$y''+\epsilon y'=\epsilon$, where $y(0)=1$, $y'(0)=0$

Question

I am trying to solve $y''+\epsilon y'=\epsilon$, where $y(0)=1$, $y'(0)=0$ using perturbation theory.

Using the substitution $y=y_{0}+y_{1}\epsilon$ I got the series $y=1+\epsilon(1+\frac{x^{2}}{2})+O(\epsilon ^{2})$.

However wolframalpha tells me the exact solution involves an exponential, (see http://m.wolframalpha.com/input/?i=y%22%2B0.1y%27%3D0.1 where I set $\epsilon=0.1$).

Am I on the right track with the solution? I'd then like to determine the validity of the solution as $x\rightarrow\infty$ but I'm not confident with this.

Answer 1

The exaxt solution has exponental

$$y''+\epsilon y'=\epsilon$$ Just integrate $$y'+\epsilon y=\epsilon x+ K_1$$ $$e^{\epsilon x}y'+e^{\epsilon x}\epsilon y=e^{\epsilon x}(\epsilon x+ K_1)$$ $$(e^{\epsilon x}y)=\int e^{\epsilon x}(\epsilon x+ K_1)dx$$ $$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+e^{-\epsilon x}\int e^{\epsilon x}\epsilon xdx)$$ $$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+ x - \frac 1 {\epsilon}$$ $$\boxed {y=K_1+e^{-\epsilon x}K_2 + x }$$ Use conditions to get the constants $$y'+\epsilon y=\epsilon x+ K_1 \text { and } y'0)=0 \to K_1=\epsilon$$ $$y=e^{-\epsilon x}K_2 + \frac {K_1}{\epsilon}+ x - \frac 1 {\epsilon} \text { and } y(0)=1 \to K_2=\frac 1 {\epsilon}$$ $$\boxed {y(x)=\frac {e^{-\epsilon x}-1}{\epsilon} + x +1}$$

Answer 2

If $$ y''+\varepsilon y'=\varepsilon, \quad y(0)=0,\,y'(0)=1 $$ then integrating the ODE over $[0,x]$ we obtain $$ y'+\varepsilon y=1+\varepsilon x $$ and hence $$ \mathrm{e}^{\varepsilon x}\big(y'+\varepsilon y\big)=\mathrm{e}^{\varepsilon x}(1+\varepsilon x) $$ or $$ (\mathrm{e}^{\varepsilon x}y)'=\frac{1}{\varepsilon^2}((\varepsilon x-(\varepsilon-1))e^{ex})' $$ and finally $$ y=\frac{1}{\varepsilon^2}\big(\varepsilon x-(\varepsilon-1)\big)+c\mathrm{e}^{-\varepsilon x} $$ and since $y(0)=0$, then $c=(\varepsilon-1)/\varepsilon^2$, i.e., $$ y=\frac{1}{\varepsilon^2}\big(\varepsilon x-(\varepsilon-1)\big)+\frac{(\varepsilon-1)\mathrm{e}^{-\varepsilon x}}{\varepsilon^2} $$

Answer 3

Zeroth order: $y_0''=0$, with these boundary conditions you get $y_0=1$.

First order: $y_1''+y_0'=1,y_1(0)=0$. (This boundary condition comes because you have already fully taken into account the boundary condition in computing $y_0$, so you want all corrections to leave the boundary conditions alone.) Thus $y_1''=1$ so $y_1=x^2/2$. This matches the first order series expansion of the exact solution, which is$x+\exp(-\varepsilon x)/\varepsilon-(1/\varepsilon-1)$.

Answer 4

I'm new to perturbation theory, but since the other answers have addressed the exact solving of the DE, I thought i'd chip in.

Taking $y=y(x,\epsilon)$, then $y\sim y(x,0)+\epsilon\frac{\partial y}{\partial \epsilon}|_{\epsilon=0}$

Solving $y(x,0)$ gives $y(x,0)=c_1x+c_2$.

Now for $\frac{\partial y}{\partial \epsilon}$, one has to solve: $\frac{\partial}{\partial \epsilon}\left( \frac{\partial^2}{\partial x^2}y+\epsilon\frac{\partial y}{\partial x}\right)\bigg|_{\epsilon=0}=\frac{\partial}{\partial \epsilon}\epsilon\bigg|_{\epsilon=0}=1\bigg|_{\epsilon=0}=1$

This gives $$\begin{align} \frac{\partial}{\partial \epsilon}\left( \frac{\partial^2}{\partial x^2}y+\epsilon\frac{\partial y}{\partial x}\right)\bigg|_{\epsilon=0}&=1\\ \frac{\partial^2}{\partial x^2}\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}+\frac{\partial}{\partial \epsilon}\epsilon\frac{\partial y}{\partial x}\bigg|_{\epsilon=0}&=1\\ \frac{\partial^2}{\partial x^2}\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}+\left(\frac{\partial y}{\partial x}+\epsilon\frac{\partial}{\partial x}\frac{\partial y}{\partial \epsilon}\right)\bigg|_{\epsilon=0}&=1\\ \frac{\partial^2}{\partial x^2}\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0}+\frac{\partial y(x,0)}{\partial x}&=1\\ \frac{\partial^2}{\partial x^2}\upsilon(x)+c_1&=1 &&\text{where }\upsilon(x)=\frac{\partial}{\partial \epsilon}y\bigg|_{\epsilon=0} \end{align}$$

Then $\upsilon(x)=\frac{1-c_1}{2}x^2+c_3x+c_4$

Imposing the initial conditions and solving for the coefficients gives: $$ y(x,\epsilon)\sim 1-c_3x+\epsilon\left(c_3x+x^2\frac{1+c_3}{2}\right) $$

Knowing the exact solution i believe $c_3=0$, so actually $y\sim 1+\frac{1}{2}\epsilon x^2$