Consider the differential operator $\left(\frac{d^{2}}{d x^{2}}+a \frac{d}{d x}+b\right)=L, a, b$ constants.

Fundamental solution $F$ of second order operator $L$ is given by $LF=\delta$.

Let $f$ and $g$ satisfy $L f=0, L g=0, f(0)=g(0), f^{\prime}(0)-g^{\prime}(0)=1$. Consider the function $$ F(x)=\left\{\begin{array}{l} f(x), x \geqslant 0 \\ g(x), x<0 \end{array}\right. $$ Then I wanted to show that $F$ is a fundamental solution for $L .$

We can write $F(x)=H(x)\left(f(x)-g(x)\right)+g(x)$ where $H(x)$ is Heaviside step function.

From linearity, one can deduce that $LF=0$ everywhere except at $x=0$.

I know that derivative of distribution generated by Heaviside function is Dirac delta function. But unable to use that here.

Any help or hint will be appreciated.


The correct mathematical way to show this is to take a test function $\varphi\in C^\infty_c(\mathbb R)$ and show that $$ \int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx = \varphi(0), $$ where $L^* = \frac{d^{2}}{d x^{2}}-a \frac{d}{d x}+b.$

Inserting the definition of $F$ into the left hand side gives $$\begin{align} \int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx &= \int_{-\infty}^{0} g(x) \, \left(\varphi''(x)-a\varphi'(x)+b\varphi(x)\right) \, dx\\ &+ \int_{0}^{\infty} f(x) \, \left(\varphi''(x)-a\varphi'(x)+b\varphi(x)\right) \, dx . \end{align}$$

Integrating by parts, moving derivatives from $\varphi$ to $g$ and $f$ we get $$\begin{align} \int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx &= \left[ g(x)\,\varphi'(x) - g'(x)\,\varphi(x) - g(x)\,\varphi(x) \right]_{-\infty}^{0} \\ &+ \int_{-\infty}^{0} \left( g''(x)+ag'(x)+bg(x)\right) \, \varphi(x) \, dx \\ &+ \left[ f(x)\,\varphi'(x) - f'(x)\,\varphi(x) - f(x)\,\varphi(x) \right]_{0}^{\infty} \\ &+ \int_{0}^{\infty} \left( f''(x) + af'(x) + bf(x) \right) \, \varphi(x) \, dx . \end{align}$$

Here the integrals vanish and the $[\ldots]$ parts vanish at $\pm\infty$ so we are left with $$\begin{align} \int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx &= \left( g(0) \, \varphi'(0) - g'(0) \, \varphi(0) - g(0) \, \varphi(0) \right) \\ &- \left( f(0) \, \varphi'(0) - f'(0) \, \varphi(0) - f(0) \, \varphi(0) \right) \\ &= \left( g(0)-f(0) \right) \varphi'(0) - \left( g'(0)-f'(0) \right) \varphi(0) - \left( g(0) - f(0) \right) \varphi(0) \\ &= 0 \varphi'(0) - (-1)\varphi(0) - 0\varphi(0) \\ &= \varphi(0). \end{align}$$


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