Integral Transform Method

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PDEs can be solved by applying the Laplace or Fourier transform. This reduces the PDE to an ODE that can be solved and then transformed back into a PDE. In this article this Integral Transformation Method is demonstrated at the example of the homogenious Diffusion Equation. \begin{align} \frac{\partial u}{\partial t}-D \frac{\partial ^2 u}{\partial x^2}=0 \end{align}

Laplace Integral Transform

One way to solve the equation involves the Laplace transform \begin{align} \int_0^{\infty} \frac{\partial u}{\partial t} e^{-st}dt = \int_0^{\infty} D \frac{\partial ^2 u}{\partial x^2} e^{-st}dt \end{align} The integral on the RHS is with respect to time, but differentiation with respect to $x$. Thus we can pull out the derivatives and first integrate, which simply gives the transform of $u(x,t)$, which we denote $\bar{u}(x,t)$. The integral on the LHS is, as the differentiation with respect to time $t$. We use partial integration to rearrange the term \begin{align} \int_0^{\infty} \frac{\partial u}{\partial t} e^{-st}dt= \left[ u(x,t) \cdot e^{-st} \right]_0^{\infty} - \int_0^{\infty} u(x,t) \cdot s \cdot e^{-st} dt= -u(x,0) - (-s) \bar{u}(x,s) \end{align} The transformed equation reads \begin{align} D \frac{\partial ^2 \bar{u}}{\partial x^2}+u(x,0)-s\bar{u}(x,s)=0 \tag{1} \end{align} We are left with an ODE with independent variable $x$. To solve the equation we need to know the initial/boundary conditions $u(x,0)$ and $u(0,t)$. Say we have tube filled with water, which at time $t_0$ is put in contact with a salt solution. As time proceeds, the salt will then diffuse through the tube according to the diffusion equation. This problem has initial condition $u(x,0)$ and the boundary condition $u(0,t)=u_0$. The solution to the homogenious equation (1) is \begin{align} \bar{u}(x,s)=A e^{\sqrt{\frac{s}{D}} x} + B e^{-\sqrt{\frac{s}{D}} x} \end{align} The boundary condition requires that $u(x,t) \rightarrow 0$ as $x \rightarrow \infty$. Hence $A$ must be zero $A=0$ and $B$ can be determined by \begin{align} \bar{u}(0,s)= \int_0^{\infty} u_0 e^{-st} = B e^{-\sqrt{\frac{s}{D}} 0} \end{align} \begin{align} \bar{u}(x,s)= \frac{u_0}{s} e^{-\sqrt{\frac{s}{D}} x} \end{align} The back-transformation of this equation is usually not an easy problem. This is the expense of this method, that allows to easily solve the transformed problem. Backtransformation can be done by contour integration, but we only state the solution here. After back-transformation we get \begin{align} u(x,t)= u_0 \left[ 1 - erf \left( \frac{x}{\sqrt{4Dt}} \right) \right] \end{align} where $erf()$ is the errorfunction $erf(x)=2/\sqrt{\pi} \int_0^x e^{-u^2}du$.

We will see that there are nicer ways to find the solution to a general diffusion problem, once we have found the Green's Function for the Diffusion Equation. We can find this Green's Function by using the Fourier Integral Transform instead of the Laplace Integral Transform.


Fourier Integral Transform

In contrast to the Laplace Transform, the Fourier Transform leaves (fuctions of) $t$ and transforms (fuctions of) $x$. \begin{align} \frac{1}{\sqrt{2 \pi}} \frac{\partial }{\partial t} \int_{-\infty}^{\infty} u e^{-ikx}dx = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} D \frac{\partial ^2 u}{\partial x^2} e^{-ikx}dx \end{align} Transformation gives the ODE \begin{align} \frac{\partial \tilde{u}(k,t)}{\partial t} = -D k^2 \tilde{u}(k,t) \end{align} with the exponential solution \begin{align} \tilde{u}(k,t)= \tilde{u}(k,0) \, e^{-Dk^2t} \end{align} Assuming an arbitrary initial condition $u(x,0)=f(x)$ on $(-\infty, \infty)$, we find \begin{align} \tilde{u}(k,0) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} u(x,0) e^{-ikx}dx = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} u(x,0) e^{-ikx}dx =\tilde{f}(k) \end{align} Using this result, we rewrite the solution for $\tilde{u}(k,t)$ as \begin{align} \tilde{u}(k,t)= \tilde{f}(k) \, e^{-Dk^2t}=\sqrt{2 \pi} \tilde{f}(k) \tilde{G}(k,t) \end{align} where we defined $\tilde{G}(k,t)=\frac{1}{\sqrt{2 \pi}} e^{-Dk^2t}$. According to the Convolution Theorem multiplication in the frequency domain is the same as convolution in the time domain. Back-transformation gives \begin{align} u(x,t)= \int_{-\infty}^{\infty} G(x-x',t)f(x')dx' \end{align} with the Green's Function for the Diffusion Equation \begin{align} G(x-x',t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{ -Dk^2t} e^{ik(x-x')} dk \end{align} which we want to integrate with respect to $k$. To this end, we write everthing in the exponent and pull out the factor in front of $k^2$ \begin{align} G(x-x',t)= \frac{1}{2 \pi} \int_{-\infty}^{\infty} exp \left[ -Dt \left( k^2 - \frac{i(x-x')}{Dt}k \right) \right] dk \end{align} then we complete the square of the exponential \begin{align} G(x-x',t)= \frac{1}{2 \pi} exp \left( - \frac{(x-x')^2}{4Dt} \right) \int_{-\infty}^{\infty} exp \left[ -Dt \left( k^2 - \frac{i(x-x')}{2Dt} \right)^2 \right] dk. \end{align} Now we can solve the Gaußian Integral and get the final version of

Green's Function for the Diffusion Equation (also called Heat Kernel) \begin{align} G(x-x',t)= \frac{1}{\sqrt{4 \pi Dt}} exp \left( - \frac{(x-x')^2}{4Dt} \right) \end{align} This gaußian function is centered about the value $x'$ and its broadness depends on the diffusion constant $D$ and the time $t$. The Fundamental Solution $G(x-x',t)$ is the solution of a Diffusion Equation with the delta function $\delta(x-a)$ - centered at $a$ - as a source.