New answers tagged pde
0
One can do this almost straightforwardly. Define
\begin{align}
\hat{f}(\mathbf{k})=\mathcal{F}^{-1}(f)=\frac{1}{(2\pi)^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{-i(k_x x+k_yy+k_zz)}}{1+x^2+y^2+z^2}dxdydz
\end{align}
Now rotate the coordinate system in $x,y,z$-space so that the direction of $\mathbf{k}=(k_x,k_y,k_z)$ ...
0
I think that "Boundary Value Problems and Green's Functions" by Stakgold and Holst is a good text.
0
I think that you can't solve this equation in $H^2_0$. If a function $u$ is in $H^2_0$ then you have $u=0$ and $\nabla u=0$ at the boundary and these are too much boundary conditions for this equation. Let me think how one can prove the nonexistence.
2
For real-valued functions $u(x,y)$ and $v(x,y)$, we say that $v$ is a harmonic conjugate of $u$ if $u+iv$ is analytic. This is not a symmetric relation between $u$ and $v$. For example, $e^x\sin y$ is a harmonic conjugate of $e^x\cos y$ because $e^x\cos y+ie^x\sin y=e^z$ is analytic, but $e^x\cos y$ is not a harmonic conjugate of $e^x\sin y$ because $e^x\sin ...
1
Let $w = a + bi$. Note that $\text{Im } (iw) = \text{Im } (-b + ai) = a = \text{Re } (w)$.
So yes, it is the same function up to a minus sign, which I don't know where it comes from - perhaps it comes from the manual's specific definition of a harmonic conjugate.
0
(This may not be a neat way to prove the assertion, but it's a proof anyway.) Let $\eta=x+t$ and $\nu=x-t$. Then $x=\eta+\nu$ and $t=\eta-\nu$ are functions of $\eta$ and $\nu$, $A=A(\eta+\nu,\eta-\nu)$ and similarly for $B$. As both $A$ and $B$ are independent of $\nu$, by the total derivative formula, we get
\begin{align*}
0 = \frac{dA}{d\nu} &= ...
0
As far as I can tell, "quasilinear" means "linear in the highest-order derivatives".
Here the highest-order derivatives are $u_{xy}$ and $u_{xx}$, and the equation is indeed linear in those. The fact that the coefficient of $u_{xx}$ involves $u$ is no problem for quasilinear. On the other hand, since this coefficient does involve $u$ rather than depending ...
0
The factor $e^{bx}$ seems to be irrelevant for the theorem.
I haven't a reference, but it's straightforward to get an estimate, using an explicit
formula for solution via Green's function:
$$
W(t,x)=\int_0^\infty G(x,y,t)F(y)\,dy.
$$
It can be assumed that $D=1$. For Dirichlet condition, say,
$$
G(x,y,t)=\Gamma(x-y,t)-\Gamma(x+y,t),
$$
where
$$
...
2
Let us fix an arbitrary $\epsilon>0$ first. Choose $R_0>0$ such that $C_\epsilon <\epsilon R_0$. From the assumption on $u$, we know that when $|z|>R_0$, $u(z)<2\epsilon |z|$. Since $u$ is harmonic, by maximum value principle, for every $R\ge R_0$,
$$\max_{|z|\le R}u(z)=\max_{|z|=R}u(z)<2\epsilon R.$$
That is to say, $2\epsilon R-u(z)$ is ...
1
Let us consider for simplicity two times, $t_1$ and $t_2$. We can evolve from the point $(0,0)$ in the 2D "time"-space to another point $(T_1,T_2)$ by different means.
One way, for example, is to use the 1st evolution equation to arrive to the point $(T_1,0)$ and then use the 2nd evolution equation to go from $(T_1,0)$ to $(T_1,T_2)$ (with fixed $T_1$!).
...
1
Since $u$ is harmonic on the whole plane and is bounded linearly, we know by Liouville's theorem that $u$ is a harmonic polynomial of order less than or equal to 1. The coefficient on the linear term must be 0, since if it were anything else, then you would have a contradiction for $\epsilon$ sufficiently small.
6
First notice $2\phi \phi' = (\phi^2)'$:
$$\phi^2(\infty) - \phi^2(0) = \int^\infty_0 2\phi(x)\phi'(x)\,dx.$$
Compactly supportedness drives $\phi\to 0$ when $x\to \infty$, so:
$$
\phi^2(0) = \left|\int^\infty_0 2\phi(x)\phi'(x)\,dx\right| \leq \int_{0}^{\infty}|2\phi(x)\phi'(x)|\,dx
$$
Lastly simply using $2ab\leq a^2 + b^2$:
$$
...
