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

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

$ -\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.


1 Answer 1


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.

  • $\begingroup$ 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 $\left(\frac{\partial C}{\partial t}+s^{2}C(s,t)-bC\right)$? $\endgroup$
    – jibo
    Oct 26, 2015 at 15:03
  • $\begingroup$ @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 $\sum_{n}\{c_n'(t)-\lambda_n c_n(t)-bc_n(t)\}e_n = 0$. $c_n'-\lambda_n c_n-bc_n = 0$. $\endgroup$ Oct 26, 2015 at 15:22

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