Homework 2 AM212
Haberman 2.3.8
Problem definition
The heat equation with homogenous boundary conditions:
\[\begin{cases} u_t = ku_{xx} - \alpha u \\ u(0, t) = 0 \\ u(L, t) = 0 \end{cases} \tag{1}\]
Part a
What are the possible equilibrium temperature distributions if \(\alpha > 0\)?
Solution
An equilibrium temperature distribution is such that the time derivative of temperature is zero (\(u_t = 0\)). Let us solve the homogenous BVP:
\[ku_{xx} - \alpha u = 0, u(0) = 0, u(L) = 0 \tag{2}\]
Before proceeding it is important to note that \(k\) corresponds to the \(\textbf{thermal diffusivity}\) of the material and is made of \(K_0\) the \(\textbf{thermal conductivity}\), \(c\) \(\textbf{specific heat capacity}\), and \(\rho\) \(\textbf{mass density}\) such that
\[k = \frac{K_0}{c\rho}\] where \(k > 0\)
Let \(u(x) = e^{rx}\), the characteristic polynomial becomes:
\(kr^2 - \alpha = 0\)
where the roots are
\(r = \pm \sqrt{\frac{\alpha}{k}}\)
the solution of positive real roots:
\[u(x) = c_1e^{\sqrt{\frac{\alpha}{k}}x} + c_2e^{-{\sqrt{\frac{\alpha}{k}}x}}\]
plugging in the boundary conditions:
\[\begin{align} &\bigg[\begin{matrix} 1 & 1 \\ e^{\sqrt{\frac{\alpha}{k}}L} & e^{-\sqrt{\frac{\alpha}{k}}L} \end{matrix}\bigg] \bigg[\begin{matrix} c_1 \\ c_2 \end{matrix}\bigg] = \bigg[\begin{matrix} 0 \\ 0 \end{matrix}\bigg] \\ &\det{\bigg(\bigg[\begin{matrix} 1 & 1 \\ e^{\sqrt{\frac{\alpha}{k}}L} & e^{-\sqrt{\frac{\alpha}{k}}L} \end{matrix}\bigg]\bigg)} = e^{-\sqrt{\frac{\alpha}{k}}L} - e^{\sqrt{\frac{\alpha}{k}}L} = -2\sinh{(\sqrt{\frac{\alpha}{k}}L)} \end{align}\]
\(c_1 = c_2 = 0\)
This leaves us with only a trivial solution \(u(x) = 0\).
Thus if \(\alpha\) is strictly positive the only possible equilibrium temperature distribution is zero everywhere.
Part b
Solve the time dependent problem with IC \(u(x, 0) = f(x)\) if \(\alpha > 0\) and analyze temperature as \(t \rightarrow \infty\) and compare with part a.
The IBVP:
\[\begin{cases} u_t = ku_{xx} - \alpha u \\ u(x, 0) = f(x) \\ u(0, t) = 0 \\ u(L, t) = 0 \end{cases} \tag{3}\]
Let
\[\begin{align} u(x, t) &= X(x)T(t) \\ u_t &= XT' \\ u_{xx} &= X''T \end{align}\]
Plugging into Equation 3
\[\begin{align} XT' &= kX''T - \alpha XT \\ \frac{T'}{kT} - \frac{\alpha}{k} &= \frac{X''}{X} = \lambda \end{align}\]
The solution of the eigen-value problem \(X''(x) = \lambda X(x), X(0) = X(L) = 0\) is
\[\begin{align} X_n(x) = \sin{(\frac{n\pi x}{L})} && \lambda_n = -\big(\frac{n\pi}{L}\big)^2 && n = 1, 2, 3, ... \end{align}\]
Solving for the IVP:
\[T' = \lambda_n kT + \alpha T\]
Let \(T_n(t) = e^{rt}\)
the characteristic polynomial:
\(r - \lambda_n k - \alpha = 0\)
with a single root:
\(r = \lambda_n k + \alpha\)
thus the solution to the BVP:
\[u(x, t) = \sum_{n=1}^{\infty} a_n\sin{(\frac{n\pi x}{L})}e^{(-k(\frac{n\pi}{L})^2 + \alpha)t}\]
Applying the initial condition:
\[\begin{align} u(x, 0) = \sum_{n=1}^{\infty} a_n\sin{(\frac{n\pi x}{L})} &= f(x) \\ \int_0^L \sum_{n=1}^{\infty} a_n\sin{(\frac{n\pi x}{L})}\sin{(\frac{m\pi x}{L})}dx &= \int_0^Lf(x)\sin{(\frac{m\pi x}{L})}dx \\ \sum_{n=1}^{\infty}\int_0^L a_n\sin{(\frac{n\pi x}{L})}\sin{(\frac{m\pi x}{L})}dx &= \int_0^Lf(x)\sin{(\frac{m\pi x}{L})}dx \end{align}\]
all terms in the integral cancel when \(n = m\) (Sturm-Louiville) due to the eigenfunctions forming an orthonormal basis.
