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For $\lambda\in \mathbb{R}$ consider the boundary value problem

$$x^2 \frac{d^2y}{dx^2}+2x\frac{dy}{dx}+\lambda y=0\;\;\;,y(1)=y(2)=0$$ called as $(P_{\lambda})$

Then i Have to prove which of following is right and which is wrong and why?

$1).$For any continuous function $f:[1,2]\to \mathbb{R}\;\;$with $f(x)\neq 0\;\;$for some $x \in[1,2]$ there $\\$ exist a solution $u$ of ($P_{\lambda}$)

$2).$There exist a $\lambda_0 \in \mathbb{R}$ such that ($P_{\lambda}$) has a nontrivial solution for any $\lambda>\lambda_0$

The solution i tried- we can write the given differntial equation in form

$$\theta(\theta-1)+2\theta+\lambda=0\\ \theta^2+\theta+\lambda=0\\$$ which is quadratic in $\theta$

after solving this i get

$$\theta =\frac{-1 \pm \sqrt {1-4\lambda}}{2}$$

after that i solved further i get

$$\displaystyle y=c_1x^{\frac{-1 + \sqrt {1-4\lambda}}{2}}+c_2x^{\frac{-1 -\sqrt {1-4\lambda}}{2}}$$

further applying the boundary conditions i get two conditions

$$c_1+c_2=0\;\; and\;\; c_1 2^{\frac{-1 + \sqrt {1-4\lambda}}{2}}+c_2 2^{\frac{-1 - \sqrt {1-4\lambda}}{2}}$$

for non trivial solution there Cofficents matrix determinent should not be zero. in end after all calculation i get

$$\lambda \neq \frac{1}{4}$$

after that i have no clue how to get to these options? how can i check which is true and why ?

Please help

Thankyou.

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  • $\begingroup$ For question (1), what is $f$ doing there? The question is about a solution in $P_\lambda$, no? $\endgroup$ Oct 15, 2019 at 10:27
  • $\begingroup$ Don't know in option it is given $\endgroup$ Oct 15, 2019 at 10:33

1 Answer 1

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Let $t:=\ln(x)$. Then, $$\frac{\text{d}y}{\text{d}x}=\exp(-t)\,\frac{\text{d}y}{\text{d}t}\,.$$ That is, $$x^2\,\frac{\text{d}^2y}{\text{d}x^2}=\exp(t)\,\frac{\text{d}}{\text{d}t}\,\left(\exp(-t)\,\frac{\text{d}y}{\text{d}t}\right)=\frac{\text{d}^2y}{\text{d}t^2}-\frac{\text{d}y}{\text{d}t}\,.$$ Therefore, the given differential equation now becomes $$\frac{\text{d}^2y}{\text{d}t^2}+\frac{\text{d}y}{\text{d}t}+\lambda\,y=0\,.$$ If $\lambda<\dfrac14$, then the solutions are given by $$\begin{align}y&=A_-\,\exp\left(\frac{-1-\sqrt{1-4\lambda}}{2}\,t\right)+A_+\,\exp\left(\frac{-1-\sqrt{1+4\lambda}}{2}\,t\right)\\&=A_-\,x^{\frac{-1-\sqrt{1-4\lambda}}{2}}+A_+\,x^{\frac{-1+\sqrt{1-4\lambda}}{2}}\,,\end{align}$$ where $A_-$ and $A_+$ are constants. If $\lambda=\dfrac14$, then the solutions are given by $$y=(a+b\,t)\,\exp\left(-\frac{1}{2}\,t\right)=\big(a+b\,\ln(x)\big)\,x^{-\frac12}\,,$$ where $a$ and $b$ are constants. If $\lambda>\dfrac14$, then the solutions are given by $$\begin{align}y&=\exp\left(-\frac{1}{2}\,t\right)\,\Biggl(C\,\cos\left(\frac{\sqrt{4\lambda-1}}{2}\,t\right)+S\,\sin\left(\frac{\sqrt{4\lambda-1}}{2}\,t\right)\Biggr) \\&=x^{-\frac{1}{2}}\,\Biggl(C\,\cos\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(x)\right)+S\,\sin\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(x)\right)\Biggr)\,,\end{align}$$ where $C$ and $S$ are constants.

For $\lambda<\dfrac14$, if $y(x=1)=0$ and $y(x=2)=0$, then $$A_-+A_+=0\text{ and }A_-+A_+\,2^{\sqrt{1-4\lambda}}=0\,.$$ It follows that $A_-=A_+=0$, so $y$ is identically zero. For $\lambda=\dfrac14$, if $y(x=1)=0$ and $y(x=2)=0$, then $$a=0\text{ and }a+b\,\ln(2)=0\,,$$ whence $a=b=0$ and so $y$ is again identically zero. For $\lambda>\dfrac14$, if $y(x=1)=0$ and $y(x=2)=0$, then $$C=0\text{ and }C\,\cos\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(2)\right)+S\,\sin\left(\frac{\sqrt{4\lambda-1}}{2}\,\ln(2)\right)=0\,.$$ This shows that $C=0$ and $S=0$, or $C=0$ and $$\frac{\sqrt{4\lambda-1}}{2}\,\ln(2)=k\pi$$ for some positive integer $k$. Thus, $$\lambda=\frac{k^2\pi^2}{\big(\ln(2)\big)^2}+\frac{1}{4}\tag{*}$$ for some positive integer $k$.

That is, there exists a nonzero solution to $P_\lambda$ if and only if $\lambda$ is given by (*). In that case, all solutions take the form $$y=S\,x^{-\frac12}\,\sin\big(k\pi\,\log_2(x)\big)$$ where $S$ is a constant. Hence, Option (2) is false. I do not understand Option (1).

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  • $\begingroup$ Thank you for such beautiful and complete solution. $\endgroup$ Oct 15, 2019 at 14:06

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