Mixing
Turning A Partial Differential Equation into an Algebraic Equation using Fourier Transforms
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It is commonly known that the Fourier transform, as an operator, can turn an Ordinary Differential Equation (ODE) into an Algebraic Equation, and a Partial Differential Equation (PDE) into an ODE.
I was wondering if one can turn a PDE (say of two variables) into an Algebraic Equation by applying the Fourier Transform twice, once applied to one variable, and the second, applied to the second variable.
For instance,
Let:
$frac{du}{dx}=frac{du}{dt}$. we can apply the Fourier transform once with respect to $x$ and thus we have: $(is)F(y)(s,t)=frac{dF(y)(s,t)}{dt}$ where $F(y)(s)$ is the Fourier of $y$ taking it from the spacial $x$ domain to the frequency $s$ domain. We can now repeat this process but taking the transform from the temporal domain to some new domain $k$.
Thus we have: $(is)F[F(y)](s,k)=(ik)F[F(y)](s,k)$. as can be seen, this is now a Algebraic Equation, and by applying the inverse Fourier twice, we could arrive at our original function.
Thanks.
functional-analysis ordinary-differential-equations fourier-analysis
edited Jan 22 at 11:19
dmtri
1,5642521
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
$endgroup$
add a comment |
$begingroup$
It is commonly known that the Fourier transform, as an operator, can turn an Ordinary Differential Equation (ODE) into an Algebraic Equation, and a Partial Differential Equation (PDE) into an ODE.
I was wondering if one can turn a PDE (say of two variables) into an Algebraic Equation by applying the Fourier Transform twice, once applied to one variable, and the second, applied to the second variable.
For instance,
Let:
$frac{du}{dx}=frac{du}{dt}$. we can apply the Fourier transform once with respect to $x$ and thus we have: $(is)F(y)(s,t)=frac{dF(y)(s,t)}{dt}$ where $F(y)(s)$ is the Fourier of $y$ taking it from the spacial $x$ domain to the frequency $s$ domain. We can now repeat this process but taking the transform from the temporal domain to some new domain $k$.
Thus we have: $(is)F[F(y)](s,k)=(ik)F[F(y)](s,k)$. as can be seen, this is now a Algebraic Equation, and by applying the inverse Fourier twice, we could arrive at our original function.
Thanks.
functional-analysis ordinary-differential-equations fourier-analysis
edited Jan 22 at 11:19
dmtri
1,5642521
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
$endgroup$
2
$begingroup$
I believe something along these lines works, these notes talk about something at least tangentially related in the first section.
$endgroup$
– Mark
Oct 20 '17 at 3:55
$begingroup$
It could be I did not understand it properly but it seemed like that section was analyzing the transformation under Fourier of a "derivitive-like' operator. There was not much talk about the repeated use of the Fourier to simplify PDE's
$endgroup$
– Yonah shmalo
Oct 20 '17 at 4:04
$begingroup$
A repeated fourier transform is nothing but a multivariate fourier transform (as long as all integrals converge absolutely and so on)
$endgroup$
– Calvin Khor
Jan 22 at 11:30
add a comment |
$begingroup$
It is commonly known that the Fourier transform, as an operator, can turn an Ordinary Differential Equation (ODE) into an Algebraic Equation, and a Partial Differential Equation (PDE) into an ODE.
I was wondering if one can turn a PDE (say of two variables) into an Algebraic Equation by applying the Fourier Transform twice, once applied to one variable, and the second, applied to the second variable.
For instance,
Let:
$frac{du}{dx}=frac{du}{dt}$. we can apply the Fourier transform once with respect to $x$ and thus we have: $(is)F(y)(s,t)=frac{dF(y)(s,t)}{dt}$ where $F(y)(s)$ is the Fourier of $y$ taking it from the spacial $x$ domain to the frequency $s$ domain. We can now repeat this process but taking the transform from the temporal domain to some new domain $k$.
Thus we have: $(is)F[F(y)](s,k)=(ik)F[F(y)](s,k)$. as can be seen, this is now a Algebraic Equation, and by applying the inverse Fourier twice, we could arrive at our original function.
Thanks.
functional-analysis ordinary-differential-equations fourier-analysis
edited Jan 22 at 11:19
dmtri
1,5642521
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
$endgroup$
It is commonly known that the Fourier transform, as an operator, can turn an Ordinary Differential Equation (ODE) into an Algebraic Equation, and a Partial Differential Equation (PDE) into an ODE.
I was wondering if one can turn a PDE (say of two variables) into an Algebraic Equation by applying the Fourier Transform twice, once applied to one variable, and the second, applied to the second variable.