Let us solve for the coefficient \(a_n\) on the LHS:
\[\begin{align} a_n\int_0^L\sin^2{(\frac{n \pi x}{L})}dx \\ \text{Let} \ \ u &= \frac{n \pi x}{L} \\ \frac{du}{dx} &= \frac{n \pi}{L} \\ dx &= \frac{L}{n \pi}du \\ \end{align}\]
\[\begin{align} a_n\frac{L}{n\pi}\int_0^{n \pi}\sin^2{(u)}du &= a_n\frac{L}{n\pi}\bigg[\frac{1}{2}(x - \sin{(x)}\cos{(x)})\bigg\vert_0^{n \pi}\bigg] \\ &= a_n \frac{L}{2} \\ a_n &= \frac{2}{L} \int_0^L f(x) \sin{(\frac{n \pi x}{L})}dx \end{align}\]
the full solution to the the IBVP of Equation 1 is thus,
\[u(x, t) = \sum_{n=1}^{\infty} a_n\sin{(\frac{n\pi x}{L})}e^{(-k(\frac{n\pi}{L})^2 + \alpha)t} \tag{4}\]
If we take the limit as (\(t \rightarrow \infty\))
\[\begin{align} \lim_{t \rightarrow \infty} \bigg[ e^{\alpha t}\sum_{n=1}^{\infty} e^{-k(\frac{n \pi}{L})^2 t}\bigg] \end{align}\]
Since \(k, L, \alpha\) are all positive real numbers we can clearly see that the limit goes to zero since the infinite sum appraoches \(e^{-\infty}\) much faster than \(e^{\alpha t}\) approaches \(e^{\infty}\)
Haberman 4.4.3
Problem definition
Consider a slightly damped vibrating string that satisfies:
\[\rho_0 \frac{d^2u}{dt^2} = T_0 \frac{d^2u}{dx^2} - \beta \frac{du}{dt} \tag{5}\]
Part a
Why must the parameter \(\beta > 0\)?
Solution
The parameter \(\beta\) must be positive, because this is the frictional coefficient. If \(\beta < 0\) the amplitude of the vibrating string would blow up to infinity as time progresses instead of converging to zero as \(t \rightarrow \infty\).
