Consider the differential equation

$$u_{t t} = c^2 u_{x x} - k u, \qquad k, u > 0$$

together with the Cauchy data

$$u(x, 0) = e^x, u_t(x, 0) = 0$$

I now want to find a solution $u: \mathbb R^2 \to \mathbb R$ for this equation which is real-analytic with respect to $t$.

Now I was given the tip that I might construct the desired solution $u$ via Taylor expansion in a point $(x, 0)$ (with respect to $t$) and see what comes out for the first few Taylor coefficients, and then prove a formula for all of them per induction. (And afterwards of course check that it actually solves the equation, but I think that will be the easier matter.)

Therefore, I started this way: let's say we write the desired solution $u$ in its Taylor expansion $u(x, t) = \sum_{n=0}^\infty \frac{u^{(n)}}{n!} t^n$ around $(x, 0)$. Then because of the Cauchy data, we would immediately have $u^{(0)} = u(x, 0) = e^x$ and $u^{(1)} = u_t(x, 0) = 0$. Putting that into our equation, we get $u^{(2)} = u_{t t}(x, 0) = c^2 e^x - k e^x = (c^2 - k) e^x$ I think. So we got the first few coefficients already.

But that's where my wisdom ends... how would I be able to get the rest of the coefficients? I don't even know how I can get the third one.

In other words, if I have the coefficients $u^{(0)}, \dots, u^{(n)}$ of the Taylor series expansion of the desired solution, how can I deduce the $n+1$-the one using only the given PDE and Cauchy data? As for a general formula for them, it would seem plausible that each $u^{(n)}$ is a product of $e^x$ and some constant term, given what the first few look like, but I'm not sure how I can actually prove that.

  • $\begingroup$ Is there any reason why you wouldn't just find the solution by means of a Fourier Transform (or Green's function, etc) and deduce the real analiticity of the solution? Because I'm pretty sure the resulting solution is going to be pretty smooth. At least then you'd know what you would expect such a taylor series expansion to look like? $\endgroup$
    – DaveNine
    Nov 8 '17 at 22:44
  • $\begingroup$ @DaveNine No particular reason, no, although my knowledge about that is a bit rusted I guess so I didn't really see that this would be possible aswell. Could you maybe give more details to how I would go about solving this equation with means of Fourier transformation? $\endgroup$
    – moran
    Nov 8 '17 at 22:56
  • $\begingroup$ Actually, come to think of it, you're going to have problems with the fourier transform method due to the blow-upness of $e^x$, so that method won't work. I've a feeling it'll take more heavy lifting to prove existence of a solution. $\endgroup$
    – DaveNine
    Nov 9 '17 at 0:04

Let $p=\dfrac{\sqrt kx}{c}$ ,

Then $u_{tt}=ku_{pp}-ku$ with $u(p,0)=e^\frac{cp}{\sqrt k}$ and $u_t(p,0)=0$

Let $q=\sqrt kt$ ,

Then $u_{qq}=u_{pp}-u$ with $u(p,0)=e^\frac{cp}{\sqrt k}$ and $u_q(p,0)=0$

Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,

Let $u(p,q)=\sum\limits_{m=0}^\infty\dfrac{q^m}{m!}\dfrac{\partial^mu(p,0)}{\partial q^m}$ ,

Then $u(p,q)=\sum\limits_{m=0}^\infty\dfrac{q^{2m}}{(2m)!}\dfrac{\partial^{2m}u(p,0)}{\partial q^{2m}}+\sum\limits_{m=0}^\infty\dfrac{q^{2m+1}}{(2m+1)!}\dfrac{\partial^{2m+1}u(p,0)}{\partial q^{2m+1}}$



Similarly, $\dfrac{\partial^{2m}u}{\partial q^{2m}}=\sum\limits_{n=0}^m(-1)^{m-n}C_n^m\dfrac{\partial^{2n}u}{\partial p^{2n}}$




Similarly, $\dfrac{\partial^{2m+1}u}{\partial q^{2m+1}}=\sum\limits_{n=0}^m(-1)^{m-n}C_n^m\dfrac{\partial^{2n+1}u}{\partial p^{2n}\partial q}$

$\therefore u(p,q)=e^\frac{cp}{\sqrt k}\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m-n}C_n^mc^{2n}q^{2m}}{k^n(2m)!}$

Hence $u(x,t)=e^x\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m-n}C_n^mc^{2n}k^{m-n}t^{2m}}{(2m)!}$


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