Characteristic curves

Question

I have to solve the initial value problem:

$$2u_{xx}(x, t)-u_{tt}(x, t)+u_{xt}(x, t)=f(x, t), x \in \mathbb{R}, t>0 \\ u(x, 0)=0, x \in \mathbb{R} \\ u_t(x, 0)=0, x \in \mathbb{R}$$

using Green's Theorem.

To do that we have to find the characteristic curves, right??

We have the equation $$2u_{xx}-u_{tt}+u_{xt}=f(x, t)$$

This is equal to $$\left (\frac{2\partial^2}{\partial{x^2}}-\frac{\partial ^2}{\partial{t^2}}+\frac{\partial ^2}{\partial{x}\partial{t}}\right )u=f$$

To find the characteristics do we solve the homogeneous equation $$\frac{2\partial^2}{\partial{x^2}}-\frac{\partial ^2}{\partial{t^2}}+\frac{\partial ^2}{\partial{x}\partial{t}}=0$$ ??

EDIT:

$$2u_{xx}-u_{tt}+u_{xt}=f \\ \Rightarrow \left (2\frac{∂^2}{∂x^2}-\frac{∂^2}{∂t^2}+\frac{∂^2}{∂x∂t}\right )u=f \\ \Rightarrow \left(\frac{∂}{∂x}+\frac{∂}{∂t}\right)·\left(2\frac{∂}{∂x}-\frac{∂}{∂t}\right)u=‌​f$$

The $g(x−t)$ and $h(x+2t)$ are solutions of the homogeneous differential equation $2u_{xx}−u_{tt}+u_{xt}=0$ for any twice differentiable functions. (Or once differentiable?? )

So, the characteristic curves are $x−t=x_0−t_0$ and $x+2t=x_0+2t_0$. Is this correct?? Is the formulation correct??

Answer 1

Find the factorization into linear factors $$ 2a^2+ab-b^2=(a+b)(2a-b). $$ Then $$ 2\frac{∂^2}{∂x^2}-\frac{∂^2}{∂t^2}+\frac{∂^2}{∂x∂t} = \left(\frac{∂}{∂x}+\frac{∂}{∂t}\right)·\left(2\frac{∂}{∂x}-\frac{∂}{∂t}\right) $$

Answer 2

we will make a change of variable $$\pmatrix{\xi\\\eta} = \pmatrix{1&2\\1&-1}\pmatrix{x\\t}, \,\pmatrix{x\\t} = \frac 13\pmatrix{1&2\\1&-1}\pmatrix{\xi\\\eta}, dx\,dt = -\frac13 d\xi \, d\eta.$$ with this change of variables, we have $$\frac d{dx}=\frac d{d\xi} + \frac d{d\eta}, \frac d{dt}=2\frac d{d\xi} -\frac d{d\eta}.$$

the line $t = 0,$ where the initial values of $u, u_t$ are given, is now $\xi = \eta.$ so that $$u_{\xi}\big|_{\eta = \xi} = 0 $$ we can transform $$\begin{align}f &= 2u_{xx}-u_{tt} + u_{xt} \\ &=2\left(u_{\xi\xi}+2u_{\xi\eta} + u_{\eta\eta}\right) - \left(4u_{\xi\xi}-4u_{\xi\eta} +u_{\eta\eta}\right) +\left(2u_{\xi\xi}+u_{\xi\eta} -u_{\eta\eta}\right) \\&= 9u_{\xi\eta}\end{align} $$

pick a point $$P = (x, t), A = (x-t, 0), B= (x + 2t, 0).$$ note that $$\begin{align} \text{ line }AP&: \eta = x- t\\ \text{ line }BP &: \xi = x+2t\\ \text{ line }AB &: \eta = \xi\\\end{align}$$

we will integrate $f = 2u_{xx}-u_{tt} + u_{xt}$ over the triangle $ABP.$ so we get $$\begin{align}\int_{\Delta ABP}f(x,t)\, dx \, dt &=\int_{\Delta ABP} \left(2u_{xx}-u_{tt} + u_{xt} \right)\, dx \, dt \\ &=-3\int_{\Delta ABP} u_{\xi \eta} \, d\xi \, d\eta \\ &= -3\int_{\xi = x-t}^{\xi=x+2t}\, d\xi \int_{\eta = x - t}^{\eta = \xi} u_{\xi \eta} \,d\eta \\ &=-3\int_{\xi = x-t}^{\xi=x+2t} u_{\xi}\big|_{\eta = x - t}^{\eta = \xi} \,d\xi \\ &= 3\int_{\xi = x-t}^{\xi=x+2t} u_{\xi}\big|_{\eta = x - t} \,d\xi \\ &= -3\left(u(P) - u(A) \right) = -3u(x,t) \end{align} $$ therefore we have $$u(x,t) = -\frac 13 \int_{\Delta ABP}f(x,t)\, dx \, dt = -\frac 13 \int_0^t ds \int_{x-t + s}^{x+2t-2s} f(y,s) \, dy $$

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example: take $f(x,t) = -2$ we know that $u = t^2$ is a solution. we will see if we get that solution.

$$\begin{align}u(x,t) &= -\frac{1}{3} \int_0^t ds \int_{x-t+s}^{x+2t-2s} -2 \, dy\\ &= \frac 23 \int_0^t (x+2t - 2s)-(x-t+s) \, ds = \int_0^t (t-s) \, ds \\ &= t^2 \end{align}$$

the solution is correct, at least in the case where $f$ is constant.