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Finding eigenfunction to this operator
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I have the operator
$$
A : -e^-2ax
fracpartialpartial x
left(e^2axfracpartialpartial xright)\
D_A = left( v in C^2[0,L] quad | quad v(0) = v(L) = 0 right)
$$
and I need to find the eigenfunction $p$. But first I have a question. How do I apply this operator to a function $u$? I see no $u$ in the formula. Is it a typo, should it be $fracpartial u partial x$ in the last partial derivative within the parenthesis? If this is the case, this is my attempt:
$$Ap = lambda p$$
$$-2ap' - p'' - lambda p = 0$$
$$p'' + 2ap' + lambda p = 0$$
$$p(x) = C_1 e^-a + sqrta^2 - lambda + C_2e^-a - sqrta^2 - lambda.$$
But this doesn't at all look like the correct answer
$$e^-axsinfrackpi xL, k=1,2,3,...$$
What am I doing wrong and how do I find the right answer (without complicating things)?
operator-theory hilbert-spaces operator-algebras eigenfunctions
edited Jul 25 at 18:37
gt6989b
30.3k22148
asked Jul 25 at 18:18
Heuristics
415111
add a comment |Â
up vote
2
down vote
favorite
I have the operator
$$
A : -e^-2ax
fracpartialpartial x
left(e^2axfracpartialpartial xright)\
D_A = left( v in C^2[0,L] quad | quad v(0) = v(L) = 0 right)
$$
and I need to find the eigenfunction $p$. But first I have a question. How do I apply this operator to a function $u$? I see no $u$ in the formula. Is it a typo, should it be $fracpartial u partial x$ in the last partial derivative within the parenthesis? If this is the case, this is my attempt:
$$Ap = lambda p$$
$$-2ap' - p'' - lambda p = 0$$
$$p'' + 2ap' + lambda p = 0$$
$$p(x) = C_1 e^-a + sqrta^2 - lambda + C_2e^-a - sqrta^2 - lambda.$$
But this doesn't at all look like the correct answer
$$e^-axsinfrackpi xL, k=1,2,3,...$$
What am I doing wrong and how do I find the right answer (without complicating things)?
operator-theory hilbert-spaces operator-algebras eigenfunctions
edited Jul 25 at 18:37
gt6989b
30.3k22148
asked Jul 25 at 18:18
Heuristics
415111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the operator
$$
A : -e^-2ax
fracpartialpartial x
left(e^2axfracpartialpartial xright)\
D_A = left( v in C^2[0,L] quad | quad v(0) = v(L) = 0 right)
$$
and I need to find the eigenfunction $p$. But first I have a question. How do I apply this operator to a function $u$? I see no $u$ in the formula. Is it a typo, should it be $fracpartial u partial x$ in the last partial derivative within the parenthesis? If this is the case, this is my attempt:
$$Ap = lambda p$$
$$-2ap' - p'' - lambda p = 0$$
$$p'' + 2ap' + lambda p = 0$$
$$p(x) = C_1 e^-a + sqrta^2 - lambda + C_2e^-a - sqrta^2 - lambda.$$
But this doesn't at all look like the correct answer
$$e^-axsinfrackpi xL, k=1,2,3,...$$
What am I doing wrong and how do I find the right answer (without complicating things)?
operator-theory hilbert-spaces operator-algebras eigenfunctions
edited Jul 25 at 18:37
gt6989b
30.3k22148
asked Jul 25 at 18:18
Heuristics
415111
I have the operator
$$
A : -e^-2ax
fracpartialpartial x
left(e^2axfracpartialpartial xright)\
D_A = left( v in C^2[0,L] quad | quad v(0) = v(L) = 0 right)
$$
and I need to find the eigenfunction $p$. But first I have a question. How do I apply this operator to a function $u$? I see no $u$ in the formula. Is it a typo, should it be $fracpartial u partial x$ in the last partial derivative within the parenthesis? If this is the case, this is my attempt:
$$Ap = lambda p$$
$$-2ap' - p'' - lambda p = 0$$
$$p'' + 2ap' + lambda p = 0$$
$$p(x) = C_1 e^-a + sqrta^2 - lambda + C_2e^-a - sqrta^2 - lambda.$$
But this doesn't at all look like the correct answer
$$e^-axsinfrackpi xL, k=1,2,3,...$$
What am I doing wrong and how do I find the right answer (without complicating things)?
