Mixing
In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal?
$begingroup$
In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal?
Comprobe the orthogonality direct by integration.
Where,
$$x^2y''+xy'+lambda y=0tag1$$
with boundary condition:
$y(1)=0,,,,y(b)=0$
My attempt
The solution of the ODE is:
$$y(x)=C_1cos(ln(x)sqrt{lambda})+C_2sin(ln(x)sqrt{x})$$
Note the eigenvalues of $(1)$ are:
$$lambda_k=frac{k^2pi^2}{ln(b)^2}$$
with $b>1$ and $k=0,1,2,...$
The minimun eigenvalue is $lambda_0=0$ and
$$lim_{krightarrowinfty}lambda_k=infty$$
Moreover, the equation $(1)$ in Sturm-Liouville form is:
$$xy''+y'+frac{lambda}{x}y=0tag2$$
Here, i'm stuck. Can someone help me?
pde
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
$endgroup$
add a comment |
$begingroup$
In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal?
Comprobe the orthogonality direct by integration.
Where,
$$x^2y''+xy'+lambda y=0tag1$$
with boundary condition:
$y(1)=0,,,,y(b)=0$
My attempt
The solution of the ODE is:
$$y(x)=C_1cos(ln(x)sqrt{lambda})+C_2sin(ln(x)sqrt{x})$$
Note the eigenvalues of $(1)$ are:
$$lambda_k=frac{k^2pi^2}{ln(b)^2}$$
with $b>1$ and $k=0,1,2,...$
The minimun eigenvalue is $lambda_0=0$ and
$$lim_{krightarrowinfty}lambda_k=infty$$
Moreover, the equation $(1)$ in Sturm-Liouville form is:
$$xy''+y'+frac{lambda}{x}y=0tag2$$
Here, i'm stuck. Can someone help me?
pde
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
$endgroup$
add a comment |
$begingroup$
In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal?
Comprobe the orthogonality direct by integration.
Where,
$$x^2y''+xy'+lambda y=0tag1$$
with boundary condition:
$y(1)=0,,,,y(b)=0$
My attempt
The solution of the ODE is:
$$y(x)=C_1cos(ln(x)sqrt{lambda})+C_2sin(ln(x)sqrt{x})$$
Note the eigenvalues of $(1)$ are:
$$lambda_k=frac{k^2pi^2}{ln(b)^2}$$
with $b>1$ and $k=0,1,2,...$
The minimun eigenvalue is $lambda_0=0$ and
$$lim_{krightarrowinfty}lambda_k=infty$$
Moreover, the equation $(1)$ in Sturm-Liouville form is:
$$xy''+y'+frac{lambda}{x}y=0tag2$$
Here, i'm stuck. Can someone help me?
pde
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
$endgroup$
In accord to Sturm-Liouville theorem, With regard to which weight are the autofunctions orthogonal?
Comprobe the orthogonality direct by integration.
Where,
$$x^2y''+xy'+lambda y=0tag1$$
with boundary condition:
$y(1)=0,,,,y(b)=0$
My attempt
The solution of the ODE is:
$$y(x)=C_1cos(ln(x)sqrt{lambda})+C_2sin(ln(x)sqrt{x})$$
Note the eigenvalues of $(1)$ are:
$$lambda_k=frac{k^2pi^2}{ln(b)^2}$$
with $b>1$ and $k=0,1,2,...$
The minimun eigenvalue is $lambda_0=0$ and
$$lim_{krightarrowinfty}lambda_k=infty$$
Moreover, the equation $(1)$ in Sturm-Liouville form is:
$$xy''+y'+frac{lambda}{x}y=0tag2$$
Here, i'm stuck. Can someone help me?
