# Thread: Green's function for a BVP

1. ## Green's function for a BVP

Dear MHF members,

I need help in finding the Green's function for the BVP
$\displaystyle y^{\prime\prime}+\frac{1}{x}y^{\prime}=-f(x)$ with $\displaystyle \lim_{x\to0^{+}}|y(x)|<\infty$ and $\displaystyle y(1)=0$.

Many thanks.
bkarpuz

Note. Typo is corrected after TheEmptySet's warning.

2. Originally Posted by bkarpuz
Dear MHF members,

I need help in finding the Green's function for the BVP
$\displaystyle y^{\prime\prime}-\frac{1}{x}y^{\prime}=-f(x)$ with $\displaystyle \lim_{x\to0^{+}}|y(x)|<\infty$ and $\displaystyle y(1)=0$.

Many thanks.
bkarpuz
I am going to assume that there is a typo in the statement of the problem otherwise the boundary condition at zero isn't needed as the solution to the homogenous problem is

$\displaystyle y=c_1+c_2x^2$

Which is bounded no matter what the constants are but if we solve the problem

$\displaystyle y''+\frac{1}{x}y'=-f(x) \iff -\frac{d}{dx}\left( xy' \right)=xf(x)$

Now this is in its self adjoint form.

Solving the homogeneous equation gives

$\displaystyle y_c=c_1+c_2\ln(x)$

Now we need to find two solutions each satisfying one of the boundary conditions.

$\displaystyle y_0=1$

other wise the log function is unbounded at zero and

$\displaystyle y_1=\ln(x)$ to give zero at 1.

So the Green's function has the form

$\displaystyle g(x,s)=\begin{cases}A\ln(x), \text{ if } s \le x \\ B\ln(s), \text{ if } s \ge x \end{cases}$

Now if we take the Wronskian we get

$\displaystyle \begin{vmatrix}A\ln(x) & B \\ \frac{A}{x} & 0 \end{vmatrix}=\frac{-AB}{x}$

Now we take the function off of the derivative of y in the self adjoint form and multiply it by the Wronskian and set it equal to negative 1 to get

$\displaystyle -AB=-1$

So we can pick any real numbers for A and B that satisfy this equation so let

$\displaystyle A=B=1$

So the Green's function is

$\displaystyle g(x,s)=\begin{cases}\ln(x), \text{ if } s \le x \\ \ln(s), \text{ if } s \ge x \end{cases}$

So the solution to the ODE is

$\displaystyle y=\ln(x)\int_{0}^{x}sf(s)ds+\int_{x}^{1}s\ln(s)f(s )ds$

Notice this satisfies the boundary conditions and that

$\displaystyle y'=\frac{1}{x}\int_{0}^{x}sf(s)ds+x\ln(x)f(x)-x\ln(x)f(x)=\frac{1}{x}\int_{0}^{x}sf(s)ds$

and

$\displaystyle y''=-\frac{1}{x^2}\int_{0}^{x}sf(s)ds-f(x)$

If you plug this back into the ODE you get

$\displaystyle -\frac{1}{x^2}\int_{0}^{x}sf(s)ds-f(x)+\frac{1}{x}\left( \frac{1}{x}\int_{0}^{x}sf(s)ds\right)=-f(x)$

