Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.