Change of variable to solve boundary value problem

Question

We are given the differential equation $$u''(x)-\lambda u(x)=0$$ (negative $\lambda$), with boundary conditions $u(0)=u(1)=0$. As the solution will involve trig functions I'd like to do a change of variable substitution to get the new boundary conditions $u(0)=u(\pi)=0$. To achieve this I consider a new variable $X=\pi x$ with a new constant $\gamma=\lambda \pi^2$. I then try to solve $$u''(X)-\gamma u(X)=0$$ I think this is correct but I'm not entirely sure what we're looking for in terms of equivalencies to be able to check so I'd be much obliged if someone could explain how I can determine myself whether this is a correct variable change that doesn't change the solution.

Using this change of variable I get a sequence of solutions of the form $$u_m(X)=A_m\sin(mX)$$ At this point I'd like to change back to the original variable, and so I think the solution should be of the form $$u_m(x)=A_m\sin\big (\frac{mx}{\pi}\big )$$ but once again I'm not sure if it's as simple as a reverse substitution, because $m$ depends on $\gamma$ as $-m^2=\gamma$ and I don't know if I need to account for this. Any advice would be appreciated.

Answer 1

When you change variables it can be a good thing to keep track of with respect to what you are differentiating. The original differential equation reads $$\frac{d^2u}{dx^2}-\lambda u=0$$ If we let $X=\pi x$ then $$\frac{du}{dx}=\frac{du}{dX}\frac{dX}{dx}=\pi\frac{du}{dX}$$ $$\frac{d^2u}{dx^2}=\frac{d}{dx}\left(\pi\frac{du}{dX}\right)=\frac{d}{dX}\left(\pi\frac{du}{dX}\right)\frac{dX}{dx}=\pi^2\frac{d^2u}{dX^2}$$ So now we get $$\pi^2\frac{d^2u}{dX^2}-\lambda u=0$$ $$\frac{d^2u}{dX^2}-\mu u=0$$ And so $\mu=\lambda/\pi^2$. Solutions for $\mu<0$ are $$u=c_1\cos\left(\sqrt{-\mu}X\right)+c_2\sin\left(\sqrt{-\mu}X\right)$$ Applying the boundary conditions we see that $c_1=0$ and when $x=1$, $X=\pi$, $\sqrt{-\mu}\pi=m\pi$, so $\mu=-m^2$, $u=\sin mX=\sin m\pi x$, $\lambda=\mu\pi^2=-m^2\pi^2$.

Answer 2

I'm going to ignore the requirement that $\lambda>0$ (since there are no solutions in this case, as we shall see) and guide you through the general procedure for this kind of problem. A change of variables is not overly helpful here - ultimately what we want to do first is find the general solution to $u''-\lambda u=0$ and then plug in our boundary conditions.

In general, a second order linear differential equation with constant coefficients of the type $au''+bu'+cu=0$ is solved by letting $u=e^{\gamma x}$ and finding the auxiliary equation, then solving for $\gamma$. The auxiliary equation is found by plugging into the equation:

$$0=a(e^{\gamma x})''+b(e^{\gamma x})'+ce^{\gamma x}=e^{\gamma x}(a\gamma^2+b\gamma+c)$$

and so $a\gamma^2+b\gamma+c=0$, which is the auxiliary equation. Once we have the two solutions $\gamma_1$ and $\gamma_2$, the general solution is found by taking linear combinations, i.e. $$u=Ae^{\gamma_1x}+Be^{\gamma_2x}$$ where $A,B$ are arbitrary constants - this is the general solution. If $\gamma_1=\alpha+i\beta$ is complex (and hence $\gamma_2=\alpha-i\beta$) we need to take $B=\bar A$. Since $A=C+Di$ there are still two arbitrary real constants, and the solution becomes $$u=e^{\alpha x}(C\cos(\beta x)+D\sin(\beta x)).$$ (There is the special case case where $\gamma_1=\gamma_2$, but I won't go into this case.) In our case, the auxiliary equation is $\gamma^2-\lambda=0$, so the solutions are $\gamma=\pm\sqrt{\lambda}$ if $\lambda\ge0$ or $\gamma=\pm i\sqrt{|\lambda|}$ if $\lambda<0$. Since the equation is linear, linear combinations of solutions are still solutions and so $$u=Ae^{\sqrt{\lambda}x}+Be^{-\sqrt{\lambda}x}$$ if $\lambda\ge0$ or $$u=C\cos(\sqrt{|\lambda|}x)+D\sin(\sqrt{|\lambda|}x)$$ if $\lambda<0$. Plugging in the boundary condition, we see that no solutions can exist when $\lambda>0$. If $\lambda=0$ then we get the trivial solution $u=0$. If $\lambda<0$, we have that $$u=A\sin(\sqrt{|\lambda|}x)$$ is a solution if and only if $\sqrt{|\lambda|}=k\pi$ for some integer $k$, that is, $\lambda=-k^2\pi^2$ for some integer $k$. ($A$ is an arbitrary constant, which makes sense since this is an eigenvalue equation and so constant multiples of a solution should still be a solution.)

As an additional remark, you can see that you must have $\lambda\le0$ by integration by parts: $$\lambda\int_0^1u^2dx=\int_0^1(\lambda u)udx=\int_0^1u''udx=u'u\big|_{x=0}^1-\int_0^1(u')^2dx=-\int_0^1(u')^2dx\le0$$ and so $\lambda\le0$. This has the advantage of generalizing to PDEs of a similar type.