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I am aware that the PDE is already solved in this thread.

But, I want to know how that person got from $u_x+u_t=g(4x+t)$ to $2u_{x'}=g(4x+t)$ by using the change of variables $x′=x+t,t′=x−t$. I don't understand the $2u_{x'}$ part. Finally, after doing all that, how do you arrive at the final solution?

sucksatmath
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1 Answers1

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I prefer not following external links unless I have to, so I hope the following might help.

Since $a^2-3 ab-4b^2=(a+b)(a-4b)$ we can (not rigorously) write: $$ \left( \frac{\partial^2}{\partial x^2} -3 \frac{\partial^2}{\partial x \partial t}-4 \frac{\partial^2}{\partial t^2}\right)u=\left( \frac{\partial}{\partial x} +\frac{\partial}{\partial t}\right)\left( \frac{\partial}{\partial x} -4\frac{\partial}{\partial t}\right)u=0$$

And obtain two 1st order PDEs:

$$\left( \frac{\partial}{\partial x} +\frac{\partial}{\partial t}\right)u_1=0$$

$$\left( \frac{\partial}{\partial x} -4\frac{\partial}{\partial t}\right)u_2=0$$

Where we are searching for the solution as:

$$u(x,t)=A u_1(x,t)+B u_2(x,t)$$

By the form of the above equations, it's clear that we should search for solutions as:

$$u_1(x,t)=f(x-t)$$

$$u_2(x,t)=g(4x+t)$$

You can check all of this by direct substitution in the original PDE.

Yuriy S
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    Thank you!. I want to confirm if $u_1$ and $u_2$ is supposed to be the be the final answer or is $u(x,t) = c_1f(x-t)+c_2g(4x+t)+c_3$ the final answer? – sucksatmath Jan 27 '18 at 15:48
  • Also, what would happen if the factor of the operator is of the form (a+b)(a+b)? Would that mean there is only one solution f(x-t)? – sucksatmath Jan 27 '18 at 15:56
  • You can certainly add a constant since the equation is homogeneous. As for the second question this would be a degenerate case which also should have two linearly independent solutions, I think they should be $$C_1 f(x-t)+C_2 (x-t) f(x-t)$$ check it out, it works though I don't remember the proof – Yuriy S Jan 27 '18 at 20:23