0
I would recommend "Partial Differential Equations with Fourier Series and Boundary Value Problems" written by Asmar. It is a text for beginners in the subject, but eventually it gets very far, to a thorough treatment of Green's functions. It does also include a lot of material on Bessels functions.
1
You're right that the eigenvalues of $A^{-1}$ are $\lambda^{-1}$ for each eigenvalue $\lambda$ of $A$. (The notation is suggestive for a reason.) Furthermore, the identity matrix commutes with every matrix; so the eigenvalues of
$qI-rA^{-1}$
are just $q-r/\lambda$ for each eigenvalue $\lambda$ of $A$. Using this, you can compute the eigenvalues of ...
0
The issue here is, the problem seems to set you up to use certain results, but those theorems are irrelevant in the problem as actually stated.
So long as the boundary conditions lie in the appropriate trace space ($H^{-\frac{1}{2}}$ of the boundary), and $f$ lies in $H^{-1}$ of the unit square, you can solve the equation on the square in the weak sense; ...
0
Conservation laws imply some notion of 'time' in which a quantity $M(t)$ like the mass $$\int_\Omega u(x,t) \,\mathrm d x$$ is conserved.
This is equivalent to proving that
$$\frac {\mathrm d}{\mathrm d t}\int_\Omega u(x,t) \,\mathrm d x = \int_\Omega \frac {\partial}{\partial t} u(x,t) \,\mathrm d x = 0$$
For example, an equation of the form
$$\frac ...
0
The concept of conservation of mass (or any other meaningful quantity) is usually applied to time dependent equations, like the heat equation
$$
u_t-\Delta u=0,
$$
or the wave equation
$$
u_{tt}-\Delta u=0.
$$
For the heat equation conservation of mass (or heat, to be more precise) means that $\int_\Omega u(x,t)\,dx$ is constant (i.e. it does not depend on ...
2
Just to continue Chris Wong's argument.
Let $\Omega = \{x\in \mathbb{R}^n: \|x\|_{\infty}= \max|x_i|<1 \}$ be an $n$-cube with side length $2$, then it can be verified that:
$$
u = \prod_{i=1}^n \sin(2\pi x_i)
$$
is in $H^1_0(\Omega)\cap H^2(\Omega)$, since on boundary of the $n$-cube, there is some $x_i = \pm 1$. However for $\partial u/\partial x_i$ ...
1
Using the definitions, you should be able to prove that $H^2_0(\Omega) \subset H_0^1(\Omega) \cap H^2(\Omega)$. The reverse inclusion does not hold; as a counterexample, try picking a really nice $\Omega$ (such as a disc) and writing down a well-behaved function that vanishes at the boundary, but whose first derivatives do not. Then such a function will be ...
1
The inequality $$|u(x) - u(y)| \leq \sup_\xi |\nabla f(\xi)| |x-y|\tag1$$ in which the supremum is taken over $\xi$ on the line segment from $x$ to $y$, has nothing to do with harmonic functions, or their mean value property. It is a corollary of the Mean Value Theorem.
As for the original question, the key point is the interior regularity of harmonic ...
1
I will assume that the initial data is smooth. First we define the appropriate energy (this depends on oyur problem, but usually it is enough to take some high order Sobolev norm). For instance we take $E[u]=\|u\|_{H^1}$. Now we obtain an a priori bound for this energy. Multiplying the equation by u and integrating by parts we get
$$
...
2
Clearly, there is no angular dependence, so the potential will be a function of $r$ and $z$. We may separate variables by letting the potential $V(r,z)= R(r)Z(z)$ The equation for the potential is then
$$\frac{1}{rR} \frac{dR}{dr} + \frac{1}{R} \frac{d^2 R}{d r^2} + \frac{1}{Z} \frac{d^2 Z}{d z^2} = 0$$
We deal with the $R$ equation first:
...
0
I have a question concerning the domain: is the time going backward? where are the boundary conditions in u(0,t)?
1
As gerw wrote, in general neither condition implies the other. However, in the context of PDE you may have extra information about the function which allows to pass from one condition to the other.
Fact. If $T\to 0$ and $T''$ is bounded, then $T'\to 0$.
I remember the following proof from Littlewood's Miscellany. Draw the graph of $T'$ (in red):
If ...
