Eigenvalues and Eigenfunctions of Integral Equation
Compute the eigenvalues and eigenfunctions of
$$
varphi(x) - lambda int_0^1 e^{x+s} varphi(s) ds = f(x)
$$
Are there functions $f$ such that the inhomogenous equation has for every real $lambda$ at least one solution?
Attempt at my solution: I have to solve
$$
int_0^1 e^{x+s} varphi(s) ds = frac{1}{lambda} varphi(x)
$$
with
$$
int_0^1 e^{x+s} varphi(s) ds = e^x int_0^1 e^s varphi(s) ds
$$
we have the eigenvalue $1/lambda = int_0^1 e^s varphi(s) ds$ and the eigenfunctions
$$
varphi(x) = C cdot e^x
$$
for some constant $C$. Is this correct? I just derived it with looking at the equation, is there some theory by which I get this result? And are that all eigenvalues/eigenfunctions? And how to solve the question for functions $f$ such that there always exists eigenfunctions for real $lambda$?
integration analysis functional-analysis pde
asked Jul 26 '14 at 23:37
StefanH
8,06652161
bumped to the homepage by Community♦ 16 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
Compute the eigenvalues and eigenfunctions of
$$
varphi(x) - lambda int_0^1 e^{x+s} varphi(s) ds = f(x)
$$
Are there functions $f$ such that the inhomogenous equation has for every real $lambda$ at least one solution?
Attempt at my solution: I have to solve
$$
int_0^1 e^{x+s} varphi(s) ds = frac{1}{lambda} varphi(x)
$$
with
$$
int_0^1 e^{x+s} varphi(s) ds = e^x int_0^1 e^s varphi(s) ds
$$
we have the eigenvalue $1/lambda = int_0^1 e^s varphi(s) ds$ and the eigenfunctions
$$
varphi(x) = C cdot e^x
$$
for some constant $C$. Is this correct? I just derived it with looking at the equation, is there some theory by which I get this result? And are that all eigenvalues/eigenfunctions? And how to solve the question for functions $f$ such that there always exists eigenfunctions for real $lambda$?
integration analysis functional-analysis pde
asked Jul 26 '14 at 23:37
StefanH
8,06652161
bumped to the homepage by Community♦ 16 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $phi(x)mapstophi(x)-lambdaint_0^1e^{x+s}phi(s)dx$?
– Daniel Robert-Nicoud
Jul 26 '14 at 23:40
I bet it is $$phi(x)to int_{0}^{1}e^{x+s}phi(s),ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $lambda=frac{2}{e^2-1}$ eigenspace.
– Jack D'Aurizio
Jul 26 '14 at 23:46
Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $int_0^1 e^{x+s} varphi(s)ds = e^x int_0^1 e^svarphi(s) ds$ which is $e^x$ times a constant, so $varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning?
– StefanH
Jul 26 '14 at 23:56
Nothing wrong, I mean "constant times $e^x$".
– Jack D'Aurizio
Jul 27 '14 at 1:42
add a comment |
Compute the eigenvalues and eigenfunctions of
$$
varphi(x) - lambda int_0^1 e^{x+s} varphi(s) ds = f(x)
$$
Are there functions $f$ such that the inhomogenous equation has for every real $lambda$ at least one solution?
Attempt at my solution: I have to solve
$$
int_0^1 e^{x+s} varphi(s) ds = frac{1}{lambda} varphi(x)
$$
with
$$
int_0^1 e^{x+s} varphi(s) ds = e^x int_0^1 e^s varphi(s) ds
$$
we have the eigenvalue $1/lambda = int_0^1 e^s varphi(s) ds$ and the eigenfunctions
$$
varphi(x) = C cdot e^x
$$
for some constant $C$. Is this correct? I just derived it with looking at the equation, is there some theory by which I get this result? And are that all eigenvalues/eigenfunctions? And how to solve the question for functions $f$ such that there always exists eigenfunctions for real $lambda$?
integration analysis functional-analysis pde
asked Jul 26 '14 at 23:37
StefanH
8,06652161
Compute the eigenvalues and eigenfunctions of
$$
varphi(x) - lambda int_0^1 e^{x+s} varphi(s) ds = f(x)
$$
Are there functions $f$ such that the inhomogenous equation has for every real $lambda$ at least one solution?
