I have to solve the nonhomogeneous classic problem

$$\left(P_{1}\right)\;\,\left\{ \begin{aligned} u_{t}\;-\; \Delta u\; &= \;f& &\textrm{on}\;\;\; \Omega\times\left(0,\,\infty\right) &\\ u\; &=\; 0 & &\textrm{on}\;\;\; \partial\Omega\times\left(0,\,\infty\right)&\\ u\; &= \;g & &\textrm{on}\;\;\; \Omega\times\left\{t\;=\;0\right\}& \end{aligned}\right.\\$$

where $\,\Omega = B(x_0,R).\,$ We assume that $\,f=f\left(x,t\right)\;$ and $\,g=g\left(x,t\right)\,$ are smooth enough.

My attempt

Following Lawrence C. Evans guidelines (page 45), I first tried to solve the homogeneous equation looking for a particular solution:


Once I assumed the solution to be radial, $\,v\left(y\right)=w\left(\left|y\right|\right)=w\left(r\right)\;$ I solved the ODE:

$$\frac{w}{2}r^n\,+\,w'r^{n-1} = A$$ $$\displaystyle w\left(r\right)=Be^{-\frac{r^2}{4}}\,+\,Ae^{-\frac{r^2}{4}}\cdot\int_{0}^{r}\frac{e^{\frac{s^2}{4}}}{s^{n-1}}ds$$

I got stuck just here. I cannot assume $\,A=0\,$ and obtain the fundamental solution $\,\Phi\,$ because it doesn't verify the BC's.

How can I go on? Any kind of help would be useful. Thanks in advance!

  • $\begingroup$ There seems to be a typo in the form of the ODE above. On pg. 46 the ODE is given as rn−1w′+12rnw=A. Evans goes on to state that as r→∞, w,w′→0 so that A=0. This can be seen from the form of the solution to the ODE. If A≠0 then w,w′ cannot tend to zero as r grows. $\endgroup$ – asd Sep 23 '18 at 13:39
  • $\begingroup$ Are you sure Problem ($P$) is your goal? It is a nonlinear PDE. Or is Problem ($P_1$) which is a linear PDE what you are really after for this question? $\endgroup$ – Hans Sep 28 '18 at 4:52
  • $\begingroup$ The whole problem is $\,(P),\,$ so that 's my goal, but the question concerns $\,(P_1)\,$ in this case. $\endgroup$ – CarlIO Sep 28 '18 at 7:41
  • $\begingroup$ It looks like you are unfamiliar with how this site operates. If you want to direct your comment to a person who is not the originator of the question, write @(name) which in this case is @Hans. The site will inform that person of your comment. Back to the question, to confirm, you only need a solution to ($P_1$), correct? $\endgroup$ – Hans Sep 28 '18 at 7:59
  • $\begingroup$ @user313212: Have you seen my answer below? $\endgroup$ – Hans Sep 29 '18 at 18:39

The spherical symmetric solution in your question provides a basis for the PDE in the infinite space $R^n$. Your boundary value problem in a sphere of a finite radius, you need to expand the solution in the series of products of the spherical Bessel function of $r$ and $n$-dimensional spherical harmonics.

The Laplace operator in $3$-dimensional spherical coordinate is

$$0=-\frac{\partial u}{\partial t}+\Delta u =\frac{\partial u}{\partial t}-\frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial u}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta} \frac{\partial^2 u}{\partial \varphi^2} \tag1$$ Suppose $u(r,\theta,\phi)=T(t)R(r)\Theta(\theta)\Phi(\varphi)$. Substitute it into Equation (1) and divide it by $f$, we have

$$0=-\frac1T\frac{\partial T}{\partial t}+\frac{1}{r^2R}\frac{\partial}{\partial r} \left(r^2 \frac{\partial R}{\partial r}\right) + \frac{1}{r^2\Theta \sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta\Phi} \frac{\partial^2 \Phi}{\partial \varphi^2} \tag2$$ We see that the first term is independent of $(r,\theta,\varphi)$ while the remaining terms are independent of $t$. The first term must then be a constant. Let us call it $-\lambda$. We have $$T'(t)=-\lambda T(t) \tag3$$ Multiplying (2) by $r^2$, we see that the third and fourth terms are independent of $r$ while the second term depends only on $r$. So we introduce another constant $-\mu$. $$\frac{1}{r^2R}\frac{d}{d r} \left(r^2 \frac{d R}{d r}\right)+\Big(\lambda-\frac\mu {r^2}\Big)R=0. \tag4$$ Apply the same trick again, we obtain $$\Phi''(\varphi)+\nu\Phi(\varphi)=0 \tag5$$ for some constant $\nu$ and for $\Theta(\theta)$ $$\Theta''+\cot\theta\,\Phi'+(\mu-\nu(\csc\theta)^2)\Theta=0 \tag6$$ for some constant $\mu$.