Part b
n Solve the IBVP:
\[\begin{align} \begin{cases} \rho_0 \frac{d^2u}{dt^2} = T_0 \frac{d^2u}{dx^2} - \beta \frac{du}{dt} \\ u(0, t) = 0 & u(L, t) = 0 \\ u(x, 0) = f(x) & \frac{d}{dt}u(x, 0) = g(x) \end{cases} \end{align}\]
Solution
Let us non-dimensionalize the problem
Let
\[\begin{align} &\hat{u} = \frac{u}{u_s} \Longrightarrow u = \hat{u}u_s &\hat{x} = \frac{x}{x_s} \Longrightarrow u = \hat{x}x_s & &\hat{t} = \frac{t}{t_s} \Longrightarrow u = \hat{t}t_s \end{align}\]
\[\begin{align} \frac{\rho_0 u_s}{t_s^2}\frac{d^2 \hat{u}}{d\hat{t}^2} + \beta \frac{u_s}{t_s} \frac{d\hat{u}}{\hat{t}} = T_0 \frac{u_s}{x_s^2} \frac{d^2 \hat{u}}{d\hat{x}^2} \\ \frac{d^2 \hat{u}}{d \hat{t}^2} + \frac{\beta}{\rho_0} t_s \frac{d\hat{u}}{d\hat{t}} = \frac{T_0}{\rho_0} \frac{t_s^2}{x_s^2} \frac{d^2\hat{u}}{d \hat{x}^2} \end{align}\]
Let
\[\begin{align} t_s = \tau = \frac{\rho_0}{\beta} \end{align}\]
where \(\tau\) is the characteristic time scale for Equation 5.
\[\begin{align} x_s = L_c = \sqrt{\frac{T_0 \rho_0}{\beta^2}} \end{align}\]
where \(L_c\) is the characteristic length scale for Equation 5.
Let us solve the non-dimensionalized IBVP
\[ \begin{cases} u_{tt} + u_t = u_{xx} \\ u(0, t) = 0 & u(L, t) = 0 \\ u(x, 0) = f(x) & u_t(x, 0) = g(x) \end{cases} \tag{6}\]
Let \(u(x, t) = X(x) T(t)\)
\[\begin{align} \frac{T'' + T'}{T} = \frac{X''}{X} = -\lambda \end{align}\]
The solution to the eigen value problem: \[\begin{align} \begin{cases} X'' = \lambda X \\ X(0) = X(L) = 0 \end{cases}\\ \Longrightarrow X_n(x) = \sin{(\frac{n \pi x}{L})} \quad n = 1, 2, 3, \cdots \end{align}\]
where \(\lambda = -\big(\frac{n \pi}{L}\big)^2\)
for the IVP:
\[\begin{align} T'' + T' = \tau^2 \lambda T \end{align}\]
the characteristic equation becomes:
\[r^2 + r + \frac{n^2\pi^2}{\hat{L}} = 0 \quad \text{where } \hat{L} = \frac{L}{L_c}\]
where the roots of the characteristic polynomial is:
\[\begin{align} r_{1, 2} = \frac{-1}{2} \pm \frac{1}{2}\sqrt{1 - \frac{4n^2\pi^2}{\hat{L}^2}} \end{align}\]
Assuming the frictional coefficient \(\beta\) is small where \(\beta^2 < \frac{4\pi^2\rho_0T_0}{L^2}\) let us plug in \(\hat{L}\) to check if the roots are complex.
