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HW 13 Mathematics 503, Mathematical Modeling,
CSUF , August 2, 2007
Nasser M. Abbasi
October 8, 2025
Contents
1 Problem 10 page 268 section 4.5 (Distributions)
problem:
Find fundamental solution associated with operator \(L\) defined by \(Lu=-x^{2}u^{\prime \prime }-xu^{\prime }+u,0<x<1,\) such that \(u\left ( x,\xi \right ) =x\) for \(0<x<\xi .\)
answer:
The fundamental solution can be written as
\[ u=\left \{ \begin {array} [c]{cc}x & 0<x<\xi \\ A\left ( \xi \right ) u_{1}\left ( x\right ) +B\left ( \xi \right ) u_{2}\left ( x\right ) & \xi <x<1 \end {array} \right . \]
And our goal is to determine \(A\left ( \xi \right ) ,B\left ( \xi \right ) \). In the above \(u_{1},u_{2}\) are the 2 independent solution to the homogenous
equation \(-x^{2}u^{\prime \prime }-xu^{\prime }+u=0\)
We start by finding \(u_{1},u_{2}.\) We try solution \(u=x^{m}\) and substitute this into the above homogenous equation, we
obtain the characteristic equation \(m^{2}=1\), hence \(m=\pm 1\) the 2 solution are \(u_{1}=x\) and \(u_{2}=x^{-1}\). Hence our fundamental solution
now looks like
\[ u=\left \{ \begin {array} [c]{cc}x & 0<x<\xi \\ A\left ( \xi \right ) x+B\left ( \xi \right ) x^{-1} & \xi <x<1 \end {array} \right . \]
Now consider the test function \(\phi \), hence
\begin{align*} \left ( Lu,\phi \right ) & =\left ( -x^{2}u^{\prime \prime }-xu^{\prime }+u,\phi \right ) \\ & =\left ( -x^{2}u^{\prime \prime },\phi \right ) -\left ( xu^{\prime },\phi \right ) +\left ( u,\phi \right ) \ \ \ \ \ \ \ \ \ \ \text {linearity of distribution}\\ & =\left ( u^{\prime \prime },-x^{2}\phi \right ) -\left ( u^{\prime },x\phi \right ) +\left ( u,\phi \right ) \ \ \ \ \ \ \ \ \ \ \text {property of distribution}\\ & =\left ( u,\left ( -x^{2}\phi \right ) ^{\prime \prime }\right ) +\left ( u,\left ( x\phi \right ) ^{\prime }\right ) +\left ( u,\phi \right ) \ \ \text {property of distribution}\\ & =\left ( u,\left ( -x^{2}\phi \right ) ^{\prime \prime }+\left ( x\phi \right ) ^{\prime }+\phi \right ) \\ & =\left ( u,L^{\ast }\phi \right ) \end{align*}
Where \(L^{\ast }\phi =\left ( -x^{2}\phi \right ) ^{\prime \prime }+\left ( x\phi \right ) ^{\prime }+\phi \)
Hence expanding the differentiation in the above and simplifying we obtain
\[ \fbox {$\left ( Lu,\phi \right ) =\left ( u,-3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) $}\]
Now take \(Lu=\delta _{\xi }\), i.e. put a point source as input, then we are looking for \(\left ( Lu,\phi \right ) =\phi \left ( \xi \right ) \) from the properties of delta
function. In other words, we are looking for
\[ \left ( Lu,\phi \right ) =\int _{0}^{1}u\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx=\phi \left ( \xi \right ) \]
Hence
\begin{align} \phi \left ( \xi \right ) & =\int _{0}^{\xi }u_{1}\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx+\int _{\xi }^{1}u_{2}\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx\nonumber \\ & =\int _{0}^{\xi }x\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx+\int _{\xi }^{1}\left ( A\left ( \xi \right ) x+B\left ( \xi \right ) x^{-1}\right ) \left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx \tag {1}\end{align}
Looking at the first integral, and perform integration by parts. In these calculations we note that
