How can I solve this:

$u_t - x u_x -x^2 u_{xx} = \ln{x}$

$u(x,0) = \sin ((\pi/2) \ln{x})$

$u(1,t) = 0 \quad u_x(e,t)=0$

What I have so far:

Since we have homogenous BC consider no forcing term to get eigenfunctions:

$u_t = x u_x + x^2u_{xx}$

Let $u(x,t) = X(x)T(t)$

$(1) \dot{T}+\lambda^2T=0$


Solving (2) we get:

$X(x)=Acos(\lambda \ln{x}) + B sin(\lambda ln{x})$

Plugging in BC we find:

$X_n(x)=\sin{(\lambda_n \ln{x})}, \lambda_n = \frac{(2n+1)\pi}{2}, n = 0,1,2,3,... \leftarrow$ this is our eigenfunction

Now I want to expand forcing term in terms of eigenfunction:

$\ln{x} = \sum_{n+1} ^\infty S_n(t) \sin{(\lambda_n \ln{x})}$ where $S_n(t) = \frac{2}{e-1}\int_1 ^e \ln{x} \sin{(\lambda_n \ln{x})} dx$

Ya so I am not sure if the last line is correct. If I could expand ln(x) then I could guess an eigenvalue expansion as the solution for u as well.

  • 1
    $\begingroup$ What have you learnt in class so far? That way we can find a suitable method for you to use. $\endgroup$ Commented Mar 8, 2015 at 9:14
  • $\begingroup$ Eigenvalue expansion or separation of variable. I prefer sep of var but can't seem to make it work with the usual u=X(x)T(t) guess. $\endgroup$
    – sci-guy
    Commented Mar 8, 2015 at 9:19
  • $\begingroup$ It is definitely a problem related to eigenfunction expansion method. Looks easy except $u(x, 0)$. Would you please confirm that. $\endgroup$
    – incognito
    Commented Mar 8, 2015 at 9:58
  • 1
    $\begingroup$ @renegade05 Separating variables should work, you should get an Euler Equation of the form $x X'' + x X' +\lambda X = 0$ (I set my separation constant to be $-\lambda$) with the conditions that $X(1) = X'(e) = 0$. If you want to see how to solve the Euler Equation, see here. $\endgroup$ Commented Mar 8, 2015 at 11:36
  • 1
    $\begingroup$ Your eigenfunction seems incorrect it should be $\sin{\left(\frac{(2n-1)\pi}{2}\ln{x}\right)}$. Please check it once. $\endgroup$
    – incognito
    Commented Mar 8, 2015 at 12:02

2 Answers 2


Well I am escaping the part you have already understood. By the eigenfunction expansion method, if we assume $$u(x,t) = \sum_{n = 1}^{\infty} T_n(t) X_n(x),$$ We substitute the above expansion in the original PDE i.e. $$\frac{1}{x}u_t - (xu_x)_x = \frac{1}{x} \ln{x},$$ Here please note that we don't change the original form of the equation. Now we make use of following: $$ \int\limits_{1}^{e} \frac{1}{x}\sin{(m \ln{x})}\sin{(n \ln{x})}dx = \frac{1}{2},$$ for $m = n$ and will be zero otherwise. Moreover as you have mentioned the $S_n(t)$ will be given by: $$S_n(t) = \frac{2}{1-e} \int\limits_{1}\frac{1}{x} \ln{x} \sin{((\frac{2n-1)\pi}{2} \ln{x})},$$ $$\implies S_n(t) = -\frac{8 (-1)^n}{(1-e)(\pi -2 \pi n)^2} = s_n (say).$$ The differential equations for $T_n(t)$ are: $$\frac{d T_n}{dt} + \lambda^2 T_n = s_n.$$ Where $\lambda$ represents the eigenvalues. The initial value $u(x,0)$ suggest that we should have following initial conditions for the ODEs for $T_n(t)$: $$T_1(0) = 1, T_2(0) = T_3(0) = \cdots = 0.$$ Form here onwards you can handle I guess. If you still have any issues with it please let me know.

  • $\begingroup$ I am sure I can. Did you end up getting a final answer so I can compare? THANKS! @incognito $\endgroup$
    – sci-guy
    Commented Mar 9, 2015 at 3:14
  • $\begingroup$ No man !! I haven't. I just typed what I had in mind. Be careful. Don't just copy !! ;-) $\endgroup$
    – incognito
    Commented Mar 9, 2015 at 3:18
  • $\begingroup$ @renegade05 Did the solution I provide work ?? $\endgroup$
    – incognito
    Commented Mar 9, 2015 at 5:17
  • $\begingroup$ I have not tried yet ... Am a tad preoccupied at the moment. Plus just looking at it, I am not sure why you didn't change the original form of the equation and how we would go about getting the final u(x,t) answer... $\endgroup$
    – sci-guy
    Commented Mar 9, 2015 at 5:22
  • $\begingroup$ Without that $\frac{1}{x}$ the orthogonality of $\sin{(m \ln{x})}$ won't work !!! $\endgroup$
    – incognito
    Commented Mar 9, 2015 at 5:27


Let $u(x,t)=\sum\limits_{n=0}^\infty A(n,t)\sin\dfrac{(2n+1)\pi\ln x}{2}$ so that it automatically satisfies $u(1,t)=0$ and $u_x(e,t)=0$ ,

Then $\sum\limits_{n=0}^\infty\dfrac{\partial A(n,t)}{\partial t}\sin\dfrac{(2n+1)\pi\ln x}{2}-x\sum\limits_{n=0}^\infty\dfrac{(2n+1)\pi A(n,t)}{2x}\cos\dfrac{(2n+1)\pi\ln x}{2}-x^2\left(-\sum\limits_{n=0}^\infty\dfrac{(2n+1)^2\pi^2A(n,t)}{4x^2}\sin\dfrac{(2n+1)\pi\ln x}{2}-\sum\limits_{n=0}^\infty\dfrac{(2n+1)\pi A(n,t)}{2x^2}\cos\dfrac{(2n+1)\pi\ln x}{2}\right)=\ln x$

$\sum\limits_{n=0}^\infty\dfrac{\partial A(n,t)}{\partial t}\sin\dfrac{(2n+1)\pi\ln x}{2}-\sum\limits_{n=0}^\infty\dfrac{(2n+1)\pi A(n,t)}{2}\cos\dfrac{(2n+1)\pi\ln x}{2}+\sum\limits_{n=0}^\infty\dfrac{(2n+1)^2\pi^2A(n,t)}{4}\sin\dfrac{(2n+1)\pi\ln x}{2}+\sum\limits_{n=0}^\infty\dfrac{(2n+1)\pi A(n,t)}{2}\cos\dfrac{(2n+1)\pi\ln x}{2}=\ln x$

$\sum\limits_{n=0}^\infty\left(\dfrac{\partial A(n,t)}{\partial t}+\dfrac{(2n+1)^2\pi^2A(n,t)}{4}\right)\sin\dfrac{(2n+1)\pi\ln x}{2}=\ln x$

$\sum\limits_{n=0}^\infty\left(\dfrac{\partial A(n,t)}{\partial t}+\dfrac{(2n+1)^2\pi^2A(n,t)}{4}\right)\sin\dfrac{(2n+1)\pi y}{2}=y$


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