operator-theory hilbert-spaces operator-algebras eigenfunctions
edited Jul 25 at 18:37
gt6989b
30.3k22148
asked Jul 25 at 18:18
Heuristics
415111
edited Jul 25 at 18:37
gt6989b
30.3k22148
edited Jul 25 at 18:37
gt6989b
30.3k22148
edited Jul 25 at 18:37
gt6989b
30.3k22148
30.3k22148
asked Jul 25 at 18:18
Heuristics
415111
asked Jul 25 at 18:18
Heuristics
415111
asked Jul 25 at 18:18
Heuristics
415111
415111
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
How do we apply an operator such as
$A = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right ) tag 0$
to a function $u$? Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:
We start with the following calculation:
$Au = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartial upartial x right )$
$=-e^-2ax left ( 2ae^2ax dfracpartial upartial x + e^2ax dfracpartial^2upartial x_2 right ) = -2adfracpartial upartial x - dfracpartial u^2partial x^2; tag 1$
if $lambda$ is an eigenvalue of this operator, then
$-e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = lambda u, tag 2$
then, via (1),
$-2adfracpartial upartial x - dfracpartial u^2partial x^2 = lambda u, tag 3$
or
$dfracpartial u^2partial x^2 + 2adfracpartial upartial x + lambda u = 0; tag 4$
since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^rho x$ where $rho$ is a root of the quadratic equation
$rho^2 + 2a rho + lambda = 0; tag 5$
we have
$rho_pm = dfrac-2a pm sqrt4a^2 - 4 lambda2 = -a pm sqrta^2 - lambda; tag 6$
thus we examine functions of the form
$u(x) = C_+ e^rho_+ x + C_- e^rho_- x tag 7$
which satisfy
$u(0) = u(L) = 0; tag 8$
we note that (7) and (8) yield
$C_+ + C_- = u(0) = 0, tag 9$
so that
$C_+ = -C_-; tag10$
we can thus take
$C_+ = C = -C_-, tag11$
and write
$u(x) = C(e^rho_+ x - e^rho_- x); tag12$
we next note that for $lambda < a^2$ in (6), we obtain two distinct real $rho_pm$, and if we apply the condition
$u(x) = C(e^rho_+ L - e^rho_- L) = u(L) = 0, tag13$
we see that with $C ne 0$ it leads to
$e^rho_+ L = e^rho_- L, tag14$
which is impossible if $rho_+ ne rho_-$; therefore (4), (5) have no solutions in $D_A$ with $lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $lambda < a^2$. If $lambda = a^2$, then we see via (6) that (5) has a repeated root $rho = rho_pm = -a$; it is well-known the solutions in this case are $e^rho x = e^-ax$ and $xe^-ax$; then
$u(x) = C_1 e^-ax + C_2 x e^-ax = (C_1 + C_2x) e^-ax, tag15$
whence
$C_1 = u(0) = 0, tag16$
and hence
$u(x) = C_2 x e^-ax, tag17$
whence
$C_2 L e^-aL = u(L) = 0 Longrightarrow C_2 = 0 Longrightarrow u(x) = 0, tag18$
eigenfunctions cannot be zero; thus we rule out the case $lambda = a^2$.
We are left only with the possibility that
$lambda > a^2; tag19$
then we have
$rho_pm = -a pm i sqrtlambda - a^2; tag20$
formulas (7), (9) and (12) still hold so we still write
$u(x) = C(e^rho_+ x - e^rho_- x); tag21$
with $rho_pm$ as in (20) we find
$e^rho_pm x = e^-ax e^pm sqrtlambda - a^2 x = e^-ax(cos sqrtlambda - a^2 x pm i sin sqrtlambda - a^2 x), tag22$
so that
$e^rho_+ x - e^rho_- x = 2i e^-ax sin sqrtlambda - a^2 x, tag23$
and
$u(x) = 2C i e^-ax sin sqrtlambda - a^2 x; tag24$
the criterion
$u(L) = 0 tag25$
forces
$sin sqrtlambda - a^2 L = 0, tag26$
or
$sqrtlambda - a^2 L = kpi, ; 0 ne k in Bbb Z tag27$
(we preclude $k = 0$ to enforce $u(x) ne 0$); then
$lambda = a^2 + dfrack^2 pi^2L^2, ; 0 ne k in Bbb Z, tag28$
and the eigenfunctions may be taken to be
$e^-ax sin sqrtlambda - a^2 x = e^-ax sin dfracx kpiL, 0 ne k in Bbb Z. tag29$
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
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2
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Hint: Your functions are
$$
p(x) = C_1 e^(-a + sqrta^2 - lambda)x + C_2e^(-a - sqrta^2 - lambda)x
$$
Consider how they behave when $lambda > a^2$. Then consider what $lambda$ and the $C_i$ need to be to satisfy the boundary conditions.
answered Jul 25 at 18:35
eyeballfrog
4,585527
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up vote
1
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How do I apply this operator to a function $u$?
Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.
And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.
Thus I believe $A$ applies like this
$$
beginalign
A u
&= -e^-2ax partial_x left(e^2ax partial_x u right) \
&= -e^-2axleft(e^2ax 2a partial_x u + e^2ax partial_x^2 uright) \
&= - (partial_x^2 u + 2a partial_x u)\
endalign
$$
to $u in D_A$.
answered Jul 25 at 18:23
mvw
30.2k22250
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
How do we apply an operator such as
$A = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right ) tag 0$
to a function $u$? Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:
We start with the following calculation:
$Au = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartial upartial x right )$
$=-e^-2ax left ( 2ae^2ax dfracpartial upartial x + e^2ax dfracpartial^2upartial x_2 right ) = -2adfracpartial upartial x - dfracpartial u^2partial x^2; tag 1$
if $lambda$ is an eigenvalue of this operator, then
$-e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = lambda u, tag 2$
then, via (1),
$-2adfracpartial upartial x - dfracpartial u^2partial x^2 = lambda u, tag 3$
or
$dfracpartial u^2partial x^2 + 2adfracpartial upartial x + lambda u = 0; tag 4$
since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^rho x$ where $rho$ is a root of the quadratic equation
$rho^2 + 2a rho + lambda = 0; tag 5$
we have
$rho_pm = dfrac-2a pm sqrt4a^2 - 4 lambda2 = -a pm sqrta^2 - lambda; tag 6$
thus we examine functions of the form
$u(x) = C_+ e^rho_+ x + C_- e^rho_- x tag 7$
which satisfy
$u(0) = u(L) = 0; tag 8$
we note that (7) and (8) yield
$C_+ + C_- = u(0) = 0, tag 9$
so that
$C_+ = -C_-; tag10$
we can thus take
$C_+ = C = -C_-, tag11$
and write
$u(x) = C(e^rho_+ x - e^rho_- x); tag12$
we next note that for $lambda < a^2$ in (6), we obtain two distinct real $rho_pm$, and if we apply the condition
$u(x) = C(e^rho_+ L - e^rho_- L) = u(L) = 0, tag13$
we see that with $C ne 0$ it leads to
$e^rho_+ L = e^rho_- L, tag14$
which is impossible if $rho_+ ne rho_-$; therefore (4), (5) have no solutions in $D_A$ with $lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $lambda < a^2$. If $lambda = a^2$, then we see via (6) that (5) has a repeated root $rho = rho_pm = -a$; it is well-known the solutions in this case are $e^rho x = e^-ax$ and $xe^-ax$; then
$u(x) = C_1 e^-ax + C_2 x e^-ax = (C_1 + C_2x) e^-ax, tag15$
whence
$C_1 = u(0) = 0, tag16$
and hence
$u(x) = C_2 x e^-ax, tag17$
whence
$C_2 L e^-aL = u(L) = 0 Longrightarrow C_2 = 0 Longrightarrow u(x) = 0, tag18$
eigenfunctions cannot be zero; thus we rule out the case $lambda = a^2$.