The contents inside of the square root:
\[\begin{align} 1 - \frac{4n^2\pi^2}{\hat{L}^2} &= 1 - \frac{4n^2\pi^2}{(\frac{L}{L_c})^2} \\ &= 1 - \frac{4n^2\pi^2}{(\frac{L}{\sqrt{\frac{T_0 \rho_0}{\beta^2}}})^2} \\ &= 1 - \frac{4n^2\pi^2\rho_0T_0}{\beta^2 L^2} \end{align}\]
since the denominator \(\beta^2 L^2\) is less than the numerator we know that the contents inside the square root is always negative. Thus we have a pair of complex roots and the general solution can be written as:
\[\begin{align} T(t) = e^{-\frac{\hat{t}}{2}}\left(a_n \cos{(\omega t)} + b_n \sin{(\omega t)}\right) \quad \text{where } \omega = \frac{1}{2} \sqrt{\frac{4n^2\pi^2}{\hat{L}} - 1} \end{align}\]
for initial conditions \(T(0) = f(x) \quad T'(0) = g(x)\) we have:
\[\begin{align} T(0) = f(x) = a_n \\ T'(0) = g(x) &= \frac{1}{2}\left(b_n2\omega - f(x)\right)\\ b_n &= \frac{2g(x) + f(x)}{2\omega} \end{align}\]
The final solution can be written as:
\[u(x, t) = e^{-\frac{\hat{t}}{2}} \sum_{n=1}^{\infty} \sin{(\frac{n \pi x}{\hat{L}})}\left(f(x) \cos{(\omega_n t)} + \frac{2g(x) + f(x)}{2\omega_n}\sin{(\omega_n t)}\right) \quad \text{where } \omega_n = \frac{1}{2} \sqrt{\frac{4n^2\pi^2}{\hat{L}} - 1}\]
Haberman 2.5.2
Problem definition
\[\begin{cases} u_{xx} + u_{yy} = 0 \\ u_x(0, y) = 0 & u_y(x, 0) = 0 \\ u_x(L, y) = 0 & u_y(x, H) = f(x) \end{cases} \tag{7}\]
Part a
Without solving this problem, explain the physical condition under which there is a solution to this problem.
Solution
The steady-state condition must be maintained along the boundaries. Along \(y\) from \([0, H]\) we have homogeneous boundary conditions such that the flow along left and right vertical sides of the rectangle are both zero. Along \(x\) we must enforce the condition that the sum of the contributions of the flow along \(x\) from \([0, L]\) also sum to zero. Thus the following solvavility criteria must be imposed in order for a solution to exist.
\[\int_0^L f(x) dx = 0\]
Part b
Solve the problem only by separation of variables. Show that the solution exists only under the condition in part a.
Solution
Let \(u(x, y) = X(x)Y(y)\) such that
\[\begin{align} -\frac{Y''}{Y} = \frac{X''}{X} = \lambda \end{align}\]
Since the non-homogenous BC are in \(y\) we can expect that \(Y(y)\) is composed of exponentials and \(X(x)\) oscillates.
The eigen value problem \(X'' = \lambda X \quad X'(0) = X'(L) = 0\) (Neumann boundary conditions) produces the following solutions:
\[X_n(x) = \cos{(\frac{n \pi x}{L})} \quad n = 1, 2, 3, ... \quad \lambda_n = -(\frac{n \pi}{L})^2\]
The ODE in the \(y\) domain
\[Y_n(y) = \frac{n^2 \pi^2}{L^2} Y_n(y)\]
The general solution:
\[Y_n(y) = c_1 \cosh{(\frac{n \pi y}{L})} + c_2 \sinh{(\frac{n \pi y}{L})}\]
Enforcing the boundary conditions at \(y = 0\)
\(\begin{align} Y'_n(y) &= c_1 \frac{n \pi}{L}\sinh{(\frac{n \pi y}{L})} + c_2 \frac{n \pi}{L}\cosh{(\frac{n \pi y}{L})} \\ Y'_n(0) &= 0 = c_2 \frac{n \pi}{L} \\ \Longrightarrow c_2 = 0 \end{align}\)
\(\begin{align} Y'_n(H) = f(x) &= c_1 \frac{n \pi}{L}\sinh{(\frac{n \pi H}{L})} \\ \frac{L}{n \pi}\int_0^L f(x) \sinh{(\frac{m \pi H}{L})} dx &= c_1 \frac{L}{2} \quad \textit{orthogonality} \\ \Longrightarrow c_1 = \frac{2}{n \pi}\int_0^L f(x) \sinh{(\frac{m \pi H}{L})} dx \end{align}\)
The full solution to the Laplace equation with 3 homogeneous boundary conditions is thus,
\[u(x, y) = \frac{2}{n \pi} \int_0^L f(x) \sinh{(\frac{n \pi H}{L})}dx \cosh{(\frac{n \pi y}{L})} \cos{(\frac{n \pi x}{L})} \quad n = 1, 2, 3, ...\]