Solving differential equation using Fourier

Question

I'd like to solve the following equation using Fourier $$ y(t)+ {\sqrt 2\over 2\pi5} {dy(t)\over dt}+({1\over2\pi5})^2 {d^2y(t)\over dt^2}=x(t) $$ where $x(t) = u(t)$ (step function)
So far i've got $$ Y(f)+{\sqrt 2j2\pi f\over 2\pi 5}Y(f)+({j2\pi f\over2\pi5})^2Y(f)=X(f) $$ $$ Y(f)=X(f)+5\sqrt 2 \pi {1\over j2\pi f}X(f)+{25\over -f^2}X(f) $$ $$ y(t)=x(t)+5\sqrt 2 \pi {\int x(t)dt} +... $$ I'm kinda stuck can you help me? (i'm not even sure if it's correct so far...) thanks

Answer 1

Let us consider the following PDE$$a\frac{d^{2}y}{dt^{2}}\left(t\right)+b\frac{dy}{dt}\left(t\right)+cy\left(t\right)=u\left(t\right).$$ One formally applies the Fourier transform$$\hat{y}\left(\omega\right)=TF\left[y\right]\left(\omega\right)=\intop y\left(t\right)e^{-i\omega t}dt$$ and obtains$$-a\omega^{2}\hat{y}\left(\omega\right)+ib\omega\hat{y}\left(\omega\right)+c\hat{y}\left(\omega\right)=\hat{u}\left(\omega\right)$$ i.e.$$\hat{y}\left(\omega\right)=\frac{\hat{u}\left(\omega\right)}{-a\omega^{2}+ib\omega+c}.$$ Now using the convolution formula, one gets$$y\left(t\right)=\left(TF^{-1}\left[\omega\mapsto\frac{1}{-a\omega^{2}+ib\omega+c}\right]*TF^{-1}\left[\hat{u}\right]\right)\left(t\right).$$ To compute $F^{-1}\left[\omega\mapsto\frac{1}{-a\omega^{2}+ib\omega+c}\right]$ , one can use the residue formula :$$TF^{-1}\left[\omega\mapsto\frac{1}{-a\omega^{2}+ib\omega+c}\right]\left(t\right)$$ $$=\frac{1}{2\pi}\intop\frac{e^{i\omega t}}{-a\omega^{2}+ib\omega+c}d\omega =\frac{1}{2\pi}\intop\frac{e^{i\omega t}}{-a\left(\omega-\omega_{-}\right)\left(\omega-\omega_{+}\right)}d\omega$$ where $\omega_{\pm}=\frac{-ib\pm\sqrt{4ac-b^{2}}}{2a}.$ One obtains$$TF^{-1}\left[\omega\mapsto\frac{1}{-a\omega^{2}+ib\omega+c}\right]=i\frac{e^{-\frac{b}{2a}t-i\frac{\sqrt{4ac-b^{2}}}{2a}t}}{\sqrt{4ac-b^{2}}}-i\frac{e^{-\frac{b}{2a}t+i\frac{\sqrt{4ac-b^{2}}}{2a}t}}{\sqrt{4ac-b^{2}}}=\frac{2e^{-\frac{b}{2a}t}}{\sqrt{4ac-b^{2}}}\sin\left(\frac{\sqrt{4ac-b^{2}}}{2a}t\right).$$ I think I made no mistake, but you may check it. Verify also that all the above calcula are allowed (why the complex contour integral converges ?). I also let you evaluate $\hat{u}\left(\omega\right) .$

Answer 2

You have $$Y(f)+C_1 fY(f)+C_2f^2Y(f)=X(f),$$ where $C_i$ are (eventually) complex constants. Their exact form depend on your conventions for Fourier transform.

Hence, $$Y(f) = \frac{X(f)}{1+C_1f+C_2f^2}.$$ Now you can take the inverse Fourier transform of both sides to reconstruct $y(t)$. As you can guess, Fourier transform for solving linear ODE's is largely non-optimal (except for carefully tailored examples).