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I'm trying to solve the equation $u_t = \alpha^2 U_{yy}$ given u(y,t) bounded y $\rightarrow\infty$ and u(0,t) = $U_o e^{iw_ot}$. Initial is u(y,0) = 0. I have gotten both separations as Y'' - $\lambda$Y=0 and T' = $\alpha^2$T, but from here I get confused what to do, I never learnded PDE and am trying to solve a model. Thanks

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Why rollback to your bad formatting? –  doraemonpaul Mar 8 at 23:10

2 Answers 2

I would like to point out that if $u(y,0)=0$, the solution is $u(y,t)=0$. This is a classical Possion problem which is derive from heat diffusion. So this equation is also called heat equation. I think a simple way to solve this equation is to use Fourier Transformation.

Let $\hat{u}(x,t)=F[u(y,t)]$ denote the corresponding spatial Fourier transformation of $u$. Assume the initial condition is not zero, say $u(y,0)=\phi(y)$. Then the equation can be written by: \begin{equation} \hat{u}_t(x,t)+\alpha^2y^2\hat{u}(x,t)=0\\ \hat{u}(x,0)=\hat{\phi}(x) \end{equation} This is a classical 1st order ODE initial problem. The solution is \begin{equation} \hat{u}(x,t)=\hat{\phi}(x)e^{-a^2x^2t} \end{equation} Then take the inverse Fourier transformation (denoted by $F^{-1}[u(x,t)]$), we obtain the solution of the original PDE: \begin{equation} u(y,t)=F^{-1}[\hat{\phi}(x)e^{-a^2x^2t}]=\phi(y)*F^{-1}[e^{-a^2x^2t}]\\ =\frac{1}{2\alpha\sqrt{\pi t}}\int_{-\infty}^{+\infty}\phi(\xi)e^{-\frac{(y-\xi)^2}{4\alpha^2t}}d\xi \end{equation}

But this mehtod is limited because it assume the spatial boundary of region of $u(y,t)$ is zero then the $u$ is an element of Schwartz space which made Fourier transformation is valid. If the spatial boundary is not zero, you can use the technique suggested by Heat Equation.

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It's not all $0$ because of the boundary condition at $y=0$. –  Robert Israel Mar 4 at 3:51
    
@RobertIsrael Yes, that is. I just give him a hint of this problem. In fact, the nonhomogeneous IBVP can be solve by fundamental solution which is base by this deduction. –  Lion Mar 4 at 4:00

I presume that should be $u_t = \alpha^2 u_{yy}$. Take the Laplace transform of the PDE with respect to $t$. If $U(y,s)$ is the Laplace transform of $u(y,t)$, you get

$$ s U(y,s) - u(y,0) = s U(y,s) = \alpha^2 \dfrac{\partial^2}{\partial y^2} U(y,s) $$

Considering this as an ODE in $y$, the general solution is

$$ U(y,s) = c_1(s) e^{-\sqrt{s} y/\alpha} + c_2(s) e^{\sqrt{s} y/\alpha}$$

Now you want $u$ (and presumably $U$) bounded as $y\to +\infty$, so you want to take $c_2 = 0$. On the other hand, for $y = 0$ you want $u(0,t) = U_0 e^{i\omega_0 t}$ which says $c_1(s) = U(0,s) = \dfrac{U_0}{s - i \omega_0}$. Thus $U(y,s) = \dfrac{U_0}{s - i \omega_0} e^{-\sqrt{s} y/\alpha}$, and $u(y,t)$ is its inverse Laplace transform, which according to Maple is

$$ u(y,t) = \dfrac{U_0 y}{2 \sqrt{\pi} \alpha} \int_0^t r^{-3/2} e^{-y^2/(4 \alpha^2 r)} e^{i \omega_0 (t - r)}\; dr $$

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