For instance,
Let:
$frac{du}{dx}=frac{du}{dt}$. we can apply the Fourier transform once with respect to $x$ and thus we have: $(is)F(y)(s,t)=frac{dF(y)(s,t)}{dt}$ where $F(y)(s)$ is the Fourier of $y$ taking it from the spacial $x$ domain to the frequency $s$ domain. We can now repeat this process but taking the transform from the temporal domain to some new domain $k$.
Thus we have: $(is)F[F(y)](s,k)=(ik)F[F(y)](s,k)$. as can be seen, this is now a Algebraic Equation, and by applying the inverse Fourier twice, we could arrive at our original function.
Thanks.
functional-analysis ordinary-differential-equations fourier-analysis
functional-analysis ordinary-differential-equations fourier-analysis
edited Jan 22 at 11:19
dmtri
1,5642521
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
edited Jan 22 at 11:19
dmtri
1,5642521
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
edited Jan 22 at 11:19
dmtri
1,5642521
edited Jan 22 at 11:19
dmtri
1,5642521
edited Jan 22 at 11:19
dmtri
1,5642521
1,5642521
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
asked Oct 20 '17 at 3:51
Yonah shmaloYonah shmalo
454
454
2
$begingroup$
I believe something along these lines works, these notes talk about something at least tangentially related in the first section.
$endgroup$
– Mark
Oct 20 '17 at 3:55
$begingroup$
It could be I did not understand it properly but it seemed like that section was analyzing the transformation under Fourier of a "derivitive-like' operator. There was not much talk about the repeated use of the Fourier to simplify PDE's
$endgroup$
– Yonah shmalo
Oct 20 '17 at 4:04
$begingroup$
A repeated fourier transform is nothing but a multivariate fourier transform (as long as all integrals converge absolutely and so on)
$endgroup$
– Calvin Khor
Jan 22 at 11:30
add a comment |
2
$begingroup$
I believe something along these lines works, these notes talk about something at least tangentially related in the first section.
$endgroup$
– Mark
Oct 20 '17 at 3:55
$begingroup$
It could be I did not understand it properly but it seemed like that section was analyzing the transformation under Fourier of a "derivitive-like' operator. There was not much talk about the repeated use of the Fourier to simplify PDE's
$endgroup$
– Yonah shmalo
Oct 20 '17 at 4:04
$begingroup$
A repeated fourier transform is nothing but a multivariate fourier transform (as long as all integrals converge absolutely and so on)
$endgroup$
– Calvin Khor
Jan 22 at 11:30
2
2
$begingroup$
I believe something along these lines works, these notes talk about something at least tangentially related in the first section.
$endgroup$
– Mark
Oct 20 '17 at 3:55
$begingroup$
I believe something along these lines works, these notes talk about something at least tangentially related in the first section.
$endgroup$
– Mark
Oct 20 '17 at 3:55
$begingroup$
It could be I did not understand it properly but it seemed like that section was analyzing the transformation under Fourier of a "derivitive-like' operator. There was not much talk about the repeated use of the Fourier to simplify PDE's
$endgroup$
– Yonah shmalo
Oct 20 '17 at 4:04
$begingroup$
It could be I did not understand it properly but it seemed like that section was analyzing the transformation under Fourier of a "derivitive-like' operator. There was not much talk about the repeated use of the Fourier to simplify PDE's
$endgroup$
– Yonah shmalo
Oct 20 '17 at 4:04
$begingroup$
A repeated fourier transform is nothing but a multivariate fourier transform (as long as all integrals converge absolutely and so on)
$endgroup$
– Calvin Khor
Jan 22 at 11:30
$begingroup$
A repeated fourier transform is nothing but a multivariate fourier transform (as long as all integrals converge absolutely and so on)
$endgroup$
– Calvin Khor
Jan 22 at 11:30
add a comment |
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2
$begingroup$
I believe something along these lines works, these notes talk about something at least tangentially related in the first section.
$endgroup$
– Mark
Oct 20 '17 at 3:55
$begingroup$
It could be I did not understand it properly but it seemed like that section was analyzing the transformation under Fourier of a "derivitive-like' operator. There was not much talk about the repeated use of the Fourier to simplify PDE's
$endgroup$
– Yonah shmalo
Oct 20 '17 at 4:04
$begingroup$
A repeated fourier transform is nothing but a multivariate fourier transform (as long as all integrals converge absolutely and so on)
$endgroup$
– Calvin Khor
Jan 22 at 11:30