I am learning this method for solving PDSs by means of power series. Since I am studying it with lecture notes and I can't find any other book that describes this method, I am going to summarize it below then I ask my question.

A general PDE can be written as: $$F(u,u_{yy},u_{xx},u_{xy},u_{x},u_{y},x,y)=0$$ where $u=u(x,y)$. So, if we able able to solve this equation for $u_{yy}$ we can write $$u_{yy}=f(u_{xx},u_{xy},u_{x},u_{y},x,y) $$ Then, we are given the following side conditions: $u(x,0)$ and $u_{y}(x,0)$, such that we are able to find the other derivatives at $(x,0)$. Then, we can use a power series around $y=0$ to find $u(x,y)$: $$u(x,y)=u(x,0)+yu_{y}(x,0)+\frac{y^2}{2!}u_{yy}(x,0)+\frac{y^3}{3!}u_{yyy}(x,0)+... $$

For example, suppose we have the PDE $u_{xx}-u_{yy}=0$ with the conditions $u(x,0)=x^2+x$ and $u_{y}(x,0)=2x+1$. It follows that $u_{x}(x,0)=2x+1$ and $u_{xx}(x,0)=2=u_{yy}(x,0)$ (this last equality is true because of the PDE). Then if we differentiate $u_{yy}(x,0)$ with respect to $y$ we get $u_{yyy}(x,0)=u_{yyyy}(x,0)=...=0$. Then, the solution as a power series is $$u(x,y)=x^2+x+y(2x+1)+y^2 $$ which satisfy the PDE and the side conditons.

Now, my question is about the following problem: $$u_{t}=u_{xx}$$ with the side conditions $u(0,t)=t^2$ and $u_{x}(0,t)=0$. In this case, I know the function along the t-axis, so I should try to apply this method by writing a power series around $x=0$. To do this, I need to find the derivatives of $u$ with respect to $x$. Since $u(0,t)=t^2$, we have $u_{t}(0,t)=2t$=$u_{xx}(0,t)$ (again, this is true because of the PDE). If I derive $u_{xx}(0,t)$ with respect to $x$, I can get higher derivatives to write the power series. But it turns out the $u_{xx}(0,t)=u_{xxx}(0,t)=...=0$ because $u_{xx}(0,t)$ only depends on $t$. Thus, we can write: $$u(x,t)=u(0,t)+xu_{x}(0,t)+\frac{x^2}{2!}u_{xx}(0,t)+\frac{x^3}{3!}u_{xxx}(0,t)+...=t^2+x^2t $$But this solution does not satisfy the PDE. What is wrong with this reasoning?

  • 1
    $\begingroup$ I used your example for demonstrating SymPy PDE solving over here. $\endgroup$
    – Foad
    Aug 2, 2018 at 11:54
  • 1
    $\begingroup$ @Foad very interesting. Thanks for bringing this up. $\endgroup$
    – Thiago
    Oct 24, 2018 at 21:21

1 Answer 1


The mistake is that $u_{xxxx}(0,t) \neq 0$. We have

$$\begin{align} u_{xxxx}(0,t) &= (u_{xx})_{xx}(0,t)\\ &= (u_t)_{xx}(0,t)\\ &= (u_{xx})_t(0,t)\\ &= u_{tt}(0,t)\\ &= 2. \end{align}$$

And therefore you get

$$u(x,t) = u(0,t) + \frac{x^2}{2}u_{xx}(0,t) + \frac{x^4}{4!}u_{xxxx}(0,t) = t^2 + tx^2 + \frac{1}{12}x^4,$$

which satisfies $u_t = u_{xx}$.

(Since $u_{ttt}(0,t) \equiv 0$, we need not convert higher derivatives wrt $x$ to derivatives wrt $t$.)


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