We will solve Equation ($4$) for with boundary condition $R(r=1)=0$ where we set the radius of the sphere at $1$.


Jumping to the conclusion first, let $$q_{\lambda,k,m}(r,\theta,\varphi):=j_k(r\sqrt\lambda)Y_{k,m}(\theta,\varphi),\ \lambda>0,$$ where $j_k$ is the spherical Bessel function with $\sqrt\lambda$ being a positive root of it, $Y_{k,m}$ is the spherical harmonic function. Expand $$g(r,\theta,\varphi)=\sum_{\lambda,|m|\le k}a_{\lambda,k,m}q_{\lambda,k,m}(r,\theta,\varphi)$$ to solve for the coefficient $a_{\lambda,k,m}$. This is facilitated by the property that the set $\{q_{\lambda,k,m}(r,\theta,\varphi)\}_{\lambda,k,m}$ is orthogonal in $L^2$.

In the case of the homogeneous PDE where $f=0$, $$u(r,\theta,\varphi,t) = \sum_{\lambda,|m|\le k}a_{\lambda,k,m}q_{\lambda,k,m}(r,\theta,\varphi)e^{-\lambda t}.$$ By Duhamel's principle, the inhomogeneous equation can be viewed as linear superposition of the homogeneous problem and $$u(r,\theta,\varphi,t) = \sum_{\lambda,|m|\le k}a_{\lambda,k,m}q_{\lambda,k,m}(r,\theta,\varphi,t)e^{-\lambda t}+\int_0^t ds \sum_{\lambda,|m|\le k}a_{\lambda,k,m}(s)q_{\lambda,k,m}(r,\theta,\varphi)e^{-\lambda (t-s)},$$ where $$f(r,\theta,\varphi,t)=\sum_{\lambda,|m|\le k}a_{\lambda,k,m}(t)q_{\lambda,k,m}(r,\theta,\varphi).$$

This is how one builds up the solution in a sphere. More details can be found in e.g. ON GREEN'S FUNCTIONS IN THE THEORY OF HEAT CONDUCTION IN SPHERICAL COORDINATES by Arnold N. Lowan and Green’s Function for the Heat Equation by Abdelgabar Adam Hassan.

Now if $g(x)$ is spherically symmetric thus is only a radial function $g(r)$, then the solution is simpler.

  • $\begingroup$ I'm new so I didn't know about the "@". Regarding your answer, honestly it doesn't help me much, because the solution was not supposed to be a series. Besides, the method should not be separation of variables. However you are the only who has thought of a proper response so... I dind't set the bounty by myself so I can't award it now, but if I accept your answer and it scores at least 2 votes you will get it anyway. $\endgroup$ – CarlIO Sep 29 '18 at 10:44
  • $\begingroup$ @CarlIO: OK. By "the solution was not supposed to be a series" do you mean any series solution is wrong or you do not like a series solution but an integral one? If you mean the former, can you show a proof that any series solution must be wrong? If you mean the latter, is this a homework problem and are you told by the professor that he wants an integral form solution because he is sure there is one? By "the method should not be separation of variables" do you mean the seperation of variables method is wrong and you can prove it is? $\endgroup$ – Hans Sep 29 '18 at 17:04
  • $\begingroup$ @CarlIO: The seperation of variable method only serves the purpose to obtain the terms used in the series solution. It can be proved that the series spans the set of continuous function compactly supported on the unit ball. So long as that is true, the seperation of variable method is correct. It is another matter, if again by "should not be separation of variable", you mean it is a homework and your professor explicitly forbid the use of seperation of variable and thus the series solution. Is this the case? $\endgroup$ – Hans Sep 29 '18 at 17:11
  • $\begingroup$ @CarlIO: I added some references on the heat equation Green's function both of which are in terms of the infinite series of the Bessel functions and the spherical harmonics. There is no avoiding it. With regard to the integral form solution, even in the case of spatial dimension 1, if the Gaussian integral is to be used, it has to be terms of an infinite series rather than a single integral, by the image method. In higher dimension, it is not clear the image method still works. $\endgroup$ – Hans Sep 29 '18 at 18:42
  • $\begingroup$ No, no way. I didn't mean series solution nor separation of variables method is wrong. But I was looking for an integral form solution. I'm sure it exists because I've been told to follow L. Evans guidelines. $\endgroup$ – CarlIO Oct 1 '18 at 14:10

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