pde
pde
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
asked Jan 2 at 23:21
Bvss12Bvss12
1,785618
1,785618
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, $lambda ne 0$ since it would result in a trivial solution (prove this yourself). So the eigenfunctions are
$$ y_k(x) = sinleft(kpi frac{ln x}{ln b}right) $$
up to a multiplicative constant, with $lambda_k = frac{k^2pi^2}{ln^2 b}$ and $k = 1,2,3,dots$
Let's compare 2 distinct eigenfunctions $y_n$ and $y_m$, with corresponding eigenvalues $lambda_nnelambda_m$. By the Sturm-Liouville form, we have the following relations:
begin{align}
frac{lambda_n}{x}y_n &= -(x{y_n}')' \
frac{lambda_m}{x}y_m &= -(x{y_m}')'
end{align}
Using integration by parts, we can show
begin{align}
int_1^b frac{lambda_n}{x}y_ny_m dx &= -int_1^b (x{y_n}')'y_m dx = int_1^b x{y_n}'{y_m}' dx \
int_1^b frac{lambda_m}{x}y_my_n dx &= -int_1^b (x{y_m}')'y_n dx = int_1^b x{y_m}'{y_n}' dx
end{align}
It follows that
$$ int_1^b frac{lambda_n}{x}y_ny_m dx = int_1^b frac{lambda_m}{x}y_my_n dx implies (lambda_n-lambda_m) int_1^b frac{1}{x}y_ny_m dx = 0 $$
But $lambda_nne lambda_m$, so the integral must be zero. This means the eigenfunctions $y_n$ and $y_m$ are orthogonal w.r.t the weighing function $w(x)=frac{1}{x}$
The second part of the question asks you to show this by direct integration, i.e
$$ int_1^b frac{1}{x} sinleft(npifrac{ln x}{ln b}right)sinleft(mpifrac{ln x}{ln b}right) dx = 0 $$
which can be done with the substitution $t = ln x$
answered Jan 3 at 6:32
DylanDylan
12.4k31026
$endgroup$
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First of all, $lambda ne 0$ since it would result in a trivial solution (prove this yourself). So the eigenfunctions are
$$ y_k(x) = sinleft(kpi frac{ln x}{ln b}right) $$
up to a multiplicative constant, with $lambda_k = frac{k^2pi^2}{ln^2 b}$ and $k = 1,2,3,dots$
Let's compare 2 distinct eigenfunctions $y_n$ and $y_m$, with corresponding eigenvalues $lambda_nnelambda_m$. By the Sturm-Liouville form, we have the following relations:
begin{align}
frac{lambda_n}{x}y_n &= -(x{y_n}')' \
frac{lambda_m}{x}y_m &= -(x{y_m}')'
end{align}
Using integration by parts, we can show
begin{align}
int_1^b frac{lambda_n}{x}y_ny_m dx &= -int_1^b (x{y_n}')'y_m dx = int_1^b x{y_n}'{y_m}' dx \
int_1^b frac{lambda_m}{x}y_my_n dx &= -int_1^b (x{y_m}')'y_n dx = int_1^b x{y_m}'{y_n}' dx
end{align}
It follows that
$$ int_1^b frac{lambda_n}{x}y_ny_m dx = int_1^b frac{lambda_m}{x}y_my_n dx implies (lambda_n-lambda_m) int_1^b frac{1}{x}y_ny_m dx = 0 $$
But $lambda_nne lambda_m$, so the integral must be zero. This means the eigenfunctions $y_n$ and $y_m$ are orthogonal w.r.t the weighing function $w(x)=frac{1}{x}$
The second part of the question asks you to show this by direct integration, i.e
$$ int_1^b frac{1}{x} sinleft(npifrac{ln x}{ln b}right)sinleft(mpifrac{ln x}{ln b}right) dx = 0 $$
which can be done with the substitution $t = ln x$
answered Jan 3 at 6:32
DylanDylan
12.4k31026
$endgroup$
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
add a comment |
$begingroup$
First of all, $lambda ne 0$ since it would result in a trivial solution (prove this yourself). So the eigenfunctions are
$$ y_k(x) = sinleft(kpi frac{ln x}{ln b}right) $$
up to a multiplicative constant, with $lambda_k = frac{k^2pi^2}{ln^2 b}$ and $k = 1,2,3,dots$
Let's compare 2 distinct eigenfunctions $y_n$ and $y_m$, with corresponding eigenvalues $lambda_nnelambda_m$. By the Sturm-Liouville form, we have the following relations:
begin{align}
frac{lambda_n}{x}y_n &= -(x{y_n}')' \
frac{lambda_m}{x}y_m &= -(x{y_m}')'
end{align}
Using integration by parts, we can show
begin{align}
int_1^b frac{lambda_n}{x}y_ny_m dx &= -int_1^b (x{y_n}')'y_m dx = int_1^b x{y_n}'{y_m}' dx \
int_1^b frac{lambda_m}{x}y_my_n dx &= -int_1^b (x{y_m}')'y_n dx = int_1^b x{y_m}'{y_n}' dx
end{align}
It follows that
$$ int_1^b frac{lambda_n}{x}y_ny_m dx = int_1^b frac{lambda_m}{x}y_my_n dx implies (lambda_n-lambda_m) int_1^b frac{1}{x}y_ny_m dx = 0 $$
But $lambda_nne lambda_m$, so the integral must be zero. This means the eigenfunctions $y_n$ and $y_m$ are orthogonal w.r.t the weighing function $w(x)=frac{1}{x}$
