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Finite Fourier Transform on a 2d wave
How does the finite Fourier transform work exactly?
The transform of f(x) is
\widetilde{f}(\lambda_{n}) =\int^{L}_{0} f(x) X_{n} dx
If I had a 3d wave equation pde and I applied Finite Fourier transform on the pde for
z(x,y,t)=X(x)Y(y)T(t)
f(x)=z_{xx}
\tilde{z_{xx}(\lambda_{n})} =\int^{L}_{0} z_{xx} X_{n}(x) dx
f(x)=z_{yy}
\tilde{z_{yy}(\lambda_{n})} =\int^{L}_{0} z_{yy} Y_{n}(y) dy
etc for T
or
is the transform the same for everything, namely:
f(x)=z_{xx}
\tilde{z_{xx}(\lambda_{n})} =\int^{L}_{0} z_{xx} X_{n}(x) Y_{n}(y) dx
or is it a double integral?
or do you apply finite Fourier twice with respect to y and x.
How does the finite Fourier transform work exactly?
The transform of f(x) is
\widetilde{f}(\lambda_{n}) =\int^{L}_{0} f(x) X_{n} dx
If I had a 3d wave equation pde and I applied Finite Fourier transform on the pde for
z(x,y,t)=X(x)Y(y)T(t)
f(x)=z_{xx}
\tilde{z_{xx}(\lambda_{n})} =\int^{L}_{0} z_{xx} X_{n}(x) dx
f(x)=z_{yy}
\tilde{z_{yy}(\lambda_{n})} =\int^{L}_{0} z_{yy} Y_{n}(y) dy
etc for T
or
is the transform the same for everything, namely:
f(x)=z_{xx}
\tilde{z_{xx}(\lambda_{n})} =\int^{L}_{0} z_{xx} X_{n}(x) Y_{n}(y) dx
or is it a double integral?
or do you apply finite Fourier twice with respect to y and x.
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