Attempt at my solution: I have to solve
$$
int_0^1 e^{x+s} varphi(s) ds = frac{1}{lambda} varphi(x)
$$
with
$$
int_0^1 e^{x+s} varphi(s) ds = e^x int_0^1 e^s varphi(s) ds
$$
we have the eigenvalue $1/lambda = int_0^1 e^s varphi(s) ds$ and the eigenfunctions
$$
varphi(x) = C cdot e^x
$$
for some constant $C$. Is this correct? I just derived it with looking at the equation, is there some theory by which I get this result? And are that all eigenvalues/eigenfunctions? And how to solve the question for functions $f$ such that there always exists eigenfunctions for real $lambda$?
integration analysis functional-analysis pde
integration analysis functional-analysis pde
asked Jul 26 '14 at 23:37
StefanH
8,06652161
asked Jul 26 '14 at 23:37
StefanH
8,06652161
asked Jul 26 '14 at 23:37
StefanH
8,06652161
asked Jul 26 '14 at 23:37
StefanH
8,06652161
asked Jul 26 '14 at 23:37
StefanH
8,06652161
8,06652161
bumped to the homepage by Community♦ 16 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 16 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $phi(x)mapstophi(x)-lambdaint_0^1e^{x+s}phi(s)dx$?
– Daniel Robert-Nicoud
Jul 26 '14 at 23:40
I bet it is $$phi(x)to int_{0}^{1}e^{x+s}phi(s),ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $lambda=frac{2}{e^2-1}$ eigenspace.
– Jack D'Aurizio
Jul 26 '14 at 23:46
Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $int_0^1 e^{x+s} varphi(s)ds = e^x int_0^1 e^svarphi(s) ds$ which is $e^x$ times a constant, so $varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning?
– StefanH
Jul 26 '14 at 23:56
Nothing wrong, I mean "constant times $e^x$".
– Jack D'Aurizio
Jul 27 '14 at 1:42
add a comment |
1
Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $phi(x)mapstophi(x)-lambdaint_0^1e^{x+s}phi(s)dx$?
– Daniel Robert-Nicoud
Jul 26 '14 at 23:40
I bet it is $$phi(x)to int_{0}^{1}e^{x+s}phi(s),ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $lambda=frac{2}{e^2-1}$ eigenspace.
– Jack D'Aurizio
Jul 26 '14 at 23:46
Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $int_0^1 e^{x+s} varphi(s)ds = e^x int_0^1 e^svarphi(s) ds$ which is $e^x$ times a constant, so $varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning?
– StefanH
Jul 26 '14 at 23:56
Nothing wrong, I mean "constant times $e^x$".
– Jack D'Aurizio
Jul 27 '14 at 1:42
1
1
Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $phi(x)mapstophi(x)-lambdaint_0^1e^{x+s}phi(s)dx$?
– Daniel Robert-Nicoud
Jul 26 '14 at 23:40
Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $phi(x)mapstophi(x)-lambdaint_0^1e^{x+s}phi(s)dx$?
– Daniel Robert-Nicoud
Jul 26 '14 at 23:40
I bet it is $$phi(x)to int_{0}^{1}e^{x+s}phi(s),ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $lambda=frac{2}{e^2-1}$ eigenspace.
– Jack D'Aurizio
Jul 26 '14 at 23:46
I bet it is $$phi(x)to int_{0}^{1}e^{x+s}phi(s),ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $lambda=frac{2}{e^2-1}$ eigenspace.
– Jack D'Aurizio
Jul 26 '14 at 23:46
Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $int_0^1 e^{x+s} varphi(s)ds = e^x int_0^1 e^svarphi(s) ds$ which is $e^x$ times a constant, so $varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning?
– StefanH
Jul 26 '14 at 23:56
Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $int_0^1 e^{x+s} varphi(s)ds = e^x int_0^1 e^svarphi(s) ds$ which is $e^x$ times a constant, so $varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning?
– StefanH
Jul 26 '14 at 23:56
Nothing wrong, I mean "constant times $e^x$".
– Jack D'Aurizio
Jul 27 '14 at 1:42
Nothing wrong, I mean "constant times $e^x$".