\[ \phi \left ( 0\right ) =\phi \left ( 1\right ) =\phi ^{\prime }\left ( 0\right ) =\phi \left ( 1\right ) =0 \]
Hence
\begin{align*} \int _{0}^{\xi }x\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx & =\int _{0}^{\xi }-3x^{2}\phi ^{\prime }-x^{3}\phi ^{\prime \prime }dx\\ & =\int _{0}^{\xi }-3x^{2}\phi ^{\prime }dx-\int _{0}^{\xi }x^{3}\phi ^{\prime \prime }dx\\ & =-3\left ( \left [ x^{2}\phi \right ] _{0}^{\xi }-\int _{0}^{\xi }2x\phi dx\right ) -\left ( \left [ x^{3}\phi ^{\prime }\right ] _{0}^{\xi }-\int _{0}^{\xi }3x^{2}\phi ^{\prime }dx\right ) \\ & =-3\left ( \left [ \xi ^{2}\phi \left ( \xi \right ) \right ] -2\int _{0}^{\xi }x\phi dx\right ) -\left ( \left [ \xi ^{3}\phi ^{\prime }\left ( \xi \right ) \right ] -3\int _{0}^{\xi }x^{2}\phi ^{\prime }dx\right ) \\ & =-3\xi ^{2}\phi \left ( \xi \right ) +6\int _{0}^{\xi }x\phi dx-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +3\int _{0}^{\xi }x^{2}\phi ^{\prime }dx \end{align*}
Now do integration by part on the last integral above
\begin{align} \int _{0}^{\xi }x\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx & =-3\xi ^{2}\phi \left ( \xi \right ) +6\int _{0}^{\xi }x\phi dx-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +3\left ( \left [ x^{2}\phi \right ] _{0}^{\xi }-\int _{0}^{\xi }2x\phi dx\right ) \nonumber \\ & =-3\xi ^{2}\phi \left ( \xi \right ) +6\int _{0}^{\xi }x\phi dx-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +3\xi ^{2}\phi \left ( \xi \right ) -6\int _{0}^{\xi }x\phi dx\nonumber \\ & =-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) \tag {2}\end{align}
Now looking at (1) above, we now do integration by part on \(\int _{\xi }^{1}\left ( Ax+Bx^{-1}\right ) \left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx\)
\[ \int _{\xi }^{1}\left ( Ax+Bx^{-1}\right ) \left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx=\int _{\xi }^{1}Ax\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx+\int _{\xi }^{1}Bx^{-1}\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx \]
Consider the first integral above in the RHS, we write
\begin{align*} \int _{\xi }^{1}Ax\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx & =A\int _{\xi }^{1}-3x^{2}\phi ^{\prime }-x^{3}\phi ^{\prime \prime }dx\\ & =A\left ( \int _{\xi }^{1}-3x^{2}\phi ^{\prime }dx-\int _{\xi }^{1}x^{3}\phi ^{\prime \prime }dx\right ) \\ & =A\left ( 3\left ( \left [ -x^{2}\phi \right ] _{\xi }^{1}-\int _{\xi }^{1}-2x\phi dx\right ) -\left ( \left [ x^{3}\phi ^{\prime }\right ] -\int _{\xi }^{1}3x^{2}\phi ^{\prime }dx\right ) \right ) \\ & =A\left ( 3\left ( \left [ -\phi \left ( 1\right ) +\xi ^{2}\phi \left ( \xi \right ) \right ] +2\int _{\xi }^{1}x\phi dx\right ) -\left ( \left [ \phi ^{\prime }\left ( 1\right ) -\xi ^{3}\phi ^{\prime }\left ( \xi \right ) \right ] -3\int _{\xi }^{1}x^{2}\phi ^{\prime }dx\right ) \right ) \\ & =A\left ( 3\left ( \xi ^{2}\phi \left ( \xi \right ) +2\int _{\xi }^{1}x\phi dx\right ) -\left ( -\xi ^{3}\phi ^{\prime }\left ( \xi \right ) -3\int _{\xi }^{1}x^{2}\phi ^{\prime }dx\right ) \right ) \\ & =A\left ( 3\xi ^{2}\phi \left ( \xi \right ) +6\int _{\xi }^{1}x\phi dx+\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +3\int _{\xi }^{1}x^{2}\phi ^{\prime }dx\right ) \end{align*}
