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Let $u_t=a^2u_{xx}+hu_x$ with the following conditions: $u(0,t)=0=u(L,t)$ and $u(0,t)=f(x)$ such that $x \in (0,L)$ and $t>0$.

I'm using the method of separation of variables. Let $u(x,t)=X(x)T(t)$. Then:

$$XT'=a^2X''T+hX'T$$

Shifting stuff around we get:

$$a^2X''+hX'-\lambda X=0\text{ and }T'-T\lambda=0.$$

What's throwing me off here is trying to apply separation of variables with the additional $u_x$ term and then moving on to finding eigenvalues.

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  • $\begingroup$ Making the change of variables $u(x,t) = v(x+ht, t)$ transforms your PDE to $$v_{t} = a^{2} v_{xx}$$ Now transform your data and solve in $v$, or solve in $v$ and back solve for $u$ then apply data. $\endgroup$ – mattos Dec 4 '16 at 4:29
  • $\begingroup$ I'm familiar with that method, but I was hoping to try to solve this problem directly using separation of variables. $\endgroup$ – emka Dec 4 '16 at 4:39
  • $\begingroup$ Then you'll need to solve the ODE $a^{2}X'' + hX' - \lambda X = 0$, with the BCs $X(0) = X(L) = 0$.. I'm not entirely sure what your question is? $\endgroup$ – mattos Dec 4 '16 at 4:49
  • $\begingroup$ Well, specifically $\lambda$ so I can find the eigenvalues and eigenfunctions. $\endgroup$ – emka Dec 4 '16 at 5:01
  • $\begingroup$ You get $\lambda$ by solving the ODE in $X$ with associated BCs.. And $\lambda$ is your eigenvalue. $\endgroup$ – mattos Dec 4 '16 at 5:31
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How to deal if there is $u_{x}$

We have : $$a^2X''+hX'-\lambda X=0 $$ let $ X=e^{mx} $ which implies : $$a^2m^2+hm-\lambda=0 \Rightarrow m=\frac{-h\pm\sqrt{h^2+4a^2 \lambda}}{2a^2}$$ You can see that the only case that gives non trivial solution is when $h^2+4a^2\lambda<0$

Now let $-\gamma^2=h^2+4a^2\lambda$ which implies : $$X=e^{\frac{-hx}{2a^2}}[c_{1}cos(\frac{\gamma x}{2a^2})+c_{2}sin(\frac{\gamma x}{2a^2})] $$ since $X(0)=X(L)=0$ we have $c_{1}=0$ and

$$ sin(\frac{\gamma L}{2a^2})=0 \Rightarrow \gamma =\frac{2a^2n\pi}{L} \Rightarrow -\gamma^2=-\frac{4a^2n^2\pi^2}{L^2}\Rightarrow \lambda_{n}=-(\frac{h^2}{4a^2}+\frac{n^2\pi^2}{L^2})$$ Now we have : $$X_{n}(x)T_n(t)=e^{\frac{-hx}{2a^2}}sin(\frac{n\pi x}{L})e^{-(\frac{h^2}{4a^2}+\frac{n^2\pi^2}{L^2})t} $$ Finally we have : $$ u(x,t)=\sum_{n=1}^{\infty}c_n e^{\frac{-hx}{2a^2}}sin(\frac{n\pi x}{L})e^{-(\frac{h^2}{4a^2}+\frac{n^2\pi^2}{L^2})t}$$ and from $u(x,0) =f(x)$ you can get $c_{n}$.

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If you want to solve directly, and have orthogonality of eigenfunctions in $x$, you'll need to write the equation for $X$ in Sturm-Liouville form. An integrating factor will help: $$ e^{hx/a^2}a^2X''+e^{hx/a^2}hX'=\lambda e^{hx/a^2}X \\ \frac{d}{dx}\left(e^{hx/a^2}a^2\frac{dX}{dx}\right)=\lambda e^{hx/a^2}X $$ The reason for doing this is to discover an orthogonality relation between eigenfunctions in $X$, which you'll need in order to expand in a series. From this you see that, if $X_1,X_2$ are solutions of the equation that vanish at both endpoints of the interval $[0,L]$, $$ \int_{0}^{L}X_1(x)X_2(x)e^{hx/a^2}dx=0. $$ This is because of the weight function multiplying $\lambda$. Then you can solve the original equation directly, and use this orthogonality relation to isolate the Fourier coefficients in the expansion with respect to the eigenfunctions of the equation in $X$. The equation $$ X''+\frac{h}{a^2}X'-\frac{\lambda}{a^2} X = 0 $$ is Euler's equation, which has known solutions. So you solve for a solution $X$ such that $X(0)=0$, $X'(0)=1$, and then you determine the set of $\lambda$ for which that solution satisfies $X(L)=0$, and you'll have the eigenvalues, along with the eigenfunctions. Then, once you have the eigenfunctions $$ X_1, X_2, X_3, \cdots $$ with corresponding eigenvalues $\lambda_k$, the general solution is $$ u(x,t) = \sum_{k=1}^{\infty}C_kX_k(x)e^{\lambda_k t}, $$ and you'll be able to solve $u(x,0)=f(x)$ using orthogonality with respect to the weight $e^{hx/a^2}$: $$ f(x) = \sum_{k=1}^{\infty}C_k X_k(x) \\ \int_{0}^{L}f(x)X_k(x)e^{hx/a^2}dx = C_k\int_{0}^{L}X_k(x)^2e^{hx/a^2}dx \\ C_k = \frac{\int_{0}^{L}f(x)X_k(x)e^{hx/a^2}dx}{\int_{0}^{L}X_k(x)^2e^{hx/a^2}dx} $$ Note: Because of the way you chose $\lambda$ instead of $-\lambda$, the eigenvalues will be mostly negative. You know this from the $T(t)=e^{\lambda t}$ solutions.

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