We are left only with the possibility that
$lambda > a^2; tag19$
then we have
$rho_pm = -a pm i sqrtlambda - a^2; tag20$
formulas (7), (9) and (12) still hold so we still write
$u(x) = C(e^rho_+ x - e^rho_- x); tag21$
with $rho_pm$ as in (20) we find
$e^rho_pm x = e^-ax e^pm sqrtlambda - a^2 x = e^-ax(cos sqrtlambda - a^2 x pm i sin sqrtlambda - a^2 x), tag22$
so that
$e^rho_+ x - e^rho_- x = 2i e^-ax sin sqrtlambda - a^2 x, tag23$
and
$u(x) = 2C i e^-ax sin sqrtlambda - a^2 x; tag24$
the criterion
$u(L) = 0 tag25$
forces
$sin sqrtlambda - a^2 L = 0, tag26$
or
$sqrtlambda - a^2 L = kpi, ; 0 ne k in Bbb Z tag27$
(we preclude $k = 0$ to enforce $u(x) ne 0$); then
$lambda = a^2 + dfrack^2 pi^2L^2, ; 0 ne k in Bbb Z, tag28$
and the eigenfunctions may be taken to be
$e^-ax sin sqrtlambda - a^2 x = e^-ax sin dfracx kpiL, 0 ne k in Bbb Z. tag29$
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
add a comment |Â
up vote
1
down vote
accepted
How do we apply an operator such as
$A = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right ) tag 0$
to a function $u$? Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:
We start with the following calculation:
$Au = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartial upartial x right )$
$=-e^-2ax left ( 2ae^2ax dfracpartial upartial x + e^2ax dfracpartial^2upartial x_2 right ) = -2adfracpartial upartial x - dfracpartial u^2partial x^2; tag 1$
if $lambda$ is an eigenvalue of this operator, then
$-e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = lambda u, tag 2$
then, via (1),
$-2adfracpartial upartial x - dfracpartial u^2partial x^2 = lambda u, tag 3$
or
$dfracpartial u^2partial x^2 + 2adfracpartial upartial x + lambda u = 0; tag 4$
since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^rho x$ where $rho$ is a root of the quadratic equation
$rho^2 + 2a rho + lambda = 0; tag 5$
we have
$rho_pm = dfrac-2a pm sqrt4a^2 - 4 lambda2 = -a pm sqrta^2 - lambda; tag 6$
thus we examine functions of the form
$u(x) = C_+ e^rho_+ x + C_- e^rho_- x tag 7$
which satisfy
$u(0) = u(L) = 0; tag 8$
we note that (7) and (8) yield
$C_+ + C_- = u(0) = 0, tag 9$
so that
$C_+ = -C_-; tag10$
we can thus take
$C_+ = C = -C_-, tag11$
and write
$u(x) = C(e^rho_+ x - e^rho_- x); tag12$
we next note that for $lambda < a^2$ in (6), we obtain two distinct real $rho_pm$, and if we apply the condition
$u(x) = C(e^rho_+ L - e^rho_- L) = u(L) = 0, tag13$
we see that with $C ne 0$ it leads to
$e^rho_+ L = e^rho_- L, tag14$
which is impossible if $rho_+ ne rho_-$; therefore (4), (5) have no solutions in $D_A$ with $lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $lambda < a^2$. If $lambda = a^2$, then we see via (6) that (5) has a repeated root $rho = rho_pm = -a$; it is well-known the solutions in this case are $e^rho x = e^-ax$ and $xe^-ax$; then
$u(x) = C_1 e^-ax + C_2 x e^-ax = (C_1 + C_2x) e^-ax, tag15$
whence
$C_1 = u(0) = 0, tag16$
and hence
$u(x) = C_2 x e^-ax, tag17$
whence
$C_2 L e^-aL = u(L) = 0 Longrightarrow C_2 = 0 Longrightarrow u(x) = 0, tag18$
eigenfunctions cannot be zero; thus we rule out the case $lambda = a^2$.
