# PDE:Fourier Transforms

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

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

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

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

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

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

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

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

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

• $u(x,y,t) = \int_0^\infty \int_0^\infty U(\lambda,\mu,t) \sin(\lambda x) \sin(\mu y)\,d\lambda\,d\mu\,$

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

• $u(x,y,t) = \int_0^\infty \int_0^\infty U(\lambda,\mu,t) \sin(\lambda x) \cos(\mu y)\,d\lambda\,d\mu\,$

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

• $u(x,y,t) = \int_0^\infty \int_0^\infty U(\lambda,\mu,t) \sin(\lambda x) \cos(\mu y)\,d\lambda\,d\mu\,$

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