3
Take $T(x) = \sin(x^2)/x$. Then, $T(x) \to 0$ as $x \to +\infty$, but $T'(x) = (2\,x^2\,\cos(x^2)+ \sin(x^2))/x^2 \not\to 0$ as $x \to +\infty$.
Also, the reverse implication is not true: take $T(x) = \ln(x)$. Then, $T'(x) \to 0$, but $T(x) \not\to 0$.
Hence, those two conditions are not related.
1
I am assuming that $E=C^2([0,1])$ and $F=C([0,1])$. Also, if $T$ is a linear transoformation, we gonna use the symbol $\langle T,x\rangle =T(x)$. Remember that the derivative of a function $F$ between two Banach spaces $X$ and $Y$ is a function $F':X\to L(X,Y)$ between $X$ and the set of all linear bounded transformations between $X$ and $Y$, $L(X,Y)$.
...
2
Mathematicians often look to other fields to gain inspiration for some of their research (the mother of all examples might be vector calculus and electromagnetism), so this is not a bad question. Lots of mathematical modelling in biology is very approximate and many models are entirely useless to the average experimentalist. An example of this are the PDEs ...
3
The book referenced by @Henrik Finsberg is OK if you'd like to learn some basic facts about PDEs (and about their applications in biology). If you feel already confident about PDEs, try to get a hold of
James D. Murray, Mathematical Biology, Vol. 2
Benoît Perthame, Transport equations in biology
Robert S. Cantrell, Chris Cosner, Spatial Ecology via ...
4
No, you do not miss anything: no explicit $\lambda$s can be found. However, the general theory tells us:
there are infinitely many lambda's
there is minimal $\lambda$ and there is no maximal one
the eigenfunctions corresponding to lambdas are orthogonal on the interval
Hence we can always write the solution as an infinite series. And if you want to ...
1
Define $$
f(t) = \left\{ \begin{array}{rl}
-t\log{t} &\mbox{ if $t\in[0,1]$} \\
0 &\mbox{ otherwise}
\end{array} \right.
$$
Note that $f$ is continuous. COnsider the problem
$$\tag{1}
\left\{ \begin{array}{rl}
-\Delta u-\lambda u=f(u) &\mbox{ in $\Omega$} \\
u=0 &\mbox{ in $\partial\Omega$}
\end{array} \right.
$$
1 - ...
3
There is a great book about this: Leah Edelstein-Keshet, Mathematical models in biology chapter 10.
1
Using separation of variables
$$
u = XY \\
u_y = XY' \\
u_{xx} = X''Y \\
XY' = X''Y - XY \\
\frac {Y'}Y = \frac {X''}X - 4
$$
Since LHS and RHS are functions of different variables, they must be equal to some constant
$$
\frac {Y'}Y = \frac {X''}X - 4 = -\lambda^2-4 \\
\frac {Y'}Y = -\lambda^2-4 \\
Y = Y_0 e^{-(\lambda^2+4) y} \\
\frac {X''}X = -\lambda^2 \\
...
1
"$\Rightarrow$" Let $h>0$ arbritrary, then clearly
$$\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta) = \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D \leq h)}_{=:p_1} + \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D > h)}_{=:p_2}$$
We have $$\begin{align*} p_1 &\leq \mathbb{P}^w \left( \sup_{t \leq h} |B_t-z| \geq \delta ...
0
In my opinion Thermodynamics is physics, chemistry and engineering.
In thermodynamics mathematical operations are used that are unwarranted and would not be acceptable anywhere else. (eg applying a Legendre transform, without solving the differential equation involved.) I would start with 'Cengel and Boles' not with Callen.
As Gibbs noted: "A mathematician ...
2
This is just chain rule, $ \phi(t) = f(tx) $ implies $$ \phi'(t) = \nabla f(tx).x = \sum_{i=1}^n\frac{\partial f}{\partial x_i}(tx)x_i $$
You have gradient defined as $$ \nabla f = \Big(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},....,\frac{\partial f}{\partial x_n}\Bigr) $$
and for any curve $\gamma $ chain rule gives you that $ (f\circ ...
1
Yes; since you're taking only one $\eta_i$ to vary everything just reduces to the case where you just had $x_k(\eta)$ in fact.
One thing that is worth stressing is that the $\partial/\partial x_k$ derivatives are holding the other $x_m$ constant, rather than the other $\eta_j$ variables held constant elsewhere. This is probably implicit, but important to ...