We are left only with the possibility that
$lambda > a^2; tag19$
then we have
$rho_pm = -a pm i sqrtlambda - a^2; tag20$
formulas (7), (9) and (12) still hold so we still write
$u(x) = C(e^rho_+ x - e^rho_- x); tag21$
with $rho_pm$ as in (20) we find
$e^rho_pm x = e^-ax e^pm sqrtlambda - a^2 x = e^-ax(cos sqrtlambda - a^2 x pm i sin sqrtlambda - a^2 x), tag22$
so that
$e^rho_+ x - e^rho_- x = 2i e^-ax sin sqrtlambda - a^2 x, tag23$
and
$u(x) = 2C i e^-ax sin sqrtlambda - a^2 x; tag24$
the criterion
$u(L) = 0 tag25$
forces
$sin sqrtlambda - a^2 L = 0, tag26$
or
$sqrtlambda - a^2 L = kpi, ; 0 ne k in Bbb Z tag27$
(we preclude $k = 0$ to enforce $u(x) ne 0$); then
$lambda = a^2 + dfrack^2 pi^2L^2, ; 0 ne k in Bbb Z, tag28$
and the eigenfunctions may be taken to be
$e^-ax sin sqrtlambda - a^2 x = e^-ax sin dfracx kpiL, 0 ne k in Bbb Z. tag29$
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
How do we apply an operator such as
$A = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right ) tag 0$
to a function $u$? Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:
We start with the following calculation:
$Au = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartial upartial x right )$
$=-e^-2ax left ( 2ae^2ax dfracpartial upartial x + e^2ax dfracpartial^2upartial x_2 right ) = -2adfracpartial upartial x - dfracpartial u^2partial x^2; tag 1$
if $lambda$ is an eigenvalue of this operator, then
$-e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = lambda u, tag 2$
then, via (1),
$-2adfracpartial upartial x - dfracpartial u^2partial x^2 = lambda u, tag 3$
or
$dfracpartial u^2partial x^2 + 2adfracpartial upartial x + lambda u = 0; tag 4$
since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^rho x$ where $rho$ is a root of the quadratic equation
$rho^2 + 2a rho + lambda = 0; tag 5$
we have
$rho_pm = dfrac-2a pm sqrt4a^2 - 4 lambda2 = -a pm sqrta^2 - lambda; tag 6$
thus we examine functions of the form
$u(x) = C_+ e^rho_+ x + C_- e^rho_- x tag 7$
which satisfy
$u(0) = u(L) = 0; tag 8$
we note that (7) and (8) yield
$C_+ + C_- = u(0) = 0, tag 9$
so that
$C_+ = -C_-; tag10$
we can thus take
$C_+ = C = -C_-, tag11$
and write
$u(x) = C(e^rho_+ x - e^rho_- x); tag12$
we next note that for $lambda < a^2$ in (6), we obtain two distinct real $rho_pm$, and if we apply the condition
$u(x) = C(e^rho_+ L - e^rho_- L) = u(L) = 0, tag13$
we see that with $C ne 0$ it leads to
$e^rho_+ L = e^rho_- L, tag14$
which is impossible if $rho_+ ne rho_-$; therefore (4), (5) have no solutions in $D_A$ with $lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $lambda < a^2$. If $lambda = a^2$, then we see via (6) that (5) has a repeated root $rho = rho_pm = -a$; it is well-known the solutions in this case are $e^rho x = e^-ax$ and $xe^-ax$; then
$u(x) = C_1 e^-ax + C_2 x e^-ax = (C_1 + C_2x) e^-ax, tag15$
whence
$C_1 = u(0) = 0, tag16$
and hence
$u(x) = C_2 x e^-ax, tag17$
whence
$C_2 L e^-aL = u(L) = 0 Longrightarrow C_2 = 0 Longrightarrow u(x) = 0, tag18$
eigenfunctions cannot be zero; thus we rule out the case $lambda = a^2$.
We are left only with the possibility that
$lambda > a^2; tag19$
then we have
$rho_pm = -a pm i sqrtlambda - a^2; tag20$
formulas (7), (9) and (12) still hold so we still write
$u(x) = C(e^rho_+ x - e^rho_- x); tag21$
with $rho_pm$ as in (20) we find
$e^rho_pm x = e^-ax e^pm sqrtlambda - a^2 x = e^-ax(cos sqrtlambda - a^2 x pm i sin sqrtlambda - a^2 x), tag22$
so that
$e^rho_+ x - e^rho_- x = 2i e^-ax sin sqrtlambda - a^2 x, tag23$
and
$u(x) = 2C i e^-ax sin sqrtlambda - a^2 x; tag24$
the criterion
$u(L) = 0 tag25$
forces
$sin sqrtlambda - a^2 L = 0, tag26$
or
$sqrtlambda - a^2 L = kpi, ; 0 ne k in Bbb Z tag27$
(we preclude $k = 0$ to enforce $u(x) ne 0$); then
$lambda = a^2 + dfrack^2 pi^2L^2, ; 0 ne k in Bbb Z, tag28$
and the eigenfunctions may be taken to be
