Researching refracted wave front equation

Question

I am studying seismology, particularly refracted wave front equation.

I know equation for distance: $$ X(z)=\int_0^z \frac{pv_0(1+\beta u)\;du}{\sqrt{1-p^2v_0^2(1+\beta u)^2}} $$ where $v_0$ is the initial speed of the wave, $p$ is the parameter of the ray, and $\beta$ is an arbitral non-zero constant.

For the time we get $$ t(z)=\int_0^z \frac{du}{v_0(1+\beta u)\sqrt{1-p^2v_0^2(1+\beta u)^2}} $$

So far I managed to obtain $$ t(z)=\frac{1}{v_0\beta}\log\frac{(1+\beta z)\left(1+\sqrt{1-p^2v_0^2}\right)}{1+\sqrt{1-p^2v_0^2(1+\beta z)^2}} $$ And now am struggling to prove the most important statement: $$ X(z)^2+\left(z-\frac{\cosh(v_0\beta\cdot t(z))-1}{\beta}\right)^2=\frac{\sinh^2(v_0\beta\cdot t(z))}{\beta^2} $$

I have tried exponentiating $t$ to get $\cosh$ from somewhere, but got no results.

Answer 1

Notice, for $\text{t}\left(\text{z}\right)$ you've to be carefull:

$$\text{t}\left(\text{z}\right)=\int_0^\text{z}\frac{1}{\mathcal{v}_0\cdot\left(1+\beta\cdot u\right)\cdot\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot u\right)^2}}\space\text{d}u$$

Substitute $x=1+\beta\cdot u$ and $\text{d}x=\beta\space\text{d}u$:

$$\text{t}\left(\text{z}\right)=\frac{1}{\mathcal{v}_0\cdot\beta}\int_1^{1+\beta\cdot\text{z}}\frac{1}{x\cdot\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot x^2}}\space\text{d}x$$

Substitute $x=\frac{\sin\left(\text{s}\right)}{\text{p}\cdot\mathcal{v}_0}$ and $\text{d}x=\frac{\cos\left(\text{s}\right)}{\text{p}\cdot\mathcal{v}_0}\space\text{d}\text{s}$:

$$\text{t}\left(\text{z}\right)=\int_{\arcsin\left(\mathcal{v}_0\cdot\text{p}\right)}^{\arcsin\left(\mathcal{v}_0\cdot\text{p}\cdot\left(1+\beta\cdot\text{z}\right)\right)}\csc\left(\text{s}\right)\space\text{d}\text{s}=-\left[\ln\left|\cot\left(\text{s}\right)+\csc\left(\text{s}\right)\right|\right]_{\arcsin\left(\mathcal{v}_0\cdot\text{p}\right)}^{\arcsin\left(\mathcal{v}_0\cdot\text{p}\cdot\left(1+\beta\cdot\text{z}\right)\right)}$$

So:

1. When $\text{s}=\arcsin\left(\mathcal{v}_0\cdot\text{p}\right)$: $$\ln\left|\cot\left(\text{s}\right)+\csc\left(\text{s}\right)\right|=\ln\left|\frac{1+\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0}}{\text{p}\cdot\mathcal{v}_0}\right|$$
2. When $\text{s}=\arcsin\left(\mathcal{v}_0\cdot\text{p}\cdot\left(1+\beta\cdot\text{z}\right)\right)$: $$\ln\left|\cot\left(\text{s}\right)+\csc\left(\text{s}\right)\right|=\ln\left|\frac{1+\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot\text{z}\right)^2}}{\text{p}\cdot\mathcal{v}_0\cdot\left(1+\beta\cdot\text{z}\right)}\right|$$

So:

$$\text{t}\left(\text{z}\right)=\ln\left|\frac{\left(1+\beta\cdot\text{z}\right)\cdot\left(1+\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0}\right)}{1+\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot\text{z}\right)^2}}\right|$$

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And, $x=1+\beta\cdot u$ and $\text{d}x=\beta\space\text{d}u$:

$$\text{X}\left(\text{z}\right)=\int_0^\text{z}\frac{\text{p}\cdot\mathcal{v}_0\cdot\left(1+\beta\cdot u\right)}{\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot u\right)^2}}\space\text{d}u=\frac{\text{p}\cdot\mathcal{v}_0}{\beta}\int_1^{1+\beta\cdot\text{z}}\frac{x}{\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot x^2}}\space\text{d}x$$

Use again a substitution $\text{s}=1-\text{p}^2\cdot\mathcal{v}^2_0\cdot x^2$ and $\text{d}\text{s}=-2\cdot\text{p}^2\cdot\mathcal{v}^2_0\cdot x$:

$$\text{X}\left(\text{z}\right)=-\frac{1}{2\cdot\beta\cdot\text{p}\cdot\mathcal{v}_0}\int_{1-\text{p}^2\cdot\mathcal{v}^2_0}^{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot\text{z}\right)^2}\frac{1}{\sqrt{\text{s}}}\space\text{d}\text{s}=-\frac{1}{\beta\cdot\text{p}\cdot\mathcal{v}_0}\cdot\left[\sqrt{\text{s}}\right]_{1-\text{p}^2\cdot\mathcal{v}^2_0}^{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot\text{z}\right)^2}=$$ $$\frac{\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0}-\sqrt{1-\text{p}^2\cdot\mathcal{v}^2_0\cdot\left(1+\beta\cdot\text{z}\right)^2}}{\beta\cdot\text{p}\cdot\mathcal{v}_0}$$