- #1
NicolaiTheDane
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- TL;DR Summary
- Using separation of variables I have found a Fourier series to solve a PDE. How do I find the coefficients?
Hey there!
I am current taking an introductory course on PDE's, and our professor hasn't really emphasized last part of solutions from separation of variables. Now its not strictly going to be on the exam, however I remember doing this with ease a few years back, but for some reason now I simply cannot recall, or translate my books explanation into action. This below is the solution
$$u(x,y,t)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}C_{m,n}\cdot e^{-(m^2+n^2)\cdot t} \cdot \sin(m \cdot x) \cdot \cos(n \cdot y)$$
with ##(x,y,t) \in (0,\pi) \times (0,\pi) \times (0,\infty)##. I need an expression for ##C_{m,n}##, which seems pretty clear to me is the Fourier Coefficients. My attempt is to use the Fourier Transform:
$$C_{m,n}=\frac{1}{\pi} \cdot \int_{0}^{\pi} \frac{1}{\pi} \cdot \int_{0}^{\pi}u(x,y,t) \cdot e^{-i \cdot m \cdot x} \cdot e^{-i \cdot n \cdot y} dydx$$
However this seems odd. Can someone enlighten me?
I am current taking an introductory course on PDE's, and our professor hasn't really emphasized last part of solutions from separation of variables. Now its not strictly going to be on the exam, however I remember doing this with ease a few years back, but for some reason now I simply cannot recall, or translate my books explanation into action. This below is the solution
$$u(x,y,t)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}C_{m,n}\cdot e^{-(m^2+n^2)\cdot t} \cdot \sin(m \cdot x) \cdot \cos(n \cdot y)$$
with ##(x,y,t) \in (0,\pi) \times (0,\pi) \times (0,\infty)##. I need an expression for ##C_{m,n}##, which seems pretty clear to me is the Fourier Coefficients. My attempt is to use the Fourier Transform:
$$C_{m,n}=\frac{1}{\pi} \cdot \int_{0}^{\pi} \frac{1}{\pi} \cdot \int_{0}^{\pi}u(x,y,t) \cdot e^{-i \cdot m \cdot x} \cdot e^{-i \cdot n \cdot y} dydx$$
However this seems odd. Can someone enlighten me?
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