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Consider the equation $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2} + a\frac{\partial u}{\partial x}$$ for a function $u(x,t)$ with initial value $$u(x,0)=f(x).$$ Let $\hat{u}(y,t)$ and $\hat{f}(y)$ denote the Fourier transform in the $x$ variable of $u$ and $f$. I want to solve for $\hat{u}$ in terms of $\hat{f}$.

Taking the Fourier transform and using the formula for Fourier transform of derivatives, I get $$\frac{\partial}{\partial t}\hat{u}(y,t)=(iy)^2\hat{u}(y,t)+aiy\hat{u}(y,t)=-y^2\hat{u}(y,t)+aiy\hat{u}(y,t)$$

I don't understand how to get $\hat{u}$ in terms of $\hat{f}$. There is no $\hat{f}$ in the equation.

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up vote 1 down vote accepted

$ \hat u(y,0) = \hat f(y) $. Solve the ODE as an equation in $t$, which happens to depend upon a parameter $y$. That is, treat $y$ as a constant while you solve the ODE.

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Observe that the original PDE, in the Fourier space, reduces to a family of ODE (indexed in $y$). Notice also that the initial data for $\hat{u}$ is $\hat{f}$. Thus, you only need to solve the ODE in the Fourier space.

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