$e^-ax sin sqrtlambda - a^2 x = e^-ax sin dfracx kpiL, 0 ne k in Bbb Z. tag29$
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
How do we apply an operator such as
$A = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right ) tag 0$
to a function $u$? Typically we work from right to left, taking the argument $u$ to occupy the first or right-most "empty slot"; bearing this in mind:
We start with the following calculation:
$Au = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = -e^-2ax dfracpartialpartial x left (e^2ax dfracpartial upartial x right )$
$=-e^-2ax left ( 2ae^2ax dfracpartial upartial x + e^2ax dfracpartial^2upartial x_2 right ) = -2adfracpartial upartial x - dfracpartial u^2partial x^2; tag 1$
if $lambda$ is an eigenvalue of this operator, then
$-e^-2ax dfracpartialpartial x left (e^2ax dfracpartialpartial x right )u = lambda u, tag 2$
then, via (1),
$-2adfracpartial upartial x - dfracpartial u^2partial x^2 = lambda u, tag 3$
or
$dfracpartial u^2partial x^2 + 2adfracpartial upartial x + lambda u = 0; tag 4$
since this is a linear, constant coefficient, second order ordinary differential equation, we may find solutions of the general form $e^rho x$ where $rho$ is a root of the quadratic equation
$rho^2 + 2a rho + lambda = 0; tag 5$
we have
$rho_pm = dfrac-2a pm sqrt4a^2 - 4 lambda2 = -a pm sqrta^2 - lambda; tag 6$
thus we examine functions of the form
$u(x) = C_+ e^rho_+ x + C_- e^rho_- x tag 7$
which satisfy
$u(0) = u(L) = 0; tag 8$
we note that (7) and (8) yield
$C_+ + C_- = u(0) = 0, tag 9$
so that
$C_+ = -C_-; tag10$
we can thus take
$C_+ = C = -C_-, tag11$
and write
$u(x) = C(e^rho_+ x - e^rho_- x); tag12$
we next note that for $lambda < a^2$ in (6), we obtain two distinct real $rho_pm$, and if we apply the condition
$u(x) = C(e^rho_+ L - e^rho_- L) = u(L) = 0, tag13$
we see that with $C ne 0$ it leads to
$e^rho_+ L = e^rho_- L, tag14$
which is impossible if $rho_+ ne rho_-$; therefore (4), (5) have no solutions in $D_A$ with $lambda < a^2$; since solutions to (4) are eigenfunctions of the operator $A$, we see it has no eigenvalues $lambda < a^2$. If $lambda = a^2$, then we see via (6) that (5) has a repeated root $rho = rho_pm = -a$; it is well-known the solutions in this case are $e^rho x = e^-ax$ and $xe^-ax$; then
$u(x) = C_1 e^-ax + C_2 x e^-ax = (C_1 + C_2x) e^-ax, tag15$
whence
$C_1 = u(0) = 0, tag16$
and hence
$u(x) = C_2 x e^-ax, tag17$
whence
$C_2 L e^-aL = u(L) = 0 Longrightarrow C_2 = 0 Longrightarrow u(x) = 0, tag18$
eigenfunctions cannot be zero; thus we rule out the case $lambda = a^2$.
We are left only with the possibility that
$lambda > a^2; tag19$
then we have
$rho_pm = -a pm i sqrtlambda - a^2; tag20$
formulas (7), (9) and (12) still hold so we still write
$u(x) = C(e^rho_+ x - e^rho_- x); tag21$
with $rho_pm$ as in (20) we find
$e^rho_pm x = e^-ax e^pm sqrtlambda - a^2 x = e^-ax(cos sqrtlambda - a^2 x pm i sin sqrtlambda - a^2 x), tag22$
so that
$e^rho_+ x - e^rho_- x = 2i e^-ax sin sqrtlambda - a^2 x, tag23$
and
$u(x) = 2C i e^-ax sin sqrtlambda - a^2 x; tag24$
the criterion
$u(L) = 0 tag25$
forces
$sin sqrtlambda - a^2 L = 0, tag26$
or
$sqrtlambda - a^2 L = kpi, ; 0 ne k in Bbb Z tag27$
(we preclude $k = 0$ to enforce $u(x) ne 0$); then
$lambda = a^2 + dfrack^2 pi^2L^2, ; 0 ne k in Bbb Z, tag28$
and the eigenfunctions may be taken to be
$e^-ax sin sqrtlambda - a^2 x = e^-ax sin dfracx kpiL, 0 ne k in Bbb Z. tag29$
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
answered Jul 26 at 2:32
Robert Lewis
36.9k22155
36.9k22155
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: Your functions are
$$
p(x) = C_1 e^(-a + sqrta^2 - lambda)x + C_2e^(-a - sqrta^2 - lambda)x
$$
Consider how they behave when $lambda > a^2$. Then consider what $lambda$ and the $C_i$ need to be to satisfy the boundary conditions.
