# PDE : Solving the damped (nonhomogeneous) heat conduction equation

$\frac{\partial v(x,t)}{\partial t}- C\frac{\partial^2 v(x,t)}{\partial x^2}=b{v(x,t)}$

$IC: v(x,0) = f(x)$

$-\infty \le x \le \infty ;$ b and C are both constants; $C > 0$

A particular solution would be

$v_p(x,t) = c_p e^{ikx-\alpha t}$

which as I understand it would solve the homogenous version of the initial PDE:

$\frac{\partial v(x,t)}{\partial t}- C\frac{\partial^2 v(x,t)}{\partial x^2}=0$

Now how would one procede to solve this taking into account for the nonhomogenous version combining with the particular solution and superposition? Help is appreciated.

The idea is to diagonalize the operator $-\frac{d^{2}}{dx^{2}}$ and then to expand $v(x,t)$ in the eigenfunctions of this operator to build the final solution: $$v(x,t)=\int_{-\infty}^{\infty}C(s,t)e^{isx}ds \\ \frac{\partial v}{\partial t}-C\frac{\partial^{2}v}{\partial x^{2}} = bv(x,t) \\ \int_{-\infty}^{\infty}\left(\frac{\partial C}{\partial t}+s^{2}C(s,t)-bC\right)e^{-isx}ds=0$$ So the coefficient function $C$ must satisfy $$\frac{\partial C}{\partial t}=-s^{2}C+bC\\ C(s,t) = K(s)e^{-s^{2}t+bt}$$ The coefficient function $K$ is determined by the initial condition $$f(x)=u(x,0)=\int_{-\infty}^{\infty}K(s)e^{isx}ds \\ K(s)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-isx}dx$$ The final solution: $$v(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\left(\int_{-\infty}^{\infty}f(y)e^{-isy}dy\right)e^{-s^{2}t+bt}\right]e^{isx}ds.$$ The quantity in square brackets is $C=K(s)e^{-is^{2}t+bt}$, while the quantity in parentheses is $K(s)$. Now you use inverse Fourier transform and convolution results to simplify.
• Thanks that definitely helps! When you say diagonalize and expand in the eigenfunctions, is that what you are doing in the 3 first steps or what exactly is happening there (how do you find that $C(s,t)$ is equal to $\left(\frac{\partial C}{\partial t}+s^{2}C(s,t)-bC\right)$?
• @BoroBorooooooooooooooooooooooo : Yes, you are writing a solution is a continuous sum of eigenfunctions of differentiation. The function $e^{isx}$ is an eigenfunction of $\frac{1}{i}\frac{d}{dx}$ with eigenvalue $s$. So this "continuous basis" diagonalizes $\frac{d^{2}}{dx^{2}}$, too, and turns that operator into multiplying by the eigenvalue. Analogous to where you have an orthonormal basis $\{e_n\}$ for a matrix $A$ and you want to solve $\frac{dx}{dt}=Ax+bx$. Then $x(t)=\sum_{n}c_n(t)e_n$ and you get $\sum_{n}\{c_n'(t)-\lambda_n c_n(t)-bc_n(t)\}e_n = 0$. $c_n'-\lambda_n c_n-bc_n = 0$. Oct 26, 2015 at 15:22