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Consider a boundary value problem $$ \frac{\mathrm{d^2}y }{\mathrm{d} x^2 } =f(x) $$ with boundary conditions $$ y(0)=y(1)=\frac{dy}{dx}(1) $$ where $f$ is real-valued continuous function on the interval $[0,1]$.

Then which of the following is true .

  • (1) Given BVP has unique solution for every $f$.
  • (2) Given BVP does not have unique solution for some $f$.
  • (3) $y(x)=\int_{0}^{x}xtf(t)dt+\int_{x}^{1}(t-x+xt)f(t)dt$ is the solution given BVP.
  • (4) $y(x)=\int_{0}^{x}(x-t+xt)f(t)dt+\int_{x}^{1}xtf(t)dt$ is the solution of given BVP.

for this i use

Theorem: Existence and Uniqueness Let $p(t)$, $q(t)$, and $g(t)$ be continuous on $[a,b]$ , then the differential equation $$y'' + p(t)y' + q(t)y = g(t)\\ y(t_0) = y_0 \\ y'(t_0) = y'_0$$ has a unique solution defined for all $t$ in $[a,b]$.

Therefore we have unique solution for every $f$ . Reject option (2 ). Option (1) is correct .

for option (4) I directly put $x=0$ and $1$ in given solution in the option. But I get $y(0)=0$ and $y(1)= \int_{0}^{1}f(t)dt$. Clearly both are not always equal. so I reject option (4).

For option (3) I differentiate given solution in option (3) i.e $$y(x)=\int_{0}^{x}xtf(t)dt+\int_{x}^{1}(t-x+xt)f(t)dt$$ and i get given BVP. So according to me answers should be (1) and (3).

please solve this problem. This is very important to me . you can just tell me the correct options .

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3
  • $\begingroup$ Please tell us what you tried and where you are stuck at? So we can help you better, $\endgroup$
    – prog_SAHIL
    Jan 2, 2018 at 15:29
  • $\begingroup$ i add my approach to the question . you can check it now .please help me now . $\endgroup$
    – Gilll
    Jan 3, 2018 at 3:36
  • $\begingroup$ it is very important to me please reply anyone $\endgroup$
    – Gilll
    Jan 3, 2018 at 3:37

1 Answer 1

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The uniqueness is obvious Now we jump to the existence:

Doing like here Stability property of Poisson equation with zero boundary values you have $$ y''(x) = f(x)\implies y'(t) = \int_0^tf(s)ds +c \\\implies y(x) = \int_0^x \left(\int_0^tf(s)ds +c\right)dt+ a$$

Now $$y(0) = a = y(1) = \int_0^1 \left(\int_0^tf(s)ds +c\right)dt+ a $$

Then $$\int_0^ 1\left(\int_0^tf(s)ds +c\right)dt=0\implies c=-\int_0^ 1\int_0^tf(s)ds dt $$

also $$y(0) =a= y'(1) = \int_0^1f(s)ds +c= \int_0^1f(s)ds-\int_0^ 1\int_0^tf(s)ds dt $$

Therefore, $$y(x) = \int_0^x \left(\int_0^tf(s)ds +c\right)dt+ a\\=\int_0^x \int_0^tf(s)dsdt -x\int_0^ 1\int_0^tf(s)ds dt + \int_0^1f(s)ds-\int_0^ 1\int_0^tf(s)ds dt $$

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  • $\begingroup$ Values of a and c can be find this means option 1 and 3 are correct .am I right sir ? $\endgroup$
    – Gilll
    Jan 3, 2018 at 12:29
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    $\begingroup$ @guronkar yes 3 and 1 are correct. Don't forget to vote and checkmark if the solution suite you $\endgroup$
    – user503348
    Jan 3, 2018 at 13:00

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