– Jack D'Aurizio
Jul 27 '14 at 1:42
add a comment |
1 Answer
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Suppose $int_{0}^{1}f(s)e^{s},ds=0$. Then your equation always has the solution $varphi = f$ which holds for all $lambda$. So this condition is sufficient. Actually, this condition is also necessary in order to have a solution for all real $lambda$, but I'll leave that to you to see. (Hint: choose $lambda = 1/int_{0}^{1}e^{2x},dx$, and show that $h(x)=varphi(x)-lambdaint_{0}^{1}varphi(s)e^{s+x},ds$ satisfies $int_{0}^{1}h(x)e^{x},dx=0$.)
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
add a comment |
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Suppose $int_{0}^{1}f(s)e^{s},ds=0$. Then your equation always has the solution $varphi = f$ which holds for all $lambda$. So this condition is sufficient. Actually, this condition is also necessary in order to have a solution for all real $lambda$, but I'll leave that to you to see. (Hint: choose $lambda = 1/int_{0}^{1}e^{2x},dx$, and show that $h(x)=varphi(x)-lambdaint_{0}^{1}varphi(s)e^{s+x},ds$ satisfies $int_{0}^{1}h(x)e^{x},dx=0$.)
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
add a comment |
Suppose $int_{0}^{1}f(s)e^{s},ds=0$. Then your equation always has the solution $varphi = f$ which holds for all $lambda$. So this condition is sufficient. Actually, this condition is also necessary in order to have a solution for all real $lambda$, but I'll leave that to you to see. (Hint: choose $lambda = 1/int_{0}^{1}e^{2x},dx$, and show that $h(x)=varphi(x)-lambdaint_{0}^{1}varphi(s)e^{s+x},ds$ satisfies $int_{0}^{1}h(x)e^{x},dx=0$.)
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
add a comment |
Suppose $int_{0}^{1}f(s)e^{s},ds=0$. Then your equation always has the solution $varphi = f$ which holds for all $lambda$. So this condition is sufficient. Actually, this condition is also necessary in order to have a solution for all real $lambda$, but I'll leave that to you to see. (Hint: choose $lambda = 1/int_{0}^{1}e^{2x},dx$, and show that $h(x)=varphi(x)-lambdaint_{0}^{1}varphi(s)e^{s+x},ds$ satisfies $int_{0}^{1}h(x)e^{x},dx=0$.)
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
Suppose $int_{0}^{1}f(s)e^{s},ds=0$. Then your equation always has the solution $varphi = f$ which holds for all $lambda$. So this condition is sufficient. Actually, this condition is also necessary in order to have a solution for all real $lambda$, but I'll leave that to you to see. (Hint: choose $lambda = 1/int_{0}^{1}e^{2x},dx$, and show that $h(x)=varphi(x)-lambdaint_{0}^{1}varphi(s)e^{s+x},ds$ satisfies $int_{0}^{1}h(x)e^{x},dx=0$.)
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
edited Jul 28 '14 at 1:50
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
answered Jul 27 '14 at 17:37
DisintegratingByParts
58.5k42579
58.5k42579
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add a comment |
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1
Wait, what is exactly the operator of which you are searching the eigenvectors? Is it $phi(x)mapstophi(x)-lambdaint_0^1e^{x+s}phi(s)dx$?
– Daniel Robert-Nicoud
Jul 26 '14 at 23:40
I bet it is $$phi(x)to int_{0}^{1}e^{x+s}phi(s),ds$$. In this case, the only eigenfunctions are the constant functions and they all belong to the $lambda=frac{2}{e^2-1}$ eigenspace.
– Jack D'Aurizio
Jul 26 '14 at 23:46
Yes Jack D'Aurizio wrote the right operator. But why the constant functions? On the RHS we have $int_0^1 e^{x+s} varphi(s)ds = e^x int_0^1 e^svarphi(s) ds$ which is $e^x$ times a constant, so $varphi(x)$ must also be $e^x$ times a constant, and not constant, or what is wrong with my reasoning?
– StefanH
Jul 26 '14 at 23:56
Nothing wrong, I mean "constant times $e^x$".
– Jack D'Aurizio
Jul 27 '14 at 1:42