at 13:46
  • $\begingroup$ The weak limit differs from the usual notion of the limit. $\endgroup$
    – user64494
    Nov 19 '21 at 13:54
  • $\begingroup$ Your "unsuccessful attempt" gives a solution, similar to your first solution in your question. But last step eps->0 is not alowed! $\endgroup$
    – Ulrich Neumann
    Nov 19 '21 at 14:25
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    $\begingroup$ Your last "personal" hint is misplaced, completely unnecessary and completely unfounded! Like many other users too in the stackexchange community I'll stop my contributions to your questions completely! $\endgroup$
    – Ulrich Neumann
    Nov 20 '21 at 9:11
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To long for a comment:

Assuming c==1 the first solution follows to

sol1=DSolve[{f''[x] - f[x] == DiracDelta[x - y] }, f ,x][[1, 1]][[2, 2]] /. {C[1] -> 0, C[2] -> 0}

Your "unsuccessful attempt" gives

eps =.
F = Function[{x, y, eps}, Evaluate[DSolve[{f''[x] -  f[x] == eps/Pi/(eps^2 + (x - y)^2), f[0] == 0 }, f , x][[1, 1]][[2, 2]] /. {C[1] -> 0,  C[2] -> 0}]]

Plot of both solutions

Plot3D[Evaluate[{sol1, Boole[x > y] F[x,y, .1]}], {x, 0, 3}, {y, 0,3}, PlotRange -> All]

enter image description here

shows quite good approximation.

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  • $\begingroup$ You wrote "shows quite good approximation". Can you elaborate it? What exactly is well approximated? $\endgroup$
    – user64494
    Nov 19 '21 at 15:46
  • $\begingroup$ Do you put eps=0.0;? The result of eps=.; is eps. $\endgroup$
    – user64494
    Nov 19 '21 at 15:51
  • $\begingroup$ I find your "too long for acomment" very similar to that answer . $\endgroup$
    – user64494
    Nov 19 '21 at 15:57
  • $\begingroup$ I added the plot command with eps=.1. $\endgroup$
    – Ulrich Neumann
    Nov 19 '21 at 15:57
  • $\begingroup$ Thank you. Still don't understand all that [[1, 1]][[2, 2]] and see "a good approximation". $\endgroup$
    – user64494
    Nov 19 '21 at 15:59
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I'm not sure why your second solution is missing the HeavisideTheta factor in the second solution, but it is fairly straightforward to show your first solution satisfies the differential equation.

eq=f''[x]-c^2*f[x]==DiracDelta[x-y]

DSolve[eq,f[x],x]//Flatten

(*   {f[x] -> C[1]*E^(c*x) + C[2]/E^(c*x) - (E^(-(c*x) - c*y)*(E^(2*c*y) - E^(2*c*x))*
      HeavisideTheta[x - y])/(2*c)}   *)

f[x_]=f[x]/.%

Plug the result into the equation to test our solution.

eq = eq//Simplify

(*   2*E^(c*(x - y))*DiracDelta[x - y] + (E^(c*(y - x)) - E^(c*(x - y)))*
    DiracDelta[x - y] + ((E^(2*c*x) - E^(2*c*y))*
     Derivative[1][DiracDelta][x - y])/(E^(c*(x + y))*(2*c)) == 
  DiracDelta[x - y]   *)

It looks like we do not have a match, but realizing that DiracDelta's only mean something inside an integral, we multiply each side by g[x] and integrate:

Integrate[eq[[1]]*g[x], {x, -Infinity, Infinity}] == Integrate[eq[[2]]*g[x], {x, -Infinity, Infinity}]
(*   ConditionalExpression[True, Element[y, Reals]]  *)

So the result satisfies the differential equation at least in the case of real y.

Doing the integrals by hand, I agree with Mathematica for the above. In the case of your second answer, doing the integrals by hand and assuming y lies inside the limits of integration I get the same result as the first solution except that it is missing the HeavisideTheta factor. The two solutions should be the same.

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  • $\begingroup$ Thank you for your work which confirms that the implementation of the $\delta$-distribution in Mathematica leaves much to be desired. $\endgroup$
    – user64494
    Nov 30 '21 at 8:26
  • $\begingroup$ Replacing eq=f''[x]-c^2*f[x]==DiracDelta[x-y] by eq=f''[x]-c^2*f[x]==DiracDelta[y-x] and executing the rest of your code without any changes, I obtain ConditionalExpression[True, y \[Element] Reals] . $\endgroup$
    – user64494
    Nov 30 '21 at 10:43
  • $\begingroup$ Moreover, ClearAll[f,x,y]; eq1 = f''[x] - c^2*f[x] == DiracDelta[x - y]; eq2 = f''[x] - c^2*f[x] == DiracDelta[y - x]; DSolve[eq2, f[x], x] // Flatten; f[x_] = f[x] /. %;eq1 = eq1 // Simplify;Integrate[eq1[[1]]*g[x], {x, -Infinity, Infinity}] == Integrate[eq1[[2]]*g[x], {x, -Infinity, Infinity}] produces ConditionalExpression[True, y \[Element] Reals], This means that the weak solution of eq2 is the weak solution of eq1. I repeat that Integrate[(E^(-c K[1]) DiracDelta[y - K[1]])/( 2 c), {K[1], 1, x}] makes no sense in traditional math. $\endgroup$
    – user64494
    Nov 30 '21 at 13:19
  • $\begingroup$ I really do not understand why you can't do that integral. It is in the standard form given in all the books covering DiracDelta integrals. Integrate[f[x] DiracDelta[a-x] dx. The value is f[a] if a is contained in the integration limits, and zero otherwise. For your integral the value is Exp[-c y]/(2 c) for y between 1 and x. If y is outside those values, the value is zero, and the user must actually use knowledge of the situation if y is either 1 or x. $\endgroup$
    – Bill Watts
    Nov 30 '21 at 21:33

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