# PDE:Fourier Transforms

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solution Find the Fourier transform of ${\displaystyle f(t)=e^{-|t|}\,}$

solution Find the Fourier transform of ${\displaystyle f(t)={\begin{cases}1&|t|<1\\0&|t|>1\end{cases}}\,}$

 solution ${\displaystyle u_{t}=ku_{xx}\,}$ ${\displaystyle u(0,t)=0\,}$${\displaystyle u(x,0)=f(x)\,}$${\displaystyle t>0,\,\,0

 solution ${\displaystyle u_{xx}+u_{yy}=0\,}$ ${\displaystyle u(0,y)=0\,}$ ${\displaystyle u(1,y)=0\,}$ ${\displaystyle u(x,0)=0\,}$ ${\displaystyle u(x,1)=Bx(1-x)\,}$${\displaystyle t>0,\,\,0

 solution ${\displaystyle u_{t}=-u_{xxxx}\,}$ ${\displaystyle u(x,0)=f(x)\,}$${\displaystyle t>0,\,\,x\in \mathbb {R} ,}$

 solution ${\displaystyle u_{tt}=c^{2}\,u_{xx}\,}$ ${\displaystyle u(x,0)=f(x)\,}$${\displaystyle u_{t}(x,0)=g(x)\,}$${\displaystyle t>0,\,\,x\in \mathbb {R} ,}$

 solution ${\displaystyle u_{xx}+u_{yy}+u_{zz}=0\,}$ ${\displaystyle u(x,y,0)=f(x,y)\,}$Auxiliary condition: ${\displaystyle u}$ is bounded. ${\displaystyle t>0,\,\,x,y\in \mathbb {R} ,\,\,\,z>0\,}$

• ${\displaystyle u(x,y,z)=\int \!\!\!\int _{\Re }e^{i\lambda x+i\mu y}B(\lambda ,\mu )e^{-{\sqrt {\lambda ^{2}+\mu ^{2}}}\,z}\,d\lambda d\mu \,}$

 [Quick Answer] Write the form of the solution: ${\displaystyle u_{tt}=c^{2}(u_{xx}+u_{yy})\,}$ ${\displaystyle u(x,0,t)=g(x)\,}$ ${\displaystyle u(0,y,t)=h(y)\,}$ ${\displaystyle u(x,y,0)=0\,}$ ${\displaystyle u_{t}(x,y,0)=f(x,y)\,}$ ${\displaystyle 00\,}$

• ${\displaystyle u(x,y,t)=\int _{0}^{\infty }\int _{0}^{\infty }U(\lambda ,\mu ,t)\sin(\lambda x)\sin(\mu y)\,d\lambda \,d\mu \,}$

 [Quick Answer] Write the form of the solution: ${\displaystyle u_{tt}=c^{2}(u_{xx}+u_{yy})\,}$ ${\displaystyle u_{y}(x,0,t)=g(x)\,}$ ${\displaystyle u(0,y,t)=h(y)\,}$ ${\displaystyle u(x,y,0)=0\,}$ ${\displaystyle u_{t}(x,y,0)=f(x,y)\,}$ ${\displaystyle 00\,}$

• ${\displaystyle u(x,y,t)=\int _{0}^{\infty }\int _{0}^{\infty }U(\lambda ,\mu ,t)\sin(\lambda x)\cos(\mu y)\,d\lambda \,d\mu \,}$

 solution${\displaystyle u_{tt}=c^{2}(u_{xx}+u_{yy})\,}$ ${\displaystyle u_{y}(x,0,t)=g(x)\,}$ ${\displaystyle u(0,y,t)=h(y)\,}$ ${\displaystyle u(x,y,0)=0\,}$ ${\displaystyle u_{t}(x,y,0)=f(x,y)\,}$ ${\displaystyle 00\,}$

• ${\displaystyle u(x,y,t)=\int _{0}^{\infty }\int _{0}^{\infty }U(\lambda ,\mu ,t)\sin(\lambda x)\cos(\mu y)\,d\lambda \,d\mu \,}$