3. Originally Posted by TheEmptySet
I am going to assume that there is a typo in the statement of the problem otherwise the boundary condition at zero isn't needed as the solution to the homogenous problem is ...
Thanks a lot TheEmptySet. Let me tell you what I have done. The problem was actually the following equation
$\displaystyle y^{\prime\prime}+\dfrac{1}{x}y^{\prime}-\dfrac{\lambda^{2}}{x^{2}}y=-f(x)$ with $\displaystyle |y(0^{+})|<\infty$ and $\displaystyle y(1)=0$,
$\displaystyle (xy^{\prime})^{\prime}-\dfrac{\lambda^{2}}{x}y=-xf(x)$
The functions $\displaystyle \phi_{1}=2\lambda x^{\lambda}$ and $\displaystyle \phi_{2}=(x^{\lambda}-x^{-\lambda})/(2\lambda)$ are respective solutions of the homogeneous equations
$\displaystyle (x y^{\prime})^{\prime}-\dfrac{\lambda^{2}}{x}y=-xf(x)$ with $\displaystyle |y(0^{+})|<\infty$
and
$\displaystyle (x y^{\prime})^{\prime}-\dfrac{\lambda^{2}}{x}y=-xf(x)$ with $\displaystyle y(1)=0$.
Then, $\displaystyle xW(\phi_{1},\phi_{2})=2\lambda$, which yields
$\displaystyle G(x,t)=\dfrac{1}{2\lambda}\begin{cases}(x/t)^{\lambda}-(tx)^{\lambda}, &x\leq t\\ (t/x)^{\lambda}-(tx)^{\lambda}, &x\geq t\end{cases}$
............$\displaystyle =\dfrac{1}{2\lambda}\big[(t/x)^{\lambda\mathrm{sgn}(x-t)}-(tx)^{\lambda}\big]$.
Thus the solution is given by
$\displaystyle \phi=\frac{1}{2\lambda}\int_{0}^{1}\big[(s/x)^{\lambda\mathrm{sgn}(x-s)}-(sx)^{\lambda}\big]sf(s)\mathrm{d}s$.

Letting here $\displaystyle \lambda\to0$ (assuming that it is possible), I get
$\displaystyle \phi=\frac{1}{2}\int_{0}^{1}\big[\mathrm{sgn}(x-s)\ln(s/x)-\ln(sx)\big]sf(s)\mathrm{d}s$
.....$\displaystyle =\int_{0}^{1}\left\{\begin{array}{c}\ln(s),\ x\leq s\\ \ln(x),\ x\geq s\end{array}\right\}sf(s)\mathrm{d}s$, which is exactly the same expression you have obtained. But the problem here is that in my case, the Wronskian tends to $\displaystyle 0$. What is the reason that the method I have applied above does not work (by considering the homogeneous equation with different boundary conditions separately)?

Thanks.
bkarpuz

4. Here is what I think is going on.

When we compare the two equations and their solutions.

$\displaystyle y''+\frac{1}{x}y'+\frac{\lambda^2}{x^2}y=0$

$\displaystyle y_c=c_1x^\lambda+c_2x^{-\lambda}$

and

$\displaystyle y''+\frac{1}{x}y'=0$

$\displaystyle y_c=d_1+d_2\ln(x)$

Now if we take the limit as

$\displaystyle \lambda \to 0$

The first ODE goes to the 2nd but

$\displaystyle \lim_{\lambda \to 0}c_1x^\lambda+c_2x^{-\lambda}=c_1+c_2$

This does not go to the complimentary solution of the 2nd equation.

5. Originally Posted by TheEmptySet
Here is what I think is going on.

When we compare the two equations and their solutions.

$\displaystyle y''+\frac{1}{x}y'+\frac{\lambda^2}{x^2}y=0$

$\displaystyle y_c=c_1x^\lambda+c_2x^{-\lambda}$

and

$\displaystyle y''+\frac{1}{x}y'=0$

$\displaystyle y_c=d_1+d_2\ln(x)$

Now if we take the limit as

$\displaystyle \lambda \to 0$

The first ODE goes to the 2nd but

$\displaystyle \lim_{\lambda \to 0}c_1x^\lambda+c_2x^{-\lambda}=c_1+c_2$

This does not go to the complimentary solution of the 2nd equation.
But if we write the solution of the first one as $\displaystyle y=c_{1}x^{\lambda}+c_{2}(x^{\lambda}-x^{-\lambda})/(2\lambda)$ it does.

6. Originally Posted by bkarpuz
But if we write the solution of the first one as $\displaystyle y=c_{1}x^{\lambda}+c_{2}(x^{\lambda}-x^{-\lambda})/(2\lambda)$ it does.
But the Wronskian of the above function is independent of lambda and non zero.

$\displaystyle \begin{vmatrix} x^\lambda & \frac{x^{\lambda} - x^{-\lambda}}{2\lambda} \\ \lambda x^{\lambda -1} & \frac{x^{\lambda-1}+x^{-\lambda-1}}{2} \end{vmatrix} = \frac{x^{2\lambda-1}+x^{-1}}{2} - \frac{ x^{2\lambda -1}- x^{-1}}{2} =\frac{1}{x} \ne 0$