0
$\begingroup$

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 .

$\endgroup$
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 '18 at 15:29
  • $\begingroup$ i add my approach to the question . you can check it now .please help me now . $\endgroup$
    – Gilll
    Jan 3 '18 at 3:36
  • $\begingroup$ it is very important to me please reply anyone $\endgroup$
    – Gilll
    Jan 3 '18 at 3:37
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Values of a and c can be find this means option 1 and 3 are correct .am I right sir ? $\endgroup$
    – Gilll
    Jan 3 '18 at 12:29
  • 1
    $\begingroup$ @guronkar yes 3 and 1 are correct. Don't forget to vote and checkmark if the solution suite you $\endgroup$
    – user503348
    Jan 3 '18 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.