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Solve the BVP $$\begin{cases} u''+a^2u=sin\pi x, \quad 0

for all $a\in\mathbb{R}$. What are the solutions in the cases $a=\pm\pi$?

Here, $'=\frac{du}{dx}$. I know I need the characteristic function, which I get $r^2+a^2=0\implies r=\pm ai$. Then the complementary solution is, $$u_c=C_1e^{ai*t}+C_2e^{-ai*t}$$

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    Is $u''=\frac{d^2u}{dx^2}$? If yes then this is an ODE. – 2017-02-21
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    Are you familiar with Laplace transforms, Green's functions, etc? – 2017-02-21
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    @CheeHan Yes, it is. – 2017-02-21
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    @Dr.MV I am not familiar with Laplace transforms, Green's functions. From what I am reading, I need to find the complementary function in the form $u(x)=Au_1(x)+Bu_2(x)+P(x)$. And I am not missing a $u''$ after the $a^2$ term. – 2017-02-21
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    If there is no $u''$ after the $a^2$ term, then you are on the right track. Solving $u''=0$ gives $u(x)=Ax+B$, you are left with finding the particular solution. – 2017-02-21
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    @ozarka Are you missing a $u$ after $a^2$?? – 2017-02-21
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    @Dr.MV Oh my goodness, I am missing $u$ after $a^2$. Thank you, I had mistyped it in latex! Edited. – 2017-02-21

2 Answers 2

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HINT:

Since the homogeneous solutions to the ODE $u''+\pi^2 u=\sin(\pi x)$ are of the form $u_h=A\sin(\pi x)+B\cos(\pi x)$, try complementary solutions of the form $u_c=Cx\sin(\pi x)+Dx\cos(\pi x)$.

Alternatively, use Laplace Transforms.

Alternatively, find the Green (or Green's) function for the problem

$$G''(x,x')+\pi^2 G(x,x')=\delta(x-x')$$

with $G(0,x')=1$, $G(1,x')=-2$, $G$ is continuous at $x=x'$, and $\left.\frac{\partial G}{\partial x}\right|_{x=x'^+}-\left.\frac{\partial G}{\partial x}\right|_{x=x'^-}=1$ and integrate.

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    I found that $r^2+a^2=0\implies r=\pm ai$. So the complementary solution is $u_c=Cx\sin(ax)+Dx\cos(ax)$. Is this correct so far? – 2017-02-21
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    Yes, you've got it now. You'll find $C=0$ and $D=1/(2\pi)$. I'm pleased to see that my hint was useful. – 2017-02-21
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    Since we are given initial conditions, would I define a particular solution, that is, $u_p=A\cos ax+B\sin ax$ and then find $u_p'\ \text{and}\ u_p''$ to plug into our given BVP? – 2017-02-21
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    No. The solution is of the form $u=u_h+u_p$, where $u_h=A\sin(\pi x)+B\cos(\pi x)$ and $u_p=Cx\sin(\pi x)+Dx\cos(\pi x)$. Now, you should be able to show that $C=0$ and $D=1/(2\pi)$. With those values of $C$ and $D$, apply the boundary conditions to $u_h+u_p$, not just $u_h$. – 2017-02-21
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    Are you solving BVP when $a=\pi$? With your help, I am getting $u=u_h+u_p=A\sin(ax)+B\cos(ax)+Cx\cos(ax)+Dx\sin(ax)$. I set $u=1$ at $x=0$ to eventually get $B=1$. Then I get $u=-2$ at $x=1$ to get $-2=(A+D)\sin(a)+(1+C)\cos(a)$. – 2017-02-21
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    First, note that $C=0$ and $D=1/(2\pi)$ (why?). At that point, only $A$ and $B$ are unresolved. Then, apply the two boundary conditions to find the two unknowns $A$ and $B$. – 2017-02-21
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    $C=0$ and $D=\frac{1}{2\pi}$ because once you found $u=u_h+u_p=A\sin(\pi x)+B\cos(\pi x)+Cx\cos(\pi x)+Dx\sin(\pi x)$, you took the second derivative and plugged into the original BVP? – 2017-02-21
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    I used the form of $u_p$ and substituted it into the original ODE (without regard for the boundary conditions). – 2017-02-21
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    so you found $u_p''$? I get like a 6 term expression. – 2017-02-21
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    $u_p''=2C\pi\cos(\pi x)-2D\pi\sin(\pi x)-\pi^2Cx\sin(\pi x)+D\pi^2x\cos(\pi x)$ – 2017-02-21
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    Yes, and adding $u"_p$ to $\pi^ u_p$ yields?? Then, exploit the linear independence of the sine and cosine functions gives what? – 2017-02-21
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    So I plug in $u_p''$ and $u_p$ to my BVP and get $$(2C\pi+D\pi^2x+a^2Cx)cos(\pi x) + (-2D\pi-C\pi^2x+a^2Dx)sin(\pi x)=sin(\pi x)$$ Then I set $(2C\pi+D\pi^2x+a^2Cx)=0$ and $(-2D\pi-C\pi^2x+a^2Dx)=1$. And then I see that $C=0$ but $D=-\frac{1}{2\pi}$. – 2017-02-21
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    $$u_p=Cx\sin(\pi x)+Dx\cos(\pi x)\implies u_p''=(-\pi ^2 C x\sin(\pi x)-\pi^2 Dx\cos(\pi x)) \,\,+2\pi C\cos(\pi x)-2\pi D\sin(\pi x)$$Now, for $a=\pm \pi$, $a^2=\pi^2$ and $$u_p''+\pi^2u_p=2\pi C\cos(\pi x)-2\pi D\sin(\pi x)=\sin(\pi x)\implies C=0,\,\,D=-1/(2\pi) $$ – 2017-02-21
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If you do not know about integral transforms and Green's functions go and learn it. But Here is the way of avoiding this stuff for your concrete problem: $$y''+a^{2}y=f(x)$$ You note that $$y''+a^{2}y=\frac{d}{dt}(y'+iay)-ia(y'+iay)$$ Now let $$\xi(x)=y'+iay$$ Hence $$\xi'-ia\xi=f(x)$$ Using the integrating factor $$\xi(x)=\frac{d}{dx}y+aiy=e^{aix}\Big\{\int_{0}^{x}f(z)e^{-aiz}dz+\xi_{0}\Big\}$$ Using the integrating factor the second time you get $$y(x)=e^{-aix}\Big\{\int_{0}^{x}e^{2aiw}\Big\{\int_{0}^{w}f(z)e^{-aiz}dz+\xi_{0}\Big\}dw+x_{0}\Big\}$$ Then do the integrals)