0
I have done this as; howerver i am embarrass when use the condition:
$$\begin{array}{l}
a = 1;b = y^2/2;c = 0 = > b^2 - ac = \frac{y^2}{4} = > \frac{dx}{dy} = \frac{b \pm \sqrt{ b^2 - ac }}{a} = \frac{y^2/2 \pm y^2/2}{1} = \left[ \begin{array}{l}
0 \\
{y^2} \\
\end{array} \right. \\
\left\{ \begin{array}{l}
\frac{{dx}}{{dy}} = 0 = > dx = ...
0
Let $v=u_y$. The equation is then 1st order in $v$:
$$y^2 v_x + v_y - \frac{2}{y} v = 0$$
$$v(x,1) = 3$$
This may be solved using the method of characteristics to determine $v(x,y)$; then use $v = u_y$, $u(x,1)=1-x$ to find $u(x,y)$.
I leave the details to the reader; I get
$$u(x,y) = 5 y - 2 y \log y - x - 4$$
ADDENDUM
The method of characteristic ...
1
Consider first order PDE depending on two independent variables. It is
linear, if it has the form
$$
a(x,y)\partial_x u+b(x,y)\partial_yu=c(x,y)u+f(x,y),
$$
Example:
$$
\partial_xu+\partial_yu=0.
$$
semi-linear if it has the form
$$
a(x,y)\partial_x u+b(x,y)\partial_yu=c(x,y,u),
$$
Example:
$$
\partial_xu+\partial_yu=u^2
$$
quasi-linear if it has the form
...
1
So you wanna "master" the following topics "really fast":
(Linear) Partial differential equations
Finite element methods
Optimization technique in numerical treatments of PDE
Not discouraging you, or keeping you from motivated, but I highly doubt it can be done within say, one or two month or something, assuming you are working towards a summer project ...
2
Changes of units, in any equation, modify the coefficients. If different coefficients are subject to different scaling exponents (which is what it means to be "dimensionally wrong"), it means that for the equation to not depend on choice of units, the coefficients must be dimensionful, and in a specific way. There is a unique set of units for the ...
4
The units are fine. $L$ is inductance per length unit and $R$ is resistance per length unit.
1
Peter Olver's Partial Differential Equations is a very nice set of lecture notes (you will need to supplement them with some problems). But hurry, very soon these lecture notes will turn into a book and probably disappear from his site.
1
The law describing the distribution of heat is the heat equation,namely
$$u_t(t,x,y) = D\Delta u(t,x,y).$$
Here the function $u(t,x,y)$ gives the temperature at position $(x,y)$ and time $t$, while $D$ is a constant representing the thermal diffusivity. You can find how to derive the equation here: http://en.wikipedia.org/wiki/Heat_equation.
Now, if the ...
2
If you set $R,c$ to constants, you obtain a massive scalar field equation (kinda like a time dependent Helmholtz equation).
This has solutions of the form
$$\psi = \exp (i\omega t - i \mathbf k \cdot\mathbf x)$$
where you can figure out the relationship needed between the parameters.
You can also linearly combine these (effectively Fourier transforms) to ...
0
I suspect what you're looking for is the theory of pseudodifferential operators on manifolds. Peter Gilkey's book has a good introduction.
1
I assume your equation is of the form
$$
u_t-a^2u_{xx}=0,\quad 0\le x\le \ell,\quad t\ge0,
$$
with initial condition $u(x,0)=u_0(x)$ and boundary conditions $u(0,t)=1$, $u(\ell,t)=0$. You cannot apply directly the method of separation of variables because of the boundary condition at $x=0$. But it is possible to solve the equation. Let $h(x)=1-x/\ell$. Then ...
0
Gage and Hamilton prove this fact in the paper "The Heat Equation Shrinking Convex Plane Curves" see page 80 of this paper.. I confess that I didn't understand the argument too..
I can't access the link you furnished..
0
Of course, see, e.g. the second chapter of book. This should be adressed in every course on the numerical treatment of PDEs.
1
This integral may be evaluated using residue theory. In this case, convert the integral over $\phi$ to a contour integral over the unit circle in the complex plane, and evaluate the residues of the poles inside the unit circle. Let $z=e^{i \phi}$, then $d\phi = dz/(i z)$, $\cos{\phi} = (z+z^{-1})/2$, $\sin{\phi} = (z-z^{-1})/(2 i)$. Then, after some ...
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