Now do integration by part on the last term in the above line
\begin{align} \int _{\xi }^{1}Ax\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx & =A\left ( 3\xi ^{2}\phi \left ( \xi \right ) +6\int _{\xi }^{1}x\phi dx-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +3\left ( \left [ x^{2}\phi \right ] _{\xi }^{1}-\int _{\xi }^{1}2x\phi dx\right ) \right ) \nonumber \\ & =A\left ( 3\xi ^{2}\phi \left ( \xi \right ) +6\int _{\xi }^{1}x\phi dx-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +3\left ( -\xi ^{2}\phi \left ( \xi \right ) -2\int _{\xi }^{1}x\phi dx\right ) \right ) \nonumber \\ & =A\left ( 3\xi ^{2}\phi \left ( \xi \right ) +6\int _{\xi }^{1}x\phi dx-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) -3\xi ^{2}\phi \left ( \xi \right ) -6\int _{\xi }^{1}x\phi dx\right ) \nonumber \\ & =A\xi ^{3}\phi ^{\prime }\left ( \xi \right ) \tag {3}\end{align}
Now we do integration by parts on \(\int _{\xi }^{1}Bx^{-1}\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx\)
\begin{align} \int _{\xi }^{1}Bx^{-1}\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx & =B\int _{\xi }^{1}-3\phi ^{\prime }-x\phi ^{\prime \prime }dx\nonumber \\ & =-B\left ( \int _{\xi }^{1}3\phi ^{\prime }dx+\int _{\xi }^{1}x\phi ^{\prime \prime }dx\right ) \nonumber \\ & =-B\left ( 3\left [ \phi \right ] _{\xi }^{1}+\left ( \left [ x\phi ^{\prime }\right ] _{\xi }^{1}-\int _{\xi }^{1}\phi ^{\prime }dx\right ) \right ) \nonumber \\ & =-B\left ( -3\phi \left ( \xi \right ) +\left ( -\xi \phi ^{\prime }\left ( \xi \right ) -\left [ \phi \right ] _{\xi }^{1}\right ) \right ) \nonumber \\ & =-B\left ( -3\phi \left ( \xi \right ) -\xi \phi ^{\prime }\left ( \xi \right ) +\phi \left ( \xi \right ) \right ) \nonumber \\ & =2B\phi \left ( \xi \right ) +B\xi \phi ^{\prime }\left ( \xi \right ) \tag {5}\end{align}
Hence, from (2),(3),(4),(5), we have
\begin{align*} \phi \left ( \xi \right ) & =\int _{0}^{\xi }x\left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx+\int _{\xi }^{1}\left ( A\left ( \xi \right ) x+B\left ( \xi \right ) x^{-1}\right ) \left ( -3x\phi ^{\prime }-x^{2}\phi ^{\prime \prime }\right ) dx\\ \phi \left ( \xi \right ) & =-\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +A\xi ^{3}\phi ^{\prime }\left ( \xi \right ) +2B\phi \left ( \xi \right ) +B\xi \phi ^{\prime }\left ( \xi \right ) \end{align*}
or
\begin{equation} \phi \left ( \xi \right ) =\phi \left ( \xi \right ) \left ( 2B\right ) +\phi ^{\prime }\left ( \xi \right ) \left [ -\xi ^{3}+A\xi ^{3}+B\xi \right ] \tag {6}\end{equation}
By looking at the coefficients on \(\phi \left ( \xi \right ) \), and compare, we see that \(2B=1\) or
\[ B=\frac {1}{2}\]
We can now use the continuity condition at \(x=\xi \) and write
\(u_{1}=u_{2}\) at \(x=\xi \), hence
\begin{align*} \xi & =Au_{1}+Bu_{2}\\ & =A\xi +B\xi ^{-1}\\ \xi & =A\xi +\frac {1}{2}\xi ^{-1}\\ 2\xi ^{2} & =2A\xi ^{2}+1 \end{align*}
Hence
\[ A=\frac {2\xi ^{2}-1}{2\xi ^{2}}\]
Therefor the fundamental solution is
\begin{align*} u & =\left \{ \begin {array} [c]{cc}x & 0<x<\xi \\ Ax+Bx^{-1} & \xi <x<1 \end {array} \right . \\ & =\left \{ \begin {array} [c]{cc}x & 0<x<\xi \\ \frac {2\xi ^{2}-1}{2\xi ^{2}}x+\frac {1}{2x} & \xi <x<1 \end {array} \right . \end{align*}
Here is a plot for few values of \(\xi \)