answered Jul 25 at 18:35
eyeballfrog
4,585527
add a comment |Â
up vote
2
down vote
Hint: Your functions are
$$
p(x) = C_1 e^(-a + sqrta^2 - lambda)x + C_2e^(-a - sqrta^2 - lambda)x
$$
Consider how they behave when $lambda > a^2$. Then consider what $lambda$ and the $C_i$ need to be to satisfy the boundary conditions.
answered Jul 25 at 18:35
eyeballfrog
4,585527
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Your functions are
$$
p(x) = C_1 e^(-a + sqrta^2 - lambda)x + C_2e^(-a - sqrta^2 - lambda)x
$$
Consider how they behave when $lambda > a^2$. Then consider what $lambda$ and the $C_i$ need to be to satisfy the boundary conditions.
answered Jul 25 at 18:35
eyeballfrog
4,585527
Hint: Your functions are
$$
p(x) = C_1 e^(-a + sqrta^2 - lambda)x + C_2e^(-a - sqrta^2 - lambda)x
$$
Consider how they behave when $lambda > a^2$. Then consider what $lambda$ and the $C_i$ need to be to satisfy the boundary conditions.
answered Jul 25 at 18:35
eyeballfrog
4,585527
answered Jul 25 at 18:35
eyeballfrog
4,585527
answered Jul 25 at 18:35
eyeballfrog
4,585527
answered Jul 25 at 18:35
eyeballfrog
4,585527
4,585527
add a comment |Â
add a comment |Â
up vote
1
down vote
How do I apply this operator to a function $u$?
Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.
And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.
Thus I believe $A$ applies like this
$$
beginalign
A u
&= -e^-2ax partial_x left(e^2ax partial_x u right) \
&= -e^-2axleft(e^2ax 2a partial_x u + e^2ax partial_x^2 uright) \
&= - (partial_x^2 u + 2a partial_x u)\
endalign
$$
to $u in D_A$.
answered Jul 25 at 18:23
mvw
30.2k22250
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
add a comment |Â
up vote
1
down vote
How do I apply this operator to a function $u$?
Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.
And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.
Thus I believe $A$ applies like this
$$
beginalign
A u
&= -e^-2ax partial_x left(e^2ax partial_x u right) \
&= -e^-2axleft(e^2ax 2a partial_x u + e^2ax partial_x^2 uright) \
&= - (partial_x^2 u + 2a partial_x u)\
endalign
$$
to $u in D_A$.
answered Jul 25 at 18:23
mvw
30.2k22250
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
How do I apply this operator to a function $u$?
Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.
And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.
Thus I believe $A$ applies like this
$$
beginalign
A u
&= -e^-2ax partial_x left(e^2ax partial_x u right) \
&= -e^-2axleft(e^2ax 2a partial_x u + e^2ax partial_x^2 uright) \
&= - (partial_x^2 u + 2a partial_x u)\
endalign
$$
to $u in D_A$.
answered Jul 25 at 18:23
mvw
30.2k22250
How do I apply this operator to a function $u$?
Your $D_A$ looks like the function space of all 2 times differentiable functions on $[0,L]$. So this seems like a function of one variable.
And therefore these seem ordinary differential operators, no partial ones, in the definition of $A$.
Thus I believe $A$ applies like this
$$
beginalign
A u
&= -e^-2ax partial_x left(e^2ax partial_x u right) \
&= -e^-2axleft(e^2ax 2a partial_x u + e^2ax partial_x^2 uright) \
&= - (partial_x^2 u + 2a partial_x u)\
endalign
$$
to $u in D_A$.
answered Jul 25 at 18:23
mvw
30.2k22250
edited Jul 25 at 18:34
answered Jul 25 at 18:23
mvw
30.2k22250
answered Jul 25 at 18:23
mvw
30.2k22250
answered Jul 25 at 18:23
mvw
30.2k22250
30.2k22250
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
add a comment |Â
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Okay yes, that's what I thought and used as well. But using it, my answer does not seem coherent with the correct answer
– Heuristics
Jul 25 at 18:25
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Then I am at a loss what kind of function a feasible $u$ is.
– mvw
Jul 25 at 18:27
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
Is my method of finding the eigenfunction correct (although somewhat incomplete)?
– Heuristics
Jul 25 at 18:28
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
And you have to apply the boundary conditions from the $D_A$ definition. This might reduce your exponential solution to that damped sine function.
– mvw
Jul 25 at 18:30
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
Have a look at Universal oscillator.
– mvw
Jul 25 at 18:37
add a comment |Â
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