beautiful partial differential equation

[logistic_guy]

Solve.

\(\displaystyle \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2} + 2\beta \frac{\partial u}{\partial t}, \ \ \ \ \ 0 < x < \pi, \ \ \ 0 < \beta < 1, \ \ \ t > 0\)

\(\displaystyle u(0,t) = 0\)
\(\displaystyle u(\pi,t) = 0\)
\(\displaystyle u(x,0) = f(x)\)
\(\displaystyle u_t(x,0) = 0\)

 

[logistic_guy]

Solve.

We have already used separation of variables in a previous thread. And we have found the following solutions:

\(\displaystyle \textcolor{blue}{\lambda = 0}\)
\(\displaystyle X(x) = c_1x + c_2\)
\(\displaystyle T(t) = \frac{c_7}{\beta}e^{-2\beta t} + c_8\)

\(\displaystyle \textcolor{blue}{\lambda > 0}\)
\(\displaystyle X(x) = c_3\cos \mu x + c_4\sin \mu x\)
\(\displaystyle T(t) = c_9e^{-\beta t}\cos \alpha t + c_{10}e^{-\beta t}\sin \alpha t\)

\(\displaystyle \textcolor{blue}{\lambda < 0}\)
\(\displaystyle X(x) = c_5\cosh \mu x + c_6\sinh \mu x\)
\(\displaystyle T(t) = c_{11}e^{-\beta t}\cosh \alpha t + c_{12}e^{-\beta t}\sinh \alpha t\)

 

[logistic_guy]

\(\displaystyle X(x) = c_1x + c_2\)

We have two boundary conditions for the first equation.

\(\displaystyle X(0) = 0\)
\(\displaystyle X(\pi) = 0\)

Let us apply the first boundary condition.

\(\displaystyle 0 = c_2\)

Then we have:

\(\displaystyle X(x) = c_1x\)

Let us apply the second boundary condition.

\(\displaystyle 0 = c_1\pi\)

\(\displaystyle 0 = c_1\)

Then, when \(\displaystyle \lambda = 0\), we have the trivial solution:

\(\displaystyle X(x) = 0\)

This means that:

\(\displaystyle u(x,t) = X(x) \ T(t) = 0 \ T(t) = 0\)

And \(\displaystyle \textcolor{red}{\lambda = 0}\) is not an eigenvalue.

 

[logistic_guy]

\(\displaystyle X(x) = c_3\cos \mu x + c_4\sin \mu x\)

We have two boundary conditions for the second equation.

\(\displaystyle X(0) = 0\)
\(\displaystyle X(\pi) = 0\)

Let us apply the first boundary condition.

\(\displaystyle 0 = c_3\)

Then, we have:

\(\displaystyle X(x) = c_4\sin \mu x\)

Let us apply the second boundary condition.

\(\displaystyle 0 = c_4\sin \mu \pi\)

If \(\displaystyle c_4 \neq 0\), then \(\displaystyle \sin \mu \pi = 0\) and we have a solution.

\(\displaystyle \sin \mu \pi = 0\) when \(\displaystyle \mu \pi = \pi, 2\pi, 3\pi, \cdots\).

Or

\(\displaystyle \mu = 1, 2, 3, \cdots\)

Or

\(\displaystyle \mu = n, \ \ \ \ n = 1,2,3,\cdots\)

We know that \(\displaystyle \lambda = \mu^2\), then the eigenvalues when \(\displaystyle \textcolor{red}{\lambda > 0}\) are:

\(\displaystyle \lambda_n = \mu^2_n = n^2, \ \ \ \ n = 1,2,3,\cdots\)

And our non-trivial solution is:

\(\displaystyle X_n(x) = c_4\sin nx\)

Then

\(\displaystyle u_n(x,t) = X_n(x) \ T_n(t) = c_4\sin nx\bigg[c_9e^{-\beta t}\cos \alpha t + c_{10}e^{-\beta t}\sin \alpha t\bigg]\)

Or

\(\displaystyle u_n(x,t) = \sin nx\bigg[Ae^{-\beta t}\cos \alpha t + Be^{-\beta t}\sin \alpha t\bigg]\)


where \(\displaystyle \alpha = \sqrt{n^2 - \beta^2} \ \ \ \ \) and \(\displaystyle \ \ \ \ n = 1,2,3,\cdots\)

 

[logistic_guy]

\(\displaystyle X(x) = c_5\cosh \mu x + c_6\sinh \mu x\)

We have two boundary conditions for the third equation.

