Solving the BVP $u_{xx}+a^2u=\sin(\pi x)$ with $u(0)=1$ and $u(1)=-2$

I am trying to solve the BVP $$u_{xx}+a^2u=\sin(\pi x), \ \ \text{for} \ \ 0 with $$u(0)=1$$ and $$u(1)=-2$$, $$\forall a\in\mathbb{R}$$.

I begin by solving the homogeneous equation $$u_{xx}+a^2u=0$$. The roots of the characteristic equation of this ODE are $$\pm ai$$ and so the general solution of the homogeneous equation is $$u_H(x)=A\cos(ax)+B\sin(ax), \ \ A,B\in\mathbb{R}.$$ Searching for a particular solution, I guessed that $$u_p(x)=x(C_1\cos(\pi x)+C_2\sin(\pi x))$$ where $$C_1,C_2\in\mathbb{R}$$. So, \begin{align} u'_p(x)&=-C_1x\pi\sin(\pi x)+C_1\cos(\pi x)+C_2x\pi\cos(\pi x)+C_2\sin(\pi x) \\ u''_p(x)&=-C_1\pi^2 x\cos(\pi x)-2C_1\pi\sin(\pi x)-C_2\pi x\sin(\pi x)+2C_2\pi\cos(\pi x). \end{align} Substitution into the original ODE gives $$C_1=-\frac{1}{2\pi}$$ and $$C_2=0$$ and so $$u_p(x)=-\frac{x}{2\pi}\cos(\pi x).$$ Thus, $$u(x)=u_H(x)+u_P(x)=A\cos(ax)+B\sin(ax)-\frac{x}{2\pi}\cos(\pi x).$$ The boundary condition $$u(0)=1$$ implies $$A=1$$. The boundary condition $$u(1)=-2$$ implies \begin{align} -2&=\cos(a)+B\sin(a)-\frac{1}{2\pi}\cos(\pi) \\ \implies B&=\frac{1}{\sin(a)}\left(-2-\frac{1}{2\pi}-\cos(a)\right). \end{align} Hence, $$u(x)=\cos(ax)+\frac{1}{\sin(a)}\left(-2-\frac{1}{2\pi}-\cos(a)\right)\sin(ax)-\frac{x}{2\pi}\cos(\pi x).$$ But the resource I found this question states that this is not the correct answer. Have I made an error in logic?

• By inspection there is a particular solution of the type $c\sin (\pi x)$. The value of $c$ is $\frac 1 {\pi^{2}-a{2}}$. What you have obtained is not a solution to the given DE. – Kabo Murphy Mar 13 at 8:52
• Have I made a mistake somewhere? – Steven Mar 13 at 9:04
• I haven't checked carefully but the last term should be a constant times $\sin (\pi x)$. The might help you to find the mistake. – Kabo Murphy Mar 13 at 9:08
• Your particular solution is only valid for $a=\pm\pi$. For all other values there is no resonance, you get the particular solution in the form of @KaviRamaMurthy. – LutzL Mar 13 at 9:26
• $u_p(x)=x(C_1\cos(\pi x)+C_2\sin(\pi x))$ is not a convenient proposal. – Cesareo Mar 13 at 9:36

In the general case $$|a|\ne \pi$$ there is no resonance in the forcing term on the right and you get the solution by the method of undetermined coefficients in the base case, that is, $$u_p(x)=C_1\cos(πx)+C_2\sin(πx).$$ This results in $$C_1=0$$ and $$(-\pi^2+a^2)C_2=1$$.

Inserting boundary conditions into $$u(x)=A\cos(ax)+B\sin(ax)+\frac1{a^2-\pi^2}\sin(πx)$$ gives $$1=A$$ and $$-2=\cos(a)+B\sin(a)$$, $$B=-\frac{2+\cos(a)}{\sin(a)}$$. Thus in general there will be no solutions for $$a=k\pi$$.

For $$a\approx \pi$$ one can write the solution as $$u(x)=\cos(ax)-\frac{\sin(a)\sin(ax)}{1+\cos(a-\pi)}-\frac{\sin(ax)-\sin(πx)}{(a-π)(a+π)}+\left(\frac1{a^2-π^2}+\frac1{\sin(a-π)}\right)\sin(ax)$$ While the first three terms have a limit for $$a\toπ$$, the last coefficient does not cancel the two poles in its terms, thus is singular. There is no solution for $$a=\pi$$.

Using Laplace transform instead

$$U(s) = \frac{s(s^2 +\pi^2)u(0)+(s^2 + \pi^2)u'(0)+\pi }{\left(s^2+\pi ^2\right) \left(a^2+s^2\right)}$$

or

$$U(s) = \frac{s(s^2 +\pi^2)C_1+(s^2 + \pi^2)C_2+\pi }{\left(s^2+\pi ^2\right) \left(a^2+s^2\right)}$$

after inverting with previous expansion we have

$$u(t) = \frac{1}{a(a^2-\pi^2)}\left(a(a^2-\pi^2)\cos(a t)C_1+(a^2-\pi^2)\sin(a t)C_2+a\sin(\pi t)-\pi\sin(a t)\right)$$

The constants are

$$\begin{cases} C_1 = 1\\ C_2 = \frac{\pi}{a^2-\pi^2}-a(2+\cos a)\csc a \end{cases}$$

• I appreciate the solution, but I haven't learnt about Laplace transforms in this sense. Is there a solution with a similar method to the one which I proposed? – Steven Mar 13 at 10:07