Mixing
Solving differential equation with delta function in nonhomogeneous part
$begingroup$
I have the following equation, which I need to solve:
$left(frac{mathrm{d}^2}{mathrm{d} x^2}-a^2right)A(x)=c delta(x-b)$.
The solution should be in form
$A(x)=C_1 e^{-x a}+C_2 e^{x a}+f(x)$,
but I am not sure with the non-homogeneous part $f(x)$. Mathematica offers the solution with Heaviside theta function
$f(x)=left[c e^{-a b - a x} (-e^{2 a b} + e^{2 a x}) Theta(-b + x)right]/(2 a)$,
however in literature I found the solution in form
$f(x)=left[c e^{-a lvert x-b rvert}right]/(2 a)$.
Are these two forms equivalent?
ordinary-differential-equations dirac-delta
asked Jun 16 '16 at 14:51
infinityinfinity
81
$endgroup$
add a comment |
$begingroup$
I have the following equation, which I need to solve:
$left(frac{mathrm{d}^2}{mathrm{d} x^2}-a^2right)A(x)=c delta(x-b)$.
The solution should be in form
$A(x)=C_1 e^{-x a}+C_2 e^{x a}+f(x)$,
but I am not sure with the non-homogeneous part $f(x)$. Mathematica offers the solution with Heaviside theta function
$f(x)=left[c e^{-a b - a x} (-e^{2 a b} + e^{2 a x}) Theta(-b + x)right]/(2 a)$,
however in literature I found the solution in form
$f(x)=left[c e^{-a lvert x-b rvert}right]/(2 a)$.
Are these two forms equivalent?
ordinary-differential-equations dirac-delta
asked Jun 16 '16 at 14:51
infinityinfinity
81
$endgroup$
1
$begingroup$
The solution Mathematica offers would vanish for $x<b$, which would seem to preclude it agreeing with the literature result which doesn't.
$endgroup$
– Semiclassical
Jun 16 '16 at 15:06
1
$begingroup$
Typo in Mathematica's solution?
$endgroup$
– Did
Jun 16 '16 at 15:09
add a comment |
$begingroup$
I have the following equation, which I need to solve:
$left(frac{mathrm{d}^2}{mathrm{d} x^2}-a^2right)A(x)=c delta(x-b)$.
The solution should be in form
$A(x)=C_1 e^{-x a}+C_2 e^{x a}+f(x)$,
but I am not sure with the non-homogeneous part $f(x)$. Mathematica offers the solution with Heaviside theta function
$f(x)=left[c e^{-a b - a x} (-e^{2 a b} + e^{2 a x}) Theta(-b + x)right]/(2 a)$,
however in literature I found the solution in form
$f(x)=left[c e^{-a lvert x-b rvert}right]/(2 a)$.
Are these two forms equivalent?
ordinary-differential-equations dirac-delta
asked Jun 16 '16 at 14:51
infinityinfinity
81
$endgroup$
I have the following equation, which I need to solve:
$left(frac{mathrm{d}^2}{mathrm{d} x^2}-a^2right)A(x)=c delta(x-b)$.
The solution should be in form
$A(x)=C_1 e^{-x a}+C_2 e^{x a}+f(x)$,
but I am not sure with the non-homogeneous part $f(x)$. Mathematica offers the solution with Heaviside theta function
$f(x)=left[c e^{-a b - a x} (-e^{2 a b} + e^{2 a x}) Theta(-b + x)right]/(2 a)$,
however in literature I found the solution in form
$f(x)=left[c e^{-a lvert x-b rvert}right]/(2 a)$.
Are these two forms equivalent?
ordinary-differential-equations dirac-delta
ordinary-differential-equations dirac-delta
asked Jun 16 '16 at 14:51
infinityinfinity
81
asked Jun 16 '16 at 14:51
infinityinfinity
81
asked Jun 16 '16 at 14:51
infinityinfinity
81
asked Jun 16 '16 at 14:51
infinityinfinity
81
asked Jun 16 '16 at 14:51
infinityinfinity
81
81
1
$begingroup$
The solution Mathematica offers would vanish for $x<b$, which would seem to preclude it agreeing with the literature result which doesn't.
$endgroup$
– Semiclassical
Jun 16 '16 at 15:06
1
$begingroup$
Typo in Mathematica's solution?
$endgroup$
– Did
Jun 16 '16 at 15:09
add a comment |
1
$begingroup$
The solution Mathematica offers would vanish for $x<b$, which would seem to preclude it agreeing with the literature result which doesn't.
