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Questions tagged [characteristics]

The method of characteristics is a way of solving certain partial differential equations by reducing them to ordinary differential equations. It is most often used for 1st order equations. Use with the (pde) tag.

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Infinite solution PDE

I want to show that the following PDE of the function $u(x,y)$ has infinitely many solutions: $$\left\{ \begin{array}{c} u_x+u_y=2xu \\ u(x,x)=e^{x^2} \\ \end{array} \right.$$ By using the ...
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How can I prove that this matrix is diagonalizable?

How can I prove that for all $a \in \mathbb{R}$ the following matrix is diagonalizable? I computed the characteristic polynomial, but I couldn't decompose it into linear factors. This is the ...
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Confusion in college book on the introduction of PDE (method of characteristics)

In my course book, the method of characteristics for solving 1st order PDE of the form $P \frac{\partial u}{\partial x} + Q\frac{\partial u}{\partial y} = R$ where $u$ is a function of $x$ and $y$ is ...
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Cauchy problem for nonlinear first order hyperbolic PDE with source via method of characteristics. (Work and characteristic plots included)

Working on the following problem. $y u_x -u u_y = x \\ u(x,x)=-2x$ I've went after this with the method of characteristics. I'm using $(t,s$) as my parametrization variables. In parameterize space I ...
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Solving 1st order PDE (Finding where solution is invalid)

I'm struggling with a PDE homework question (see below) Solve the equation: $au_x + bu_y + u = 0$ With Cauchy data $u(s) = e^{−s}$ along the straight lines $(x(s), y(s)) = (cs, ds)$. The solution ...
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Dealing with the logarithm of absolute value when solving $u_t+xu_x=1$ by method of characteristics

I was trying to come up with a solution to the PDE IVP problem $u_t + x u_x =1$, $u(x,0) = f(x)$. Here's how I went about it: We can the following relations by applying the Method of Characteristics: ...
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Method of characteristic for PDE $xu_x + yu_y = u$

I'm studying method of characteristic. This is a homework. I'm very new to this topic so hope anyone can give a direction. Given a PDE : $xu_x + yu_y = u$. a) Solve it, where the initial curve ...
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How to show that characteristic curves for a PDE fill the plane?

In a question I found my characteristics curves for my PDE to be $y=-\cos(x)+C$. When I sketched some of the curves I could see that I would fill the $x,y$. But the question asked explain why your ...
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Method of characteristics - finding the particular solution using initial conditions

I am trying to use the method from my previous question to solve this PDE: $$3u_x + 2u_t = \cos x$$ with initial condition $u(x,0) = x^2$. So I need to solve these: \begin{align} \frac{dx}{ds} &...
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Method of characteristics - eliminating variables

I am trying to follow a guide for the method of characteristics; quoting the first example: We use the method of characteristics to solve the problem $2u_x - u_y = 0, \;\; u(x, 0) = f(x)$ ...
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Analyze : $u_t-u^2u_x +cu =0, u(x,0)=g(x)$

Analyze : $u_t-u^2u_x +cu =0$, $u(x,0)=g(x)$. From This we have following \begin{align} \frac{dt}{ds} &=1 \\ \frac{du}{ds} &=c \\ \frac{dx}{ds} &=-u^2 \end{align} then how to ...
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The Burgers equation $u_y + u u_x = 1$ with $u=0$ on the parabola $y^2=2x$

For the PDE $u_y + u u_x = 1$, sketch a plot of $\Gamma$ and a few representative curves, including the envelope curve. Conditions: $u=0$ on the curve $y^2=2x$, and $y,x>0$. Express $u$ as a ...
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integral of solution between two characteristic curves

Suppose we are given a pde: $\frac{\partial u}{\partial t} +\frac{\partial {(b(x)u)}}{\partial x}=0$. Let $u(t,x)\in C^1(\mathbb{R}^2)$ be a solution and $x=X_1(t)$ and $x=X_2(t)$ be two ...
Is there a solution to this unidirectional wave equation, with initial value $v=f(x)$ and $x=t^2$
unidirectional wace equation: $$\frac{du}{dt}+c\frac{du}{dx}=0$$ The initial value $u=f(x)$ is given on the parabola $x=t^2$. Is there a solution to this problem, discuss why the solution is unique ...
Show that the PDE has vertical lines as a family of characteristic curves for $y=0$.
Show that the PDE given by $\dfrac{\partial ^2u}{\partial y^2}-y\dfrac{\partial ^2u}{\partial x^2}$ has vertical lines as a family of characteristics curves for $y=0$. Taking $S=0,R=-y,T=1$ we ...