Here is the Problem: Solve $\frac{\partial T(x,t)}{\partial t} = \frac{\partial^{2} T(x,t)}{\partial x^{2}} +2xe^{-t} $ with the following boundary conditions $T(0,t)=10, and \frac{\partial T}{\partial x} (1,t)= 0 $ and initial condition $T(x,0)=10$

The attempt:

We need to make the PDE homogeneous and the Boundary Conditions homogeneous by using the eigenfunction expansion method. Assume $T(x,t) = v(x,t) - r(x,t)$ which $r(x,t)$ is found by the equilibrium temperature solution. What I have is $r(x,t) = T_e(x) = -\frac{1}{3}x^{3} e^{-t}+xe^{-t} +10$, which the PDE, BC's and IC becomes

$\frac{\partial v}{\partial t} = \frac{\partial^{2} v}{\partial x^{2}} - \frac{1}{3} x^{3} e^{-t} + xe^{-t}$

$v(0,t) = 0$

$\frac{\partial v}{\partial x} (1,t) = 0$

$v(x,0) = 1/3 x^{3} e^{-t} -xe^{-t} $

The Eigenvalues and eigenfunctions for the homogeneous PDE are

$\lambda_n= \big(\frac{(2n-1)\pi }{2}\big)^{2} $

$\phi_n(x) = \sin(\frac{(2n-1) \pi x}{2})$

From the Generalized Principle of Superposition, this would mean that

$v(x,t) = \sum_{n=1}^{\infty} h_n(t) \phi_n(x)$ which $h_n(t) = C_n e^{-\lambda_n t}$

Now I did solve for $h_n(0)$ but it is too hard to type it on this software.

So I do plug in the sub into the PDE and the way to find $q_n(t)$ is by getting an ODE from this. How do I do that in this scenario? Does this look okay so far?

Thank you very much. I really appreciate your thoughts and inputs.

  • $\begingroup$ The procedure in this post is not correct. $\endgroup$ – Juan Ospina Dec 10 '14 at 0:25
  • $\begingroup$ Okay I see where I went wrong. My equilibrium temperature needs to change. I need to make the equilibrium temperature to be a function which makes the boundary conditions homogeneous. I having trouble with that though. $\endgroup$ – user179766 Dec 10 '14 at 0:47
  • $\begingroup$ In this case there is not an "equilibrium temperature" because it will simply 10. You need a particular solution of the PDE with the form $T_{p}(x,t)=f(x)e^{-t}$ with the boundary conditions $T_{p}(0,t) = 0$ and $dT_{p}/dx =0$ at $x=1$. $\endgroup$ – Juan Ospina Dec 10 '14 at 0:51
  • $\begingroup$ Other option: define $v(x,t)=T(x,t)-10$ then you will have homogeneous boundary conditions: $v(0,t)=0$, $dv/dx =0$ at $x=1$; and the initial condition $v(x,0)= 0$. $\endgroup$ – Juan Ospina Dec 10 '14 at 0:56
  • $\begingroup$ Okay I do not understand what you got for your first case but in the second case, that is exactly what I got and what I was thinking. I will keep working on this. Thank you very much, I really appreciate your help and inputs. $\endgroup$ – user179766 Dec 10 '14 at 2:06

The problem statement is \begin{cases} \begin{align} u_t(x,t) &= u_{xx}(x,t) + 2x e^{-t} \\ u(0,t) &= 10 \\ u_x(1,t) &= 0 \\ u(x,0) &= 10. \end{align} \end{cases}

As it stands, separation of variables may not be used on this system since the boundary conditions are not homogeneous. However, both can be made homogeneous by the simple change $$ w(x,t) = u(x,t) - 10. $$ The system for $w$ is then \begin{cases} \begin{align} w_t(x,t) &= w_{xx}(x,t) + 2x e^{-t} \\ w(0,t) &= 0 \\ w_x(1,t) &= 0 \\ w(x,0) &= 0. \end{align} \end{cases} The condition that $w(x,0) = 0$ is not necessary, only homogeneous boundary conditions is necessary for this approach.

To find the solution, we need to find the eigenfunctions associated with the homogeneous heat equation $$ v_t - v_{xx} = 0, $$ with the same conditions as $w$. Separating the variables and equating to a common constant $-\lambda$, we get the system of ODEs \begin{cases} X'' + \lambda X = 0, &\text{with}& X(0) = 0, & X'(1) = 0 \\ T' + \lambda T = 0 \end{cases}

We know the operator on $X$ is symmetric with the given conditions, so the eigenvalues $\lambda$ are real, and any solutions $X$ identically zero are of no use. With this in mind, $\lambda = 0$ is automatically eliminated from consideration.