$endgroup$
– Semiclassical
Jun 16 '16 at 15:06
1
$begingroup$
Typo in Mathematica's solution?
$endgroup$
– Did
Jun 16 '16 at 15:09
1
1
$begingroup$
The solution Mathematica offers would vanish for $x<b$, which would seem to preclude it agreeing with the literature result which doesn't.
$endgroup$
– Semiclassical
Jun 16 '16 at 15:06
$begingroup$
The solution Mathematica offers would vanish for $x<b$, which would seem to preclude it agreeing with the literature result which doesn't.
$endgroup$
– Semiclassical
Jun 16 '16 at 15:06
1
1
$begingroup$
Typo in Mathematica's solution?
$endgroup$
– Did
Jun 16 '16 at 15:09
$begingroup$
Typo in Mathematica's solution?
$endgroup$
– Did
Jun 16 '16 at 15:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The two forms are equivalent, but with different coefficients. The boundary conditions have to be taken into account in the relationships between the respective coefficients.
$$A''(x)-a^2A(x)=cdelta(x-b)$$
Let : $A(x)=e^{ax}Y(x) quadtoquad Y''(x)+2aY'(x)=c delta(x-b)e^{-ax}$
$Y'(x)=e^{-2ax}y(x) quadtoquad y'(x)=cdelta(x-b)e^{ax}$
$$y(x)=int cdelta(x-b)e^{ax}dx = c:e^{ab}theta(x-b)+C_0$$
$theta$ is the Heaviside step function.
$Y'(x)=e^{-2ax}y(x) = c:e^{ab}e^{-2ax}theta(x-b)+Ce^{-2ax}$
$Y(x)= ce^{ab}int e^{-2ax}theta(x-b)dx+C_1e^{-2ax}+C_2$
$int e^{-2ax}theta(x-b)dx= frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)$
$$Y(x)= ce^{ab} frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-2ax}+C_2$$
$A(x)=e^{ax}Y(x)=ce^{ab} frac{1}{2a}e^{ax}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$
$$A(x)=cfrac{1}{2a}left(e^{a(x-b)}-e^{-a(x-b)} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$$
This result is the same as the Mathematica result.
This is equivalent to : $quad A(x)=begin{cases}
C_1e^{-ax}+C_2e^{ax} quadtext{if } x<b\
C_3e^{-ax}+C_4e^{ax} quadtext{if } x>bend{cases}$
because both are function of $e^{ax}$ and $e^{-ax}$ but of course with different constant coefficients depending on the boundary conditions.
Comparison to the expected result :
$$quad A(x)=cfrac{1}{2a}e^{-a|x-b|} +c_1e^{-ax}+c_2e^{ax} $$
$$quad A(x)= begin{cases}
C'_1e^{-ax}+C'_2 e^{ax} quadtext{if } x<b\
C'_3e^{-ax}+C'_4 e^{ax} quadtext{if } x<b
end{cases}$$
because both are function of $e^{ax}$ and $e^{-ax}$ but with different coefficients depending on the boundary conditions.
Thus, the two forms of results (one from mathematica, the other from the literature) are equivalent, insofar the coefficients are related one to another according to the boundary conditions.
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
$endgroup$
add a comment |
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$begingroup$
The two forms are equivalent, but with different coefficients. The boundary conditions have to be taken into account in the relationships between the respective coefficients.
$$A''(x)-a^2A(x)=cdelta(x-b)$$
Let : $A(x)=e^{ax}Y(x) quadtoquad Y''(x)+2aY'(x)=c delta(x-b)e^{-ax}$
$Y'(x)=e^{-2ax}y(x) quadtoquad y'(x)=cdelta(x-b)e^{ax}$
$$y(x)=int cdelta(x-b)e^{ax}dx = c:e^{ab}theta(x-b)+C_0$$
$theta$ is the Heaviside step function.
$Y'(x)=e^{-2ax}y(x) = c:e^{ab}e^{-2ax}theta(x-b)+Ce^{-2ax}$
$Y(x)= ce^{ab}int e^{-2ax}theta(x-b)dx+C_1e^{-2ax}+C_2$
$int e^{-2ax}theta(x-b)dx= frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)$
$$Y(x)= ce^{ab} frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-2ax}+C_2$$
$A(x)=e^{ax}Y(x)=ce^{ab} frac{1}{2a}e^{ax}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$
$$A(x)=cfrac{1}{2a}left(e^{a(x-b)}-e^{-a(x-b)} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$$
This result is the same as the Mathematica result.