\(\displaystyle X(0) = 0\)
\(\displaystyle X(\pi) = 0\)

Apply the first condition.

\(\displaystyle 0 = c_5\), then we have:

\(\displaystyle X(x) = c_6\sinh \mu x\)

Apply the second condition.

\(\displaystyle 0 = c_6\sinh \mu \pi\)

\(\displaystyle \sinh \mu \pi \neq 0\), then \(\displaystyle c_6\) must be zero and we have the trivial solution

\(\displaystyle X(x) = 0\) when \(\displaystyle \textcolor{indigo}{\lambda < 0}\)

 

[logistic_guy]

So, the only non-trivial solution happens when \(\displaystyle \lambda > 0\).

\(\displaystyle u_n(x,t) = \sin nx\bigg[A_ne^{-\beta t}\cos \alpha_n t + B_ne^{-\beta t}\sin \alpha_n t\bigg]\)

This solution can be written with summation symbol.

\(\displaystyle u(x,t) = \sum_{n=1}^{\infty}\sin nx\bigg[A_ne^{-\beta t}\cos \alpha_n t + B_ne^{-\beta t}\sin \alpha_n t\bigg]\)

Now to find the coefficients \(\displaystyle A_n\) and \(\displaystyle B_n\), we need to apply the initial conditions.

Let us apply the first initial condition.

\(\displaystyle u(x,0) = f(x) = \sum_{n=1}^{\infty}\sin nx\bigg[A_n\bigg]\)

\(\displaystyle f(x)\sin mx = \sum_{n=1}^{\infty}\sin nx\sin mx\bigg[A_n\bigg]\)

\(\displaystyle \int_{0}^{\pi} f(x)\sin mx \ dx = \sum_{n=1}^{\infty}A_n \int_{0}^{\pi}\sin nx\sin mx \ dx\)

The integral on the right will not be zero only when \(\displaystyle n = m\), then

\(\displaystyle \int_{0}^{\pi} f(x)\sin mx \ dx = A_m \int_{0}^{\pi}\sin mx\sin mx \ dx\)

\(\displaystyle \int_{0}^{\pi} f(x)\sin mx \ dx = \frac{\pi}{2}A_m\)

\(\displaystyle A_m = \frac{2}{\pi}\int_{0}^{\pi} f(x)\sin mx \ dx\)

Or

\(\displaystyle A_n = \frac{2}{\pi}\int_{0}^{\pi} f(x)\sin nx \ dx\)

 

[logistic_guy]

Apply the second initial condition.

\(\displaystyle u_t(x,0) = 0 = \sum_{n=1}^{\infty}\sin nx\bigg[-\beta A_n + \alpha_nB_n\bigg]\)


\(\displaystyle B_n = \frac{\beta A_n}{\alpha_n}\)

Then, the final solution is:

\(\displaystyle u(x,t) = \textcolor{blue}{\sum_{n=1}^{\infty}A_n\sin nx\bigg[e^{-\beta t}\cos \alpha_n t + \frac{\beta}{\alpha_n}e^{-\beta t}\sin \alpha_n t\bigg]}\)

where

\(\displaystyle A_n = \textcolor{red}{\frac{2}{\pi}\int_{0}^{\pi} f(x)\sin nx \ dx}\)
And
\(\displaystyle \alpha_n = \textcolor{green}{\sqrt{n^2 - \beta^2}}\)