(i) $\lambda < 0$ \begin{align} X(x) &= c_1 e^{\sqrt{-\lambda} x} + c_2 e^{-\sqrt{ -\lambda} x} \\ &= \tilde{c}_1 \sinh \sqrt{-\lambda} x. \\ X'(x) &= \tilde{c}_1 \sqrt{-\lambda} \cosh \sqrt{-\lambda} x, \end{align} and $X'(1) = 0$ iff $\tilde{c}_1 = 0$ since $\lambda < 0$ and $\cosh y > 0$ for all $y$.

(ii) $\lambda > 0$ \begin{align} X(x) &= c_1 \cos \sqrt{\lambda} x + c_2 \sin \sqrt{\lambda} x \\ &= c_2 \sin \sqrt{\lambda}x. \\ X'(x) &= c_2 \sqrt{\lambda} \cos \sqrt{\lambda} x \\ X'(1) &= c_2 \sqrt{\lambda} \cos \sqrt{\lambda} \\ \implies X_n(x) &= c_n \sin \left(n\pi - \frac{\pi}{2} \right)x, \ \ \ n \in \mathbb{N^+}. \end{align} $X_n(x)$ are the eigenfunctions, and furthermore, $\phi_n(x)$ are the normalized eigenfunctions such that $\left<\phi_n(x), \phi_n(x)\right> = 1.$ $$ \phi_n(x) = \sqrt{2} \sin \left(n\pi - \frac{\pi}{2} \right)x. $$

Now returning to the problem for $w$, let \begin{align} w(x,t) &= \sum_{n = 1}^\infty w_n(t) \phi_n(x) & w_n(t) &= \int_0^1 w(x,t)\phi_n(x) \, dx \\ 2x e^{-t} &= \sum_{n = 1}^\infty g_n(t) \phi_n(x) & g_n(t) &= \int_0^1 2x e^{-t} \phi_n(x) \, dx. \end{align} Now multiplying the nonhomogeneous heat equation for $w$ by $\phi_n(x)$ and integrating over the domain of $x$, \begin{align} \int_0^1 w_t(x,t) \phi_n(x) \, dx &= \int_0^1 w_{xx}(x,t) \phi_n(x) \, dx + \int_0^1 2x e^{-t} \phi_n(x) \, dx \\ \frac{d}{dt} \int_0^1 w(x,t) \phi_n(x) \, dx &= \int_0^1 w(x,t) \phi_n{''}(x) \, dx + g_n(t), \end{align} where here the Lagrange-Green identity was used. We now have a well-known ordinary differential equation (5) for $w_n(t)$. \begin{align} w_n{'}(t) &= -\lambda_n w_n(t) + g_n(t) \\ \implies w_n(t) &= w_n(0) e^{-\lambda_n t} + e^{-\lambda_n t} \int_0^t g_n(s) e^{\lambda_n s} ds \\ &= e^{-\lambda_n t} \int_0^t g_n(s) e^{\lambda_n s} ds, \end{align} since $w(x,0) = 0$ and $\phi_n(x) \not\equiv 0.$ $$ w_n(t) = \frac{32 \sqrt{2}}{\pi^2} \frac{e^{-\frac{\pi^2}{4}(2n-1)^2 t} - e^{-t}}{(-1)^{-n}(2n-1)^2(\pi^2(2n-1)^2 - 4)}. $$ The solution $u$ is now known. $$ u(x,t) = 10 + \frac{64}{\pi^2}\sum_{n = 1}^\infty \frac{\big(e^{-\frac{\pi^2}{4}(2n-1)^2 t} - e^{-t}\big)\sin \left(n\pi - \frac{\pi}{2} \right)x}{(-1)^{-n}(2n-1)^2(\pi^2(2n-1)^2 - 4)}. $$

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I am obtaining the following solution

enter image description here

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  • $\begingroup$ Okay. What I have so far is that if I substitute in the sum into the PDE, I would get $\sum_{n=1}^{\infty} \Big[\frac{\partial h_n(t)}{\partial t} + \lambda_n h_n(t) \Big] \phi_n(x) = -\frac{1}{3} x^{3} e^{-t} + xe^{-t}$. For the $Q(x,t)$ term , we need it to equal to $sum_{n=1)^{\infty} q_n(t) \phi_n(x)$. Is this what you have so far? $\endgroup$ – user179766 Dec 9 '14 at 23:33
  • $\begingroup$ I am showing the final complete solution. Speaking with all true I do not understand your procedure. Please check it . Please fix some typos in your post. $\endgroup$ – Juan Ospina Dec 9 '14 at 23:41
  • $\begingroup$ Confirmed. You have an error, your $r(x,t)$ is not correct. Please note that it is not possible to obtain $r(x,t)$ making $dT/dt = 0$ because the resulting equation depends on $t$. $\endgroup$ – Juan Ospina Dec 9 '14 at 23:50

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