This is equivalent to : $quad A(x)=begin{cases}
C_1e^{-ax}+C_2e^{ax} quadtext{if } x<b\
C_3e^{-ax}+C_4e^{ax} quadtext{if } x>bend{cases}$
because both are function of $e^{ax}$ and $e^{-ax}$ but of course with different constant coefficients depending on the boundary conditions.
Comparison to the expected result :
$$quad A(x)=cfrac{1}{2a}e^{-a|x-b|} +c_1e^{-ax}+c_2e^{ax} $$
$$quad A(x)= begin{cases}
C'_1e^{-ax}+C'_2 e^{ax} quadtext{if } x<b\
C'_3e^{-ax}+C'_4 e^{ax} quadtext{if } x<b
end{cases}$$
because both are function of $e^{ax}$ and $e^{-ax}$ but with different coefficients depending on the boundary conditions.
Thus, the two forms of results (one from mathematica, the other from the literature) are equivalent, insofar the coefficients are related one to another according to the boundary conditions.
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
$endgroup$
add a comment |
$begingroup$
The two forms are equivalent, but with different coefficients. The boundary conditions have to be taken into account in the relationships between the respective coefficients.
$$A''(x)-a^2A(x)=cdelta(x-b)$$
Let : $A(x)=e^{ax}Y(x) quadtoquad Y''(x)+2aY'(x)=c delta(x-b)e^{-ax}$
$Y'(x)=e^{-2ax}y(x) quadtoquad y'(x)=cdelta(x-b)e^{ax}$
$$y(x)=int cdelta(x-b)e^{ax}dx = c:e^{ab}theta(x-b)+C_0$$
$theta$ is the Heaviside step function.
$Y'(x)=e^{-2ax}y(x) = c:e^{ab}e^{-2ax}theta(x-b)+Ce^{-2ax}$
$Y(x)= ce^{ab}int e^{-2ax}theta(x-b)dx+C_1e^{-2ax}+C_2$
$int e^{-2ax}theta(x-b)dx= frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)$
$$Y(x)= ce^{ab} frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-2ax}+C_2$$
$A(x)=e^{ax}Y(x)=ce^{ab} frac{1}{2a}e^{ax}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$
$$A(x)=cfrac{1}{2a}left(e^{a(x-b)}-e^{-a(x-b)} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$$
This result is the same as the Mathematica result.
This is equivalent to : $quad A(x)=begin{cases}
C_1e^{-ax}+C_2e^{ax} quadtext{if } x<b\
C_3e^{-ax}+C_4e^{ax} quadtext{if } x>bend{cases}$
because both are function of $e^{ax}$ and $e^{-ax}$ but of course with different constant coefficients depending on the boundary conditions.
Comparison to the expected result :
$$quad A(x)=cfrac{1}{2a}e^{-a|x-b|} +c_1e^{-ax}+c_2e^{ax} $$
$$quad A(x)= begin{cases}
C'_1e^{-ax}+C'_2 e^{ax} quadtext{if } x<b\
C'_3e^{-ax}+C'_4 e^{ax} quadtext{if } x<b
end{cases}$$
because both are function of $e^{ax}$ and $e^{-ax}$ but with different coefficients depending on the boundary conditions.
Thus, the two forms of results (one from mathematica, the other from the literature) are equivalent, insofar the coefficients are related one to another according to the boundary conditions.
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
$endgroup$
add a comment |
$begingroup$
The two forms are equivalent, but with different coefficients. The boundary conditions have to be taken into account in the relationships between the respective coefficients.
$$A''(x)-a^2A(x)=cdelta(x-b)$$
Let : $A(x)=e^{ax}Y(x) quadtoquad Y''(x)+2aY'(x)=c delta(x-b)e^{-ax}$
$Y'(x)=e^{-2ax}y(x) quadtoquad y'(x)=cdelta(x-b)e^{ax}$
$$y(x)=int cdelta(x-b)e^{ax}dx = c:e^{ab}theta(x-b)+C_0$$
$theta$ is the Heaviside step function.
