Fourier/Laplace transform for PDE

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SUMMARY

The discussion focuses on finding the fundamental solution to the diffusion equation \(\frac{\partial c}{\partial t} = D\frac{\partial ^2 c}{\partial x^2}\) with the initial condition \(c(x,0)=\delta (x)\). The Laplace transform of \(\frac{\partial c}{\partial t}\) is expressed as \(u\hat c (x,u) -c(x,0)\), while the Fourier transform of \(\frac{\partial ^2 c}{\partial x^2}\) is represented as \(-q^2 \tilde c (q,t)\). The resulting equation \(\hat c (q,u) = \frac{c(q,0)}{u+Dq^2}\) leads to the Green function for the diffusion equation, which is essential for solving more complex problems involving boundary and initial conditions.

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vladimir69
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hello
i am trying to find the fundamental solution to
\frac{\partial c}{\partial t} = D\frac{\partial ^2 c}{\partial x^2}
where c=c(x,t)
with initial condition being c(x,0)=\delta (x)
where \delta (x) is the dirac delta function.
i have the solution and working written out in front of me.
first off its got the laplace transform of \frac{\partial c}{\partial t} as
u\hat c (x,u) -c(x,0)
and the Fourier transform of \frac{\partial ^2 c}{\partial x^2} as
-q^2 \tilde c (q,t)
and then out of nowhere we get
\hat c (q,u) = \frac{c(q,0)}{u+Dq^2}
once that bit of magic is done and a leap of faith is taken then i can see how the rest of it falls into place but can anyone explain the above steps to me?
 
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To get the Fourier and laplace transform of the first derivative you put df/dx into the definitions, integrate by parts and use, in the Fourier case, that f(x) must vanish as x goes to infinity in order for the the Fourier transform of f(x) to exist. And this you easily can generalize to higher derivatives.

If you set c(q,0)=1 you get the Green function for the diffusion equation in the transformed space. The Green function is very useful for solving more complicated problems. A general solution to the diffusion equation is then integrals over the Green function G(r,r';t,t') times the source term, the boundary conditions, the initial condition, respectively.
 

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