The second part of the question asks you to show this by direct integration, i.e
$$ int_1^b frac{1}{x} sinleft(npifrac{ln x}{ln b}right)sinleft(mpifrac{ln x}{ln b}right) dx = 0 $$
which can be done with the substitution $t = ln x$
answered Jan 3 at 6:32
DylanDylan
12.4k31026
$endgroup$
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
add a comment |
$begingroup$
First of all, $lambda ne 0$ since it would result in a trivial solution (prove this yourself). So the eigenfunctions are
$$ y_k(x) = sinleft(kpi frac{ln x}{ln b}right) $$
up to a multiplicative constant, with $lambda_k = frac{k^2pi^2}{ln^2 b}$ and $k = 1,2,3,dots$
Let's compare 2 distinct eigenfunctions $y_n$ and $y_m$, with corresponding eigenvalues $lambda_nnelambda_m$. By the Sturm-Liouville form, we have the following relations:
begin{align}
frac{lambda_n}{x}y_n &= -(x{y_n}')' \
frac{lambda_m}{x}y_m &= -(x{y_m}')'
end{align}
Using integration by parts, we can show
begin{align}
int_1^b frac{lambda_n}{x}y_ny_m dx &= -int_1^b (x{y_n}')'y_m dx = int_1^b x{y_n}'{y_m}' dx \
int_1^b frac{lambda_m}{x}y_my_n dx &= -int_1^b (x{y_m}')'y_n dx = int_1^b x{y_m}'{y_n}' dx
end{align}
It follows that
$$ int_1^b frac{lambda_n}{x}y_ny_m dx = int_1^b frac{lambda_m}{x}y_my_n dx implies (lambda_n-lambda_m) int_1^b frac{1}{x}y_ny_m dx = 0 $$
But $lambda_nne lambda_m$, so the integral must be zero. This means the eigenfunctions $y_n$ and $y_m$ are orthogonal w.r.t the weighing function $w(x)=frac{1}{x}$
The second part of the question asks you to show this by direct integration, i.e
$$ int_1^b frac{1}{x} sinleft(npifrac{ln x}{ln b}right)sinleft(mpifrac{ln x}{ln b}right) dx = 0 $$
which can be done with the substitution $t = ln x$
answered Jan 3 at 6:32
DylanDylan
12.4k31026
$endgroup$
First of all, $lambda ne 0$ since it would result in a trivial solution (prove this yourself). So the eigenfunctions are
$$ y_k(x) = sinleft(kpi frac{ln x}{ln b}right) $$
up to a multiplicative constant, with $lambda_k = frac{k^2pi^2}{ln^2 b}$ and $k = 1,2,3,dots$
Let's compare 2 distinct eigenfunctions $y_n$ and $y_m$, with corresponding eigenvalues $lambda_nnelambda_m$. By the Sturm-Liouville form, we have the following relations:
begin{align}
frac{lambda_n}{x}y_n &= -(x{y_n}')' \
frac{lambda_m}{x}y_m &= -(x{y_m}')'
end{align}
Using integration by parts, we can show
begin{align}
int_1^b frac{lambda_n}{x}y_ny_m dx &= -int_1^b (x{y_n}')'y_m dx = int_1^b x{y_n}'{y_m}' dx \
int_1^b frac{lambda_m}{x}y_my_n dx &= -int_1^b (x{y_m}')'y_n dx = int_1^b x{y_m}'{y_n}' dx
end{align}
It follows that
$$ int_1^b frac{lambda_n}{x}y_ny_m dx = int_1^b frac{lambda_m}{x}y_my_n dx implies (lambda_n-lambda_m) int_1^b frac{1}{x}y_ny_m dx = 0 $$
But $lambda_nne lambda_m$, so the integral must be zero. This means the eigenfunctions $y_n$ and $y_m$ are orthogonal w.r.t the weighing function $w(x)=frac{1}{x}$
The second part of the question asks you to show this by direct integration, i.e
$$ int_1^b frac{1}{x} sinleft(npifrac{ln x}{ln b}right)sinleft(mpifrac{ln x}{ln b}right) dx = 0 $$
which can be done with the substitution $t = ln x$
answered Jan 3 at 6:32
DylanDylan
12.4k31026
answered Jan 3 at 6:32
DylanDylan
12.4k31026
answered Jan 3 at 6:32
DylanDylan
12.4k31026
answered Jan 3 at 6:32
DylanDylan
12.4k31026
12.4k31026
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
add a comment |
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer as always, +1.
$endgroup$
– Mattos
Jan 3 at 7:49
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
A great answer, very clear all! One last question: I don't see very clear why: $begin{align} frac{lambda_n}{x}y_n &= -(x{y_n}')' \ frac{lambda_m}{x}y_m &= -(x{y_m}')' end{align}$ Can you explain thas step a little more? Thanks for all.
$endgroup$
– Bvss12
Jan 4 at 0:29
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
Oh, now i see because that. Thanks for all :D
$endgroup$
– Bvss12
Jan 4 at 1:31
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
$begingroup$
If you find my answer helpful, feel free to "tick" it to close the question
$endgroup$
– Dylan
Jan 4 at 7:06
add a comment |
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