Finding the fundamental solution of an ODE

Question

I want to find the fundamental solution of this ODE: $$- u^{\prime\prime} + k^2 u=0, -\infty<x<\infty, k\neq 0$$ I know that it is: $$\Gamma(x,\epsilon) = \frac{e^{k|x-\epsilon|}}{2k}$$, but I don’t know how to obtain it.

We know that for $x\neq \epsilon$ $$-\Gamma^{\prime\prime}(x,\epsilon) + k^2 \Gamma(x,\epsilon) =0$$, so we can say that for $x\neq \epsilon$ We have: $$ \Gamma(x,\epsilon) = Ae^{k(x-\epsilon)} + Be^{-k(x-\epsilon)}$$ Can anyone please show me how can I compute A and B?

Thanks.

Answer 1

Write

$$u'u''=k^2uu'$$ and integrate to get

$$u'^2=k^2u^2\pm c^2.$$

This is a separable equation,

$$\frac{u'}{\sqrt{\dfrac{k^2}{c^2}u^2\pm1}}=c^2$$

which integrates as

$$\frac ck\text{arcosh}\frac{ku}c=\pm cx+c'$$ or

$$\frac ck\text{arsinh}\frac{ku}c=\pm cx+c'.$$

Finally,

$$u=\frac ck\cosh(\pm kx+c'')$$ or

$$u=\frac ck\sinh(\pm kx+c'').$$

You can convince yourself that this covers all cases of

$$ae^{kx}+be^{-kx}.$$

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Needless to say, by the theory of linear ODE with constant coefficients, the solution is an arbitrary linear combination of exponentials with parameters equal to the roots of the characteristic equation

$$\lambda^2-k^2=0.$$

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As you are looking for the fundamental solution, $a$ and $b$ remain free parameters.

Answer 2

This does look like an exercise from an homework... anyway, just some tips: the reasoning you do is correct, $\Gamma(x,\epsilon)$ indeed is a linear combination of $e^{k(x-\epsilon)}$ and $e^{-k(x-\epsilon)}$ away from $x=\epsilon$. However, one must remember that $\Gamma$ is not differentiable for $x=\epsilon$, so you should expect the general form of $\Gamma$ to be $$ \Gamma(x,\epsilon)= \begin{cases} A e^{k(x-\epsilon)}+B e^{-k(x-\epsilon)} & x<\epsilon\,, \\ C e^{k(x-\epsilon)}+D e^{-k(x-\epsilon)} & x>\epsilon\,. \end{cases} $$ To find the coefficients $A,B,C,D\in\mathbb{R}$, you need to use the definition of fundamental solution. Namely, for any test function $\varphi\in\mathcal{C}_c^\infty(\mathbb{R})$ it must hold \begin{equation} \varphi(\epsilon)\,=\,\int_{-\infty}^{\infty}\Gamma(x-\epsilon)\left[-\varphi''(x)+k^2\varphi(x)\right]\,dx\,. \end{equation} Split the integral on the right hand side into two parts (from $-\infty$ to $\epsilon-\delta$ and from $\epsilon+\delta$ to $+\infty$, with $\delta>0$ small number that you will eventually send to zero) and integrate by parts a couple of time. You will be left with some boundary terms. Impose that these boundary terms equal $\varphi(\epsilon)$ at the limit $\delta\to 0$ and you will find some conditions on your coefficients $A,B,C,D$. As LutzL mentioned, at the end you will find that these conditions force $\Gamma\,'(x-\epsilon)$ to have a jump of height $1$ at $x=\epsilon$. Also, the coefficients $A,B,C,D$ will not (and cannot) be completely fixed, but you will have some freedom in their choice, due to the fact that the fundamental solution is not uniquely determined (if $\Gamma$ is a fundamental solution, then $\Gamma+h$ is also a fundamental solution for any function $h$ such that $-h''+k^2 h=0$).

Good luck!