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Thanks for looking at my question. I'm working through/self-studying the second edition of Partial Differential Equations: An Introduction by Walter A. Strauss.

On page three, example two, he says

"Solve the PDE $u_{xx} + u = 0$. Again, its really an ODE with an extra variable y. We know how to solve the ODE, so the solution is $u = f(y)cos(x) + g(y)sin(x)$, where again $f(y)$ and $g(y)$ are two arbitrary functions of $y$. You can easily check this formula by differentiating twice to verify that $u_{xx} = -u$."

What I don't understand is how he gets $u = f(y)cos(x) + g(y)sin(x)$.

He says it's basically just an ODE with an extra variable y, but I'm not quite seeing that. I was able to understand Example 1 before it and Example 3 after it, and I can sort of see that this PDE is similar to the ODE form $y'' + y = 0$, but it's just been a hot minute since I've solved an ODE like this. I see it's homogenous, and one could use the method of integrating factors for it, but since this is a PDE I'm not sure how to solve this. My initial guess of $C_{1}e^{r_{1}t} + C_{2}e^{{r_2}t}$ didn't work, so I'm not sure how they got what they got for $u$. I get that when you integrate with respect to $x$ the constant you get is a function of $y$, but that's all I understand about this problem. Could someone show me how Walter got his solution for $u$, please? Thanks.

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  • $\begingroup$ You're usually given boundary conditions. $\endgroup$ – Shogun Jun 3 at 21:23
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We want to solve $\frac{\partial^2}{\partial x^2} u(x,y)+u(x,y)=0.$ As you said, we can solve this like an ODE. Let us consider a guess $u(x,y)=c(y)e^{rx}.$ We make $c$ depend on $y$ because $c(y)$ is still constant in $x$, which is the variable that we're differentiating in. Plugging in such a guess, we get the characteristic equation $$r^2+1=0,$$ which has imaginary roots at $\pm i$, which yields the solution $$u(x,y)=c(y)\cos x+b(y)\sin x.$$ So, the mistakes in your interpretation were that your guess should depend on $y$ and that your guess was a perfectly valid one, but you must consider complex roots here.

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  • $\begingroup$ Thanks for the answer, it is definitely helpful and nicely written. This is a dumb question, but I get where the $r^2$ comes from, but why do you keep the $r$ term? Shouldn't that correspond to the first derivative of a function? Something like $y'' + y' = 0$? Why is it not some constant term, like $a$? Wait, would it correspond to $1$ which creates the complex roots $+i$ and $-i$ which is where the solution comes from? Is that it? $\endgroup$ – Matthew Graham Jun 3 at 18:21
  • $\begingroup$ I made a typo, fixed! $\endgroup$ – cmk Jun 3 at 18:23
  • $\begingroup$ Ah, I see it now, yeah, makes sense now. Is there a way to solve for $c(y)$ and $b(y)$ or would we need initial conditions for those? I was just thinking of Euler's formula which looks similar to the solution here. $\endgroup$ – Matthew Graham Jun 3 at 18:26
  • $\begingroup$ You're correct, we need (boundary) condition to find our coefficients. As for Euler's formula, this is what allows us to get from our solution $c(y)e^{rx}$, with $r$ imaginary, to solutions of the form of $\sin x$ and $\cos x$. $\endgroup$ – cmk Jun 3 at 18:28
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Fix an arbitrary value for $y$, say $y_0$. Then let $f(x) =u(x,y_0)$. You can immediately see that $f''+f=0$. This is very common ODE and most people just know that $\sin$ and $\cos$ are solutions, but if you're unhappy with that you can get there with $\exp(\omega x)$ solutions and some algebra.

Now we know that $f(x) = A \sin(x) + B \cos(x)$ for an arbitrary $y_0$. The only thing that can change when we change $y_0$ are the values of constants $A,B$.

This finally leads to conclusion $u(x,y) = A(y)\sin(x) + B(y)\cos(x)$ and the $A,B$ now functions of $y$ are determined by boundary conditions.

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  • $\begingroup$ Thanks for the answer. I can see it now. $\endgroup$ – Matthew Graham Jun 3 at 18:29
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There are a few facts that are always treated as "obvious" in ODEs (because, presumably, you've been thinking about them non-stop since Calc I):

  • The derivative of a constant (with respect to any independent variable) is zero.
  • The derivative of a polynomial (with respect to its variable) reduces the degree by one.
  • The derivative of $f(x) = \mathrm{e}^{kx}$ with respect to $x$ is a constant multiple of $f$. (Precisely, $\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^{kx} = k \mathrm{e}^{kx}$.)
  • The second derivatives of $g(x) = \sin kx$ and $h(x) = \cos kx$ with respect to $x$ are constant multiples of $g$ and $h$, respectively. (Precisely, $\dfrac{\mathrm{d}^2}{\mathrm{d}x^2} \sin kx = -k^2 \sin x$, and similarly for cosine.) Additionally, odd order derivatives of $g$ and $h$ swap them.

So, when you see "the second derivative of $y$ is the negative of $y$", you should be thinking "sine and cosine" pretty much immediately.

The bullet point about sine and cosine can be rolled into the one about exponentials, so you could have gotten there with your characteristic equation method, but you need to recall what exponentiation does to complex numbers. In particular, you recall $\mathrm{e}^{\mathrm{i}x} = \cos x + \mathrm{i} \sin x$. So what would happen with your characteristic equation is, from $$ y'' + y = 0 $$ you have the characteristic equation $$ x^2 + 1 = 0 \text{.} $$ Then the characteristic roots are $\pm \mathrm{i}$, so the solutions (with arbitrary constants $c_1$ and $c_2$) are $c_1 \mathrm{e}^{\mathrm{i} x}$ and $c_2 \mathrm{e}^{-\mathrm{i} x}$. These are sines and cosines in disguise: \begin{align*} \mathrm{e}^{\mathrm{i} x} &= \cos x + \mathrm{i} \sin x \\ \mathrm{e}^{-\mathrm{i} x} &= \cos x - \mathrm{i} \sin x \text{.} \end{align*}

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  • $\begingroup$ Ahh, yes, thanks for the answer. I appreciate the thoroughness and how well written it is. Yeah, those facts are definitely at play here. It all makes sense now though. $\endgroup$ – Matthew Graham Jun 3 at 18:28

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