$Y'(x)=e^{-2ax}y(x) = c:e^{ab}e^{-2ax}theta(x-b)+Ce^{-2ax}$
$Y(x)= ce^{ab}int e^{-2ax}theta(x-b)dx+C_1e^{-2ax}+C_2$
$int e^{-2ax}theta(x-b)dx= frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)$
$$Y(x)= ce^{ab} frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-2ax}+C_2$$
$A(x)=e^{ax}Y(x)=ce^{ab} frac{1}{2a}e^{ax}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$
$$A(x)=cfrac{1}{2a}left(e^{a(x-b)}-e^{-a(x-b)} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$$
This result is the same as the Mathematica result.
This is equivalent to : $quad A(x)=begin{cases}
C_1e^{-ax}+C_2e^{ax} quadtext{if } x<b\
C_3e^{-ax}+C_4e^{ax} quadtext{if } x>bend{cases}$
because both are function of $e^{ax}$ and $e^{-ax}$ but of course with different constant coefficients depending on the boundary conditions.
Comparison to the expected result :
$$quad A(x)=cfrac{1}{2a}e^{-a|x-b|} +c_1e^{-ax}+c_2e^{ax} $$
$$quad A(x)= begin{cases}
C'_1e^{-ax}+C'_2 e^{ax} quadtext{if } x<b\
C'_3e^{-ax}+C'_4 e^{ax} quadtext{if } x<b
end{cases}$$
because both are function of $e^{ax}$ and $e^{-ax}$ but with different coefficients depending on the boundary conditions.
Thus, the two forms of results (one from mathematica, the other from the literature) are equivalent, insofar the coefficients are related one to another according to the boundary conditions.
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
$endgroup$
The two forms are equivalent, but with different coefficients. The boundary conditions have to be taken into account in the relationships between the respective coefficients.
$$A''(x)-a^2A(x)=cdelta(x-b)$$
Let : $A(x)=e^{ax}Y(x) quadtoquad Y''(x)+2aY'(x)=c delta(x-b)e^{-ax}$
$Y'(x)=e^{-2ax}y(x) quadtoquad y'(x)=cdelta(x-b)e^{ax}$
$$y(x)=int cdelta(x-b)e^{ax}dx = c:e^{ab}theta(x-b)+C_0$$
$theta$ is the Heaviside step function.
$Y'(x)=e^{-2ax}y(x) = c:e^{ab}e^{-2ax}theta(x-b)+Ce^{-2ax}$
$Y(x)= ce^{ab}int e^{-2ax}theta(x-b)dx+C_1e^{-2ax}+C_2$
$int e^{-2ax}theta(x-b)dx= frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)$
$$Y(x)= ce^{ab} frac{1}{2a}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-2ax}+C_2$$
$A(x)=e^{ax}Y(x)=ce^{ab} frac{1}{2a}e^{ax}left(e^{-2ab}-e^{-2ax} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$
$$A(x)=cfrac{1}{2a}left(e^{a(x-b)}-e^{-a(x-b)} right)theta(x-b)+C_1e^{-ax}+C_2e^{ax}$$
This result is the same as the Mathematica result.
This is equivalent to : $quad A(x)=begin{cases}
C_1e^{-ax}+C_2e^{ax} quadtext{if } x<b\
C_3e^{-ax}+C_4e^{ax} quadtext{if } x>bend{cases}$
because both are function of $e^{ax}$ and $e^{-ax}$ but of course with different constant coefficients depending on the boundary conditions.
Comparison to the expected result :
$$quad A(x)=cfrac{1}{2a}e^{-a|x-b|} +c_1e^{-ax}+c_2e^{ax} $$
$$quad A(x)= begin{cases}
C'_1e^{-ax}+C'_2 e^{ax} quadtext{if } x<b\
C'_3e^{-ax}+C'_4 e^{ax} quadtext{if } x<b
end{cases}$$
because both are function of $e^{ax}$ and $e^{-ax}$ but with different coefficients depending on the boundary conditions.
Thus, the two forms of results (one from mathematica, the other from the literature) are equivalent, insofar the coefficients are related one to another according to the boundary conditions.
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
answered Jun 20 '16 at 8:18
JJacquelinJJacquelin
45.2k21856
45.2k21856
add a comment |
add a comment |
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$begingroup$
The solution Mathematica offers would vanish for $x<b$, which would seem to preclude it agreeing with the literature result which doesn't.
$endgroup$
– Semiclassical
Jun 16 '16 at 15:06
1
$begingroup$
Typo in Mathematica's solution?
$endgroup$
– Did
Jun 16 '16 at 15:09