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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.

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1 Answer 1

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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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